# Part II Applications of Quantum Mechanics Lent 2012 Contents Prof. R.R. Horgan

Part II Applications of Quantum Mechanics Lent 2012 Prof. R.R. Horgan March 8, 2012 Contents 1 The 1.1 1.2 1.3 1.4 Variational Principle Estimating E0 . . . . . Accuracy . . . . . . . . Evaluation of hψ|H|ψi Further Results . . . . . . . . 1 1 2 2 3 . . . . . . 5 6 7 8 9 12 15 3 Classical Scattering in D = 3 3.1 The Scattering Cross-section . . . . . . . . . . . . . . . . . . . . . 16 16 4 Quantum Scattering in 3D 4.1 Partial wave analysis and phase shifts . . . . . . . . 4.1.1 Partial wave expansion for eikz . . . . . . . . 4.1.2 Partial wave expansion for scattering states 4.1.3 The optical theorem . . . . . . . . . . . . . . . 4.1.4 Solving for the asymptotic scattered wave 4.1.5 Low-energy scattering . . . . . . . . . . . . . . 4.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.1 Higher partial waves in the square well . . . 4.3 Influence of bound states . . . . . . . . . . . . . . . . 4.4 Resonance Poles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 20 . . . . . . . . . 21 23 24 24 25 26 28 28 30 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Bound States and Scattering States in D = 1 2.1 Scattering for particle incident from left . 2.2 Scattering for particle incident from right 2.3 The S-matrix . . . . . . . . . . . . . . . . . . 2.4 Bound states and the S-matrix . . . . . . . 2.5 Resonance states and the S-matrix . . . . 2.6 Properties of S(k) . . . . . . . . . . . . . . . i . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . CONTENTS 4.5 ii Green’s Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5.1 Integral equations . . . . . . . . . . . . . . . . . . . . . . . . 4.5.2 The Born series . . . . . . . . . . . . . . . . . . . . . . . . . . 5 Periodic Structures 5.1 Crystal Structure in 3D . . . . . . . . . . . 5.2 Discrete Translation Invariance . . . . . . . 5.3 Periodic Functions . . . . . . . . . . . . . . . 5.4 Reciprocal Vectors and Reciprocal Lattice 5.5 Bragg Scattering on a rigid Lattice . . . . 31 33 34 . . . . . 36 38 40 41 41 43 . . . . . . 46 47 48 51 51 53 53 . . . . . 59 59 60 60 61 63 . . . . . . 63 64 65 65 67 68 68 9 Elastic Vibrations of the Lattice – Phonons 9.1 Quantization of Harmonic Vibrations . . . . . . . . . . . . . . . 9.2 Unequal Masses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3 Debye-Waller Factor in 1D . . . . . . . . . . . . . . . . . . . . . . . 69 70 73 75 10 Quantum Particle in a Magnetic Field 10.1 Classical Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . . . . 77 77 . . . . . . . . . . . . . . . . . . . . 6 Particle in a One-Dimensional Periodic Potential 6.1 Band Structure . . . . . . . . . . . . . . . . . . . . . 6.1.1 The Tight Binding Model . . . . . . . . . . 6.1.2 Bloch’s Theorem and Bloch Waves in 1D 6.1.3 The Floquet Matrix . . . . . . . . . . . . . . 6.1.4 Brillouin Zones . . . . . . . . . . . . . . . . . 6.1.5 The Nearly-Free Electron Model . . . . . . 7 Particle in a 3-Dimensional Periodic Potential 7.1 Bloch’s Theorem . . . . . . . . . . . . . . . . 7.2 Ambiguities in k . . . . . . . . . . . . . . . . 7.3 The Nearly Free Electron Model in 3D . . 7.4 The Tight-Binding Model in 3D . . . . . . 7.4.1 Energy Contours . . . . . . . . . . . . 8 Electronic Properties 8.1 The Effect of Temperature . . . . 8.2 Insulators and Conductors . . . . 8.2.1 Dynamics . . . . . . . . . . 8.3 Semiconductors . . . . . . . . . . . 8.4 More on Extended Zone Scheme 8.4.1 Overlapping Bands . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . CONTENTS iii 10.2 Quantum Hamiltonian . . . . . . . . . . . . . . 10.2.1 Approximate Hamiltonian . . . . . . . . 10.3 Charged Particle in Uniform Magnetic Field 10.4 Degeneracy of Landau Levels . . . . . . . . . . 10.5 Effects of Spin . . . . . . . . . . . . . . . . . . . 10.6 Level Filling Effects . . . . . . . . . . . . . . . . 10.7 Lowest Landau Level and Complex Variables 10.8 Aharonov-Bohm Effect . . . . . . . . . . . . . . 10.8.1 Stationary State . . . . . . . . . . . . . . 10.8.2 Double Slit Experiment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78 79 81 82 82 83 83 83 84 84 BOOKS • D.J. Griffiths Introuction to Quantum Mechanics, 2nd edition, Pearson Education 2005. • N.W Ashcroft and N.D. Mermin Solid-State Physics, Holt-Saunders 1976. • L.D. Landau and E.M. Lifshitz Quantum Mechanics (Course in Theoretical Physics Vol 3), Butterworth-Heinemann 1976. • L. Schiff Quantum Mechanics, McGraw-Hill Book Company. • B. Dutta-Roy Elements of Quantum Mechanics, New-Age Science Limited. Other books may be recommeneded through the course. Also, consult particular topics in library books since many books are good in one area but poor in another. 1 THE VARIATIONAL PRINCIPLE 1 1 The Variational Principle A quantum system has a hermitian Hamiltonian H with energy eigenvalues E0 ≤ E1 ≤ E2 ≤ . . ., which are assumed to be discrete, and corresponding stationary states (eigenfunctions) ψ0 , ψ1 , ψ2 , . . . . (1.1) which are orthogonal and chosen to be normalized to unity. Let ψ be a general normalized state. Its energy expectation value is hE i = hψ|H|ψi . (1.2) We can always expand ψ on the complete basis of states in eqn. (1.1) as ψ = ∞ X an ψn ∞ X with n=0 |an |2 = 1 . (1.3) n=0 Then hEi = X = X = X a∗n am hψn |H|ψm i n,m a∗n am Em δnm n,m |an |2 En . (1.4) n Clearly, the minimum of hEi is E0 and occurs when a0 = 1, an = 0, n > 0; i.e., when ψ = ψ0 . This implies the variational definition of E0 : The ground state energy E0 is the minimum of hψ|H|ψi with respect to the choice hψ|H|ψi of ψ, or equivalently of if ψ is not normalized; this is the Rayleigh-Ritz hψ|ψi quotient 1.1 Estimating E0 This is a useful application of the variation approach when ψ0 and E0 cannot be easily or accurately computed. Consider a trial wavefunction ψ which depends on a few parameters α. Calculate hEi(α) = hψ|H|ψi(α) (1.1.1) and minimize with respect to α. The minimum value of hEi is an upper bound on E0 since, by expanding ψ as in eqn. (1.3) and using eqn. (1.4), we may always write X X X hEi = |an |2 En = E0 |an |2 + |an |2 (En − E0 ) ≥ E0 . (1.1.2) n This follows since 1. n P n n |an |2 = 1 and the last term is non-negative since En > E0 , n ≥ 1 THE VARIATIONAL PRINCIPLE 1.2 2 Accuracy Write 1 (ψ0 + εψ⊥ ) , (1.2.1) 1 + ε2 where ψ⊥ is a linear superposition of ψ1 , ψ2 , . . ., so hψ0 |ψ⊥ i = 0 =⇒ hψ0 |H|ψ⊥ i = 0. Then 1 hψ0 + εψ⊥ |H|ψ0 + εψ⊥ i hEi = 1 + ε2 1 2 = E + ε hψ |H|ψ i 0 ⊥ ⊥ 1 + ε2 1 2 E + ε Ẽ where Ẽ ≥ E1 = 0 1 + ε2 = E0 + ε2 (Ẽ − E0 ) + . . . . (1.2.2) ψ=√ We deduce that an O(ε) error in ψ gives O(ε2 ) error in estimate for energy E0 . In many cases can get estimate that is very close to E0 . 1.3 Evaluation of hψ|H|ψi For H=− ~2 2 ∇ + V (x) 2m (1.3.1) have ~2 2 hψ|H|ψi = ψ (x) − ∇ ψ(x) + V (x)ψ(x) dx 2m Z 2 ~ ∗ ∗ = ∇ψ (x) · ∇ψ(x) + ψ (x)V (x)ψ(x) dx 2m = hT i + hV i . Z ∗ (1.3.2) This R is applicable in all dimensions. Note that we have used the divergence theorem ( by parts in 1D) and so the surface term must vanish. Example Consider the quartic oscillator in scaled units: H=− d2 + x4 dx2 Can solve for the eigenstates numerically. In 1D use Given’s procedure, for example, which is basic approach of the CATAM project 5.3. I found the energies, in ascending 1 THE VARIATIONAL PRINCIPLE 3 order, to be (2 decimal places) 1.06, 3.80, 7.46, 11.64, . . .. Since the potential is symmetric The states are alternately even and odd parity, with the ground state even (this is always the case). Use normalized trial wavefunction a 1/4 2 ψ(x) = e−ax /2 (1.3.3) π and get Z hEi = dψ dx = 3 a + 2 . 2 4a 2 ! + x4 ψ 2 dx (1.3.4) The minimum of hEi occurs at a = 31/3 giving the estimate for the ground-state energy E0 = 1.08, which is less than 2% larger (it is always larger) than the true value. The variational (solid) and true (dotted) wavefunctions for ψ0 , ψ1 are 1.4 Further Results (1) To estimate E1 we can use the fact that ψ1 has odd parity and so use trial wavefunction ψ= 4a3 π 1/4 xe−ax 2 /2 . (1.4.1) This is already a bit long-winded to work out; I used MAPLE and tried other trial wavefunctions as well, but none so good as the oscillator. The estimate is E1 = 3.85, which is rather good. The variational (solid) and true (dotted) wavefunctions are shown above. The reason this works is that the expansion of ψ on the complete set of states only includes those with odd parity: X ψ(x) = an ψn (x), (1.4.2) n odd and the variational principle gives the lowest energy in this odd-parity sector. This is only possible if parity is a good quantum number which can be used to guarantee that the ground state does not appear in this expansion. In general, it is hard to pick 1 THE VARIATIONAL PRINCIPLE 4 out the excited states by clever choice of ψ. Here, we used parity symmetry to get E1 because V (x) is symmetric under reflection. Even then it is not obvious how to get E2 etc. We need, for example, to choose ψ with even parity and know that we exactly have hψ|ψ0 i = 0. (2) Lower bound on E0 . Clearly, h|H|i > min V , so min V is a lower bound on E0 . (3) Virial Theorem. Let the potential V (x) be a homogeneous function of degree n; i.e., V (λx) = λn V (x), in D dimensions. (1.4.3) Let ψ0 (x) be the true ground state, not necessarily normalized, and assumed real. Then Z 2 ~ 2 ∇ψ0 (x) · ∇ψ0 (x) + V (x)ψ0 (x) dD x 2m Z E0 = ψ02 (x)dD x = hT i0 + hV i0 . (1.4.4) Now consider ψ(x) = ψ0 (αx) as a trial wavefunction. Then hEi = α2 hT i0 + α−n hV i0 . (1.4.5) Check: Z hV i = V Z (x)ψ02 (αx) Z = α−n Z D d x = α −n ψ02 (αx)dD x V (αx)ψ02 (αx) dD x Z ψ02 (αx)dD x V (y)ψ02 (y) dD y Z = α−n hV i0 , 2 D ψ0 (y)d y (1.4.6) and similarly for the kinetic term, where the kinetic operator has n = −2. The minimum of hEi is clearly at α = 1 where the result is the exact ground state energy value. Hence, must have 2αhT i0 − nα−n−1 hV i0 = 0 at α = 1 , (1.4.7) and thus we find 2hT i0 = nhV i0 The Virial Theorem. (1.4.8) Equivalently, E0 = Examples: 2 1+ n hT i0 = n 1+ hV i0 . 2 (1.4.9) 2 BOUND STATES AND SCATTERING STATES IN D = 1 5 (i) The harmonic oscillator has n = 2, and so hV i0 = hT i0 = 1 E0 . 2 (1.4.10) (ii) The coulomb potential has n = −1, and so hV i0 = − 2hT i0 ⇒ E0 = − hT i0 negative and so a bound state . (1.4.11) (iii) The potential V (r) = 1/r3 , which has n = −3. This is relevant in D = 3 when the orbital angular momentum is zero: l = 0. Then find 2 E0 = 1− hT i0 > 0 . (1.4.12) 3 This is a contradiction since we assumed that there was a bound state; i.e., a state with normalizable wavefunction. In fact, there are no bound states and the particle falls to the centre. 2 Bound States and Scattering States in D = 1 Consider the stationary Schrödinger equation in D = 1: ~2 d2 ψ + V (x)ψ = Eψ , − 2m dx2 (2.1) where V (x) → 0 as x → ±∞. V(x) x The time-dependence of a stationary state of energy E is e−iωt with ω = E/~. A bound state has a negative energy E = −~2 κ2 /2m, κ > 0 and it is described by a normalizable wavefunction a e−κx x→∞ ψ(x) ∼ . (2.2) κx ae x → −∞ To obtain the bound state solutions we impose this asymptotic behaviour as a boundary condition. Generically, a solution that is exponentially growing from the left will continue to grow exponentially to the right; bound state solutions are special and discrete. The particle is localized in the neighbourhood of the origin and the probability for finding it at position x when |x| → ∞ is exponentially small. A scattering state in the same potential has positve energy E = ~2 k 2 /2m, k > 0, The asymptotic behaviour of a scattering state can be classified for x → ±∞ as follows: 2 BOUND STATES AND SCATTERING STATES IN D = 1 6 x→∞ ψ(x) ∼ a eikx a e−ikx right moving, outgoing/emitted wave left moving, incoming/incident wave. (2.3) right moving, incoming/incident wave, left moving, outgoing/emitted wave. (2.4) x → −∞ ψ(x) ∼ a eikx a e−ikx The momentum operator is p = −i~d/dx and so the momenta for e±ikx are ±~k. Really, the particles are to be thought of as wave packets which are made of a superposition of waves with wave vectors taking values in (k − ∆k, k + ∆k); the particles are then localized. This does not materially affect the analysis using the idealized system of pure plane waves. Why are we concerned with the properties of the scattering wavefunction as |x| → ∞? It is because we fire a projectile from far away (far enough away for the influence of the potential to be negligible) into the region where the potential is non-zero. We then observe how the particle is deflected by observing the emitted particle trajectory long after the scattering (long on the time-scale of the interaction) and so again far from the region of influence of the potential. If we think of the neighbourhood of the origin as a black box whose properties we probe using projectiles and studying how they scatter in this way, then we can probe the properties of the system (i.e., the potential) located in the region of the origin. This, after all, is how Rutherford discovered the nucleus at the centre of the atom and how we know that quarks exist as point-like objects in elementary particles such as the proton, neutron, pion etc. Indeed, quarks were originally a group-theoretical mathematical device for building the “periodic table” of the many elementary (actually not so elementary) particles. It was only in the 1960s that it was mooted (by Dalitz and others) that they were real and then in the 1970s that their effects were manifest in scattering experiments. 2.1 Scattering for particle incident from left For each k there is a solution for ψ with no incoming wave from the right; this a standard solution familiar in 1D calculations for reflection and transmission from a potential. A second, independent, solution has no incoming wave from the left (see later). In the former case we have ikx e + r e−ikx x → −∞ ψ(x) ∼ (2.1.1) t eikx x→∞ The reflection probability is The transmission probability is R = |r|2 T = |t|2 (2.1.2) To recap, recall that from the Schrödinger equation and its complex conjugate we can show, for real V and E (i.e., a stationary state), that ~ dψ(x) dψ ∗ (x) dj(x) ∗ = 0, j(x) = − i ψ (x) − ψ(x) dx 2m dx dx ⇒ j(x) is constant, (2.1.3) 2 BOUND STATES AND SCATTERING STATES IN D = 1 7 where j(x) is the probability current. This is equivalent to realizing that the Wronskian for the two independent solutions Re (ψ) and Im (ψ) is constant since there is no firstorder derivative term in the Schrödinger equation. For large -ve x j(x) = (e−ikx + r∗ eikx ) = ~k ikx (e − r e−ikx ) + complex conjugate 2m ~k (1 − |r|2 ) m (2.1.4) For large +ve x j(x) = ~k 2 |t| m (2.1.5) Thus, 1 − |r|2 = |t|2 ⇒ R + T = 1. (2.1.6) This shows that probability is conserved: the scattering is unitary. Note that r, t, R, T are functions of k and hence also of E. 2.2 Scattering for particle incident from right For each k there is a solution for ψ with no incoming wave from the left. This is the parity reversed situation to the familiar one discussed above. We need not solve again from scratch; indeed we can recover the solution from what has already been done. The Schrödinger equation is real, and so can take the complex conjugate of the solution ψ(x) of the previous section (2.1) and get another solution with asymptotic behaviour −ikx e + r∗ eikx x → −∞ ∗ ψ (x) ∼ (2.2.1) ∗ −ikx te x→∞ Yet another solution is ψ̃(x) = ψ ∗ (x) − r∗ ψ(x) with asymptotic behaviour (1 − |r|2 )e−ikx x → −∞ ψ̃(x) ∼ t∗ e−ikx − r∗ t eikx x→∞ (2.2.2) Since tt∗ = 1 − |r|2 , I can write this scattering state normalized to standard form as −ikx te x → −∞ ψL (x) ∼ (2.2.3) −ikx 0 ikx e +re x→∞, This has the particle incident from the right, i.e. Left-moving, and then reflected and transmitted with amplitudes for reflection and transmission (use subscript ”L” on ψL and prime superscript to signify quantities associated with scattering of incident Left moving particle.) r0 = − 1 − |r|2 T r∗ t 0 , t = = ∗ = t ∗ ∗ t t t the transmission amplitudes are the same. (2.2.4) 2 BOUND STATES AND SCATTERING STATES IN D = 1 8 Then we find the reflection and transmission probabilities are |r∗ t|2 = |r|2 = R same as for particle incident from left, |t|2 (2.2.5) = T also same as for particle incident from left. R0 = T0 Clearly, r0 and r differ only by an overall phase factor, and t0 = t. Note that the potential V need not be symmetric about the origin (symmetric here means an even function). If V is symmetric then V commutes with the parity operator P f (x) = f (−x): [P, V ] = 0 and hence so does the Hamiltionian H = p2 /2m + V (x): [P, H] = 0, and the eigenstates of H, including the scattering states discussed here, can be chosen as simultaneous eigenstate of H and P . The eigenvalues of P are ±1 since (a) P is hermitian and so has real eigenvalues, (b) P ψ(x) = λψ(x) and P 2 ψ(x) = ψ(x) ⇒ λ2 = 1. Thus, the simultaneous eigenvalues are (E = k 2 /2m, λ = ±1). In the case that [P, V ] = 0 it is clear that r0 = r and t0 = t since ψ(−x) of section (2.1) is also a solution but with left and right interchanged. Note that the equation for r0 can be written as r0 t∗ + r∗ t = 0 , (2.2.6) and for a symmetric potential this is rt∗ + r∗ t = 0 ⇒ Re (rt∗ ) = 0 . 2.3 (2.2.7) The S-matrix In the previous two sections I defined two different scattering wavefunctions which I shall call ψR , ψL signifying that the incident particle is Right-moving and Left-moving, respectively. These are two independent complex solutions for the Schrödinger equation with positive energy. Define Incoming, Right-moving wave: IR (k, x) = eikx , x → −∞ eikx Incoming, Left-moving wave: IL (k, x) = e−ikx , x → ∞ e-ikx Outgoing, Right-moving wave: OR (k, x) = eikx , x → ∞ eikx Outgoing, Left-moving wave: OL (k, x) = e−ikx , x → −∞ e-ikx 2 BOUND STATES AND SCATTERING STATES IN D = 1 We can then write our two solutions in matrix form ψR IR t r OR = + . 