# D18L.

Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission. Foundations of Quantum Mechanics Dr. H. Osborn1 Michælmas 1997 1 A LT EXed by Paul Metcalfe – comments and corrections to [email protected] Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission. T O C LAIRE T HANKS FOR YOUR PATIENCE Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission. Contents Introduction 1 v Basics 1.1 Review of earlier work . . . . . . . . . . . . 1.2 The Dirac Formalism . . . . . . . . . . . . . 1.2.1 Continuum basis . . . . . . . . . . . 1.2.2 Action of operators on wavefunctions 1.2.3 Momentum space . . . . . . . . . . . 1.2.4 Commuting operators . . . . . . . . 1.2.5 Unitary Operators . . . . . . . . . . . 1.2.6 Time dependence . . . . . . . . . . . . . . . . . . . 1 1 2 3 5 6 7 8 8 2 The Harmonic Oscillator 2.1 Relation to wavefunctions . . . . . . . . . . . . . . . . . . . . . . . 2.2 More comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 10 11 3 Multiparticle Systems 3.1 Combination of physical systems . . . 3.2 Multiparticle Systems . . . . . . . . . 3.2.1 Identical particles . . . . . . . 3.2.2 Spinless bosons . . . . . . . . 3.2.3 Spin 21 fermions . . . . . . . 3.3 Two particle states and centre of mass 3.4 Observation . . . . . . . . . . . . . . . . . . . . . 13 13 14 14 15 16 17 17 4 Perturbation Expansions 4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Non-degenerate perturbation theory . . . . . . . . . . . . . . . . . . 4.3 Degeneracy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 19 19 21 5 General theory of angular momentum 5.1 Introduction . . . . . . . . . . . . . 5.1.1 Spin 21 particles . . . . . . . 5.1.2 Spin 1 particles . . . . . . . 5.1.3 Electrons . . . . . . . . . . 5.2 Addition of angular momentum . . . 5.3 The meaning of quantum mechanics 23 23 24 25 25 25 26 iii . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission. iv CONTENTS Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission. Introduction These notes are based on the course “Foundations of Quantum Mechanics” given by Dr. H. Osborn in Cambridge in the Michælmas Term 1997. These typeset notes have been produced mainly for my own benefit but seem to be officially supported. Recommended books are discussed in the bibliography at the back. A word or two about the philosophy of these notes seem in order. They are based in content on the lectures given, but I have felt free to expand and contract various details, as well as to clarify explanations and improve the narrative flow. Errors in content are (hopefully) mine and mine alone but I accept no responsibility for your use of these notes. Other sets of notes are available for different courses. At the time of typing, these courses were: Discrete Mathematics Further Analysis Fluid Dynamics 1 Geometry Foundations of QM Methods of Math. Phys Waves (etc.) Dynamical Systems Probability Analysis Quantum Mechanics Quadratic Mathematics Dynamics of D.E.’s Electrodynamics Fluid Dynamics 2 Applications of QM Statistical Physics They may be downloaded from http://pdm23.trin.cam.ac.uk/˜pdm23/maths/ http://www.damtp.cam.ac.uk/ or or you can email me on [email protected] to get a copy of the sets you require. Even if you download them please email me to let me know, so that I can keep you up to date with the errata and new note sets. The other people who have contributed time and effort to these note sets are: Richard Cameron Claire Gough Kate Metcalfe Hugh Osborn Malcolm Perry David Sanders Analysis Proofreading Probability Proofreading Accomodation Proofreading Although these notes are free of charge anyone who wishes to express their thanks could send a couple of bottles of interesting beer to Y1 Burrell’s Field, Grange Road. Paul Metcalfe 15th December 1997 v Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission. vi INTRODUCTION Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission. Chapter 1 The Basics of Quantum Mechanics Quantum mechanics is viewed as the most remarkable development in 20th century physics. Its point of view is completely different from classical physics. Its predictions are often probabilistic. We will develop the mathematical formalism and some applications. We will emphasize vector spaces (to which wavefunctions belong). These vector spaces are sometimes finite-dimensional, but more often infinite dimensional. The pure mathematical basis for these is in Hilbert Spaces but (fortunately!) no knowledge of this area is required for this course. 