# B9s.

Mathematical Tripos Part IB Quantum Mechanics Michaelmas Term 2015 Dr. J.M. Evans Some Constants and Units • Planck’s Constant: h̄ = 1.05×10−34 Js , or h = 2πh̄ = 6.63×10−34 Js • Speed of light: c = 3.00×108 ms−1 ; Wavelength of visible light (approx) 4×10−7 m to 7×10−7 m • Unit of electric charge: e = 1.60×10−19 C ; Unit of energy: electron-volt, 1 eV = 1.60×10−19 J • Fine structure constant: α = e2 /4πǫ0 h̄c ≈ 1/137 (dimensionless) Copyright © 2015 University of Cambridge. Not to be quoted or reproduced without permission. • Electron mass: me = 9.11×10−31 kg ; Proton mass: mp = 1.67×10−27 kg • Bohr radius: r1 = 4πǫ0 h̄2 /me e2 = h̄/me cα = 5.29×10−11 m Wave Behaviour We will refer to any (real or complex valued) function with some kind of periodicity in time and/or space as a wave. The following remarks summarise a few useful definitions and ideas. • A function of time t obeying f (t+T ) = f (t) has period T , frequency ν = 1/T , and angular or circular frequency ω = 2πν = 2π/T (in practice, ω is usually just called the frequency, provided this leads to no confusion). Familiar examples are f (t) = cos ωt, sin ωt or exp ±iωt. A function of position x (in one dimension) obeying f (x+λ) = f (x) has wavelength λ and wavenumber k = 2π/λ Examples are f (x) = cos kx, sin kx or exp ±ikx. The analogous functions of a position vector x with periodicity in three dimensions are f (x) = exp ik · x where k is the wave vector , and the wavelength is then λ = 2π/|k|. • The wave equation in one dimension for a function f (x, t) is 2 ∂2f 2∂ f − c =0 ∂t2 ∂x2 where c is some constant. This has solutions which are periodic in both position and time: f± (x, t) = A± exp( ±ikx − iωt ) (∗) provided that the wavelength and frequency are related by ω = ck or λν = c . Such solutions represent waves which move or propagate with speed c to the right or left, according to the sign in (∗) (assuming ω, k > 0). The constant A± is the amplitude of the wave. In electromagnetic (EM) waves, the field components obey the three dimensional wave equation, obtained by replacing ∂ 2 /∂x2 by ∇2 above. This has solutions of an analogous form f (x, t) = A exp( ik·x − iωt ) with ω = c|k| . Such a wave propagates in the direction of k, with speed c, now the speed of light. • Other kinds of waves arise as solutions of other governing equations which may differ significantly from the standard wave equation. A function does not have to satisfy the standard wave equation in order to be usefully thought of as a wave! The Schrödinger Equation is one example of an alternative governing equation; it is the central equation in QM and we will study it in some depth. (In other physical applications, e.g. waves in real fluids, we should expect the wave equation to be modified by friction or dissipative terms.) Many different governing equations give rise to propagating solutions of the form (∗), provided the frequency is chosen to be a suitable function of the wavenumber, ω(k). Moreover, if the governing equation is linear in f , then any solutions f1 and f2 can be combined to give a new solution: Copyright © 2015 University of Cambridge. Not to be quoted or reproduced without permission. f = f1 + f2 This is the Principle of Superposition and it is responsible for much behaviour we tend to think of as wave-like. • Interference or diffraction occurs when waves from different sources merge, or when parts of a wave recombine after passing around or through some obstacle. When a number of such waves are superposed, they may interfere constructively, increasing the size of the amplitude, or destructively, diminishing the amplitude. The result is an interference or diffraction pattern which depends on the sources or the obstacles. When light is passed through a number of narrow slits, the resulting diffraction pattern provides