0 ψL IL r t OL The matrix in this equation is the S-matrix: t r S = . r0 t 9 (2.3.1) (2.3.2) This is in the basis of right and left moving incoming waves denoted (R, L) basis. Conservation of probability implies that S is unitary: 2 ∗ 0∗ 1 0 |t| + |r|2 tr0∗ + rt∗ t r t r † = , (2.3.3) = SS = 0 1 r0 t∗ + tr∗ |r0 |2 + |t|2 r∗ t∗ r0 t where results from the earlier section have been used including eqn. (2.2.6). When V = 0 then S = I, the unit matrix. The solutions then are ikx ψR IR + OR e = = , ψL IL + OL e−ikx (2.3.4) i.e., unscattered right- and left-moving plane waves described by r = r0 = 0, t = 1. Another way to write the solutions in this (R, L) basis is to emphasize the departure from these non-scattering solutions which is caused by the scattering. After all, this is what we detect as being the non-trivial effect of the scattering potential. We rewrite eqn. (2.3.1) as ikx OR ψR e + (S − I) . (2.3.5) = −ikx e OL ψL It is usual to define the T-matrix by T = S − I, and then T = 0 means there is no scattering: t−1 r T = . (2.3.6) r0 t−1 2.4 Bound states and the S-matrix It is useful to reformulate the expression for the scattering states discussed above in order to highlight different features that are physically important. This will act as a paradigm for the way in which scattering in D = 3 is analyzed and allows the basic concepts to be demonstrated. In particular, we shall generally assume that there are one or more conserved quantum numbers other than E. In 1D this will be parity which is relevant for symmetric potentials V (x). We shall consider incoming and outgoing waves that are eigenstates of parity, labelled by ± for even/odd parity, respectively, as well as the wavevector k and consequently also by energy E. This is just a different basis choice and is a linear combination of the (R, L) basis states: 2 BOUND STATES AND SCATTERING STATES IN D = 1 10 Incoming, +ve parity: I+ (k, x) = e−ik|x| eikx e-ikx Incoming, −ve parity: I− (k, x) = sgn(x)e−ik|x| -eikx e-ikx Outgoing, +ve parity: O+ (k, x) = eik|x| e-ikx eikx Outgoing, −ve parity: O− (k, x) = −sgn(x)eik|x| e-ikx -eikx We have that IR 1 1 I+ , = IL I− −1 1 | {z } A O+ O− = 1 1 OR OL −1 1 (2.4.1) If there is no scattering potential the incoming particles pass through the origin without suffering any modification and become outgoing particles. The +ve(−ve) parity steadystate wave functions that describes this process are thus ψ+ (k, x) = I+ (k, x) + O+ (k, x) = eikx + e−ikx , ψ− (k, x) = I− (k, x) + O− (k, x) = − eikx + e−ikx . (2.4.2) When a scattering potential V (x) is present the incoming wave is unaffected, it has yet to encounter the scattering centre, but the outgoing wave is modified. Indeed, if V (x) is not symmetric then an incoming wave of given parity can scatter and produce both +ve and −ve parity outgoing waves. We can then write the scattering wavefunctions in matrix form using the S-matrix: ψ+ I+ (P ) O+ = + S , (2.4.3) ψ− I− O− where S (P ) is the S-matrix in the parity basis: S++ S+− t + (r + r0 )/2 (r − r0 )/2 (P ) −1 S = = ASA = . S−+ S−− (r0 − r)/2 t − (r + r0 )/2 (2.4.4) The unitarity of S (P ) follows directly from the unitarity of S; the relation between them is a similarity transformation. The important case is when V is symmetric: [P, V ] = 0. Then r0 = r and S (P ) is diagonal: t+r 0 (P ) S = , (2.4.5) 0 t−r and parity is conserved in the scattering process; a +ve(−ve) incoming state scatters only into a +ve(−ve) outgoing state. Clearly, unitarity of S (P ) in this case means |S++ |2 = |S−− |2 = 1 ⇒ S++ = t + r = e2iδ+ , S−− = t − r = e2iδ− . (2.4.6) 2 BOUND STATES AND SCATTERING STATES IN D = 1 11 Here δ± encode the phase shifts of the outgoing waves O± with respect to the case of no scattering. Thus, δ± 6= 0 signifies the occurrence of scattering and they are all we need to know to describe and quantify the process. Consider the example of scattering from the square well potential −V0 |x| < a/2 V (x) = (2.4.7) 0 |x| > a/2 The transmission (t) and reflection (r) amplitudes are given by t = 2ikq e−ika , D(k) r = (k 2 − q 2 ) sin qa e−ika D(k) where D(k) = 2ikq cos qa + (k 2 + q 2 ) sin qa , q 2 = k 2 + 2mV0 . ~2 (2.4.8) We set α = qa/2 and then can write the denominator D as D(k) = 2ikq( cos 2 α − sin 2 α) + (k 2 + q 2 )2 sin α cos α = 2 cos 2 α (q tan α + ik)(q − ik tan α) (2.4.9) Now S++ = t + r and S−− = t − r and the numerators of these expressions are N± (k) = (2ikq ± (k 2 − q 2 ) sin qa)e−ika = (2ikq ± 2(k 2 − q 2 ) sin α cos α)e−ika = 2 cos 2 α (q tan α ∓ ik)(ik tan α ∓ q)e−ika . (2.4.10) Thus, after all the algebra we find that f ∗ (k) (q tan α − ik) ≡ + , (q tan α + ik) f+ (k) f−∗ (k) −ika (q + ik tan α) ≡ . = e (q − ik tan α) f− (k) S++ = −e−ika (2.4.11) S−− (2.4.12) These expressions are both unitary since they are the ratios of a complex expression and its complex conjugate. [Side remark: f+ (k) and f− (k), properly chosen, are related to Jost functions in scattering theory] The important feature is that, as a function of k, both S++ (k) and S−− (k) have a pole at imaginary value of k and an associated zero in the complex congugate position. We examine the meaning using the example of scattering of a +ve parity wave on the well. We have ψ+ (k, x) = I+ (k, x) + S++ (k)O+ (k, x) . (2.4.13) We now observe that we can analytically continue I+ and O+ in the complex k-plane and note for k = −iκ that I+ (−iκ, x) decays exponentially in both left and right regions, and similarly for O+ (iκ, x): I+ (−iκ, x) = O+ (iκ, x) = e−κ|x| 2 BOUND STATES AND SCATTERING STATES IN D = 1 12 However, the state ψ+ (k, x) is still a solution of the Schrödinger equation for these values of k : k = ∓iκ. In the special case that S++ in eqn. (2.4.13) has a zero at k = −iκ then the first term, the term with I+ (−iκ, x), is all that is left and the state has only the exponentially decaying contribution. ψ+ (−iκ, x) ∼ e−κ|x| . (2.4.14) Because of the form of S++ (k) we expect that if S++ has a zero at k = −iκ then it will have a pole at k = iκ. Then in eqn. (2.4.13) the second term, the term with O+ (iκ, x), dominates the first term and, for the properly normalized state, again ψ+ (iκ, x) ∼ e−κ|x| . (2.4.15) In the case here the zero(pole) are at k = −iκ(iκ) where from eqn. (2.4.11) κ = q tan qa/2 remember α = qa/2 , q 2 + κ2 = 2mV0 . ~2 (2.4.16) The solutions for κ are indeed the solutions for the +ve parity bound states of the square well. A similar result holds for the −ve parity states and S−− . These are certainly bound state solutions and so we have the very important result for potentials V (x) that vanish sufficiently fast as |x| → ∞: Let V (x) be a potential such that V (x) → 0 for |x| → ∞. Then to each bound state state of V , φ(κ, x) ∼ exp(−κ|x|) |x| → ∞, there is an associated zero of the S-matrix S(k) on the negative imaginary k axis at k = −iκ, and a corresponding pole on the positive imaginary k-axis at k = iκ. Caveat: the inverse statement is not always true. Not every zero(pole) of S corresponds to a bound state. This will only be the case if the potential V (x) is localized, by which is meant essentially that, as |x| → ∞, V falls off faster than any exponential. This is true for any potential which is zero outside some neighbourhood of the origin or, for example, for a potential decaying like a Gaussian function at large |x|. This is true for the foregoing analysis where we have only demonstrated the result for localized V ; we could then identify the exact solutions to be plane-waves outside the range of V . This needs more effort if V is not locaized. The analytic properties of S(k) in the complex k-plane are important in the analysis of scattering. 2.5 Resonance states and the S-matrix We will look at another important scattering result: the effect of states that are quasistable/metastable. We examine this by an example. Consider the potential consisting of two delta functions of equal strength, V0 , at x = ±1. V (x) = V0 δ(x − 1) + V0 δ(x + 1) . (2.5.1) 2 BOUND STATES AND SCATTERING STATES IN D = 1 V0 V0 eikx S++ e e A cos kx -ikx 13 -1 -ikx S++ eikx +1 Here is shown the scattering of the +ve parity partial wave: I+ (k, x) scatters into O+ (k, x). A similar diagram for the negative parity waves has ψ(x) = A sin kx in −1 < x < 1. The boundary conditions are that ψ is continuous and, defining U0 = 2mV0 /~2 , dψ dψ = U0 ψ , Disc = U0 ψ . (2.5.2) Disc dx x=−1 dx x=1 Carry out the matching calculation in the usual way but now only need to match at one boundary, x = 1 say, since parity ensures the boundary conditions are satisfied at the other boundary. (see example sheet). We consider only the +ve parity waves to demonstrate the result. We get ik −ik −2ik (2k − iU0 )e − iU0 e . (2.5.3) S++ = e (2k + iU0 )e−ik + iU0 eik This is clearly unitary as it is the ratio of a function and its conjugate. (1) For V0 → ∞ with k finite we find S++ → −e−2ik . (2.5.4) This is to be expected since it corresponds to pure reflection off an impenetrable barrier. The “−” sign is the standard change of phase on reflection and the phase factor is there because the wave is reflected rather than passing through and hence has travelled a distance ∆x = 2 less far. (2) Consider the pole positions of S++ . These are at 2ik 2ik e =− 1− . U0 (2.5.5) For k/U0 → 0 (i.e. V0 → ∞) the solutions are k = (n + 1/2)π. These correspond to the symmetric bound states in the interior region −1 < x < 1. Consider just n = 0 from now on. This state looks like: 2 BOUND STATES AND SCATTERING STATES IN D = 1 V0 V0 -1 +1 14 Now consider the pole position for U0 finite but still U0 /k >> 1. We expect it to be close to the bound state pole positions. Take example with n = 0 and propose π k = + α − iγ . (2.5.6) 2 We expect α, γ to vanish as V0 → ∞ and will find that a power series in 1/U0 will suffice. We need to solve 2ik 2ik iπ 2(iα+γ) 2(iα+γ) e e =− 1− ⇒ e = 1− . (2.5.7) U0 U0 After a little algebra find α=− π π + , 2U0 2U02 γ= π2 . 4U02 (2.5.8) [Note, it is easier to solve after realizing that γ is O(1/U02 ).] Thus, the pole is at a complex value of momentum and lies in the lower-half of the complex k-plane. Write k = p − iγ with p = π/2 + α and γ > 0. There is, of course, a zero at the conjugate point. What does all this signify? (i) The state inside the potential region is no longer bound but is quasistable and, once established, only leaks out very slowly as long as V0 is large. (ii) The wavefunction continued to the pole value for k is dominated by O+ (k) for |x| > 1; It corresponds to analytic continuation (in k) to the boundary condition with only outgoing waves: ψ+ (k, x) ∼ eip|x| eγ|x| . (2.5.9) I.e., at a fixed time the wavefunction increases with |x|. (C.f. to investigate bound states we continued to exponentially decaying boundary conditions.) What does this mean? Suppose we have put the particle into −1 < x < 1 and it occupies a state of energy E. This is not an eigenstate since it eventually escapes by tunnelling. However, if V0 is large it will be close to one of the stationary states (here we chose the ground state). Suppose this happened at t = −T ; it is like an excited atom which will eventually decay. The particle is said to be in a metastable or quasi-stationary state. Radioactive decay is described by similar systems. As t increases the particle leaks out to ∞ and at t = 0 the wavefunction looks like 2 BOUND STATES AND SCATTERING STATES IN D = 1 -1 15 +1 vT Here v is the particle velocity. The wavefunction is bigger at larger |x| since there was a larger particle current at early times and it has travelled further by t = 0. The wavefunction, of course, is zero for |x| > vT and the whole state is still normalized. The complex energy is given by Ec = ~2 k 2 /2m and we find 1 Ec ∼ E0 − iΓ , 2 where ~2 p 2 E0 = and is very close to the bound state energy, 2m 2~2 pγ and Γ > 0 since γ > 0 . m The time dependence of the state is then Γ = e−iEc t/~ = e−iE0 t/~ e−Γt/2~ . (2.5.10) (2.5.11) (2.5.12) I.e., at fixed x the state decays with characteristic time τ = 1/Γ; this controls the half-life of the state and Γ is called the width of the state. (Note that the probability is proportional to |ψ 2 | and so decays with exp(−Γt).) As a function of energy then S++ will have the form S++ E − E0 − 21 iΓ . ∼ E − E0 + 21 iΓ (2.5.13) The quasi-stationary state is also called a resonance and in general the S-matrix has this characteristic form when E is near to E0 . We will look at the consequences of resonant structure in 3D scattering. 2.6 Properties of S(k) (1) SS † = I : probability conservation means S is unitary. If S is diagonal then Sαβ (k) = e2iδα (k) δαβ (do not confuse notation using “δ” twice here). (2) Complex conjugation ∗ and k → −k have the same effect on the incoming and outgoing states: I+ ↔ O+ , I− ↔ −O− . Thus, ψ ∗ (k) = ψ(−k) =⇒ S ∗ (k) = S(−k) . (2.6.1) For diagonal S this implies δα (−k) = −δα (k) ; the phase shifts are odd functions of k. 3 CLASSICAL SCATTERING IN D = 3 3 16 Classical Scattering in D = 3 Scattering of particles is used to probe the general structure of matter. It reveals the constituent particles composing matter (such as atoms, nuclei and quarks), determines their sizes, the characteristic length scales of structures (such as the lattice spacing in a crystal) and the nature of the forces which act. In the previous section we have established most of the mathematical features of scattering but in 3D there are generalizations of these ideas. 3.1 The Scattering Cross-section First we define the impact parameter b and the scattering angle θ: 2 b scattering centre The scattering angle is the angle between the asymptotic trajectories at late and early times and the impact parameter is the closest distance of approach if the particle were to suffer no scattering. We will ignore recoil and assume azimuthal symmetry at least. The trajectory is then planar. We shall calculate or measure θ(b) or, better, b(θ). Example of hard sphere: " 2 " b " R 3 CLASSICAL SCATTERING IN D = 3 17 Here b = R sin α , θ = π − 2α =⇒ π θ θ b = R sin − = R cos . 2 2 2 (3.1.1) There is no scattering if b ≥ R. Idea of the Cross-section: The cross-section is a measure of the size of the scattering centre presented to the incoming beam. As its name suggests it is in units of (metres)2 although the natural unit in atomic/nuclear/particle physics is the barn = 10−28 m2 and a more usual version is the nanobarn (nb) = 10−9 b. For classical scattering this is a simple idea but in quantum mechanics the scattering cross-section generally varies strongly with the energy of the incoming beam and can be particularly large for those energies corresponding to resonance poles as were discussed in the previous section. The full calculation also takes into account diffraction effects which are absent classically. The picture which explains how the cross-section is defined for both classical and quantum scattering is d5\* d *\ v - >- -- ' / p_ bf lU,-ad ) /er& .t Consider infinitesimal cross-sectional area dσ of incident beam that scatters into infinitesimal solid angle dΩ. Geometrically dσ = b db dφ , dΩ = sin θ dθ dφ . Note, dφ is the same in both cases if have azimuthal symmetry. Then dσ db b . the differential cross-section is D(θ) = = dΩ sin θ dθ D(θ) has units of area. (3.1.2) (3.1.3) 3 CLASSICAL SCATTERING IN D = 3 18 Hard sphere: db R θ = − sin =⇒ dθ 2 2 θ R θ R cos 2 R2 2 sin 2 D(θ) = = . sin θ 4 The total cross-section, σ, is Z Z σ = D(θ)dΩ = 2π (3.1.4) 1 D(θ)d cos θ . (3.1.5) −1 σ is the total area of the beam that is scattered. For the hard sphere scattering R2 σ= 4π = πR2 , 4 (3.1.6) i.e., the geometrical area presented to the beam. Coulomb (Rutherford) scattering: 2 charge q1 b Heavy charge q2 Beam particles have kinetic energy E. Orbit theory gives b= q1 q2 θ cot , 8πε0 E 2 and hence the differential cross-section is 2 dσ q 1 q2 1 D(θ) = = . dΩ 16πε0 E ( sin θ/2)4 (3.1.7) (3.1.8) The total cross-section is infinite. This arises because the coulomb, inverse square, force is of infinite range and is special. Indeed, it has to be dealt with very carefully in general. The infinity arises because the beam is idealized as being of infinite transverse extent whereas, in practice, no beam is like this; in QM the particles are in wavepackets. This is one case where the idealization goes wrong. However, it is only relevant to very small scattering angles which are never observed in practice. 