1.1 Review of earlier work This is a brief review of the salient points of the 1B Quantum Mechanics course. If you anything here is unfamiliar it is as well to read up on the 1B Quantum Mechanics course. This section can be omitted by the brave. A wavefunction ψ(x) : R3 7→ C is associated with a single particle in three dimensions. ψ represents the state of a physical system for a single particle. If ψ is normalised, that is Z 2 2 kψk ≡ d3 x |ψ| = 1 2 then we say that d3 x |ψ| is the probability of finding the particle in the infinitesimal region d3 x (at x). Superposition Principle If ψ1 and ψ2 are two wavefunctions representing states of a particle, then so is the linear combination a1 ψ1 + a2 ψ2 (a1 , a2 ∈ C). This is obviously the statement that wavefunctions live in a vector space. If ψ ′ = aψ (with a 6= 0) then ψ and ψ ′ represent the same physical state. If ψ and ψ ′ are both normalised then a = eıα . We write ψ ∼ eıα ψ to show that they represent the same physical state. For two wavefunctions φ and ψ we can define a scalar product Z (φ, ψ) ≡ d3 x φ∗ ψ ∈ C. 1 CHAPTER 1. BASICS 2 This has various properties which you can investigate at your leisure. Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission. Interpretative Postulate Given a particle in a state represented by a wavefunction ψ (henceforth “in a state 2 ψ”) then the probability of finding the particle in state φ is P = |(φ, ψ)| and if the wavefunctions are normalised then 0 ≤ P ≤ 1. P = 1 if ψ ∼ φ. We wish to define (linear) operators on our vector space — do the obvious thing. In finite dimensions we can choose a basis and replace an operator with a matrix. For a complex vector space we can define the Hermitian conjugate of the operator A to be the operator A† satisfying (φ, Aψ) = (A† φ, ψ). If A = A† then A is Hermitian. Note that if A is linear then so is A† . In quantum mechanics dynamical variables (such as energy, momentum or angular momentum) are represented by (linear) Hermitian operators, the values of the dynamical variables being given by the eigenvalues. For wavefunctions ψ(x), A is usually a differential operator. For a single particle moving in a potential V (x) we get the ~2 ∇2 + V (x). Operators may have either a continuous or disHamiltonian H = − 2m crete spectrum. If A is Hermitian then the eigenfunctions corresponding to different eigenvalues are orthogonal. We assume completeness — that any wavefunction can be expanded as a linear combination of eigenfunctions. The expectation value for A in a state with wavefunction ψ is hAiψ , defined to be P 2 2 2 i λi |ai | = (ψ, Aψ). We define the square deviation ∆A to be h(A − hAiψ ) iψ which is in general nonzero. Time dependence This is governed by the Schrödinger equation ı~ ∂ψ = Hψ, ∂t where H is the Hamiltonian. H must be Hermitian for the consistency of quantum mechanics: ∂ ı~ (ψ, ψ) = (ψ, Hψ) − (Hψ, ψ) = 0 ∂t if H is Hermitian. Thus we can impose the condition (ψ, ψ) = 1 for all time (if ψ is normalisable). If we consider eigenfunctions ψi of H with eigenvalues Ei we can expand a general wavefunction as X ıEi ψ(x, t) = ai e− ~ t ψi (x). i 2 If ψ is normalised then the probability of finding the system with energy Ei is |ai | . 1.2 The Dirac Formalism This is where we take off into the wild blue yonder, or at least a more abstract form of quantum mechanics than that previously discussed. The essential structure of quantum Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission. 