conclusive evidence that light is a wave. Passing higher energy waves, such as X-rays, through matter gives a way of determining the crystalline arrangement of atoms from the resulting diffraction patterns. A Few Historical Highlights • 1801-03: Interference/diffraction experiments by Young show that light is a wave • 1862-4: Maxwell identifies light as an EM (electromagnetic) wave • 1897: Thompson discovers the electron, the first elementary particle • 1900: Planck introduces the energy-frequency relation, with h as a new physical constant, and derives the black body spectrum (the distribution of energy with frequency for EM radiation in thermal equilibrium) • 1905: Einstein imparts clearer physical meaning to photons, using them to explain the photoelectric effect, and other experimental results • 1911: Based on scattering experiments, Rutherford proposes a model of the atom with most of its mass concentrated in a small, compact nucleus • 1913: Bohr proposes an atomic model with electrons orbiting a nucleus and with quantisation of their angular momentum, using this to derive observed line spectra • 1923: Compton scattering of X-rays on electrons confirms that photons are relativistic particles of zero rest mass • 1923-24: de Broglie proposes wave-particle duality for matter, as for radiation • 1925-30: The emergence of Quantum Mechanics, through work of Heisenberg, Born, Jordan, Dirac, Pauli, Schrödinger, and others. • 1927-28: Diffraction experiments of Davisson, Germer and Thompson confirm that electrons behave as waves as well as particles Mathematical Tripos Part IB Quantum Mechanics Michaelmas Term 2015 Dr. J.M. Evans Gaussian Wavepackets • Consider a normalised Gaussian wavefunction (at time t = 0): ψ0 (x) = (1/απ)1/4 exp(−x2 /2α) Copyright © 2015 University of Cambridge. Not to be quoted or reproduced without permission. with α a real positive constant. The expectation values of position and momentum in this state are Z ∞ Z ∞ d 2 ψ0 (x)∗ ψ0 (x) dx = 0 . x |ψ0 (x)| dx = 0 and hp̂i = −ih̄ hx̂i = dx −∞ −∞ The first integral vanishes because ψ0 is an even function of x, and the second because the integrand is a total derivative, since ψ0 is a real function in this case. • The uncertainty in x and p are calculated using the standard formulas (∆x)2 = hx̂2 i − hx̂i2 = Z ∞ x2 |ψ0 (x)|2 dx = −∞ 1 (απ)1/2 Z ∞ x2 α x2 exp − dx = α 2 −∞ and 2 2 2 (∆p) = hp̂ i − hp̂i = Z ∞ h̄2 h̄ |ψ 0 (x)| dx = (απ)1/2 −∞ 2 ′ 2 Z ∞ −∞ x2 x2 h̄2 dx = exp − . α2 α 2α The wavefunction can be interpreted as a particle which is localised around x = 0, on a length scale α1/2 . The combined uncertainty in position and momentum is as small as possible in this example, since the bound given by the Uncertainty Principle is saturated: (∆x)(∆p) = 12 h̄ . • There are similar wavefunctions with non-zero expectation values for position or momentum. For ψ0 ( x−c ) = (1/απ)1/4 exp −(x−c)2 /2α we have hx̂i = c and hp̂i = 0, representing a particle localised around x = c (any real constant). For ψu (x) = (1/απ)1/4 exp( i(mu/h̄)x ) exp( −x2 /2α ) we find hx̂i = 0 but hp̂i = mu, where u is a real constant which corresponds to a velocity. This shows that ψu (x) represents a moving particle (in some sense) even before considering the evolution of the wavefunction (and hence the probability distribution) in time. Solving the Schrödinger Equation with Zero Potential Consider the time-independent Schrödinger Equation (SE) for a wavefunction Ψ(x, t) for a free particle (zero potential), subject to an initial condition: − h̄2 ∂ ∂ Ψ = ih̄ Ψ 2 2m ∂x ∂t with Ψ(x, 0) = ψ(x) . (∗) This can be solved by expanding the initial wavefunction in terms of momentum eigenstates, otherwise known as a Fourier representation: Z ∞ Z ∞ 1/2 ikx 1/2 ψ(x) = (h̄/2π) ψ̃(k) e dk where ψ̃(k) = (1/2πh̄) ψ(x) e−ikx dx . −∞ −∞ (The constants, including