4 QUANTUM SCATTERING IN 3D 4 19 Quantum Scattering in 3D The target is described by the fixed potential V (r), V → 0 as r → ∞ (r = |r|) and the scattering is described by the wavefunction ψ(r) satisfying the Schrödinger equation ~2 2 − ∇ + V (r) ψ(r) = Eψ(r) , (4.1) 2m with E > 0. The analysis follows a similar path to that already discussed for scattering in 1D but with some important differences. We look for a stationary (time-independent) description of the scattering visualized as ttere Sca e ikz d Wave incident beam This is a steady state with continuous incident beam and an outgoing scattered spherical wave. The appropriate boundary condition for ψ(r) as r → ∞ is ψ(r) ∼ ikz + e|{z} incident wave eikr f (r̂) . r | {z } scattered wave (4.2) The probability current is j(r) = −i ~ (ψ ∗ ∇ψ − ψ∇ψ ∗ ) , 2m with ∇ · j = 0 . (4.3) For ~k =v. m Note, one particle per unit volume in the beam. For ψ = ψin = eikz : ψ = ψscatt = f (r̂) j scatt = eikr : r j in = ∇ψ ∼ ikf (r̂) ~k 1 |f (r̂)|2 r̂ + O(r−3 ) m r2 (4.4) eikr r̂ + O(r−2 ) =⇒ r (4.5) Interference terms (not shown) between the incident and scattered wave arise as a consequence of the idealization of using plane waves rather than wavepackets. Can construct a beam of finite transverse extent using superposition without materially 4 QUANTUM SCATTERING IN 3D 20 affecting the scattering amplitude; the incident beam and scattered particles are then separated and no interference at large r occurs. So: The scattered flux into element dΩ of solid angle The incident flux per unit area ~k |f (r̂)|2 dΩ m ~k . = m = (4.6) We then have dσ = scattered flux into dΩ =⇒ incident flux per unit area dσ = |f (θ, φ)|2 . dΩ (4.7) f (θ, φ) is called the scattering amplitude. If V is spherically symmetric, V (r) = V (r), then f is independent of φ. 4.1 Partial wave analysis and phase shifts We have seen many of the ideas earlier. In the 1D case we had partial waves I± and O± labelled by parity, and for symmetric V the S-matrix was parametrized by phase-shifts, δ± for each partial wave. In 3D, for spherically symmetric V , the angular momentum l = 0, 1, 2, . . . is a good quantum number. The partial waves will be labelled by l and the S-matrix parametrized by δl . Wish to solve ~2 2 ∇ + V (r) ψ(r) = Eψ(r) (4.1.1) − 2m subject to eikr ψ(r) ∼ eikz + f (θ) , r Make the partial wave expansion ψ(r) = ∞ X r → ∞. Rl (r)Pl ( cos θ) . (4.1.2) (4.1.3) l=0 The Legendre polynomials, Pl (w) satisfy d d (1 − w2 ) Pl (w) + l(l + 1)Pl (w) = 0 , dw dw (4.1.4) and are angular momentum wavefunctions; this PDE is simply L2 Pl = ~2 l(l + 1)Pl . (4.1.5) d2 l(l + 1) 2 (rRl ) + k − − U (r) rRl = 0 , dr2 r2 (4.1.6) Then Schrödinger’s equation is 4 QUANTUM SCATTERING IN 3D 21 where 2m ~2 k 2 , U (r) = 2 V (r) . (4.1.7) 2m ~ For a short-range potential (the only kind we shall consider for the time being) as r → ∞ we get d2 (rRl ) + k 2 (rRl ) = 0 . (4.1.8) 2 dr This equation has two independent solutions for Rl (r): E= e−ikr r and eikr , r (4.1.9) i.e., respectively, ingoing and outgoing spherical waves. These are the 3D analogues of the ingoing and outgoing plane waves in the 1D case. 4.1.1 Partial wave expansion for eikz When U = 0, ψ = eikz ≡ eikr cos θ is also a solution, and the corresponding radial equation ∀ r is d2 l(l + 1) (ρRl ) + 1 − ρRl = 0 , (4.1.1.1) dρ2 ρ2 with ρ = kr. Thus can write eikr cos θ = X (2l + 1)ul (kr)Pl ( cos θ) , (4.1.1.2) l where the factor of (2l + 1) is for convenience, and ul (ρ) satisfies eqn (4.1.1.1). Now Z 1 2 δlm , (4.1.1.3) Pl (w)Pm (w)dw = (2l + 1) −1 and hence 1 Z 2ul (ρ) = eiρw Pl (w)dw . (4.1.1.4) −1 ρ→∞ Integrate by parts: " eiρw Pl (w) 2ul (ρ) = iρ #1 −1 1 − iρ Z 1 eiρw Pl0 (w)dw . (4.1.1.5) −1 Have Pl (1) = 1 , Pl (−1) = (−1)l and so ul (ρ) = eiρ − (−1)l e−iρ + O(ρ−2 ) , 2iρ (4.1.1.6) where a further integration by parts shows that the integral is O(ρ−2 ). This may be rewritten as 1 l sin ρ − 2 lπ ul (ρ) ∼ i ρ→∞. (4.1.1.7) ρ 4 QUANTUM SCATTERING IN 3D 22 Then for ρ → ∞ eikz = ∞ X l=0 1 lπ sin kr − 2 Pl ( cos θ) . (2l + 1) il kr (4.1.1.8) ρ→0 Since Pn (w) is a polynomial of degree n we can write the monomial wl = X l! Pl (w) + cm Pm (w) . (2l − 1)!! m<l (4.1.1.9) Then Z 1 (iρw)l 1 + ... + . . . Pl (w)dw l! {z } | 2ul (ρ) = −1 exp(iρw) l = eikz ∼ (iρ) il ρl 2 l! + O(ρl+2 ) =⇒ ul (ρ) ∼ l! (2l + 1)!! (2l + 1)!! ∞ X l=0 il (kr)l Pl ( cos θ) . (2l − 1)!! =⇒ (4.1.1.10) The solutions to eqn. (4.1.1.1) are the spherical Bessel functions jl (ρ), nl (ρ) with jl (ρ) regular at ρ = 0. From the previous analysis we identify ul (ρ) = il jl (ρ) which defines jl (ρ). Because there is no first-order term in eqn. (4.1.1.1) the Wronskian is a constant and we define W ρjl (ρ), ρnl (ρ) = 1. This determines nl (ρ) and its asymptotic behaviour. We then have ρ→∞ sin ρ − 12 lπ jl (ρ) ∼ , ρ cos ρ − 21 lπ nl (ρ) ∼ − . ρ (4.1.1.11) ρ→0 jl (ρ) ∼ ρl , (2l + 1)!! The full solutions are l 1 d sin x l jl (x) = (−x) , x dx x nl (ρ) ∼ −(2l − 1)!! ρ−(l+1) . l nl (x) = − (−x) 1 d x dx (4.1.1.12) l cos x . x (4.1.1.13) In particular, sin x , x cos x n0 (x) = − , x j0 (x) = sin x cos x − , 2 x x cos x sin x n1 (x) = − 2 − . x x j1 (x) = (4.1.1.14) 4 QUANTUM SCATTERING IN 3D 4.1.2 23 Partial wave expansion for scattering states For U (r) = 0 there is no scattering and for r → ∞ we have " # ∞ ikr −ikr X e 2l + 1 e − (−1)l Pl ( cos θ) . eikz ∼ 2ik r r l=0 (4.2.1.1) In a given summand the first(second) term is an outgoing(ingoing) spherical wave of angular momentum l; these are the partial waves each labelled by l. They are the 3D analogues of the parity partial waves in 1D scattering. We can view the plane wave as a superposition of ingoing and outgoing spherical waves. The waves interfere to produce the unscattered beam. For each l we see an ingoing spherical wave which passes through the origin and emerges as an outgoing spherical wave; it is analogous to sum of the parity ingoing and outgoing waves of given parity in 1D. For U (r) 6= 0 there is scattering: (i) We expect only the outgoing waves to be modified; the ingoing waves have not yet encountered the target. (ii) Since l is a good quantum number it is not changed by the scattering; each partial wave scatters independently. (iii) Conservation of probability (particle number) implies that each outgoing wave of given l is modified by a phase only. This is identical to the situation with the symmetric potential in 1D. Thus expect, as r → ∞, # ikr −ikr e e ψ(r) ∼ e2iδl − (−1)l Pl ( cos θ) 2ik r r l=0 ∞ X eikr 2l + 1 2iδl ikz = e + e − 1 Pl ( cos θ) . 2ik r l=0 ∞ X 2l + 1 " (4.2.1.2) Note here that the S-matrix, S, has entries Sl l0 for a general potential but that in the spherically symmetric case it is diagonal as is, therefore, the T-matrix T = S − I: Sl l0 = e2iδl δl,l0 , Tl l0 = (e2iδl − 1)δl,l0 , (4.2.1.3) 4 QUANTUM SCATTERING IN 3D 24 where δl is a function of k. By comparing with eqn. (4.2) we identify the scattering amplitude ∞ f (θ) = 1X (2l + 1)fl Pl ( cos θ) , k l=0 fl = eiδl sin δl . (4.2.1.4) The fl are the partial wave scattering amplitudes. From eqn. (4.7) we have 1 X dσ = 2 (2l + 1)(2l0 + 1)fl fl∗0 Pl ( cos θ)Pl0 ( cos θ) =⇒ dΩ k l l0 4π X σ = (2l + 1) sin 2 δl . k2 l 4.1.3 (4.2.1.5) (4.2.1.6) The optical theorem From eqn. (4.2.1.4) we see immediately that Im f (0) = 1X (2l + 1) sin 2 δl . k l (4.3.1.1) Thus, we obtain the optical theorem σ = 4.1.4 4π Im f (0) . k (4.3.1.2) Solving for the asymptotic scattered wave From eqn. (4.2.1.2), for r → ∞ ψ(r) ∼ ∞ X sin (kr − 21 lπ + δl ) (2l + 1) il eiδl Pl ( cos θ) . kr l=0 (4.4.1.1) The task is to solve the Schrödinger equation, identify this asymptotic form and hence calculate the δl . Must therefore solve d2 l(l + 1) 2 (rRl ) + k − − U (r) rRl = 0 , (4.4.1.2) dr2 r2 where U (r) = 2mV (r)/~2 , with boundary conditions that (i) Rl (r) is regular at r = 0; (ii) rRl (r) ∼ Bl sin (kr − 12 lπ + δl ) as r → ∞. We suppose that V (r) is negligible beyond a finite range r = a (in our examples we shall choose it to be zero for large enough r), although the formalism is fine as long as V → 0 faster than 1/r as r → ∞. We note the asymptotic behaviours of the spherical Bessel functions jl (r) and nl (r) discussed earlier, see eqn. (4.1.1.11). Then for r > a we will have Rl (r) = Bl jl (ρ) cos δl (k) − nl (ρ) sin δl (k) . (4.4.1.3) 4 QUANTUM SCATTERING IN 3D 25 The boundary condition is that Rl (r) and dRl (r)/dr are continuous at r = a. Solve the Schrödinger equation d2 l(l + 1) 2 (rRl ) + k − − U (r) rRl = 0 , (4.4.1.4) dr2 r2 for r < a and U = 2mV /~2 . Define γl (ka) = aRl0 (a)/Rl (a) for r − a = 0− (i.e., just inside the well): this is the logarithmic derivative and is dimensionless. Then the matching condition is ka jl0 (ka) cos δl − n0l (ka) sin δl = γl (ka) =⇒ (4.4.1.5) jl (ka) cos δl − nl (ka) sin δl γl (ka)jl (ka) − ka jl0 (ka) . tan δl (k) = (4.4.1.6) γl (ka)nl (ka) − ka n0l (ka) If V (r) does not vanish for r > a then eqn. (4.4.1.4) must be solved for two solutions which behave as r → ∞, respectively, as ingoing and outgoing spherical waves. The solution for ψ(r) will be a linear combination of the two solutions whose boundary condition is that it behaves asymptotically as in eqn. (4.2.1.2). Then can determine the phase-shifts δl (k). 4.1.5 Low-energy scattering In the low energy limit we take k small. By this mean, for example, ka → 0. In this limit we expect γl (ka) → constant and in eqn. (4.4.1.6) we can replace the Bessel functions by their approximate behaviour for small argument (see eqn. (4.1.1.12)) nl (x) ∼ x−(l+1) jl (x) ∼ xl , =⇒ tan δl (k) ∼ (ka)2l+1 . (4.5.1.1) We thus deduce that δl (k) ∼ (ka)2l+1 1 + O(k 2 ) (4.5.1.2) It is instructive to understand physically why behaviour of this sort is to be expected. In eqn. (4.4.1.4) the potential is made up of two pieces: U (r) and the centrifugal barrier −l(l + 1)/r2 which acts to repel penetration of the particle to the origin where U (r) is appreciable. For low energy E ∼ k 2 the particle simply does not have enough energy to tunnel significantly to the region where it encounters U (r) and so the scattering is weak and weakens as l increases. l(l+1) r2 Rl(r) V(r) E= k2 2m r 4 QUANTUM SCATTERING IN 3D 26 Another way to see this from a classical perspective is that the angular momentum is ~k b where b is the impact parameter. Then, l ∼ kb =⇒ b ∼ l/k, and b is related to the closest distance of approach. If b > a then the particle never (classically) enters the region of influence of V (r) and does not scatter; this corresponds to ka < l. For small k the l = 0, S-wave, contribution is the only one that is not negligible and its contribution dominates the cross-section. We write δ0 = − as k + O(k 3 ) , (4.5.1.3) and from the formula for σ, eqn. (4.2.1.6), we find that the low-energy cross-section is σ ≈ σ0 = 4πa2s . (4.5.1.4) as is called the scattering length and can be positive, negative and under certain circumstances it can diverge as a function of parameters in the potential. This expansion is valid for k such that kas is small enough. We must have rR0 (r) → 0 as r → 0, and for r > a we have (exactly – see section 4.1.4) R0 (r) = B0 sin (kr − kas ) . r (4.5.1.5) Clearly, this function has a zero at r = as . So rR0 (r) looks like rR (r) 0 a r 8 as range of V(r) This shows that the sign of as is related to the curvature of the wavefunction just inside r < a. 4.2 Examples 1. Scattering off a hard sphere. The potential is U (r) = ∞ 0 r<a, r>a, (4.2.1) with boundary conditions ψ(r) = 0 continuous r≤a r=a. (4.2.2) 4 QUANTUM SCATTERING IN 3D 27 For r > a, Rl (r) takes the form in eqn. (4.4.1.3) and so the boundary condition at r = a means jl (ka) Rl (a) = Bl jl (ka) cos δl − nl (ka) sin δl = 0 =⇒ tan δl = . (4.2.3) nl (ka) This determines the δl and hence dσ/dΩ. For long wavelengths ka 1 and then find δl (k) ≈ − (ka)2l+1 , (2l + 1)!!(2l − 1)!! (4.2.4) with the exact result δ0 = − ka . (4.2.5) Note that we have chosen δ(0) = 0 but, of course, tan δ = 0 has solutions δ = nπ and the choice is a matter of convention. Thus, for k → 0 X (2l + 1) f (θ) ≈ δl Pl ( cos θ) k l X k 2l a2l+1 Pl ( cos θ) = − [(2l − 1)!!]2 l = −a − k 2 a3 cos θ + O(k 4 ) . (4.2.6) From eqn. (4.2.5) we see that the scattering length is as = a and therefore that the low-energy cross-section is σ = 4πa2 . This is to be compared with the classical result derived earlier of σ = πa2 . The increase is due to diffraction effects. 1. S-wave scattering on a square well. 2mV0 2mV (r) , γ2 = . (4.2.7) 2 ~ ~2 For l = 0 have wavefunction ψ(r) = u(r)/r where the radial Schrödinger equation is V (r) = −V0 0 r<a r>a. U (r) = u00 + k 2 u = 0 r > a, 00 2 u + κ u = 0 r < a, 2 2 ~k E = κ2 = k 2 + γ 2 . 2m (4.2.8) (4.2.9) (4.2.10) r < a: ψ(r) regular at origin =⇒ u(r) = A sin κr. r > a: u(r) = B sin (kr + δ0 ) where δ0 is the S-wave (l = 0) phase shift. dψ 1 du , is continuous at r = a. The boundary condition is that 1 , or equivalently u ψ dr dr This gives B k cos (ka + δ0 ) A κ cos κa = =⇒ A sin κa B sin (ka + δ0 ) κ cot κa = k cot (ka + δ0 ) or tan (ka + δ0 ) tan κa = . ka κa (4.2.11) 4 QUANTUM SCATTERING IN 3D 28 k→0 k γ cot γa k ≈ =⇒ γ cot γa ka(1 − γa cot γa) ≈ + O(k 3 ) . γa cot γa tan (ka + δ0 ) ≈ tan ka + tan δ0 1 − tan ka tan δ0 tan δ0 (4.2.12) Then δ0 = −as k + O(k 3 ) (4.2.13) gives as = − (1 − γa cot γa) a. γa cot γa (4.2.14) for the scattering length, with σ ≈ 4πa2s . 4.2.1 Higher partial waves in the square well We use the theory developed in section 4.1.4. We need to compute γl = aRl0 (a)/Rl (a) for the wavefunction in the well to perform the matching given in eqn. (4.4.1.6). The radial Schrödinger equation in the well, r < a, is l(l + 1) 00 2 (rRl ) + k − (rRl ) = 0 , (4.1.2.1) r2 and the solution is Rl (r) = A jl (κr) since aRl → 0 as r → 0. Then the matching condition from eqn. (4.4.1.6) is tan δl (k) = γl (ka)jl (ka) − ka jl0 (ka) , γl (ka)nl (ka) − ka n0l (ka) γl = κa jl0 (κa) . jl (κa) (4.1.2.2) As k → 0 have γl → const, jl (ka) ∼ (ka)l , nl (ka) ∼ (ka)−(l+1) and, as before, get δl ∼ (ka)2l+1 . Thus the contribution to the cross-section at low k from the l-th partial wave is sin 2 δl ∼ k 4l . (4.1.2.3) σl ∼ 2 k 4.3 Influence of bound states Solve for the bound states in the square well: A sin κr u(r) = B e−αr r<a r>a α>0 (4.3.1) Find the usual condition −κa cot κa = αa = (γ 2 a2 − κ2 a2 )1/2 with E = − I.e., α2 = γ 2 − κ2 . ~2 α 2 2mV0 , γ2 = . 2m ~2 (4.3.2) 4 QUANTUM SCATTERING IN 3D B B 2 3B 2 29 2B 6a A new bound state appears whenever γa = (n + 12 )π where cot γa = 0, and then for γa = (n + 21 )π ± ε, ε > 0 as = − (1 − γa cot γa) ε→0 a −→ ± ∞ . γa cot γa (i) as large and negative: new bound state not quite appeared; (4.3.3) (ii) as large and positive: new, weakly bound, state has just appeared. σ = 4πa2s (iii) For k such that kas is large must solve for δ0 directly, and from (4.2.12) find tan δ0 → ∓∞ =⇒ δ0 = (n + 12 )π. In both cases get a maximum contribution to σ from l = 0. σ0 = 4π/k 2 . (4.3.4) Now ka ≡ f (k) =⇒ κa cot κa 1 + if (k) . = e2iδ0 = e−2ika 1 − if (k) tan (ka + δ0 ) = S0 (4.3.5) S0 is the diagonal entry in the S-matrix for l = 0. Thus S0 has a pole when f (k) = −i and has the associated zero at f (k) = i. Write k = iα, α > 0, and then value of α at the pole is given by the bound state formula eqn. (4.3.2). As in 1D, corresponding to each bound state, the S-matrix has a pole on the positive imaginary axis and associated zero on the negative imaginary axis for complex k. A simple model for S is iα + k S(k) = = e2iδ0 . (4.3.6) iα − k Then for k → 0 find δ0 ≈ −as k with as = 1/α. If α is small and negative then the well is not quite deep enough for a new bound state to appear (need α > 0 for exponential decay as |x| → ∞) and as is large and negative. If α small and positive then have bound state near top of well and as large and positive; In either case the S-wave cross-section is very big: σ0 ∼ 4πa2s . An example is low energy proton-neutron scattering where the deuteron (n-p) bound state is only lightly bound. 4 QUANTUM SCATTERING IN 3D 4.4 30 Resonance Poles As in the example in 1D scattering, the S-matrix can have poles at complex energies which correspond to quasistable states or resonances. The pole will be at E = E0 − i 12 Γ and the state has the properties: (i) it was prepared by some mechanism at a time far in the past when the particle was localized inside a high potential barrier. In D > 1 this barrier can be created by the angular momentum term in the radial wave equation for l 6= 0, where the effective potential is (see eqn. (4.4.1.4)) W (r) = l(l + 1) r2 } | {z + U (r) . (4.4.1) centrifugal barrier For a nucleus the potential U (r) is short-range and strongly attractive whereas the centrifugal barrier is repulsive. Then W (r) looks like W(r) l(l+1) r2 r (ii) the state is not a stationary state and so the particle tunnels through the barrier and “leaks” out to ∞ as an outgoing spherical wave; (iii) The time-dependence of the state at fixed x, and especially inside the barrier, is 1 e−iE0 t/~ e− 2 Γt/~ (4.4.2) giving a decay-rate Γ for the probability of finding the particle at any fixed x; (iv) These states can also be excited quark-model states in the scattering of two elementary particles or excited atomic states in the scattering of an electron on an atom. Then E − E0 − i 12 Γ S(E) = e2iδ = E − E0 + i 12 Γ (4.4.3) and so (E − E0 )2 − 14 Γ2 cos 2δ = =⇒ (E − E0 )2 + 41 Γ2 Γ 4π Γ2 | sin δ| = p =⇒ σ ≈ 2 k 4(E − E0 )2 + Γ2 4(E − E0 )2 + Γ2 (4.4.4) 4 QUANTUM SCATTERING IN 3D 31 The cross-section has a bump at E = E0 of width Γ/2 much like a damped harmonic oscillator near to the resonant frequency. Note that the solution at resonance is δ(E0 ) = (n + 1/2)π. As E increases, each time E “passes through” a resonance at E = E0 , E1 , E2 , . . ., then n increases by unity. In this case the S-matrix can be modelled as a product of factors of the form shown in (4.4.3). This is how resonances and excited states are detected. * (E) 5B 2 3B 2 B 2 E1 E2 E3 E An argand plot of of the partial wave amplitude fl (E) = eiδ(E) sin δ(E) vs E is a circle of unit radius centre 0.5i. Here is a plot showing recent experimental P-wave (l = 1) data. The first time the curve is on the unitarity circle and is the ∆ resonance of mass 1230MeV. For other trajectories probability has ”leaked” away into undetected channels and so are within the circle but can still pinpoint δ1 = (n + 1/2)π in each case. It is rather messy in real life, though. Im(fl ) E 0.5 0.0 -0.6 * -0.5 4.5 Re(fl ) 0.5 -o.2 ReA o. 2 0.6 rh Fig. 2. Argand plots for P-partial-wave amplitudes threshold (1080MeV) to W * 2.5 GeV. Green’s Functions Use the Green function to analyse scattering processes. Will look for scattering state solutions to the Schrödinger equation ~2 2 − ∇ + V (r) ψ(r) = Eψ(r) . (4.5.1) 2m Then ∇2 + k 2 ψ(r) = U (r)ψ(r) , U (r) = 2m ~2 k 2 V (r) , E = . ~2 2m FJ (4.5.2) 4 QUANTUM SCATTERING IN 3D 32 Define the free or non-interacting Green function G0 by ∇2 + k 2 G0 (k; r, r 0 ) = δ(r − r 0 ) . (4.5.3) By translation invariance we have G0 (k; r, r 0 ) = G0 (k; r − r 0 ) and so can define the Fourier transform by Z G̃0 (k; q) = d3 x e−iq · x G0 (k; x) =⇒ Z d3 q iq · x G0 (k; x) = (4.5.4) e G̃(k, q) . (2π)3 Then eqn. (4.5.3) becomes (−q 2 + k 2 )G̃0 (k; q) = 1 =⇒ 1 G̃0 (k; q) = − 2 , q − k2 (4.5.5) where q 2 = |q|2 . Hence, Z 0 G0 (k; r − r ) = − 0 d3 q eiq · (r − r ) . (2π)3 q2 − k2 (4.5.6) This is a singular integral because of the poles on the real axis at q = k. This problem is resolved by introducing i into the denominator, where > 0 is infinitesimal. This is a choice of boundary condition for the Green function G0 . Remember that the Fourier transform of the Θ (or Heavyside) functions Θ(±t) differ simply by the sign of an in the denominator: Z dω eiωt . Θ(±t) = ± 2π iω ± Then define 0 Z d3 q eiq · (r − r ) (±) 0 G0 (k; r − r ) = − . (4.5.7) (2π)3 q 2 − k 2 ∓ i We are making a choice of how we must complete the contour: whether in UHP or LHP. This is determined by Jordan’s lemma (remember Θ(±t) example). We get that 0 (±) G0 (k; r 1 e±ik|r − r | −r) = − . 