1.2. THE DIRAC FORMALISM 3 mechanics is based on operators acting on vectors in some vector space. A wavefunction ψ corresponds to some abstract vector |ψi, a ket vector. |ψi represents the state of some physical system described by the vector space. If |ψ1 i and |ψ2 i are ket vectors then |ψi = a1 |ψ1 i + a2 |ψ2 i is a possible ket vector describing a state — this is the superposition principle again. We define a dual space of bra vectors hφ| and a scalar product hφ|ψi, a complex number.1 For any |ψi there corresponds a unique hψ| and we require hφ|ψi = hψ|φi∗ . We require the scalar product to be linear such that |ψi = a1 |ψ1 i + a2 |ψ2 i implies hφ|ψi = a1 hφ|ψ1 i + a2 hφ|ψ2 i. We see that hψ|φi = a∗1 hψ1 |φi + a∗2 hψ2 |φi and so hψ| = a∗1 hψ1 | + a∗2 hψ2 |. We introduce linear operators Â|ψi = |ψ ′ i and we define operators acting on bra vectors to the left hφ|Â = hφ′ | by requiring hφ′ |ψi = hφ|Â|ψi for all ψ. In general, in hφ|Â|ψi, Â can act either to the right or the left. We define the adjoint Â† of Â such that if Â|ψi = |ψ ′ i then hψ|Â† = hψ ′ |. Â is said to be Hermitian if Â = Â† . If Â = a1 Â1 + a2 Â2 then Â† = a∗1 Â†1 + a∗2 Â†2 , which can be seen by appealing to the definitions. We also find the adjoint of B̂ Â as follows: Let B̂ Â|ψi = B̂|ψ ′ i = |ψ ′′ i. Then hψ ′′ | = hψ ′ |B̂ † = hψ|Â† B̂ † and the result follows. Also, if hψ|Â = hφ′ | then |φ′ i = Â† |φi. We have eigenvectors Â|ψi = λ|ψi and it can be seen in the usual manner that the eigenvalues of a Hermitian operator are real and the eigenvectors corresponding to two different eigenvalues are orthogonal. We assume completeness — that is any |φi can be expanded in terms of the basis P a |ψ i where Â|ψi i = λi |ψi i and ai = hψi |ψi. If |ψi is ket vectors, |φi = i i i normalised — hψ|ψi = 1 — then the expected value of Â is hÂiψ = hψ|Â|ψi, which is real if Â is Hermitian. P The completeness relation for eigenvectors of Â can be written as 1̂ = i |ψi ihψi |, which gives (as before) |ψi = 1̂|ψi = X i |ψi ihψi |ψi. P We can also rewrite Â = i |ψi iλi hψi | and if λj 6= 0 ∀j then we can define P hψ |. Â−1 = i |ψi iλ−1 i i We now choose an P orthonormal basis {|ni} with hn|mi P = δnm and the completeness relation 1̂ = n |nihn|. We can thus expand |ψi = n an |ni with an = hn|ψi. P P We now consider a linear operator Â, and then Â|ψi = n an Â|ni = m a′m |mi, P with a′m = hm|Â|ψi = n an hm|Â|ni. Further, putting Amn = hm|Â|ni we get P a′m = n Amn an and therefore solving Â|ψi = λ|ψi is equivalent to solving the matrix P equation Aa = λa. AmnPis called the matrix representation of Â. We also have hψ| = n a∗n hn|, with a′n ∗ = m a∗m A†mn , where A†mn = A∗nm gives the Hermitian conjugate matrix. This is the matrix representation of Â† . 1.2.1 Continuum basis In the above we have assumed discrete eigenvalues λi and normalisable eigenvectors |ψi i. However, in general, in quantum mechanics operators often have continuous 1 bra ket. Who said that mathematicians have no sense of humour? CHAPTER 1. BASICS Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission. 4 spectrum — for instance the position operator x̂ in 3 dimensions. x̂ must have eigenvalues x for any point x ∈ R3 . There exist eigenvectors |xi such that x̂|xi = x|xi for any x ∈ R3 . As x̂ must be Hermitian we have hx|x̂ = xhx|. We define the vector space required in the Dirac formalism as that spanned by |xi. For any state |ψi we can define a wavefunction ψ(x) = hx|ψi. We also need to find some normalisation criterion, which uses the 3 dimensional Dirac delta function to get hx|x′ i = δ 3 (x − x′ ). Completeness gives Z d3 x|xihx| = 1. We can also recover the ket vector from the wavefunction by Z |ψi = 1̂|ψi = d3 x|xiψ(x). Also hx|x̂|ψi = xψ(x); the action of the operator x̂ on a wavefunction is multiplication by x. Something else reassuring is Z hψ|ψi = hψ|1̂|ψi = d3 xhψ|xihx|ψi Z 2 = d3 x |ψ(x)| . The momentum operator p̂ is also expected to have continuum eigenvalues. We can similarly define states |pi which satisfy p̂|pi = p|pi. We can relate x̂ and p̂ using the commutator, which for two operators Â and B̂ is defined by h i Â, B̂ = ÂB̂ − B̂ Â. The relationship between x̂ and p̂ is [x̂i , p̂j ] = ı~δij . In one dimension [x̂, p̂] = ı~. We have a useful rule for calculating commutators, that is: h i h i h i Â, B̂ Ĉ = Â, B̂ Ĉ + B̂ Â, Ĉ . This can be easily £ ¤proved simply by expanding the right hand side out. We can use this to calculate x̂, p̂2 . £ ¤ x̂, p̂2 = [x̂, p̂] p̂ + p̂ [x̂, p̂] = 2ı~p̂. It is easy to show by induction that [x̂, p̂n ] = nı~p̂n−1 . We can define an exponential by − ıa~p̂ e µ ¶n ∞ X ıap̂ 1 − = . n! ~ n=0 1.2. THE DIRAC FORMALISM h i ıap̂ We can evaluate x̂, e− ~ by Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission. h − ıa~p̂ x̂, e i 5 µ ¶n # ∞ X ıap̂ 1 − = x̂, n! ~ n=0 · µ ¶n ¸ ∞ X 1 ıap̂ = x̂, − n! ~ n=0 = = " ∞ X 1 ³ ıa ´n − [x̂, p̂n ] n! ~ n=0 ∞ X ³ ıa ´n 1 − ı~p̂n−1 (n − 1)! ~ n=1 =a ∞ ³ X ıa ´n−1 n−1 p̂ − ~ n=1 = ae− ıap̂ ~ and by rearranging this we get that x̂e− ıap̂ ~ = e− ıap̂ ~ (x̂ + a) ıap̂ and it follows that e− ~ |xi is an eigenvalue of x̂ with eigenvalue x + a. Thus we ıap̂ see e− ~ |xi = |x + ai. We can do the same to the bra vectors with the Hermiıap̂ ıap̂ tian conjugate e ~ to get hx + a| = hx|e ~ . Then we also have the normalisation hx′ + a|x + ai = hx′ |xi. ıap̂ ıap We now wish to consider hx + a|pi = hx|e ~ |pi = e ~ hx|pi. Setting x = 0 gives ıap ha|pi = e ~ N , where N = h0|pi is independent of x. We can determine N from the normalisation of |pi. ′ ′ Z da hp′ |aiha|pi Z ıa(p−p′ ) 2 = |N | da e ~ δ(p − p) = hp |pi = 2 = |N | 2π~ δ(p′ − p) So, because we are free to choose the phase of N , we can set N = ¡ 1 ¢ 12 ıxp e ~ . We could define |pi by thus hx|pi = 2π~ |pi = Z dx |xihx|pi = µ 1 2π~ ¶ 12 Z dx |xie ıxp ~ ¡ 1 2π~ ¢ 12 and , but we then have to check things like completeness. 1.2.2 Action of operators on wavefunctions We recall the definition of the wavefunction ψ as ψ(x) = hx|ψi. We wish to see what operators (the position and momentum operators discussed) do to wavefunctions. CHAPTER 1. BASICS 6 Now hx|x̂|ψi = xhx|ψi = xψ(x), so the position operator acts on wavefunctions by multiplication. As for the momentum operator, hx|p̂|ψi = Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission. = Z Z dp hx|p̂|pihp|ψi dp phx|pihp|ψi ¶ 12 Z ıxp 1 dp pe ~ hp|ψi = 2π~ Z d = −ı~ dp hx|pihp|ψi dx d d = −ı~ hx|ψi = −ı~ ψ(x). dx dx £ ¤ d The commutation relation [x̂, p̂] = ı~ corresponds to x, −ı~ dx = ı~ (acting on ψ(x)). µ 1.2.3 Momentum space |xi 7→ ψ(x) = hx|ψi defines a particular representation of the vector space. It is sometimes useful to use a momentum representation, ψ̃(p) = hp|ψi. We observe that Z ψ̃(p) = dx hp|xihx|ψi = µ 1 2π~ ¶ 21 Z dx e− ıxp ~ ψ(x). In momentum space, the operators act differently on wavefunctions. It is easy to d see that hp|p̂|ψi = pψ̃(p) and hp|x̂|ψi = ı~ dp ψ̃(p). We convert the Schrödinger equation into momentum space. We have the operator p̂2 equation Ĥ = 2m + V (x̂) and we just need to calculate how the potential operates on the wavefunction. hp|V (x̂)|ψi = Z dx hp|V (x̂)|xihx|ψi ¶ 12 Z ıxp 1 dx e− ~ V (x)hx|ψi 2π~ ZZ ıx(p′ −p) 1 = dx dp′ V (x)ψ̃(p′ )e ~ 2π~ Z = dp′ Ṽ (p − p′ )ψ̃(p′ ), = where Ṽ (p) = 1 2π~ R dp e− µ ıxp ~ V (x). Thus in momentum space, p2 ψ̃(p) + Hp ψ̃(p) = 2m Z dp′ Ṽ (p − p′ )ψ̃(p′ ). 1.2. THE DIRAC FORMALISM 1.2.4 7 Commuting operators Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission. h i Suppose Â and B̂ are Hermitian and Â, B̂ = 0. Then Â and B̂ have simultaneous eigenvectors. Proof. Suppose Â|ψi = λ|ψi and the vector subspace Vλ is the span of the eigenvectors of Â with eigenvalue λ. (If dim Vλ > 1 then λ is said to be degenerate.) As Â and B̂ commute we know that λB̂|ψi = ÂB̂|ψi and so B̂|ψi ∈ Vλ . If λ is non-degenerate then B̂|ψi = µ|ψi for some µ. Otherwise we have that B̂ : Vλ 7→ Vλ and we can therefore find eigenvectors of B̂ which lie entirely inside Vλ . We can label these as |λ, µi, and we know that Â|λ, µi = λ|λ, µi B̂|λ, µi = µ|λ, µi. These may still be degenerate. However we can in principle remove this degeneracy by adding more commuting operators until each state is uniquely labeled by the eigenvalues of each common eigenvector. This set of operators is