h̄, are conventional for Fourier transforms in QM.) Now, an eigenstate of momentum, with eigenvalue h̄k, is also an eigenstate of the Hamiltonian, with energy E = h̄2 k 2 /2m (because the potential vanishes); so exp(ikx) exp(−ih̄k 2 t/2m) is a solution of the SE for any k. Since the SE is linear, we can then write down the solution to (∗) in the form Z ∞ 2 1/2 ψ̃(k) eikx e−ih̄k t/2m dk . (∗∗) Ψ(x, t) = (h̄/2π) −∞ Copyright © 2015 University of Cambridge. Not to be quoted or reproduced without permission. Time Evolution of Wavepackets • Taking the initial wavefunction in (∗) to be ψ0 (x), we have Z ∞ x2 αk 2 1 1 α 1/4 ψ̃0 (k) = exp − exp − exp( −ikx ) dx = 2α 2 (2πh̄)1/2 (απ)1/4 −∞ h̄1/2 π which follows by completing the square in the exponential: 1 2 1 1 x + ikx = ( x + iαk )2 + αk 2 2α 2α 2 and shifting the variable of integration. (Such an imaginary shift in the real variable x can be justified using Cauchy’s Theorem; see e.g. Complex Methods/Analysis.) Substituting into (∗∗) gives a solution Z αk 2 h̄k 2 1 α 1/4 ∞ exp − exp −i t exp( ikx ) dk Ψ0 (x, t) = 2 2m (2π)1/2 π −∞ α 1/4 1 x2 h̄t exp − where γ(t) = α + i . = 1/2 π 2γ(t) m γ(t) The integral over k has again been evaluated by completing a square, 1 1 2 ix 2 h̄t 2 1 + x α+i k − ikx = γ k − 2 m 2 γ 2γ and shifting the variable of integration. From the solution, we find (∆x)2 = h̄ |γ(t)|2 2α and (∆p)2 = h̄ . 2α • Taking the initial wavefunction in (∗) to be ψu (x) (corresponding to a moving wavepacket) we have mu ψ̃u (k) = ψ̃0 k − . h̄ The solution Ψu (x, t) can then be found from (∗∗) in a similar way: we need only complete squares in the exponentials before using standard results for Gaussian integrals (although the algebra is now slightly more complicated). The result is h̄k 2 mu x exp −i t . Ψu (x, t) = Ψ0 ( x−ut, t ) exp i h̄ m It can also be verified, by substitution, that Ψu (x, t) is a solution of the time-dependent SE if Ψ0 (x, t) is. (This is an expression of Galilean invariance for the non-relativistic SE.) Standard Gaussian Integrals: Z ∞ 2π 1/2 , exp( − 21 βx2 ) dx = β −∞ Z ∞ x2 exp( − 21 βx2 ) dx = −∞ 2π 1/2 β3 , for Re(β) > 0 . Mathematical Tripos Part IB Quantum Mechanics Michaelmas Term 2015 Dr. J.M. Evans Commuting Operators and Simultaneous Measurements Consider a quantum system with space of states V . We assume throughout that the eigenstates of any observable, or hermitian operator, provide a basis for the space on which the operator acts. Recall also (section 5.2) that eigenvalues of an observable are real, that eigenstates with different eigenvalues are orthogonal, and that these results underpin the measurement axioms. • Let A and B be observables. A state χ is a simultaneous or joint eigenstate of A and B if Copyright © 2015 University of Cambridge. Not to be quoted or reproduced without permission. Aχ = λχ and Bχ = µχ . If the system is in state χ and measurements of A or B are made in rapid succession, in any order, then the results obtained will be λ or µ, respectively, with probability 1 each time. (The time intervals that elapse between successive measurements must be small enough that the evolution of the state in time can be ignored.) In light of this, observables A and B are said to be simultaneously measurable if V has a basis of joint or simultaneous eigenstates χn : Aχn = λn χn , Bχn = µn χn with (χm , χn ) = δmn . (An operator is also said to be diagonalisable if there exists a basis of eigenstates, and A and B are simultaneously diagonalisable if there exists a basis of joint eigenstates.) • A necessary and sufficient condition for observables A and B to be simultaneously measurable is that they commute: [ A, B ] = AB − BA = 0 . This was stated in section 5.5 and it can be established as follows. Necessity. For any joint eigenstate, ABχn = BAχn = λn µn χn ⇒ [ A, B ] χn = 0 . Since