4π |r − r 0 | 0 (4.5.8) The ambiguity occurs because we had not specified the boundary conditions G0 must have. For scattering theory we require solutions with outgoing spherical waves (+) and so choose G0 . Check: For R 6= 0 ∇2 ikR eikR 1 d2 eikR 2e = R = − k R R dR2 R R =⇒ (∇2 + k 2 ) eikR = 0. R (4.5.9) 4 QUANTUM SCATTERING IN 3D 33 At R = 0 have the usual result: ∇2 1 = −4πδ(R) . R (4.5.10) eikR = − 4πδ(R) . R (4.5.11) So, putting the two together we get (∇2 + k 2 ) Can also use divergence theorem for sphere of infinitesimal radius about the origin. 4.5.1 Integral equations (+) Using G0 we can replace the Schrödinger p.d.e together with the boundary conditions by a single integral equation which automatically incorporates the boundary conditions. We write eqn. (4.5.2) as Z (+) ik · r d3 r0 G0 (k; r − r 0 )U (r 0 )ψ(k; r 0 ) (4.1.5.1) ψ(k; r) = e| {z } + {z } homog. | particular integral soln. Check: 2 Z 2 (∇ + k )ψ(k; r) = 0 + d3 r0 δ(r − r 0 )U (r 0 )ψ(k; r 0 ) = U (r)ψ(k; r) . QED (4.1.5.2) Now ψ(k; r) takes the form of the required scattering solution, namely ψ(k; r) = eik · r + outgoing waves . (4.1.5.3) To see this consider r → ∞ such that |r| |r 0 |. Can always do this if U (r 0 ) vanishes sufficiently fast as r 0 → ∞. Then we can expand under the integral sign. Have |r − r 0 | = (r2 + r02 − 2 r · r 0 )1/2 ≈ r − r̂ · r 0 . (4.1.5.4) Define k0 = kr̂, and then (+) G0 ≈ − 1 eikr −ik0 · r 0 e , 4π r (4.1.5.5) and so Z 0 0 1 eikr ik · r ψ(k; r) ≈ e − d3 r0 e−ik · r U (r 0 )ψ(k; r 0 ) . 4π r (4.1.5.6) Comparing with earlier definition in eqn. (4.2) we identify the scattering amplitude as Z 0 0 1 0 f (k, k ) = − d3 r0 e−ik · r U (r 0 )ψ(k; r 0 ) , (4.1.5.7) 4π where k · k0 = k 2 cos θ (i.e., k · r = kr cos θ and k0 = kr̂.) 4 QUANTUM SCATTERING IN 3D 4.5.2 34 The Born series Solve the integral equation by iteration Z (+) ik · r ψ(k; r) = e + d3 r0 G0 (k; r − r 0 )U (r 0 )ψ(k; r 0 ) . (4.2.5.1) Define Z φn+1 (r) = (+) d3 r0 G0 (k; r − r 0 )U (r 0 )φn (r 0 ) , with φ0 (r) = eik · r . (4.2.5.2) Then the formal solution is ψ(k; r) = ∞ X φn (r) . (4.2.5.3) n=0 This is the Born series for the wavefunction. Then have Z 0 0 1 X 0 f (k, k ) = − d3 r0 e−ik · r U (r 0 )φn (r 0 ) . 4π n (4.2.5.4) In compressed notation the iterated series has the form Z Z Z Z Z Z ψ = φ0 + G0 U φ0 + G0 U G0 U φ0 + G0 U G0 U G0 U φ0 + Z Z . . . G0 . . . U φ0 + . . . . (4.2.5.5) ... + The integrals are convolutions of the form explicitly shown above (I have dropped the ”+” on G0 for ease of notation). Remark: think of U has containing a small parameter. Then will find φn ∼ O(λn ). The Born approximation consists of keeping just the n = 0 term: Z 0 0 0 1 0 f0 (k, k ) = − d3 r0 e−ik · r U (r 0 ) eik · r 4π 1 m = − Ũ (k0 − k) ≡ − Ṽ (k0 − k) . (4.2.5.6) 2 4π 2π~ This gives the differential scattering cross-section m 2 dσ ≈ |f0 |2 = |Ṽ (k0 − k)|2 . 2 dΩ 2π~ (4.2.5.7) The next approximation is Z Z 0 0 00 1 0 3 0 −ik · r 0 f1 (k, k ) = − dr e U (r ) d3 r00 G0 (k; r 0 − r 00 )U (r 00 )eik · r 4π Z Z 0 0 1 d3 q = − d3 r0 e−i(k − q) · r U (r 0 ) × 3 4π (2π) Z 00 (−1) d3 r00 e−i(q − k) · r U (r 00 ) 2 2 q − k − i Z 1 d3 q (−1) = − Ũ (k0 − q) 2 Ũ (q − k) . (4.2.5.8) 3 4π (2π) q − k 2 − i 4 QUANTUM SCATTERING IN 3D 35 Can be extended to fn (k, k0 ) and can give diagrammatic representation: k' f (k, k0 ) = k'-k q k' k + k'-q q-k q' k' k q k + + ..... (4.2.5.9) p q factor of Ũ (p) , factor of −1 q 2 − k 2 − i R d3 q 1 . for each internal momentum q , and overall factor of − 4π 3 (2π) This is similar to Feynman diagrams for perturbation theory. It is assumed that U is weak in some sense and so the series converges. Example of Born approximation Choose the Yukawa potential V (r) = λ e−µr r 4πλ . + µ2 (4.2.5.10) 4mE (1 − cos θ) , ~2 (4.2.5.11) =⇒ Ṽ (q) = q2 Here 1/µ is the range of the force. We have q = k − k0 =⇒ q 2 = 2k 2 − 2k 2 cos θ = and so Ṽ (q) = 4π~2 λ 4π~2 λ = . ~2 µ2 + 4mE(1 − cos θ) ~2 µ2 + 8mE( sin θ/2)2 (4.2.5.12) Using eqn. (4.2.5.7) we get m 2 16π 2 ~4 λ2 dσ = dΩ 2π~2 [~2 µ2 + 8mE( sin θ/2)2 ]2 4m2 λ2 = . [~2 µ2 + 8mE( sin θ/2)2 ]2 (4.2.5.13) For Rutherford scattering on an electric charge e have µ = 0, and λ = e2 /4πε0 ≡ α~c (α = 1/137 is fine structure constant). So we get 2 dσ λ2 α~c 1 = = . (4.2.5.14) 2 4 dΩ 16E ( sin θ/2) 4E ( sin θ/2)4 This is the classical result but there are now higher orders in λ2 from the Born series. These have higher powers of ~ as well. 5 PERIODIC STRUCTURES 5 36 Periodic Structures Many solids have crystalline structure: atoms are arranged in regular arrays, e.g. N aCl, metals, semiconductors such as silicon and germanium. The physical properties are determined by (i) the underlying periodic structure; (ii) impurities and small deviations from regularity; (iii) the motion of electrons through the crystal, e.g. the ability to conduct; (iv) the vibrations of atoms about their equilibrium positions in the periodic structure. The effects are strongly temperature dependent. An example is the simple cubic lattice in any dimension. In 2D have a2 a 1 Arrays are not cubic in general but there always exists a regularity or periodicity. This means that there is a discrete translation invariance. We can find three, linearly independent, vectors a1 , a2 , a3 such that a displacement r → r + l1 a1 + l2 a2 + l3 a3 , l1 , l2 , l3 integers (li ∈ Z) (5.1) leaves us in the same relation to the crystal as before. The choice of basis vectors ai is not unique: a2 a' 1 a 1 5 PERIODIC STRUCTURES 37 Can use (a1 , a2 ) or (a01 , a2 ). A given choice is a choice of the unit cell for the lattice; the three vectors a1 , a2 , a3 lie along the edges of the unit cell. The lattice Λ is then Λ = {l : l = l1 a1 + l2 a2 + l3 a3 li ∈ Z} , (5.2) and the volume of the unit cell is V = a1 · (a2 ∧ a3 ) (we assume a right-handed basis set). Λ is made from packing together unit cells. Note in this example, that V is independent of the particular basis chosen. Defined in this way Λ is called a Bravais lattice; the Bravais lattice defines the translation symmetry. The unit cell is not unique; indeed some may be larger in volume than others. Hence, the basis vectors chosen are not unique either. There is, however, a particular choice called the primitive unit cell which is the smallest unit of the lattice which, by translation, will tessellate the space. It is a unit cell of the smallest volume. The primitive cell is not unique either but all such cells have the same, and smallest, volume; in the example of the simple cubic lattice above both cells shown, with bases (a1 , a2 ) and (a01 , a2 ), are candidates. We can count the number of lattice sites per unit cell and for the primitive unit cell there is exactly one site per cell. In the face-centred cubic lattice the unit cell with basis (a1 , a2 ) has two sites per cell, each corner site contributes 1/4 plus 1 for the centre site. The primitive cell has basis (ap1 , ap2 ) and has one site per cell. When the basis for a primitive unit cell is used then the position, l, of every lattice site can be written using integers only: l = li ai , li ∈ Z i = 1 . . . D. This is useful. ap 2 ap 1 a2 a 1 In two-dimensions there are 5 distinct Bravais lattices and in three-dimensions there are 14 distinct Bravais lattices. We shall come back to some 3D examples shortly. Consider a general discrete set of lattice sites in RD . The Voronoi tessellation of RD is a set of cells filling the space where all the points in the interior of a given cell are closer to one particular lattice site that any other lattice site. We shall only be interested in sites in D = 1, 2, 3 which form a Bravias lattice but the idea is more general. For a Bravais lattice the Voronoi cells about each point are identical; the Voronoi cells of lattices are also called Wigner-Seitz cells. For a lattice the Voronoi or Wigner-Seitz unit cell, Γ, consists of all those points in RD for which the origin 0 is the nearest point of Λ: Γ = {x : |x − l| > |x|, x ∈ R3 , l ∈ Λ, l 6= 0} . (5.3) 5 PERIODIC STRUCTURES 38 In D = 3 there are 5 distinct Voronoi cells. The restriction is that they are parallelohedra since they tessellate the whole space when placed face to face in parallel displacement. Examples in 2D for cubic and hexagonal lattices are shown below: ' ' ' The Wigner-Seitz cell is a canonical choice for the primitive unit cell. The basis vectors are defined so that translations of Γ by l = li ai , li ∈ Z i = 1 . . . D tessellate the space RD . 5.1 Crystal Structure in 3D There are 14 Bravais lattices in 3D and we shall not detail their properties here but briefly mention some examples. The Bravais lattice is a regular lattice of sites in R3 , but the physical unit that occupies each site in a crystal structure can be complex. For example, there may be one atom or molecule of given species at each site, or there may be two or more atoms forming the unit associated with each site. A crystal structure is a Bravais lattice with a basis of the same physical unit, made of one or more atoms, at each lattice site. The crystal looks the same when viewed from each lattice site. The resulting lattice of atoms might not be a Bravais lattice but, for example, look like two or more interlaced Bravais lattices. The lattices considered here are simple-cubic, face-centred cubic (FCC) and bodycentred cubic (BCC). They are all based on a cubic structure and the latter two with decorations. a Cubic a FCC a BCC The difference in shading does not signify different atoms but is to help guide the eye. 5 PERIODIC STRUCTURES 39 (i) Cubic. The basis vectors of the primitive unit cell are a1 = ai , a2 = aj , a3 = ak , V = a3 . (5.1.1) (ii) FCC. The basis vectors of the primitive unit cell are a1 = a a a (j + k) , a2 = (i + k) , a3 = (i + j) , 2 2 2 V = a3 . 4 (5.1.2) V = a3 . 2 (5.1.3) (iii) BCC. The basis vectors of the primitive unit cell are a1 = a a (i − j + k) , a2 = (i + j + k) , a3 = ak , 2 2 There is one site per unit cell as can be seen from the volume, V , of each cell. Of course, these choices are still not unique. We could use the Voronoi cells instead – a good choice, but they do not look simple. Here is a picture from Wannier of the Voronoi cell for a BCC lattice. Fig. r.3r. Wigner-Seitz ce\Lfor the body-centred cubic lattice: a truncated octahedron. An example of a physical crystal is the diamond lattice which can be thought of as a FCC lattice with two carbon atoms at each site placed at a l and l + d, with d = (i + j + k) , l = li ai . (5.1.4) 4 The physical unit is a pair of atoms with displacement vector d. Alternatively, it looks like two identical interlaced FCC lattices each with one carbon atom per site and displaced by the vector d with respect to each other. This is also the structure of silicon and gallium-arsenide which are two important semi-conductor materials. Salt, N aCl, is another example. This is a FCC lattice with the unit N a − Cl at each site. Again it looks like two interlaced FCC lattices: one for the N a atoms and one for the Cl atoms. X X X X X a2 X a X 1 X The crosses are the sites of the FCC Bravais lattice with primitive basis vectors (a1 , a2 ). 5 PERIODIC STRUCTURES 5.2 40 Discrete Translation Invariance Consider the motion of an electron in a periodic potential; this is a model for the effect of the crystal lattice on the individual electron. Because the potential is periodic we have V (r) = V (r + l) l ∈ Λ . (5.2.1) Hence, if V (r) is known in the unit cell it is known everywhere. The Hamiltonian is, as usual, ~2 2 H = − ∇ + V (r) , (5.2.2) 2m and let ψ(r) be an eigenfunction of H: Hψ(r) = Eψ(r). It follows from periodicity that ψl (r) = ψ(r + l), l ∈ Λ, is also an eigenfunction with the same energy, E. Thus the map Tl , where T l ψl (r) = Tl ψ(r) = ψ(r + l) ψ(r) −→ (5.2.3) maps the eigenspace of H onto itself. Since Tl leaves the normalization and orthogonality of the wavefunctions unchanged then it is a unitary map: Z Z ∗ 3 ψ (r)Tl ψ(r)d x = ψ ∗ (r)ψ(r + l)d3 x Z = ψ ∗ (r − l)ψ(r)d3 x (by shift of origin) Z = [T−l ψ(r)]∗ ψ(r)d3 x =⇒ Tl† = T−l . (5.2.4) But Tl−1 = T−l =⇒ Tl† = Tl−1 . (5.2.5) I.e., Tl is unitary. Also, if l, l0 ∈ Λ then Tl Tl0 = Tl+l0 = Tl0 Tl , (5.2.6) Tl = Tl1 a1 +l2 a2 +l3 a3 = (Ta1 )l1 · (Ta2 )l2 · (Ta3 )l3 . (5.2.7) and hence, Translations form an abelian group with generators Tai , i = 1, 2, 3. Since Tai is unitary there exists an Hermitian operator Pi such that Tai = eiPi . (5.2.8) Tl H ψ(r) = E ψ(r + l) , (5.2.9) H Tl ψ(r) = H ψ(r + l) = E ψ(r + l) , (5.2.10) Now have and also since ψ(r + l) is an eigenstate of H with energy E because of periodicity. Thus [H, Tl ] = 0 , ∀ l∈Λ =⇒ [H, Pi ] = 0 , i = 1, 2, 3 . (5.2.11) 5 PERIODIC STRUCTURES 41 From eqn. (5.2.6) we infer that also [Pi , Pj ] = 0 , i, j = 1, 2, 3. Thus, the Pi are conserved operators, i.e. they correspond to constants of the motion, and from above we have that Tl = ei(l1 P1 + l2 P2 + l3 P3 ) ≡ eil · P . (5.2.12) The eigenvalues of the Pi can be used to classify the eigenstates of H as follows. Let Hψ = Eψ and Pi ψ = k i ψ . (5.2.13) Then the quantum numbers (E, k) classify the states and form part of the complete set. We have Tl ψ(r) = ψ(r + l) = eil · P ψ(r) = eil · k ψ(r) , (5.2.14) with k = (k1 , k2 , k3 ). 5.3 Periodic Functions It is straightforward to construct functions which are periodic on a lattice. By periodic we mean that F (r) is periodic if F (r + l) = F (r) ∀ l ∈ Λ. (5.3.1) A periodic function can be constructed from a general function φ(r), not necessarily periodic, as follows: X X φl (r) , (5.3.2) φ(r − l) = F (r) = l∈Λ l∈Λ where we have introduced the notation φl (r) = φ(r − l). 5.4 Reciprocal Vectors and Reciprocal Lattice The vectors bi reciprocal to the basis vectors ai are defined by 2π (a2 ∧ a3 ) etc , V = a1 · (a2 ∧ a3 ) , V 2π 1 i.e., bi = εijk aj ∧ ak . (5.4.1) V 2 V is the volume of the unit cell of Λ. Although not at all necessary, it is useful to use the basis for the primitive unit cell in what follows. Then b1 = bi · aj = 2πδij . (5.4.2) The basis set {b} is dual to the basis set {a} and allows a simple construction for the inner product even when the basis vectors are not orthonormal although they are, of course, independent. We also have V 1 2π 1 εijk bj ∧ bk = ∗ εijk bj ∧ bk , 2 (2π) 2 V 2 3 (2π) = b1 · (b2 ∧ b3 ) = . V ai = with V ∗ (5.4.3) 5 PERIODIC STRUCTURES 42 The reciprocal lattice is the set of points Λ∗ = n1 b1 + n2 b2 + n3 b3 : ni ∈ Z , (5.4.4) and V ∗ is the volume of the unit cell of Λ∗ . It is interesting to note that if Λ is FCC than Λ∗ is BCC and vice-versa. The construction is useful because it enables the Fourier representation of wavefunctions in a periodic lattice to be clearly and easily formulated. Suppose x = xi ai and q = qi bi , then x · q = 2π xi qi . (5.4.5) In particular if x ∈ Λ and q ∈ Λ∗ then x · q = 2π m , m∈Z. (5.4.6) As an example, consider the sum over Λ: ∆(Q) = X eiQ · a . (5.4.7) a∈Λ We can always write Q = qi bi and with a = li ai , li ∈ Z, we get X X X ∆(Q) = e2πiq1 l1 e2πiq2 l2 e2πiq3 l3 . l1 l2 (5.4.8) l3 Define σ(q) = N/2 X l=−N/2 sin (N + 1)πq e2πiql = . sin πq (5.4.9) Note that there are N + 1 terms where N is taken even, and that σ(q) is real. We will only be interested in the large N limit. σ(n) = N + 1 , width of peak ∼ N −1 But ∆(Q) = σ(q1 )σ(q2 )σ(q3 ) and hence ∆(Q) is sharply peaked for qi = ni , ni ∈ Z; i.e., for Q ∈ Λ∗ , the reciprocal lattice. As N → ∞ the area under the curve of σ(q) for −ε < q < ε , ε 1/N tends to 1. 5 PERIODIC STRUCTURES 43 2g 2g 1 R ∞ sin x dx = 1. To see this write sin πq ∼ πq in the denominator and use π x −∞ Thus, in the limit N → ∞, ε → 0, εN 1 we have X σ(q) = δ(q − n) =⇒ n∈Z ∆(Q) = X δ(q1 − n1 ) or ∆(Q) = V X b∈ δ(q2 − n2 ) X δ(q3 − n3 ) , n3 ∈ Z n2 ∈ Z n1 ∈ Z ∗ X δ(Q − b) . (5.4.10) Λ∗ V ∗ (V ) is volume of unit cell of Λ∗ (Λ), and V V ∗ = (2π)3 . 