called a complete commuting set. This isn’t so odd: for a single particle in 3 dimensions we have the operators x̂1 , x̂2 and x̂3 . These all commute, so for a single particle with no other degrees of freedom we can label states uniquely by |xi. We also note from this example that a complete commuting set is not unique, we might just as easily have taken the momentum operators and labeled states by |pi. To ram the point in more, we could also have taken some weird combination like x̂1 , x̂2 and p̂3 . For our single particle in 3 dimensions, a natural set of commuting operators involves the angular momentum operator, L̂ = x̂ ∧ p̂, or L̂i = ǫijk x̂j p̂k . We can find commutation relations between L̂i and the other operators we know. These are summarised here, proof is straightforward. i h • L̂i , x̂l = ı~ǫilj x̂j h i • L̂i , x̂2 = 0 i h • L̂i , p̂m = ı~ǫimk p̂k h i • L̂i , p̂2 = 0 i h • L̂i , L̂j = ı~ǫijk L̂k h i • L̂i , L̂2 = 0 If we have a Hamiltonian Ĥ = p̂2 2m h i + V (|x̂|) then we can also see that L̂, Ĥ = 0. We choose as a commuting set Ĥ, L̂2 and L̂3 and label states |E, l, mi, where the eigenvalue of L̂2 is l(l + 1) and the eigenvalue of L̂3 is m. CHAPTER 1. BASICS 8 Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission. 1.2.5 Unitary Operators An operator Û is said to be unitary if Û † Û = 1̂, or equivalently Û −1 = Û † . Suppose Û is unitary and Û |ψi = |ψ ′ i, Û |φi = |φ′ i. Then hφ′ | = hφ|Û † and ′ ′ hφ |ψ i = hφ|ψi. Thus the scalar product, which is the probability amplitude of finding the state |φi given the state |ψi, is invariant under unitary transformations of states. For any operator Â we can define Â′ = Û ÂÛ † . Then hφ′ |Â′ |ψ ′ i = hφ|Â|ψi and matrix elements are unchanged under unitary transformations. We also note that if Ĉ = ÂB̂ then Ĉ ′ = Â′ B̂ ′ . The quantum mechanics for the |ψi, |φi, Â, B̂ etc. is the same as for |ψ ′ i, |φ′ i, Â′ , ′ B̂ and so on. A unitary transform in quantum mechanics is analogous to a canonical transformation in dynamics. † Note that if Ô is Hermitian then Û = eıÔ is unitary, as Û † = e−ıÔ = e−ıÔ . 1.2.6 Time dependence This is governed by the Schrödinger equation, ı~ ∂ |ψ(t)i = Ĥ|ψ(t)i. ∂t Ĥ is the Hamiltonian and we require it to be Hermitian. We can get an explicit solution of this if Ĥ does not depend explicitly on t. We set |ψ(t)i = Û (t)|ψ(0)i, ıĤt where Û (t) = e− ~ . As Û (t) is unitary, hφ(t)|ψ(t)i = hφ(0)|ψ(0)i. If we measure the expectation of Â at time t we get hψ(t)|Â|ψ(t)i = a(t). This description is called the Schrödinger picture. Alternatively we can absorb the time dependence into the operator Â to get the Heisenberg picture, a(t) = hψ|Û † (t)ÂÛ (t)|ψi. We write ÂH (t) = Û † (t)ÂÛ (t). In this description the operators are time dependent (as opposed to the states). ÂH (t) is the Heisenberg picture time dependent operator. Its evolution is governed by ı~ which is easily proven. For a Hamiltonian Ĥ = for the operators x̂H and p̂H h i ∂ ÂH (t) = ÂH (t), Ĥ , ∂t 1 2 2m p̂(t) + V (x̂(t)) we can get the Heisenberg equations d 1 x̂H (t) = p̂H (t) dt m d p̂H (t) = −V ′ (x̂H (t)). dt These ought to remind you of something. Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission. Chapter 2 The Harmonic Oscillator In quantum mechanics there are two basic solvable systems, the harmonic oscillator and the hydrogen atom. We will examine the quantum harmonic oscillator using algebraic methods. In quantum mechanics the harmonic oscillator is governed by the Hamiltonian 1 2 1 Ĥ = p̂ + 2 mω 2 x̂2 , 2m with the condition that [x̂, p̂] = ı~. We wish to solve Ĥ|ψi = E|ψi to find the energy eigenvalues. We define a new operator â. ³ mω ´ 12 µ ¶ ıp̂ â = x̂ + 2~ mω ¶ µ ³ mω ´ 12 ıp̂ † . â = x̂ − 2~ mω â and â† are respectively called the£ annihilation and creation operators. We can ¤ easily obtain the commutation relation â, â† = 1̂. It is easy to show that, in terms ¡ † ¢ of the annihilation and creation operators, the Hamiltonian Ĥ = 21 ~ω ââ i + â† â , h i h ¢ ¡ which reduces to ~ω â† â + 12 . Let N̂ = â† â. Then â, N̂ = â and â† , N̂ = −â† . ³ ³ ´ ´ Therefore N̂ â = â N̂ − 1 and N̂ â† = â† N̂ + 1 . Suppose |ψi is an eigenvector of N̂ with eigenvalue λ. Then the commutation relations give that N̂ â|ψi = (λ − 1) â|ψi and therefore unless â|ψi = 0 it is an eigenvalue of N̂ with eigenvalue λ − 1. Similarly N̂ â† |ψi = (λ + 1) â† |ψi. But for any |ψi, hψ|N̂ |ψi ≥ 0 and equals 0 iff â|ψi = 0. Now suppose we have an eigenvalue λ ∈ / {0, 1, 2, . . . }. Then ∃n such that ân |ψi is an eigenvector of N̂ with eigenvalue λ − n < 0 and so we must have¡λ ∈ {0, ¢ 1, 2, . . . }. Returning to the Hamiltonian we get energy eigenvalues En = ~ω n + 12 , the same result as using the Schrödinger equation for wavefunctions, but with much less effort. We define |ni = Cn â†n |0i, where Cn is such as to make hn|ni = 1. We can take Cn ∈ R, and evaluate h0|ân â†n |0i to find Cn . 9 CHAPTER 2. THE HARMONIC OSCILLATOR 10 1 = hn|ni = Cn2 h0|ân â†n |0i Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission. = Cn2 h0|ân−1 ââ† â†n−1 |0i = Cn2 hn − 1|ââ† |n − 1i 2 Cn−1 = Cn2 hn − 1|N̂ + 1|n − 1i 2 Cn−1 = Cn2 (n − 1 + 1)hn − 1|n − 1i 2 Cn−1 =n Cn2 . 2 Cn−1 √ −1 We thus require Cn = Cn−1 / n and as C0 = 1 we get Cn = (n!) 2 and so we have the normalised eigenstate (of N̂ ) |ni = √1n! â† |0i (with eigenvalue n). |ni is also ¢ ¡ an eigenvector of Ĥ with eigenvalue ~ω n + 21 . The space of states for the harmonic oscillator is spanned by {|ni}. We also need to ask if there exists a non-zero state |ψi such that â† |ψi = 0. Then 0 = hψ|ââ† |ψi = hψ|ψi + hψ|â† â|ψi ≥ hψ|ψi > 0. So there exist no non-zero states |ψi such that â† |ψi = 0. 2.1 Relation to wavefunctions We evaluate 0 = hx|â|0i = ³ mω ´ 12 µ 2~ ~ d x+ mω dx ¶ hx|0i and we see that ψ0 (x) = hx|0i satisfies the differential equation µ ¶ d mω + x ψ0 (x) = 0. dx ~ 1 mω 2 This (obviously) has solution ψ0 (x) = N e− 2 ~ x for some normalisation constant N . This is the ground state wavefunction which has energy 12 ~ω. For ψ1 (x) = hx|1i = hx|â† |0i we find ³ mω ´ 12 ³ ı ´ p̂ |0i hx| x̂ − 2~ mω ¶ µ ³ mω ´ 12 ~ d = ψ0 (x) x− 2~ mω dx µ ¶1 2mω 2 = xψ0 (x). ~ ψ1 (x) = 2.2. MORE COMMENTS 11 2.2 More comments Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission. Many harmonic oscillator problems are simplified using the creation and annihilation operators.1 For example hm|x̂|ni = µ ~ 2mω = µ ~ 2mω = µ ~ 2mω ¶ 12 ¶ 12 ¶ 12 hm|â + â† |ni √ ¢ ¡√ n hm|n − 1i + n + 1 hm|n + 1i √ ¢ ¡√ n δm,n−1 + n + 1 δm,n+1 . This is non-zero only if m = n ± 1. We note that x̂r contains terms âs â†r−s , where 0 ≤ s ≤ r and so hm|x̂r |ni can be non-zero only if n − r ≤ m ≤ n + r. Ĥt Ĥt It is easy to see that in the Heisenberg picture âH (t) = eı ~ âe−ı ~ = e−ıωt â. Then using the equations for x̂H (t) and p̂H (t), we see that x̂H (t) = x̂ cos ωt + 1 mω p̂ sin ωt. Also, Ĥâ†H (t) = â†H (t)(Ĥ + ~ω), so if |ψi is an energy eigenstate with eigenvalue E then â†H (t)|ψi is an energy eigenstate with eigenvalue E + ~ω. 1 And such problems always occur in Tripos papers. You have been warned. Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission. 12 CHAPTER 2. THE HARMONIC OSCILLATOR Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission. Chapter 3 Multiparticle Systems 3.1 Combination of physical systems In quantum mechanics each physical system has its own vector space of physical states and operators, which if Hermitian represent observed quantities. If we consider two vector spaces V1 and V2 with bases {|ri1 } and {|si2 } with r = 1 . . . dim V1 and s = 1 . . . dim V2 . We define the tensor product V1 ⊗ V2 as the vector space spanned by pairs of vectors {|ri1 |si2 : r = 1 . . . dim V1 , s = 1 . . . dim V2 }. We see that dim(V1 ⊗ V2 ) = dim V1 dim V2 . We also write the basis vectors of V1 ⊗V2 as |r, si. We can define a scalar product on V1 ⊗V2 in terms of the basis vectors: hr′ , s′ |r, si = hr′ |ri1 hs′ |si2 . We can see that if {|ri1 } and {|si2 } are orthonormal bases for their respective vector spaces then {|r, si} is an orthonormal basis for V1 ⊗V2 . Suppose Â1 is an operator on V1 and B̂2 is an operator on V2 we can