this holds for all n, and χn form a basis for V , we deduce that [A, B] = 0. Sufficiency. For any eigenvalue λ of A, consider the eigenspace Vλ = {ψ : Aψ = λψ}, i.e. the subspace of V containing all the corresponding eigenstates. If [A, B] = 0, then Aψ = λψ ⇒ A(Bψ) = B(Aψ) = B(λψ) = λ(Bψ) . Hence, B maps Vλ to itself, for any λ. Now V is the direct sum of the eigenspaces Vλ over all possible eigenvalues λ (since V is spanned by eigenstates of A). Furthermore, any choice of basis for each Vλ gives a choice of basis for the entire space V . But since B is hermitian on V , it is also hermitian as an operator on each subspace Vλ . It follows that Vλ has a basis of eigenstates of B, all with a common eigenvalue λ under A, by definition, and so all of which are joint eigenstates. Since this holds for every Vλ , this provides a basis of joint eigenstates for V , as required. A simple special case of the argument for sufficiency applies if the eigenvalues of A are non-degenerate. Then each Vλ is one-dimensional and, since B maps Vλ to itself, we must have Bψ = µψ, for some µ. • The fact that observables must commute if they are to be simultaneously measureable is also reflected in the generalised uncertainty relation (∆A)(∆B) ≥ 12 | h [A, B] i | . The proof of this is set as an exercise on Example Sheet 3. Angular Momentum Eigenfunctions • The operators L3 and L2 are hermitian and satisfy [L3 , L2 ] = 0 implying that they have simultaneous or joint eigenstates Y . It is convenient to write the eigenvalues so that L2 Y = h̄2 ℓ(ℓ+1) Y . L3 Y = h̄m Y , • In spherical polar coordinates r, θ, φ, L3 = −ih̄ ∂ , ∂φ L2 = −h̄2 1 ∂ ∂ 1 ∂2 sin θ + , sin θ ∂θ ∂θ sin2 θ ∂φ2 and simultaneous eigenstates can be found in separable form: Copyright © 2015 University of Cambridge. Not to be quoted or reproduced without permission. Y (θ, φ) = Φ(φ) P (u) ⇒ dΦ = imΦ , dφ − d dP m2 (1 − u2 ) + P = ℓ(ℓ + 1)P , du du 1 − u2 where u = cos θ. The first of these separated ODEs has solutions Φm (φ) = eimφ where m must be an integer for the eigenfunction to be single-valued. The second ODE is the associated Legendre Equation. When m = 0 it reduces to the usual Legendre Equation with solutions Pℓ (u), the Legendre polynomials of degree ℓ; these are the solutions which are well-behaved at u = ±1 or θ = 0, π (see IB Methods). For more general m, the well-behaved solutions are Pℓm (u) = (1 − u2 )|m|/2 d|m| Pℓ (u) , du|m| |m| ≤ ℓ , known as associated Legendre functions, with ℓ a non-negative integer. • In conclusion, the joint eigenfunctions of L3 and L2 are the spherical harmonic functions: Yℓm (θ, φ) = (const) eimφ Pℓm (cos θ) with ℓ = 0, 1, 2, . . . , m = 0, ±1, ±2, . . . , ±ℓ . Spherically Symmetric Potentials • If the Hamiltonian H of a quantum system satisfies [ L3 , H ] = [ L2 , H ] = 0 (∗) then there exist energy eigenstates ψℓm (x) that are also joint eigenstates of L3 and L2 : H ψℓm = E ψℓm , L2 ψℓm = h̄2 ℓ(ℓ+1) ψℓm , L3 ψℓm = h̄m ψℓm . • For a particle of mass µ moving in a spherically symmetric potential V (r) the Hamiltonian is H = 1 2 h̄2 h̄2 1 ∂ 2 1 p̂ + V = − ∇2 + V (r) = − r + L2 + V (r) 2µ 2µ 2µ r ∂r2 2µr2 (the L2 term is the rotational kinetic energy and is proportional to the angular part of ∇2 ). The condition (∗) follows immediately from the expressions above for L3 and L2 (it is also easy to verify using Cartesian coordinates). Joint eigenstates of H, L2 and L3 can then be found as separable wavefunctions: h̄2 h̄2 1 ∂ 2 ψℓm (x) = R(r)Yℓm (θ, φ) with − (rR) + ℓ(ℓ + 1) + V (r) R = ER . 2µ r ∂r2 2µr2 The equation satisfied by χ(r) = rR(r) is known as the radial Schrödinger equation. The energy eigenvalues E will depend on ℓ, in general, but will be independent of m as a consequence of the rotational invariance of H. Comments to: [email protected]

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