5.5 Bragg Scattering on a rigid Lattice Scatter beam of particles (typically electrons, neutrons or X-ray photons) on a crystal consisting of atoms sites at points of a lattice Λ. The wavefunction of the incident plane wave is eik · r . For scattering off an atom at r = 0 the scattering wavefunction has the asymptotic form eikr ψ0 (r) ∼ eik · r + f (r̂) , r r→∞. (5.5.1) For scattering of the same plane wave off an atom situated at r = a we refer to an origin at r = a to implement the usual scattering boundary condition. We write the unscattered plane wave as 0 eik · r = eik · a eik · r , r0 = r − a . (5.5.2) So it looks like a plane wave w.r.t. the new origin but shifted by a familiar translation phase factor eik · a . Then as r → ∞ the scattering wavefunction from the site at r = a is ! ikr0 0 0 e ik · a ik · r ψa (r) = e e + f (r̂ ) 0 . (5.5.3) r | {z } correct asymptotic form w.r.t. origin at r = a 5 PERIODIC STRUCTURES 44 (i) The scattered wave takes the standard form relative to an origin at r = a and the initial phase factor translates the wavefunction to coordinates with origin at 0. (ii) The incident wave is the same, independent of the position of the scattering centre. We thus have 0 eikr 0 ik · r ik · a ψa (r) ∼ e + f (r̂ ) e , r0 r0 = r − a , r → ∞ . (5.5.4) Now, as r → ∞ r̂ 0 ∼ r̂ , (5.5.5) 1 ψscatt (r) ∼ f (r̂) eik · a eik(r − r̂ · a) , r (5.5.6) r0 = |r − a| ∼ r − r̂ · a , Thus the outgoing scattered wave is asymptotically where, to leading order in 1/r, have replaced r0 by r in the denominator. We define k0 = kr̂ since the outgoing particle has wavevector of magnitude k and is in the direction r̂. Then we finally get 0 eikr ψa (r) ∼ eik · r + ei(k − k ) · a f (r̂) . r (5.5.7) The total amplitude for scattering is the sum over the lattice of the individual amplitudes (ignoring multiple scattering). Thus, X fΛ (r̂) = eiQ · a f (r̂) , Q = k − k0 , (5.5.8) a∈Λ or fΛ (r̂) = ∆(Q) f (r̂) . (5.5.9) ∆(Q) is very small unless Q ∈ Λ∗ when it is sharply peaked. P Although in the approximation of an infinite exact lattice we have that ∆(Q) = b ∈ Λ∗ δ(Q − b), because of atomic vibrations etc., ∆(Q) is smeared out and is strongly peaked function for Q ∈ Λ∗ . We deduce from this result the Bragg condition which states that scattering occurs only when the momentum transfer, ~Q satisfies ~Q ≡ ~(k − k0 ) = ~b For the beam k = p b ∈ Λ∗ . (5.5.10) 2mE/~2 , |k0 | = k. The Bragg condition then becomes k − k0 = b =⇒ 2k 2 (1 − cos θ) = |b|2 2k sin(θ/2) = |b| , k= 2π . λ =⇒ (5.5.11) Only get scattering at angles θ satisfying this condition. The interpretation is that each b0 ∈ Λ∗ defines a set of parallel planes in Λ: a · b0 = 2πn , a ∈ Λ, n ∈ Z. Each plane, labelled by n, consists of atoms at the sites a. The separation between successive planes is clearly d = 2π/|b0 |. 5 PERIODIC STRUCTURES 45 d 1/2 1/2 x The path difference between beams reflected off successive planes is 2x = 2d sin(θ/2). Constructive interference at the detector implies that 2π 0 2d sin(θ/2) = nλ =⇒ 2k sin(θ/2) = n|b | ≡ |b| , k= (5.5.12) λ Note that b = nb0 ∈ Λ∗ if b0 ∈ Λ∗ . The formula for the differential scattering cross-section from a rigid lattice is then dσ = ∆2 (Q) |f (k, k0 )|2 , dΩ (5.5.13) where I have written f (r̂) = f (k, k0 ) since for an incoming wave with wavevector k then r̂ = k0 /|k| defines scattering direction. This follows the notation from Green function scattering theory. Consider a large but finite lattice consisting of N sites (N large). It is convenient to think of the lattice as having periodic boundary conditions but this is not a necessity. Then we have X 0 eiQ · (a − a ) ∆2 (Q) = a,a0 ∈ Λ = N X eiQ · a = N ∆(Q) , (5.5.14) a∈Λ where the minus sign on a0 in the exponent just means that in the sum over a0 ∈ Λ the lattice sites are counted in the opposite order from that for the sum over a; nevertheless, all terms in the summation are present but have been labelled more usefully. The result follows by translation symmetry where the site at a0 is considered as the origin; the lattice description is independent of choice of origin. Of course, for a small lattice there would be end effects which are, in practice, negligible. Thus have dσ = N ∆(Q) |f (k, k0 )|2 . dΩ (5.5.15) A practical example of a Bragg scattering experiment is where a powder is made of small crystals in which the crystals are randomly oriented. For a given incident 6 PARTICLE IN A ONE-DIMENSIONAL PERIODIC POTENTIAL 46 direction there are always many crystals in exactly the correct orientation for each set of planes to satisfy the Bragg condition. Thus, electrons (or photons as X-rays) scatter strongly and show all possible scattering angles compatible with the Bragg condition. This figure is courtesy of the Materials Science Department in Cambridge: Of course, the scattered beams will lie on cones determined by the Bragg angle, and the intensity will depend on |f (k, k0 )|2 . Typically, the scattering will be stronger for small n since the scattering angle is smaller for a given set of planes. Indeed, since |f (k, k0 )|2 will typically decrease as |k − k0 | increases, we expect that the intensity of scattering will decrease as the scattering angle, θ, increases. In the Born approximation we can derive fΛ (r̂) ≡ fΛ (k, k0 ) in a different way. In this case, for scattering on the atom at the origin with potential U (r) have f (k, k0 ) = − m m 0 hk |U |ki = − Ũ (Q) . 2π~2 2π~2 (5.5.16) The scattering amplitude from the atom at site a with potential Ua (r) = U (r − a) is then Z m m 0 0 3 −ik0 · r U (r − a) eik · r . (5.5.17) fa (k, k ) = − hk |U |ki = − d r e a 2π~2 2π~2 Changing variable to r − a get m iQ · a fa (k, k ) = − e 2π~2 0 Z d3 r eiQ · r U (r) = eiQ · a f (k, k0 ) . (5.5.18) The total scattering amplitude from the crystal is the sum of the individual amplitudes from each scattering site: X fΛ (r̂) ≡ fΛ (k, k0 ) = fa (k, k0 ) = ∆(Q) f (k, k0 ) , (5.5.19) a∈Λ as derived generally earlier, see eqn. (5.5.8). 6 Particle in a One-Dimensional Periodic Potential We apply the ideas and definitions of the previous section to the properties and dynamics of a particle moving in a 1D periodic potential. An important example is that 6 PARTICLE IN A ONE-DIMENSIONAL PERIODIC POTENTIAL 47 of an electron moving in a 1D crystal lattice consisting of an atom at each lattice site. The potential due to the atoms is then periodic and we must solve the associated Schrödinger equation for the system dynamics. There is only one Bravais lattice in 1D Λ = {x = na : n ∈ Z} , (6.1) and it looks like a with lattice spacing a which is also the basis vector for the unit cell. The translation group is simple; Tl ψ(x) = ψ(x + l) l = na, n ∈ Z . (6.2) For a periodic function, V (x), have Tl V (x) = V (x + l) = V (x). Let ψ(k, x) be an eigenfunction of the generator of translations, P , with P ψ(k, x) = k ψ(k, x). Then Tl ψ(k, x) = eikl ψ(k, x) . (6.3) A general periodic function in 1D can be constructed as ∞ X F (x) = φ(x − na) . (6.4) n=−∞ The reciprocal lattice is straightforward Λ∗ = {x = nb : b = 2π/a, n ∈ Z} , (6.5) with basis vector of length b = 2π/a, so that b satisfies the 1D version of eqn. (5.4.2), namely ba = 2π. In what follows we shall always consider the lattice to have N sites but consider the limit N → ∞ with appropriate boundary conditions. For example, periodic boundary conditions would modify the definitions so that integers are restricted to n ∈ ZN . 6.1 Band Structure If an atom is isolated then the electrons are bound to the nucleus and are localized to its neighbourhood. If a number of atoms are brought close enough together the quantum-mechanical state of the system becomes complicated very quickly, but one thing is clear that an electron formerly bound to one nucleus can migrate by tunnelling throughout the system. If the atoms are arranged at the sites of a regular lattice then the electrons formerly associated with particular atoms can tunnel arbitrarily far in the lattice to other atomic sites; they are no longer bound but become delocalized and can propagate through the lattice. The energy levels of the electrons lie in allowed 6 PARTICLE IN A ONE-DIMENSIONAL PERIODIC POTENTIAL 48 bands much like the allowed states that they can occupy in an atom; indeed, we shall see that in the case of the tight-binding model the two systems are closely related. Band structure for electrons in crystals is the basis for the rich electrical phenomena of regular solids; for example, they may be conductors, semi-conductors or insulators. To illustrate the important phenomenon of band structure we consider some model systems. 6.1.1 The Tight Binding Model In this model we assume that the electron is tightly bound to its associated atom when in isolation and that the proximity of atoms on neighbouring sites does not distort the atomic potential very strongly. However, the potential experienced by an electron is now periodic albeit strongly attractive near to each site. Consider a 1D array of N atoms and concentrate on the same atomic level in each atom. Consider an electron in that level for a given atom. Although this electron may initially be thought of as bound to the site, there is a probability that given enough time it will be found near a far-away site. What happens is that the electron tunnels through the barrier between the atoms and can thus migrate arbitrarily far from the original site. A simple picture is one where tunnelling to nearest-neighbour sites dominates all other processes: electron potential a Rn-1 Rn R n+1 n-1 n n+1 -A -A The lattice spacing is a . Consider the n-th atom where the state of the electron is represented by ψn . The Hamiltonian for this atom in isolation is H0 with H0 = − ~2 d2 + U (x − na) , 2m dx2 H0 ψn = E0 ψn , (6.1.1.1) where U (x − na) is the potential due to the atom centred at x = na. The full Hamiltonian is H = H0 + V where X V (x) = U (x − pa), p6=n (ψn , V ψn ) = α , (ψn , V ψm ) = (ψn , V ψn±1 ) = − A , 0 , m 6= n, n ± 1 . (6.1.1.2) 6 PARTICLE IN A ONE-DIMENSIONAL PERIODIC POTENTIAL 49 So H0 is the Hamiltonian for the n-th atom in isolation and ψn is the wavefunction for the chosen level on this atom. The potential V (x) is due to the proximity of atoms on all other sites. Note that ψn (x) = ψ0 (x − na) , (6.1.1.3) i.e., all wavefunctions are related by translation of the origin – it is the same level in each atom. We will also make the approximation that (ψn , ψm ) = δmn . From eqn. (6.1.1.3) this is not strictly true but is correct up to exponentially small corrections. The potential V is due solely to the presence of the neighbouring atoms and the matrix element −A is the tunnelling amplitude between neighbouring sites. We want to solve the full problem: HΨ = EΨ , (6.1.1.4) and we use a variational ansatz and write X cp ψp . Ψ = (6.1.1.5) p This is not the same as expanding on a complete basis since we have ignored other atomic wavefunctions of the atom with energies different from E0 ; these would need to be included in a full basis expansion. Then using (6.1.1.4) X X X cp ψp . (6.1.1.6) cp ψp = E cp ψp = (H0 + V ) H p p p Take the inner product of both sides with ψn and use eqns. (6.1.1.2): X cp (ψn , (H0 + V )ψp ) = Ecn p (E0 + α) cn − A(cn−1 + cn+1 ) = Ecn . | {z } (6.1.1.7) E00 [This kind of equation occurs in the classical dynamics of a set of equally spaced particles in 1D interconnected by springs. The solution to that problem is that waves can travel freely up and down the chain. As a classical model of a 1D crystal such waves are simply sound waves. We shall consider such systems later when discussing lattice vibrations and phonons.] This is true, of course, for all n. We need to solve this difference equation for the {cn } and so try the substitution: ⇒ cn = cn±1 = eikna , e±ika cn , normal mode solution (6.1.1.8) where we remember that a is the spacing and that the n-th atom is situated at xn = na . Substituting in eqn. (6.1.1.7) we get E ≡ E(k) = E00 − 2A cos ka . (6.1.1.9) The N atomic levels which were originally degenerate at E0 are replaced by N new levels which are spread out in energy according to eqn. (6.1.1.9). The old state labelling, n, was by atomic position. The new labelling is by the wavevector, k, which 6 PARTICLE IN A ONE-DIMENSIONAL PERIODIC POTENTIAL 50 contains no reference to position at all: for each value of k there is an eigenstate with energy E(k) . To get the counting of states right we must think about boundary conditions at the ends of the chain of atoms. We impose periodicity and so we must have cN = c0 =⇒ eikN a = 1 . (6.1.1.10) What happens at the ends cannot affect the bulk properties (c.f., the boundary conditions in a box that we used). Define L = N a, the chain length. Taking N even, this implies that k = 2πm/L , m an integer : − N N < m ≤ . 2 2 (6.1.1.11) So k is discrete, taking one of N values which is consistent with starting with having N states in the first place. It is important to recognize that we may write k expanded on the basis vector of the reciprocal lattice b = 2π/a k = m b N m an integer : − N N < m ≤ . 2 2 (6.1.1.12) The important features so far are (i) the eigenstates are no longer degenerate in energy but are labelled by a wavevector k and have energy E(k) which lies in a band of allowed energies determined by eqn. (6.1.1.9) called the dispersion relation. The band has width 4A. This structure will occur for each atomic level. For the lower, more tightly bound, levels A is small and the bands are narrow. The higher bands are wider and may even overlap in energy; (ii) the electron is no longer localized on a given site but has a non-zero probability of being found anywhere, i.e., they are extended: X Pk (x) ∝ |Ψk (x)|2 = | eikna ψn (x) |2 6= 0 ∀ x . n Schematically, the physical outcome of what we have solved is shown in the figure. 6 PARTICLE IN A ONE-DIMENSIONAL PERIODIC POTENTIAL 6.1.2 51 Bloch’s Theorem and Bloch Waves in 1D The stationary state is thus Ψk (x) = C X eikna ψn (x) . (6.2.1.1) n Using eqn. (6.1.1.3), ψn (x) = ψ0 (x − na), consider x → x + a: X X Ψk (x + a) = C eikna ψn (x + a) = C eikna ψn−1 (x) , n = n eika Ψk (x) . (6.2.1.2) Notice we have labelled the state by k, the wave vector. Thus we can always write Ψk (x) = uk (x) eikx , (6.2.1.3) where the function uk (x) is defined by this relation and satisfies uk (x + a) = uk (x) . (6.2.1.4) I.e., uk (x) is periodic on the lattice. That Ψk (x) can always be written in the form in eqn. (6.2.1.3) where uk (x) is a periodic function is the statement in 1D of Bloch’s theorem. The explicit expression for uk (x) can be deduced from eqn. (6.2.1.1) and we find X e−ik(x − na) ψ(x − na) , (6.2.1.5) uk (x) = n which is clearly a periodic function of the form given in eqn. (5.3.2) specialized to 1D. It is, however, also the form of the solution that should be emphasized. Displaying the time-dependence explicitly, the stationary state is Ψk (x, t) = uk (x) ei(kx − ωt) Bloch Wave Solution. (6.2.1.6) I have defined ω = E(k)/~. Bloch’s theorem can be generalized to the 3D case and we shall look at the general derivation later. This solution represents the basic form of the stationary state for all problems involving periodic potentials and especially crystal lattices. It shows that the energy eigenstates are travelling waves which are extended. In particular, the electron can now carry charge from one end to the other in the form of an electric current. 6.1.3 The Floquet Matrix There is another way to think of the wavefunction in a periodic potential in 1D. Suppose ψ1 (x) and ψ2 (x) are two real independent solutions of the Schrödinger equation with energy eigenvalue E. Since a second-order ODE has only two independent solutions then all solutions are expressible as a linear combination of the independent 6 PARTICLE IN A ONE-DIMENSIONAL PERIODIC POTENTIAL 52 ones. For a periodic potential ψ1 (x + a) and ψ2 (x + a) are also solutions and so must have ψ1 (x + a) c11 c12 ψ1 (x) = , (6.3.1.1) ψ2 (x + a) c21 c22 ψ2 (x) | {z } C where the matrix C is a function of E. Since the Wronskian is constant, we choose W (ψ1 , ψ2 ) = 1, we find that |C(E)| = 1. Our 1D example of Bloch’s theorem suggests that we look for solutions with the property that ψ1 (x + a) = λ1 ψ1 (x) , ψ2 (x + a) = λ2 ψ2 (x) , (6.3.1.2) for some λ1 , λ2 which are clearly the eigenvalues of the translation operator Ta . We have argued that we should be able to classify the states by such quantum numbers. It follows that λ1 , λ2 are the eigenvalues of C, the Floquet matrix, and the determinant condition |C| = 1 =⇒ λ1 λ2 = 1; they are the roots of the characteristic quadratic equation λ2 − b(E)λ + 1 = 0 , b(E) = trace(C(E)) . (6.3.1.3) The solutions depend on whether b(E)2 − 4 ≶ 0. (i) b(E)2 < 4 The roots are complex conjugates of unit magnitude. So we can write λ1 = eika , λ2 = e−ika . (6.3.1.4) These solutions correspond to a delocalized state since |ψi (x + na)| = |ψi (x)| n ∈ Z and so the probability of finding the particle in the neighbourhood of any site is the same. This is the Bloch wave solution which is a travelling wave. (ii) b(E)2 > 4 The roots are real and the reciprocals of each other. So we can write ψ1 (x + a) = eµa ψ1 (x) , ψ2 (x + a) = e−µa ψ2 (x) , (6.3.1.5) with µ real. These are not acceptable wavefunctions as they are unbounded in −∞ < x < ∞. Thus we find that a band structure naturally appears and corresponds to allowed energies, E, determined by the condition on the Floquet matrix that trace(C(E))2 < 4. All E for which trace(C(E))2 > 4 are forbidden. (Some small extra amount of analysis is needed for the degenerate case b(E)2 = 4 where λ1 = λ2 = ±1. These correspond to the lower and upper energies of the allowed band.) To determine C(E) in a particular circumstance, the wavefunctions ψi (x + a) must be matched to ψi (x) using continuity conditions appropriate to the potential. 