define an operator Â1 × B̂2 on V1 ⊗ V2 by its operation on the basis vectors: ´ ´³ ´ ³ ³ Â1 × B̂2 |ri1 |si2 = Â1 |ri1 B̂2 |si2 . We write Â1 × B̂2 as Â1 B̂2 . Two harmonic oscillators We illustrate these comments by example. Suppose Ĥi = p̂2i + 1 mωx̂2i 2m 2 i = 1, 2. We have two independent vector spaces Vi with bases |nii where n = 0, 1, . . . and âi and â†i are creation and annihilation operators on Vi , and ¢ ¡ Ĥi |nii = ~ω n + 12 |nii . For the combined system we form the tensor product V1 ⊗ V2 with basis |n1 , n2 i P and Hamiltonian Ĥ = i Ĥi , so Ĥ|n1 , n2 i = ~ω (n1 + n2 + 1) |n1 , n2 i. There are N + 1 ket vectors in the N th excited state. 13 CHAPTER 3. MULTIPARTICLE SYSTEMS 14 The three dimensional harmonic oscillator follows similarly. In general if Ĥ1 and Ĥ2 are two independent Hamiltonians which act on V1 and V2 respectively then the Hamiltonian for the combined system is Ĥ = Ĥ1 + Ĥ2 acting on V1 ⊗ V2 . If {|ψr i} and {|ψs i} are eigenbases for V1 and V2 with energy eigenvalues {Er1 } and {Es2 } respectively then the basis vectors {|Ψir,s } for V1 ⊗ V2 have energies Er,s = Er1 + Es2 . Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission. 3.2 Multiparticle Systems We have considered single particle systems with states |ψi and wavefunctions ψ(x) = hx|ψi. The states belong to a space H. Consider an N particle system. We say the states belong to Hn = H1 ⊗ · · · ⊗ HN and define a basis of states |ψr1 i1 |ψr2 i2 . . . |ψrN iN where {|ψri ii } is a basis for Hi . A general state |Ψi is a linear combination of basis vectors and we can define the N particle wavefunction as Ψ(x1 , x2 , . . . , xN ) = hx1 , x2 , . . . , xN |Ψi. The normalisation condition is Z 2 hΨ|Ψi = d3 x1 . . . d3 xN |Ψ(x1 , x2 , . . . , xN )| = 1 if normalised. 2 We can interpret d3 x1 . . . d3 xN |Ψ(x1 , x2 , . . . , xN )| as the probability density that particle i is in the volume element d3 xi at xi . We can obtain the probability density for one particle by integrating out all the other xj ’s. ∂ |Ψi = Ĥ|Ψi, where Ĥ is an operator For time evolution we get the equation ı~ ∂t N on H . If the particles do not interact then Ĥ = N X Ĥi i=1 where Ĥi acts on Hi but leaves Hj alone for j 6= i. We have energy eigenstates in each Hi such that Ĥi |ψr ii = Er |ψr ii and so |Ψi = |ψr1 i1 |ψr2 i2 . . . |ψrN iN is an energy eigenstate with energy Er1 + · · · + ErN . 3.2.1 Identical particles There are many such cases, for instance multielectron atoms. We will concentrate on two identical particles. “Identical” means that physical quantities are be invariant under interchange of particles. For instance if we have Ĥ = H(x̂1 , p̂1 , x̂2 , p̂2 ) then this must equal the permuted Hamiltonian H(x̂2 , p̂2 , x̂1 , p̂1 ) if we have identical particles. We introduce Û such that Û x̂1 Û −1 = x̂2 Û x̂2 Û −1 = x̂1 Û p̂1 Û −1 = p̂2 Û p̂2 Û −1 = p̂1 . We should also have Û Ĥ Û −1 = Ĥ and more generally if Â1 is an operator on particle 1 then Û Â1 Û −1 is the corresponding operator on particle 2 (and vice versa). Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission. 3.2. MULTIPARTICLE SYSTEMS 15 Note that if |Ψi is an energy eigenstate of Ĥ then so is Û |Ψi. Clearly Û 2 = 1̂ and we require Û to be unitary, which implies that Û is Hermitian. In quantum mechanics we require |Ψi and Û |Ψi to be the same states (for identical particles). This implies that Û |Ψi = λ|Ψi and the requirement Û 2 = 1̂ gives that λ = ±1. In terms of wavefunctions this means that Ψ(x̂1 , x̂2 ) = ±Ψ(x̂2 , x̂1 ). If we have a plus sign then the particles are bosons (which have integral spin) and if a minus sign then the particles are fermions (which have spin 12 , 23 , . . . ).1 The generalisation to N identical particles is reasonably obvious. Let Ûij inter−1 change particles i and j. Then Ûij Ĥ Ûij = Ĥ for all pairs (i, j). The same physical requirement as before gives us that Ûij |Ψi = ±|Ψi for all pairs (i, j). If we have bosons (plus sign) then in terms of wavefunctions we must have Ψ(x̂1 , . . . , x̂N ) = Ψ(x̂p1 , . . . , x̂pN ), where (p1 , . . . , pN ) is a permutation of (1, . . . , N ). If we have fermions then Ψ(x̂1 , . . . , x̂N ) = λΨ(x̂p1 , . . . , x̂pN ), where λ = +1 if we have an even permutation of (1, . . . , N ) and −1 if we have an odd permutation. Remark for pure mathematicians. 