6 PARTICLE IN A ONE-DIMENSIONAL PERIODIC POTENTIAL 6.1.4 53 Brillouin Zones Clearly, the range of k is −π/a < k ≤ π/a . This is called the first Brillouin zone B1 , or often just the Brillouin zone, in k space. Clearly, the wavevector kn = 2nπ/a + k with k ∈ B1 labels the same state as does k; it corresponds to the same energy because of periodicity and also to the same eigenstate of the translation operator. Thus the k-values labelling distinct states may be confined to the first Brillouin zone. It is often convenient to define the n-th Brillouin zone, which can be done in either of the following ways: (1) The Brillouin zone boundaries (in 1D) are at k = nπ/a, n ∈ Z. These boundaries bound the Brillouin zones and, for n > 1, Bn shares a common boundary with Bn±1 . (2) For all k ∈ Bn the origin (k = 0) is the n-th nearest point of the reciprocal lattice. These definitions generalize to higher dimensions. We shall see that all k ∈ Bn can be translated by some b ∈ Λ∗ to lie in B1 . I.e., For k ∈ Bn ∃ b ∈ Λ∗ such that k1 ≡ k − b ∈ B1 . (6.4.1.1) Note that k1 is equivalent to k because they label the same Bloch eigenfunction. The graph of E(k) versus k identifying the 1st and 2nd Brillouin zones is E(k) - B a - 2B a 2nd zone B a 1st Brillouin zone 2B a k 2nd zone Clearly, the k values in any one Brillouin zone are a complete set of allowed quantum numbers labelling the Bloch states whose energy eigenvalues are given as a function of k by the dispersion relation in eqn. (6.1.1.9). It is important to notice that the first Brillouin zone is identical to the Voronoi or Wigner-Seitz cell of the reciprocal lattice. The reciprocal lattice in this case is generated by the basis vector b = 2π/a and the Voronoi cell consists of all points {x : −b/2 < x < b/2}. We shall see that for the generalization to D = 2, 3 this result still holds. Thus the Brillouin zone is another name for the Voronoi cell of the reciprocal lattice. 6.1.5 The Nearly-Free Electron Model We shall consider an electron moving in a weak periodic potential V of period a. When the potential V = 0 the electron is free but we can still ask how we describe its plane 6 PARTICLE IN A ONE-DIMENSIONAL PERIODIC POTENTIAL 54 wave states from the lattice perspective. Then as above Λ = {x : x = na, n ∈ Z} , Λ∗ = {k : k = nb, b = 2π/a, n ∈ Z} . (6.5.1.1) Have p2 ~2 k 2 , ψk (x) = eikx , E(k) = . 2m 2m We can view the plot of E(k) versus k in two ways. H = (6.5.1.2) (i) Extended zone scheme E(k) k - 3B a - B a - 2B a 3rd zone B a 2nd zone 1st Brillouin zone 2B a 2nd zone 3B a 3rd zone Partition the k axis into Brillouin zones but otherwise the plot is standard. The full, and expected, range of k is associated with the set of all Brillouin zones and E(k) is a single-valued function of k. (ii) Reduced zone scheme n E(k) n 2 -2 -1 1 1 -1 0 0 - B a 4th band 3rd band 2nd band 1st band B a 1st Brillouin zone The wavevector k is restricted to the 1st Brillouin zone and E(k) is a multivalued function of k. The different curves are labelled by n ∈ Z with 2 ~2 2πn π π E(k) = k+ − <k≤ . (6.5.1.3) 2m a a a 6 PARTICLE IN A ONE-DIMENSIONAL PERIODIC POTENTIAL 55 Then for each curve labelled by n ψn,k (x) = ei(k + 2πn/a)x = eikx un (x) with u (x) = e2inπx/a . n (6.5.1.4) Here we see the explicit reformulation in terms of Bloch states. In this simple case un is independent of k. The reduced zone scheme generalizes to higher dimensions and general lattices. Here we see that each curve is labelled by a vector from the reciprocal lattice q = nb ≡ 2nπ/a and then Eq (k) = ~2 (k + q)2 . 2m (6.5.1.5) [Note: if we use the band label r = 1, 2, 3, . . . then ψr,k (x) = eikx ur,k (x) with ( e±2ipπx/a k≷0 r = 2p + 1, p = 0, 1, 2, . . . ur,k (x) = (6.5.1.6) ∓2ipπx/a e k≷0 r = 2p, p = 1, 2, . . . ] We now take V to be non-zero but weak. The electron is considered to be ”nearly” free, and we investigate the energy eigenvalues and eigenfunctions of the system but treating V as a periodic perturbation of the Hamiltonian for the free electron and so use a perturbation theory analysis. Consider a large (1D) crystal of N unit cells and with lattice spacing a: N N Λ = l : l = na, n = − + 1 . . . 0 . . . . (6.5.1.7) 2 2 Impose periodic boundary conditions on the wavefunction ψk (x + N a) = ψk (x) , (6.5.1.8) and since ψk (x + l) = eikl ψk (x) we must have k = 2πm/N a, m ∈ Z; the first Brillouin zone is −N/2 < m ≤ N/2. Have H0 = p2 , 2m V (x + na) = V (x) , n ∈ Z . (6.5.1.9) Eigenstates of H0 are denoted by |ki with wavefunction φk (x) = √ 1 ikx ~2 k 2 e , E0 (k) = , 2m Na with 1 hk |ki = Na 0 Z Na 0 dx ei(k − k )x = δk,k0 . (6.5.1.10) (6.5.1.11) 0 For given k the unperturbed energy is E0 (k). The perturbed energy E(k) is given by X hk|V |k 0 ihk 0 |V |ki E(k) = E0 (k) + hk|V |ki + + .... E0 (k) − E0 (k 0 ) k0 6=k (6.5.1.12) 6 PARTICLE IN A ONE-DIMENSIONAL PERIODIC POTENTIAL 56 We need to calculate hk 0 |V |ki. Because V (x) is periodic we have X V (x) = Vq eiqx , V−q = Vq∗ , q ∈ Λ∗ , q 1 = a Vq Z a dx e−iqx V (x) . (6.5.1.13) 0 Then have hk|V |ki = hk 0 |V |ki = = = Z Na 1 dx e−ikx V (x) eikx = V0 , Na 0 Z Na 0 1 dx ei(k − k )x V (x) Na 0 Z Na X 0 1 Vq dx ei(k − k + q)x Na 0 q 0 k0 − k ∈ / Λ∗ 0 Vq k − k = q ∈ Λ∗ . (6.5.1.14) If k 0 − k = q ∈ Λ∗ then also have hk|V |k 0 i = Vq∗ . Thus, explicitly to second order in perturbation theory, have X |Vq |2 + .... (6.5.1.15) E(k) = E0 (k) + hk|V |ki + E (k) − E (k + q) 0 0 ∗ q∈Λ q6=0 When does this perturbation series break down? Look at the second-order term. It breaks down at those values of k for which q ∈ Λ∗ E0 (k) = E0 (k + q) , q 6= 0 . (6.5.1.16) This means that V can cause a transition between the degenerate states |ki and |k+qi. In the extended scheme, appropriate to this calculation, since we are perturbing about the free particle plane wave states, we have the degeneracy points: E(k) k - 3B a - B a - 2B a 3rd zone B a 2nd zone 1st Brillouin zone 2B a 2nd zone 3B a 3rd zone In the reduced scheme the degeneracy points are at the edges and the centre of the Brillouin zone. For this model the degeneracy points are for states |ki and |k 0 i with k = − 21 q , k 0 = 1 q 2 , E = ~2 q 2 , 8m q ∈ Λ∗ . (6.5.1.17) 6 PARTICLE IN A ONE-DIMENSIONAL PERIODIC POTENTIAL 57 Degenerate perturbation theory. To resolve the degeneracy problem we must use degenerate perturbation theory. In the subspace of degenerate states, we must choose that basis for which matrix elements of V are diagonal when restricted to that subspace. In this case, the offending terms in second-order perturbation theory do not appear. Consider the unperturbed state | 21 q + εi, q ∈ Λ∗ , where ε is small. This will mix strongly with | − 12 q + εi since the energies of both states are nearly equal. Write the perturbed state as X0 (6.5.1.18) |ψi = α+ | 12 q + εi + α− |− 12 q + εi + αb | 21 q + b + εi , b∈Λ∗ P where 0 means exclude the two nearly degenerate states from the sum over b ∈ Λ∗ ; i.e., exclude b = 0, −q. Using eqn. (6.5.1.14) we recognize that this expansion contains all the states that can appear since the wavevectors of all the states that mix must differ by b ∈ Λ∗ ; we have simply selected the case where two states are almost degenerate and then highlighted them. The coefficients αb are O(V /∆E) (where ∆E is a non-zero energy denominator) and so, in the spirit of perturbation theory, they are small and calculable using the non-degenerate perturbation theory above. We shall omit them from now on. Denote |+i = | 21 q + εi, |−i = |− 21 q + εi, (0) E± = ~2 q 2 ~q ±ε . 8m 2m (6.5.1.19) Then H|ψi = E|ψi =⇒ (H0 + V ) (α+ |+i + α− |−i) = E (α+ |+i + α− |−i) . (6.5.1.20) h+| (0) = Eα+ (0) = Eα− . E+ α+ + h+|V |+iα+ + h+|V |−iα− E− α− + h−|V |−iα− + h−|V |+iα+ h−| (6.5.1.21) But h+|V |+i = h−|V |−i = V0 , h+|V |−i = Vq , h−|V |+i = Vq∗ . Then (0) E+ | ! + V0 − E Vq α+ (0) ∗ α− Vq E− + V0 − E {z } M (E) (6.5.1.22) = 0. (6.5.1.23) Then E is determined by det(M (E)) = 0. This gives a quadratic equation with roots r 2 (0) (0) (0) (0) E+ + E− ± E+ − E− + 4|Vq |2 E± = + V0 . (6.5.1.24) 2 6 PARTICLE IN A ONE-DIMENSIONAL PERIODIC POTENTIAL 58 Important: (0) E− → E− . (0) E− < E− . As V → 0 : E+ → E+ , For V 6= 0 : E+ > E+ , (0) (0) (6.5.1.25) I.e., the effect of the potential V is that the levels move apart (always the case): + + _ _ E (0) E Consider two cases: (0) (0) (i) E+ − E− 2|Vq | r (0) (0) E+ − E− 2 + 4|Vq |2 = (0) (0) E+ − E− 1 + 2|Vq |2 (0) (0) E+ − E− 2 + . . . , (6.5.1.26) or (0) E± = E± + V0 ± |Vq |2 (0) (0) E+ − E− + .... (6.5.1.27) I.e., just recover non-degenerate perturbation theory result. (0) (0) (ii) E+ − E− 2|Vq |. I.e., ε → 0 corresponding to degeneracy. (0) E± (0) E + E− ≈ + + V0 ± |Vq | . 2 (6.5.1.28) Or ~2 q 2 + V0 ± |Vq | . 8m The two degenerate states split and the energy gap is E± ≈ ∆Eq = E+ − E− = 2|Vq | . (6.5.1.29) (6.5.1.30) A gap opens up at the points of degeneracy. The features to remember are (a) The degenerate states have wavevectors which differ by a reciprocal lattice vector. ( The potential, being periodic, can have non|ki |k + qi, zero matrix elements only between such states. 2 2 ∗ k = (k + q) , q ∈ Λ , The condition for degeneracy. (6.5.1.31) 7 PARTICLE IN A 3-DIMENSIONAL PERIODIC POTENTIAL 59 (b) The consequence of (a) is that degenerate states lie at the edges of the Brillouin zones in the extended scheme, or at the edges and centre of the Brillouin zone in the reduced scheme. E(k) E(k) 4th band 3rd band 2nd band k - 3B a - B a - 2B a 3rd zone 2nd zone B a 1st Brillouin zone 2B a 2nd zone 1st band - 3B a - B a 3rd zone B a 1st Brillouin zone The gap between the n-th and (n + 1)-th band is ∆En with ∆En = 2|Vqn | , 7 qn = 2nπ ∈ Λ∗ . a (6.5.1.32) Particle in a 3-Dimensional Periodic Potential Much of what we have understood about a particle in a 1D periodic potential generalizes to three dimensions. 7.1 Bloch’s Theorem Let Let l ∈ Λ, l = li ai , li ∈ Z i = 1, 2, 3 . k = ki bi , bi basis vectors of Λ∗ i = 1, 2, 3 . (7.1.1) Note that k is a vector in reciprocal space. The energy eigenfunction ψ(x) is labelled by k1 , k2 , k3 or equivalently k. Then ψk (x + l) = eik · l ψk (x) . (7.1.2) Bloch’s theorem states that in a system which is translation invariant under {x → x + l : l ∈ Λ}, the energy eigenfunctions can be written in the form ψk (x) = eik · x uk (x) , where Pi ψk (x) = ki ψk (x) , k = ki bi , and uk (x + l) = uk (x) : uk (x) is a periodic function Proof: (7.1.3) k 7 PARTICLE IN A 3-DIMENSIONAL PERIODIC POTENTIAL 60 We are always free to define uk (x) by and then ψk (x) = eik · x uk (x) , (7.1.4) ψk (x + l) = eik · l eik · x uk (x + l) . (7.1.5) But, from above ψk (x + l) = eik · l ψk (x) = eik · l eik · x uk (x) uk (x + l) = uk (x) . QED 7.2 =⇒ (7.1.6) Ambiguities in k We have seen already that in 1D the distinct values of the wavevector k can be taken to be those in the first Brillouin zone or, equivalently, the Wigner-Seitz or Voronoi cell of the reciprocal lattice. This generalizes to D > 1 as follows. Suppose q ∈ Λ∗ , q = qi bi , qi ∈ Z and let k0 = k + q. Then 0 ψk0 (x + l) = eik · l ψk0 (x) = eik · l eiq · l ψk0 (x) . (7.2.1) But eiq · l = 1 since q · l = 2πm, m ∈ Z. Hence, ψk0 (x + l) = eik · l ψk0 (x) , (7.2.2) and so ψk0 (x) could equally well be labelled by the vector k instead of k0 . It follows that for classification purposes k and k0 are equivalent if k − k0 ∈ Λ∗ . This ambiguity is resolved by restricting k to lie in a representative set, and a standard choice is to restrict k to lie in the Wigner-Seitz, or Voronoi, cell of the reciprocal lattice. For any wavevector k there exists q ∈ Λ∗ such that k0 = k − q lies in the Voronoi cell. k0 and k are equivalent labels (or quantum numbers) for the energy eigenstate. This is the first Brillouin zone and defines the reduced zone scheme. 7.3 The Nearly Free Electron Model in 3D Armed with the definition of the Brillouin zone as the Voronoi or Wigner-Seitz cell of Λ∗ we can easily extend the band structure calculation for the D = 1 nearly free electron model to D = 3. (1) V (x + l) = V (x) , l∈Λ =⇒ V (x) = X Vq eiq · x . (7.3.1) q∈Λ∗ (2) Non-degenerate perturbation theory is adequate unless E(k) = E(k + q) , q ∈ Λ∗ and Vq 6= 0 . (7.3.2) 7 PARTICLE IN A 3-DIMENSIONAL PERIODIC POTENTIAL 61 In the nearly free particle model we get k 2 = |k + q|2 2k · q + q 2 = 0 . =⇒ (7.3.3) In the extended zone scheme the wavevectors k satisfying this condition lie on the boundaries between zones. Indeed, this is the condition that can be used to define the zone boundaries; it is the generalization of condition (2) that we saw in 1D which becomes: For all k ∈ Bn the origin (k = 0) is the n-th nearest point of the reciprocal lattice. In the reduced zone scheme the wavevectors k satisfying this condition lie on the boundary of the zone and at its centre. Can see this in the 1D model. (3) The degeneracy is two-fold on the zone surface but can be, and generally is, higher on zone edges and corners. For N-fold degeneracy the determinantal condition is applied to an N × N matrix and so the details of the energy gap will be different depending on the nature of the degeneracy and dimension, N , of the space of degenerate states. (4) The condition for Bragg scattering from k to k0 is E(k) = E(k0 ) and k0 = k + q , q ∈ Λ∗ . (7.3.4) Hence, the energy gaps occur for just those k which satisfy the Bragg condition and hence for which strong scattering occurs. Electrons with these values of k cannot move as plane waves through the lattice since they are continually rescattering and suffering interference. E.g., in D = 1 the eigenvectors at the degeneracy point are 1 √ | 12 qi ± |− 12 qi , 2 (7.3.5) with eigenfunctions r 2 cos ( 12 qx) , V r 2 sin ( 21 qx) . V I.e., standing waves. These carry no current as they are real: j = 7.4 (7.3.6) ~ Im (ψ ∗ ψ 0 ) m = 0. The Tight-Binding Model in 3D As in 1D we consider electrons tightly bound to atomic centres at the sites l of the lattice Λ. Suppose the Hamiltonian for the electron in an isolated atom at the origin is H0 (x) with potential U (x), and its energy eigenfunction for a chosen state is ψ(x). Have ~2 2 H0 ψ(x) ≡ − ∇ + U (x) ψ(x) = E0 ψ(x) . (7.4.1) 2m The Hamiltonian for the electron in the lattice is then H = H0 + V (x) , (7.4.2) 7 PARTICLE IN A 3-DIMENSIONAL PERIODIC POTENTIAL 62 where V (x) is the potential due to the proximity of atoms at all lattice sites other than the origin l = 0. Of course, I can choose the origin w.l.o.g., but could have also singled out any other site; the full expression for H would be the same: H = − X ~2 2 ∇ + U (x − l) . 2m l∈Λ (7.4.3) The potential here is clearly periodic. We consider the variational choice for the electron wavefunction in the same manner as before X Ψk (x) = eik · l ψl (x) , (7.4.4) l∈Λ where ψl (x) = ψ(x − l). We also assume that the only significant non-zero matrix elements are hψl |V |ψl i = α , hψl |V |ψl+a i = − γa a ∈ nearest neighbours. (7.4.5) In particular we take hψl |ψl0 i = δl,l0 (we could relax this condition but it has no significant effect on the outcome). It is clear from translation invariance that γ−a = γa∗ . Then have X ik · l HΨk (x) = E(k)Ψk (x) = H0 + V (x) e ψl (x) . (7.4.6) l∈Λ W.l.o.g. can take matrix elements by multiplying by ψ0∗ (x) and integrating. We find 1 X ik · a ∗ −ik · a γa e + γa e . E(k) = (E0 + α) − 2 a ∈ n.n. (7.4.7) I have deliberately double counted in the sum (hence factor of 1/2) to show explicitly that the contribution is real (note: a ∈ n.n. ⇒ − a ∈ n.n.). Clearly, as discussed earlier, only wavevectors in the first Brillouin zone are inequivalent since E(k + q) = E(k), q ∈ Λ∗ . The first Brillouin zone is, to reiterate, the Voronoi cell of the reciprocal lattice. (i) The original N degenerate atomic states were localized but the new states Ψk (x) are extended. (ii) Each energy level in the atom gives rise to a band as discussed in 1D. These bands can be labelled by the originating level quantum numbers and can be plotted in the reduced zone scheme just as is done in the nearly free model. As an example, consider a cubic lattice with unit cell of side a. Then by symmetry, for the n-th level, we have γa ≡ γn for a joining nearest neighbours. Take γn to be real and then the dispersion relation for the n-th band is En (k) = En − 2γn ( cos k1 a + cos k2 a + cos k3 a) , with band width 12γn . (7.4.8) 8 ELECTRONIC PROPERTIES 7.4.1 63 Energy Contours The plot of E(k) versus k is generally complicated and is best thought of as surfaces of constant E in 3D k-space, and restricted to the first Brillouin zone in the reduced scheme. As we have seen, this is a Voronoi cell which is can be a complicated shape. Contours of constant energy for a cubic reciprocal lattice and for both tight-binding and nearly free models in 2D are B a B a k2 k2 -B a -B a k1 B a -B a -B a k1 B a Note how the different degeneracies on edges and corners causes diferent energy splitting and so affects the energy values there. 8 Electronic Properties Electrons in the crystal obey the Pauli Exclusion Principle and at T = 0 they fill the lowest available energy eigenstates in the bands up to energy EF , the Fermi energy. All states with E < EF are filled and those with E > EF are empty. We can count the number of states in each band. Consider a finite size crystal with N sites and one atom per unit cell, and so N atoms in total. (a) In the tight-binding model each atomic energy shell corresponds to an energy band in the crystal. The N copies of a given shell combine, because of tunnelling, into a band of states. Suppose there are s energy levels in a shell, then the band will contain 2sN states, where the factor of 2 is for spin degeneracy ↑, ↓. (b) In the nearly free model and for a crystal of length N a in 1D we can impose periodic boundary conditions on the Bloch waves. Then the allowed values of k satisfy 2πm eikN a = 1 =⇒ k = , Na − N N <m≤ . 2 2 (8.1) I.e., there are N distinct Bloch states labelled by m in the band and so 2N states after accounting for spin. For a partially filled band in D > 1 there is an energy surface contour E = EF in the corresponding Brillouin zone. At T = 0 it separates the filled states from the empty states and is called the Fermi surface. 8 ELECTRONIC PROPERTIES B a 64 Fermi surface k2 -B a 8.1 B a k2 -B a -B a B a k1 -B a k1 B a The Effect of Temperature At T = 0 the electrons filling the bands occupy the lowest possible states consistent with the Pauli Principle. The probability that a given state is occupied is thus either 1 or 0: P (E < EF ) = 1, P (E > EF ) = 0. In practice, always have T > 0 and so there is a probability that the energy of an electron can fluctuate by amounts typically of order kB T where kB is Boltzmann’s constant. The filling probability is now changed but, in practice, not by very much. Typical values for the energies concerned are EF ∼ 2−4 eV , kB T ∼ 8.617 10−5 ×300O K ∼ 0.026 eV =⇒ kB T EF . (8.1.1) EF is typical of the energy scales in the isolated atom. We expect the occupation probability as a function of T for reasonable values to look like P(E) T= 0 T> 0 EF E k BT The effect of temperature is exaggerated here so as to display the difference from T = 0 more clearly. Thus for most purposes we can ignore effects due to non-zero T . The functional form for P (E) is the Fermi-Dirac distribution P (E) = 1 1+ e(E−EF )/kB T . (8.1.2) 8 ELECTRONIC PROPERTIES 8.2 65 Insulators and Conductors empty EF full valence band full valence band (i) Insulators contain atoms in which the electrons occupy closed shells. The atoms in an insulator have just the right number of electrons to fill all the shells with no electrons left over. The corresponding bands are completely filled. Typically materials with valency zero or divalent. EF lies in the gap above the highest filled band. The situation can change if bands overlap but always insulator = filled bands. [Note that for T > 0, P (EF ) = 1/2. This condition conversely fixes the value of EF .] INSULATOR EF conduction band partially filled full valence band full valence band (ii) Conductors contain atoms in which the lower-lying energy shells are closed but the last (i.e., highest energy) occupied shell is not closed. Correspondingly, the lowlying bands are full but the top band is only partially filled. Typically these materials are metals in which, for example, the outer shell with l = 0 has room for two electrons of spin ↑, ↓ but is occupied by one electron only. In this case the uppermost, or conduction band is half full and EF is in the middle of the band. Typically valency one. CONDUCTOR The lower bands, which are full, are called valence bands. Why is this? 8.2.1 Dynamics When a particle has plane–wave stationary states it has equal probability of being found anywhere. To localize it we form a wavepacket. Consider superposing two plane waves ∆k and with momenta ~(k ± ∆k) with energy dispersion E(k) = ~ω(k). Let ∆ω = dω dk then the wavefunction is: r 1 (i(k + ∆k)x − i(ω + ∆ω)t) (i(k − ∆k)x − i(ω − ∆ω)t) ψ(x, t) = e + e . 2V (8.1.2.1) The observable quantity is |ψ(x, t)|2 which is clearly given by |ψ(x, t)|2 = 1 2 1 + cos 2(∆k x − ∆ω t) = cos 2 (∆k x − ∆ω t). V V (8.1.2.2) The velocity of this wave–packet is given by following points of constant |ψ(x, t)|2 as t changes. From above this means ∆k x − ∆ω t = constant, i.e., keep the phase 8 ELECTRONIC PROPERTIES 66 constant. Thus the velocity is given by dx ∆ω dω(k) = = . The group velocity. (8.1.2.3) dt ∆k dk The group velocity is the observable velocity of particles. In a band, the group velocity of the electron in state labelled by wavenumber k is 1 dE(k) dω(k) = . (8.1.2.4) v = dk ~ dk This is the velocity at which the charge carried by the electron is transported. Consider an applied electric field E. The force on the electron of charge q is F = qE, and so the change in energy of the electron is v = dE = F dx = qEv dt . (8.1.2.5) Substituting the group velocity for v, we get q dE(k) dE = E dt . ~ dk We can reinterpret this result in terms of a time-dependent k : (8.1.2.6) q dk dE(k) = E . (8.1.2.7) dk ⇒ dt ~ dk This is what we expect classically since the momentum is ~k and d momentum = applied force . dt Since k changes with time the electron must be continually making the transition to states of more positive k (taking E > 0 w.l.o.g.). From above d 1 d dE 1 dE 1 d2 E q d2 E v̇ = k̇ = = k̇ 2 = 2 2 E , (8.1.2.8) ~ dt dk ~ dk dk ~ dk ~ dk dE = using k̇ from eqn. (8.1.2.7). Can write as an effective Newton’s law 2 −1 dE ∗ ∗ 2 , the effective mass. (8.1.2.9) m v̇ = qE , with m = ~ dk 2 Thus we can treat many calculations as classical as long as we use an effective mass, m∗ , for the electron. For any band the major features are: (i) m∗ > 0 at bottom of band and m∗ < 0 at top of band. (ii) Under an applied field the velocity of an electron starting from k = 0, v = 0 increases to its maximum at k = π/2a, and then it decreases again reaching v = 0 at the zone boundary. By periodicity, k re-enters the zone at k = −π/a. In general, symmetry requires v = dE/dk = 0 at the top and bottom of the band. E(k) m* > 0 m* < 0 m* = 4 k B 2a Now consider a simple filling of energy levels in the Brillouin zone in 1D: B a 8 ELECTRONIC PROPERTIES 67 (a) When there is no applied electric field the electrons move equally in all directions and there is no nett transport of charge. (b) When an electric field is applied it accelerates some electrons and decelerates others to generate a flow asymmetry. Electrons must vacate some states to occupy other, originally vacant, states. (c) If the band is full there can be no such asymmetry since all states are occupied at all times. Such a system is in the same state whether or not an electric field is applied and so there is no nett charge transport: the system is an insulator. (d) If a band is only partially filled there are vacant states nearby in energy to which an an electron can move and the state of the system changes when a field is applied (see figure): the system is a conductor. (e) In a perfect crystal there is no steady state for electrons in a constant applied electric field. They migrate through the Brillouin zone and their velocity oscillates giving an oscillating charge flow. However, this effect can only be seen in very large and perfect crystals at low temperature. In a realizable material lattice vibrations and impurities and imperfections in the crystal cause random jumps in k, and a steady state current current arises from detailed balance of accelerating and randomizing forces. 8.3 Semiconductors conduction band empty thermal excitation )E EF valence band full full T=0 full valence band T>0 A semiconductor is an insulator but with a small energy gap between the uppermost filled band and the vacant band above it: ∆E ∼ 0.6 eV. (a) At T = 0 all bands are filled and acts as insulator. 8 ELECTRONIC PROPERTIES 68 (b) For T > 0 significant numbers of electrons are thermally excited into states at bottom of conduction band leaving holes (i.e. vacant states) at top of valence band. The electrons are carriers of negative charge. The holes act as carriers of positive charge since the dynamics of electrons moving in a nearly full band can be described by the dynamics of the holes; absence of a negative charge, negative mass particle behaves like a positive charge, positive mass particle: a hole. Neither band is completely full and both kinds of carrier contribute to conduction. (c) The number of carriers is strongly T -dependent and from the Fermi-Dirac distribution behaves exponentially ∼ e−∆E/kB T . This is why need ∆E small. (d) The number of conduction electrons can be increased by doping with donor impurity atoms which donate electrons to the conduction band by becoming positively ionized. Conversely, the number of conduction holes can be increased by doping with acceptor impurities which capture electrons from the valence band leaving holes behind. These semiconductors are known as n-type and p-type, respectively. 8.4 More on Extended Zone Scheme Most applicable for nearly free model. Consider a 2D example with rectangular reciprocal lattice. The 1st and 2nd Brillouin zones are 2c 2d 1 2a 2b 2b 2a 2c 2d Extended Scheme 8.4.1 Reduced Scheme Overlapping Bands Consider an element with closed shells which should be an insulator. The energy contours for the nearly free model are B A C (i) (ii) 9 ELASTIC VIBRATIONS OF THE LATTICE – PHONONS 69 Consider the respective energies of states EA , EB , EC at points A, B, C, where A and C are in the first zone and B in the second zone. In crossing from A to B there is a discontinuity in energy of an amount ∆E = EB − EA > 0. There are now two possibilities regarding EB and EC : (i) EB > EC . All energies in the first zone are lower than any energy in the second zone and so there are just enough electrons to completely fill all the states in the first zone and the corresponding band is full; all higher ones are empty. The material is indeed an insulator. This is generally true when ∆E is large. (ii) EB < EC . The lowest energy state in the second zone, EB lies below the top of the highest of the first zone, EC : the bands overlap. The filling is now different and is shown in the figure above. It is clear that the lower band is not full and there are electrons in the upper band. The material is a conductor albeit probably a poor one. This occurs generally when ∆E is small. 9 Elastic Vibrations of the Lattice – Phonons Consider simple 1D picture of a lattice in the approximation where inter-atomic forces are harmonic; i.e., small displacements from equilibrium. a rn In equilibrium rn = na. Let displacement from equilibrium be xn and use periodic boundary conditions: xn = rn − na, xn+N = xn . Then T = X1 n 2 mẋ2n , V = X1 n 2 λ(xn − xn−1 )2 , pn = mẋn . (9.1) The classical Hamiltonian is X p2 1 n + λ(xn − xn−1 )2 , 2m 2 n (9.2) mẍn = −λ (2xn − xn+1 − xn−1 ) . (9.3) H= and the equations of motion are Discrete version of vibrating string with periodic boundary conditions, and similar to tight binding model. Choose N even (can also choose N odd, of course) and look for solutions of the form 1 X 2πl N N xn = √ αl (t)eikl na , kl = , − <l≤ . (9.4) Na 2 2 N l 9 ELASTIC VIBRATIONS OF THE LATTICE – PHONONS 70 Then mα̈l = −λ 2 − eikl a − e−ikl a αl ⇒ kl a 2 αl ⇒ mα̈l = −4λ sin 2 r λ kl a ±iω t l αl (t) = αl (0)e where ωl = 2 sin . m 2 (9.5) For large N we can consider k as continuous and write the dispersion relation as r ka λ ω(k) = 2 sin , (9.6) m 2 with, as is usual, k restricted to the 1st Brillouin zone, B1 ; it is ambiguous up to k → k + q, q ∈ Λ∗ . Restriction to B1 means equivalently −N/2 < l ≤ N/2 (N even). T(k) k B a B a r dw λ =a . For small k these are sound waves with velocity c = dk k=0 m 9.1 Quantization of Harmonic Vibrations Because xn (t) is real we can write the solution in the form i 1 X h −i(ωl t − kl na) † i(ωl t − kl na) √ xn (t) = X0 (t) + αl e + αl e , N l6=0 (9.1.1) with pn (t) = mẋn (t), and total momentum P0 = N mẊ0 . Note that the contribution when l = 0 corresponds to X0 (t), the centre-of-mass (CoM) coordinate. We quantize in the Heisenberg picture (operators carry the time dependence and states are timeindependent) by imposing the equal-time canonical commutation relations [xn (t), pn0 (t)] = i~δn,n0 . (9.1.2) Note: in the Schrödinger picture x and p are time-independent and we quantize using [x, p] = i~. To convert to the Heisenberg picture we have x(t) = eiHt/~ x e−iHt/~ , p(t) = eiHt/~ p e−iHt/~ , (9.1.3) 9 ELASTIC VIBRATIONS OF THE LATTICE – PHONONS 71 which leads directly to the equal-time commutation relations used here. Consider l 6= 0. Then can write 1 X 1 X † −i(kl0 + kl )na 0 − kl )na −ik na i(k l l xn (0)e = αl0 e + αl0 e N n N n,l0 6=0 = αl + αl†0 kl0 + kl ∈ Λ∗ , kl0 ∈ B1 . (9.1.4) (9.1.5) This means choose l0 such that l0 + l = 0 (mod N ) and kl0 ∈ B1 . (i.e. − N/2 < l0 ≤ N/2 ). We find from the dispersion relation and l0 + l = 0 (mod N ) that ωl0 = ωl or equivalently ω(kl0 ) = ω(kl ) . (9.1.6) Now also pn (t) = i 1 1 Xh αl ωl e−i(ωl t − kl na) − αl† ωl ei(ωl t − kl na) . P0 (t) − im √ N N l6=0 (9.1.7) By the same method as above we find, for l 6= 0, 1 X pn (0)e−ikl na = −imωl (αl − αl†0 ) N n l0 + l = 0 (mod N ), kl0 ∈ B1 , (9.1.8) where we have used ωl0 = ωl . Thus, we find X 1 e−ikl na (mωl xn (0) + ipn (0)) , 2mωl N n X 1 = eikl na (mωl xn (0) − ipn (0)) . 2mωl N n αl = αl† (9.1.9) The second follows as hermitian conjugate of the first. From the canonical commutation relations (9.1.2) evaluated at t = 0 we then find the commutation relations for αl and αl† to be ~ [αl , αl0 ] = [αl† , αl†0 ] = 0 , [αl , αl†0 ] = δl,l0 . (9.1.10) 2mωl From (9.1.1) and (9.1.7) we have X0 (t) = 1 X xn (t) , N n P0 (t) = X pn (t) , (9.1.11) n and using the commutation relations (9.1.2) we find [X0 (t), P0 (t)] = i~. r ~ Writing αl = al and similarly for αl† , we get 2mωl r h i X ~ al e−i(ωl t − kl na) + a†l ei(ωl t − kl na) , xn (t) = X0 (t) + 2N mωl l (9.1.12) 9 ELASTIC VIBRATIONS OF THE LATTICE – PHONONS 72 with [al , a†l0 ] = δl,l0 , [al , al0 ] = [a†l , a†l0 ] = 0 , pn = mẋn . (9.1.13) These are now harmonic oscillator annihilation and creation operators for independent oscillators labelled by l and with frequency ωl . Need to find H. Can substitute directly in expression but it is tedious algebra. Instead, from the Heisenberg equations of motion we have i dxn = [H, xn ] ∀ n. dt ~ (9.1.14) Substitute for xn (t) and get − iωl al = i [H, al ] =⇒ [H, al ] = −~ωl al , [H, a†l ] = ~ωl a†l . ~ (9.1.15) By susbstition in eqn. (9.2) H is clearly at most quadratic in al and a†l and so, using the commutation relations eqn. (9.1.13), we find the solution gives the quantum Hamiltonian: X † P02 H = + al al + cl ~ωl , (9.1.16) 2M l6=0 where the cl are constants, M = N m the total mass of the lattice, and P0 = M Ẋ0 the total momentum of the lattice. The first term does come from direct substitution for xn in eqn. (9.1.13) and is the CoM energy. We shall consider only eigenstates with P0 = 0 (note: in the limit M → ∞ the crystal has E = P02 /2M → 0). This is a sum over the N − 1 independent oscillators labelled by (non-zero) kl , ωl plus free motion of CoM; their excitations describe collective motions of the atoms which are travelling waves. From direct substitution we find cl = 1/2 as we would expect. However, this gives an infinite (zero-point energy) contribution. We are free to define the origin of energy such that the ground state has energy zero and this choice is cl = 0; in general only energy differences matter but sometimes must be careful with this approach (e.g., Casimir effect). The ground state |0i is defined by al |0i = 0 ∀ l, P0 |0i = 0 . (9.1.17) A general state is Y (a† )nl √l |0i . |ψi = nl ! l6=0 Here, the l-th oscillator is excited to the nl -th excited state. With H = (9.1.18) X ~ωl a†l al l6=0 (ensuring H|0i = 0) we find that the energy of state |ψi is X Eψ = ~ nl ωl . (9.1.19) l6=0 We see that a†l creates a phonon of energy ~ωl with wavevector kl ; the state it creates is a travelling wave labelled with this wave vector. In |ψi there are nl phonons each of energy ~ωl for each l. 9 ELASTIC VIBRATIONS OF THE LATTICE – PHONONS 9.2 73 Unequal Masses a r2n r2n+1 + - m M The mass on even sites is m, charge +, and on odd sites is M , charge −, with M > m. I.e., a unit cell with spacing 2a and the basis of two atoms of masses m and M associated with each site (c.f., N aCl). Take N atoms of each species with periodic boundary conditions. The equations of motion are mẍ2n = −λ(2x2n − x2n+1 − x2n−1 ) M ẍ2n+1 = −λ(2x2n+1 − x2n − x2n+2 ) . (9.2.1) Let 1 x2n (t) = α e−iωt √ e2ikna , N and define the vector displacement 1 x2n+1 = β e−iωt √ e2ikna , N X n (t) = Then find (9.2.2) x2n (t) . x2n+1 (t) (9.2.3) α m 0 α −λ 1 + e−2ika = ω2 . −λ 1 + e2ika 2λ β 0 M β 2λ (9.2.4) The allowed frequencies are determined by setting the appropriate determinant to zero and we find ω12 1/2 λ (M + m) ± (M − m)2 + 4M m cos 2 ka . (9.2.5) = mM 2 ω2 For N → ∞ have that k is continuous with −π/2a < k ≤ π/2a. We see that there are two branches ω1 and ω2 with ω1 > ω2 . The dispersion curves are T(k) ω1 is the optical branch T 1 ω2 is the acoustic branch There are two bands with band gap (evaluated at k = π/2a) √ 1 1 ∆E = ~ 2λ √ − √ . (9.2.6) m M )E T 2 k B 2a B 2a 9 ELASTIC VIBRATIONS OF THE LATTICE – PHONONS 74 In fact this is the reduced zone scheme for the original unit cell with spacing a. With m = M we have no gap and we can draw the extended and reduced schemes: B a B a B 2a B 2a For M 6= m a gap opens up at the zone boundary in the reduced scheme. Denote as es (k) the eigenvectors associated with the eigenfrequencies ωs (k), s = 1, 2. We may impose the orthonormality condition 0 † m es (k) es0 (k) = δs,s0 . (9.2.7) 0 M | {z } M See this by noting that equation (9.2.4) is of the form Ωes = ωs2 M es where Ω is Hermitian and M is diagonal. Then use the usual method to show the eigenvectors obey this orthogonality rule. The es (k) are called the polarization vectors. We can write the classical Fourier expansion 1 X X n (t) = √ αs (k)es (k) e−i(ωs (k)t − 2kna) + hermitian conjugate , N k,s π π < k ≤ , s = 1, 2 . − (9.2.8) 2a 2a To quantize we impose, as usual, the canonical commutation relations. We have p2n = mẋ2n , p2n+1 = M ẋ2n+1 and then [x2n (t), p2n0 (t)] = i~δn,n0 [x2n+1 (t), p2n0 +1 (t)] = i~δn,n0 . (9.2.9) This leads to the quantized displacement s i h X ~ as (k)es (k) e−i(ωs (k)t − 2kna) + a†s (k)e∗s (k) ei(ωs (k)t − 2kna) , X n (t) = 2N ωs (k) k,s (9.2.10) where we have the harmonic oscillator commutation relations [as (k), a†s0 (k 0 )] = δs,s0 δk,k0 , [a†s (k), a†s0 (k 0 )] = [as (k), as0 (k 0 )] = 0 . (9.2.11) 9 ELASTIC VIBRATIONS OF THE LATTICE – PHONONS 75 [To see this first note that P n (t) = M Ẋ n (t). Then use Fourier analysis to show that s ! N X 1 as (k) = es (k)† · e−2ikna [ωs (k) M X n (0) + iP n (0)] , 2N ~ωs (k) n=1 s ! N X 1 a†s (k) = e2ikna ωs (k) X Tn (0)M − iP Tn (0) · es (k) . 2N ~ωs (k) n=1 (9.2.12) The commutation relations for as (k) and a†s (k) then follow directly from eqn. (9.2.9). Best to use index notation for the matrix algebra to implement these.] We now have two kinds of creation (annihilation) operators: one set that creates acoustic phonons and the other that creates optical phonons. The difference between the two kinds is best understood by looking at small k. In this case we find that M 1 e1 ∝ , e2 ∝ . (9.2.13) −m 1 For the optical phonons the neighbouring particles are moving in opposite directions and, since they are oppositely charged, this generates an electric dipole of frequency ω1 (k) which can radiate or absorb energy from EM radiation at these frequencies, e.g. infrared. For the acoustic branch the particles move in the same direction and the dipole is negligible at low frequencies. These phonons are sound waves where the velocity of sound is r r 2λ 2λa2 dω2 (k) ka =⇒ c = . (9.2.14) , ω ∼ c = 2 dk k=0 M +m M +m 9.3 Debye-Waller Factor in 1D There are only two Bragg scattering directions in 1D: forward (θ = 0) and backward (θ = π). Consider backward scattering: a -k k Before, we assumed that the atoms were rigidly fixed at the lattice sites, but now we must allow for their position to fluctuate and we assume that the harmonic approximation holds. The result for the Bragg scattering amplitude for a rigid crystal from earlier was X fΛ (π) = ∆(Q)f (θ = π) , with ∆(Q) = eiQna , Q = 2k . (9.3.1) n 9 ELASTIC VIBRATIONS OF THE LATTICE – PHONONS 76 The factor now becomes ∆DW (Q) = X eiQ rn , with rn = na + xn (t) . (9.3.2) n This is a QM operator and so must define the initial and final states of the crystal phonons as well. We take both the inital and final states to be the ground state for all oscillators; an approximation valid for low-T for which the contributions of phonon excitations are not significant. The result is independent of time and is also translation invariant and so we can evaluate at t = 0 and site position n = 0. We also take h0|X0 |0i = 0; the crystal CoM is at the origin. Then X (9.3.3) ∆DW (Q) = eiQna h0| eiQxn (0) |0i = ∆(Q)h0| eiQx0 (0) |0i . {z } | n independent of n Using the expression for the displacement field xn (t = 0) from section 9.1 eiQx0 (0) = exp(iA + iA† ) , A = Q X l6=0 Now r ~ al , 2N mωl A† = Q X r l6=0 ~ a† . 2N mωl l † 1 † † e(iA + iA ) = eiA eiA e 2 [A , A] . (9.3.4) (9.3.5) This is the Baker-Campbell-Hausdorf formula for the case where the commutator commutes, itself, with the operators involved. We have X Q2 ~ 1 X Q2 ~ 1 † † 0 [A , A] = [a , al ] = − . √ 2 2 l,l0 6==0 2N m ωl ωl0 l 4N mωl l6=0 (9.3.6) Thus h0| eiQx0 (0) |0i = e−W (Q) h0| exp(iA† ) exp(iA) |0i , X Q2 ~ . W (Q) = 4N mωl l6=0 (9.3.7) This method of organizing all creation operators to the left of all annihilation operators is called normal ordering. By expanding the exponentials we see that the only surviving term is the unit operator; all others annihilate the ground state. The result is that the Bragg scattering amplitude is modified by the Debye-Waller factor: fΛ (π) −→ (Q) e|−W {z } fΛ (π) . DW factor (9.3.8) Exponential suppression of the amplitude and hence the intensity. Can go on to study phonon excitation in the scattering. In this case the final state would be one with phonon excitations. This is a model for radiation where here the radiation is a phonon beam. 10 QUANTUM PARTICLE IN A MAGNETIC FIELD 10 77 Quantum Particle in a Magnetic Field First review the classical dynamics of a charged particle. 10.1 Classical Hamiltonian The vacuum Maxwell equations are ∇·E = 0, ∇·B = 0, ∇ ∧ E = −Ḃ , 1 ∇ ∧ B = 2 Ė . c (10.1.1) Here “dot” means time derivative. Introduce scalar potential φ and vector potential A so that B = ∇∧A, E = −∇φ − Ȧ . (10.1.2) The 2nd and 3rd Maxwell equations are automatically satisfied and the 1st and 4th become wave equations for φ and A. There is a gauge transformation which leaves E and B unchanged: A → Ã = A + ∇f , φ → φ̃ = φ − f˙ . (10.1.3) Scalar and vector potentials related by gauge transformations describe the same physics and are equivalent. The classical Hamiltonian (i.e., the energy function) for a particle of charge −e in the background of these fields is H = 1 (p + eA) · (p + eA) − eφ , 2m (10.1.4) where A and φ are evaluated at the position x of the particle. Hamilton’s equations tell us how to derive the equations of motion (IB Variational Principles, IIC Classical Dynamics): ∂H ∂H ẋi = , ṗi = − . (10.1.5) ∂pi ∂xi Then get ẋi = ṗi = ⇒ mẍi = = = ⇒ mẍ = 1 (pi + eAi ) , m 1 − (pj + eAj )e∂i Aj + e∂i φ , m ṗi + eȦi + eẋj ∂j Ai [recall A = A(x(t), t)] −eẋj ∂i Aj + e∂i φ + eȦi + eẋj ∂j Ai −eẋj (∂i Aj − ∂j Ai ) − eEi −e(E + ẋ ∧ B) , the Lorentz force law . (10.1.6) 10 QUANTUM PARTICLE IN A MAGNETIC FIELD 78 In a constant uniform magnetic field B with E = 0 the simple motion is in circular orbits orthogonal to B with a drift parallel to B: a helical path. Choosing B = B k̂, B > 0, we substitute the trial solution x = a cos ωB t , a sin ωB t , bt (10.1.7) into the Lorentz law and find that a and b are arbitrary constants of the motion and ωB = eB , m the cyclotron frequency . (10.1.8) The orbit is anti-clockwise about the positive k̂ axis, B where b > 0 has been chosen. 10.2 Quantum Hamiltonian Replace p → −i~∇ and then for a spinless particle the Hamiltonian is derived from the classical one above: H0 = 1 (−i~∇ + eA)2 − eφ , 2m (10.2.1) and acts on the wavefunction Ψ(x, t) where A, φ generally depend on x, t. For a particle of spin 21 there is a spin operator where σ1 = 0 1 , 1 0 s = 1 ~σ , 2 σ2 = 0 −i i 0 (10.2.2) , σ3 = 1 0 , 0 −1 (10.2.3) are the Pauli matrices. There is then an additional contribution to the Hamiltonian Hs = − µB · s ≡ − µ~ B·σ , 2 (10.2.4) coupling the magnetic moment µ = µs directly to B. The full Hamiltonian is then H = H0 I + Hs , (10.2.5) 10 QUANTUM PARTICLE IN A MAGNETIC FIELD 79 ψ1 where I is the unit 2×2 matrix. The wavefunction now has two components Ψ = . ψ2 Unless B points in a uniform direction, this is a genuine matrix Hamiltonian. The Schrödinger equation is i~ ∂ 1 Ψ = (−i~∇ + eA)2 Ψ − eφ Ψ − µB · s Ψ . ∂t 2m Can rewrite as 2 ie ~2 ie ∂ − φ Ψ = − ∇ + A Ψ − µB · s Ψ . i~ ∂t ~ 2m ~ (10.2.6) (10.2.7) Under a gauge transformation A and φ are replaced by Ã = A + ∇f and φ̃ = φ − f˙ . (10.2.8) If we also replace Ψ by Ψ̃ = e−ief /~ Ψ, then Ψ̃ obeys the transformed equation: 2 ∂ ~2 ie ie i~ − φ̃ Ψ̃ = − ∇ + Ã Ψ̃ − µB · s Ψ̃ . (10.2.9) ∂t ~ 2m ~ I.e., the same Schrödinger equation as eqn. (10.2.6) but in terms of the transformed fields; the equation is form invariant under gauge transformations. We also say that the equation is gauge invariant or, alternatively, gauge covariant. (i) Ψ̃ is the gauge transform of Ψ. ∂ ie (ii) ∇ + ie ~ A and ∂t − ~ φ are called gauge covariant derivatives. These are the only operators that have a physical meaning; p = −i~∇ has no simple properties under a gauge transformation and is now only one part of the momentum operator which is −i~(∇+ ie~ A). The meaning is that the total momentum is now stored in the mechanical motion of the particle and in the electromagnetic field itself; the EM field carries momentum e.g., in a laser beam. A gauge transformation changes the partition between the mechanical and EM field parts. From now on just consider time-independent fields B and E =⇒ A(x) and φ(x) are independent of time. 10.2.1 Approximate Hamiltonian Useful for external field B acting over a small region such as an atom. We shall assume B is small in some sense and slowly varying on the atomic scale. Of course, |B| → 0 as |x| → ∞ but on a scale much larger than the atomic system. Ignore spin for the moment. Now 2 ie ie e2 ∇ + A ψ = ∇2 ψ + (∇ · (Aψ) + A · ∇ψ) − 2 |A|2 ψ . (10.1.2.1) ~ ~ ~ 10 QUANTUM PARTICLE IN A MAGNETIC FIELD 80 We can choose a gauge where ∇ · A = 0 (Coulomb gauge) and A vanishing at origin (subtract a constant). For constant and uniform magnetic field B = (0, 0, B) then find a solution 1 1 A = B ∧ x = B(−y, x, 0) . (10.1.2.2) 2 2 The statement that B is small on the scale of the system, R, means e |B|R2 1 ⇒ 1.6 × 1015 R2 |B| 1, ~ (10.1.2.3) where B is in Tesla and R in metres. For |B| small in this sense we can then ignore |A|2 term in eqn. (10.1.2.1); keep only the terms linear in A in this case. So, in Coulomb gauge, we then get (constant uniform B) ie ∇+ A ~ 2 ψ ≈ ∇2 ψ + 2ie A · ∇ψ . ~ (10.1.2.4) Then (ignoring spin for the moment) ~2 2 ∇ − eφ + 2m ~2 2 = − ∇ − eφ + 2m ~2 2 = − ∇ − eφ + 2m H ≈ − e (B ∧ x) · p 2m e B · L (L = x ∧ p is orbital angular momentum) 2m e BL3 . (10.1.2.5) 2m Here −eφ is the usual potential V , and if V is spherically symmetric the bound states are labelled by n, the radial quantum number, and quantum numbers l, l3 corresponding to eigenvalues ~2 l(l + 1) for L2 , and ~l3 for L3 with −l ≤ l3 ≤ l. e BL3 in 1st order perturbation theory and then states acquire We can treat the term 2m an additional energy of e~ EB = B l3 . (10.1.2.6) |2m {z } ∆EB This gives the Zeeman splitting of atomic levels: l3 2 1 ) EB 0 -1 -2 l=0 l=1 l=2 10 QUANTUM PARTICLE IN A MAGNETIC FIELD 10.3 81 Charged Particle in Uniform Magnetic Field Study the Landau levels for a free particle of charge −e, mass m, in uniform field B = (0, 0, B), B > 0 and with potential φ = 0. Use exact Schrödinger equation and ignore spin. 2 ~2 ie − ∇ + A Ψ = EΨ . (10.3.1) 2m ~ We can choose the gauge to simplify the algebra; any choice of gauge must give the same answer to a physical question. Choose A = (−By, 0, 0), and equation becomes − (∂x − iby)2 + ∂y2 + ∂z2 Ψ = εΨ , Let Ψ(x, y, z) = ψ(x, y)eikz z , b= eB 2mE , ε= . ~ ~2 ε = kz2 + ε⊥ . (10.3.2) (10.3.3) This works because the z-dependence is unaffected by B (c.f., the classical solution). Then − (∂x − iby)2 + ∂y2 ψ = ε⊥ ψ . (10.3.4) The solution is of the form ψ(x, y) = eikx χ(y) where −∂y2 χ + (k − by)2 χ = ε⊥ χ . (10.3.5) This is a standard harmonic oscillator centred at y0 = k/b with frequency b. The wavefunctions and energy levels are √ 2 1 b (y − y0 ) , ε⊥ = (2n + 1)b , n = 0, 1, 2, . . . (10.3.6) χn = e− 2 b(y − y0 ) Hn where Hn is the nth Hermite polynomial. ε⊥ is called the energy of the n-th Landau level. In practice, particles are often constrained to move in the 2D plane orthogonal to B and there is no z-motion, and so we take ∂Ψ/∂z = 0. Then ε = (2n + 1)b. The physical energies in this case are En = (2n + 1) ~eB 1 = (n + )~ωB , 2m 2 (10.3.7) where ωB = eB/m, the classical cyclotron frequency. The z-motion adds ~2 kz2 /2m. We will concentrate on the 2D picture from now on. Note, the wavefunction will not be gauge invariant. We could have chosen different gauges, e.g. 1 A = (0, Bx, 0) [x ↔ −y] , A = B(−y, x, 0) . (10.3.8) 2 Using any gauge will give an equivalent set of wavefunctions. A given energy level is infinitely degenerate and so the relationship between gauge-equivalent (degenerate) sets is not trivial but it will always be possible to expand a wavefunction of one set as a linear combination of those in another. 10 QUANTUM PARTICLE IN A MAGNETIC FIELD 10.4 82 Degeneracy of Landau Levels Each Landau level is infinitely degenerate as k, and hence y0 = k/b, takes any value. We can make a realistic estimate of the degeneracy for particles restricted to area A = Lx Ly using periodic boundary conditions. Then k takes quantized values with eikLx = 1 , ⇒ k= 2πn , Lx n∈Z. (10.4.1) The spacing of k is 2π/Lx . The centre y0 must be in the range 0 ≤ y0 ≤ Ly and so, since y0 = k/b, we also find 0 ≤ k ≤ bLy . Thus, the number of allowed k values is Interpretation: (10.4.2) bLx Ly eAB = ; this is the degeneracy. 2π 2π~ • AB is the total magnetic flux through A; • 2π~ is a unit of magnetic flux for particles of charge ±e; e • Degeneracy of Landau level N0 = AB = Number of flux quanta passing through A. 2π~/e • Whilst the degenercy of LL is a gauge invariant quantity, the description of the individual LL does depend on the choice of gauge; physical observables/consequences are always gauge invariant. 10.5 Effects of Spin e~B In a uniform field B an electron acquires an additional energy ± for spin ± 21 , 2m respectively. The Landau levels now have energies En = n~ωB , n = 0, 1, 2, . . .. All levels have spin degeneracy 2 except n = 0 which is spin − 12 only: E n spin degeneracy 5 2 4 2 3 2 2 2 1 2 0 1 spin down only The Lowest Landau Level (LLL) has exactly zero energy for ideal 2D Hamiltonian. 10 QUANTUM PARTICLE IN A MAGNETIC FIELD 10.6 83 Level Filling Effects A physical 2D system has a large number, N , of approximately non-interacting electrons. At low T the Fermi-Dirac distribution (or Pauli Principle) implies they fill the lowest distinct one-particle states. Just as in band theory the physics of the system is completely different when all Landau levels are completely filled from the case when there is a partially filled level. The useful feature is that the filling of levels is adjusted by varying B, which is much easier than trying to vary N . Of course the energies will vary as well. • At some strong field BL the lowest level, n = 0, is exactly filled and the rest are empty: N = N0 . • At field strength 1 B 2p+1 L the n = 0, 1, . . . , p levels are exactly filled: N = (2p + 1)N0 . Many observables such as conductivity, magnetic moment etc., change markedly when plotted against B. In particular, the Quantum Hall EFfect. 10.7 Lowest Landau Level and Complex Variables This is best seen generally in the gauge A = 12 B(−y, x, 0) but the analysis is involved. So consider the states of the LLL which are of the form 2 0 1 b(y − k/b) − ikx ψ(x, y; k) = e e 2 H0 trivial 1 2 1 0 by − ikx ky ∝ e 2 e e 1 2 1 0 by − ik(x − iy) (10.7.1) ∝ e 2 e . 1 Can take linear combinations of Fourier form Z 2 1 dk ˜ 0 − by Φ(x, y) = f (k)ψ(x, y; k) = e 2 f (x − iy) , 1 2π (10.7.2) where f is an arbitrary anti-holomorphic function (i.e., function of z̄ = x − iy). 10.8 Aharonov-Bohm Effect A quantum particle restricted to a region outside the location of any magnetic flux can nevertheless detect the fractional part of the flux when it is thought of as quantized in units of 2π~/e. A classical charge would be unaffected. 10 QUANTUM PARTICLE IN A MAGNETIC FIELD 10.8.1 84 Stationary State Flux M 0 ≤ x ≤ 2πR R x O A potential V restricts the particle to move on circle of radius R, e.g. in a gold ring. Use 1D Schrödinger equation and ignore curvature: 1 d (−i~ + eAx )2 ψ = Eψ . 2m dx (10.1.8.1) Set Φ Ax = , 2πR Z ⇒ 2πR Ax dx = Φ as reqd. by Stokes’ theorem. (10.1.8.2) 0 It is important to impose the correct boundary conditions. In this case we take ψ to be single-valued on circle. Try ψ = einx/R and find 2 2 1 eΦ ~n ~2 eΦ E= + = n+ , n∈Z. (10.1.8.3) 2m R 2πR 2mR2 2π~ The spectrum depends on the fractional part of Φ ; 2π~/e the integer part can be absorbed into a shift of n. 10.8.2 Double Slit Experiment interference fringes P 1 M electron beam P 2 10 QUANTUM PARTICLE IN A MAGNETIC FIELD 85 In addition to the usual phase difference due to difference in path lengths, there is an extra phase shift along each path due to the vector potential A, even though the magnetic flux is zero everywhere on the particle paths. This causes a shift in the Φ . The Schrödinger equation for fringes which depends on the fractional part of 2π~/e an otherwise free particle in the presence of A is 1 (−i~∇ + eA)2 ψ(x) = Eψ(x) . 2m (10.2.8.1) Define φ(x) by the substitution e ψ(x) = exp −i ~ Z x 0 A(x ) · dx 0 φ(x). (10.2.8.2) P The line integral is, in principle, over any path, P , ending at x but it is convenient to choose the optical paths P = P1 or P2 shown in the figure. Then Schrödinger’s equation is simply ~2 2 ∇ φ(x) = Eφ(x) . (10.2.8.3) − 2m The important fact is that φ is independent of A and is the wavefunction of a freely propagating particle such as is used to construct the rays/classical paths in the usual slit experiment. We must, however, now use ψ instead, but all that happens is that the phase associated with the path P is augmented by the phase Z e x A(x0 ) · dx0 . (10.2.8.4) − ~ P The change in phase between P1 and P2 is then Z Z e x e x 0 0 − A(x ) · dx + A(x0 ) · dx0 ~ P1 ~ P2 Z Z e e e 0 0 = A(x ) · dx = ∇ ∧ A · dσ = Φ ~ C ~ Σ ~ (10.2.8.5) where C is the closed curve encircling the flux in a anti-clockwise direction made from combining P1 and P2 with appropriate sign, and Σ is a surface bounded by C. Thus there is an additional phase shift whose relevant part is 2π × fractional part of Φ . 2π~/e (10.2.8.6) We see again that there is a quantum of flux 2π~/e that controls the shift in the fringes due to the presence of the magnetic field. There are other examples and all are verified by experiment. The surprising result is that although the electron never encounters the physical magnetic field and only the unobservable vector potential A is non-zero on its path, nevertheless there is an observable effect on the electron behaviour. This is topological in nature since, in all cases, there is a path that must wind around the flux line and is therefore not contractable to zero (homotopy).

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