1 and {±1} are the two possible one dimensional representations of the permutation group. 3.2.2 Spinless bosons (Which means that the only variables for a single particle are x̂ and p̂.) Suppose we have two identical non-interacting bosons. Then Ĥ = Ĥi + Ĥ2 and we have Ĥ1 |ψr ii = Er |ψr ii . The general space with two particles is H1 ⊗ H2 which has a basis {|ψr i1 |ψs i2 }, but as the particles are identical the two particle state space is (H1 ⊗ H2 )S where we restrict to symmetric combinations of the basis vectors. That is, a basis for this in terms of the bases of H1 and H2 is o n |ψr i1 |ψr i2 ; √12 (|ψr i1 |ψs i2 + |ψs i1 |ψr i2 ) , r 6= s . The corresponding wavefunctions are ψr (x1 )ψr (x2 ) and √1 2 (ψr (x1 )ψs (x2 ) + ψs (x1 )ψr (x2 )) 1 and the corresponding eigenvalues are 2Er and Er +Es . The factor of 2− 2 just ensures normalisation and √1 (1hψr ′ |2hψs′ | 2 + 1hψs′ |2hψr′ |) √12 (|ψr i1 |ψs i2 + |ψs i1 |ψr i2 ) evaluates to δrr′ δss′ + δrs′ δr′ s . P For N spinless bosons with Ĥ = Ĥi the appropriate completely symmetric states are ´ ³ √1 i + permutations thereof if ri 6= rj i . . . |ψ |ψ rN N r1 1 N! 1 Spin will be studied later in the course. CHAPTER 3. MULTIPARTICLE SYSTEMS 16 Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission. 3.2.3 Spin 1 2 fermions In this case (which covers electrons, for example) a single particle state (or wavefunction) depends on an additional discrete variable s. The wavefunctions are ψ(x, s) or ψs (x). The space of states for a single electron H = L2 (R3 ) ⊗ C2 has a basis of the form |xi|si ≡ |x, si and the wavefunctions can be written ψs (x) = hx, s|ψi. A basis of wavefunctions is {ψrλ (x, s) = ψr (x)χλ (s)}, where r and λ are labels for the basis. λ takes two values and it will later be seen µ to be natural to take λ = ± 12 . ¶ χλ (1) , in which case two possible basis We can also think of the vector χλ = χλ (2) µ ¶ µ ¶ 1 0 vectors are and . Note that χ†λ′ χλ = δλλ′ . 0 1 The scalar product is defined in the obvious way: hφr′ λ′ |φrλ i = hψr′ |ψr ihχλ′ |χλ i, which equals δrr′ δλλ′ if the initial basis states are orthonormal. The two electron wavefunction is Ψ(x1 , s1 ; x2 , s2 ) and under the particle exchange operator Û we must have Ψ(x1 , s1 ; x2 , s2 ) 7→ −Ψ(x2 , s2 ; x1 , s1 ). The two particle states belong to the antisymmetric combination (H1 ⊗ H2 )A . For N electrons the obvious thing can be done. Basis for symmetric or antisymmetric 2 particle spin states There is only one antisymmetric basis state ´ 1 ³ χA (s1 , s2 ) = √ χ 1 (s1 )χ− 1 (s2 ) − χ− 1 (s1 )χ 1 (s2 ) , 2 2 2 2 2 and three symmetric possibilities: χ 1 (s )χ 1 (s ) 2 ³1 2 2 ´ χS (s1 , s2 ) = √12 χ 1 (s1 )χ− 1 (s2 ) + χ− 1 (s1 )χ 1 (s2 ) 2 2 2 2 χ (s )χ (s ). − 21 1 − 12 s1 6= s2 2 We can now examine two non-interacting electrons, with Ĥ = Ĥ1 + Ĥ2 and take Hi independent of spin. The single particle states are |ψi i|χs i. The two electron states live in (H1 ⊗ H2 )A , which has a basis |ψr i1 |ψr i2 |χA i; 1 √ (|ψr i1 |ψs i2 + |ψs i1 |ψr i2 ) |χA i; r 6= s 2 1 √ (|ψr i1 |ψs i2 − |ψs i1 |ψr i2 ) |χS i; r 6= s, 2 with energy levels 2Er (one spin state) and Er + Es (one antisymmetric spin state and three symmetric spin states). We thus obtain the Pauli exclusion principle: no two electrons can occupy the same state (taking account of spin). As an example we can take the helium atom with Hamiltonian Ĥ = 2e2 2e2 e2 p̂2 p̂21 + 2 − − + . 2m 2m 4πǫ0 |x̂1 | 4πǫ0 |x̂2 | 4πǫ0 |x̂1 − x̂2 | 3.3. TWO PARTICLE STATES AND CENTRE OF MASS 17 If we neglect the interaction term we can analyse this as two hydrogen atoms and glue the results back together as above. The hydrogen atom (with a nuclear charge 2e) has 2e2 En = − 8πǫ 2 , so we get a ground state for the helium atom with energy 2E1 with 0n no degeneracy and a first excited state with energy E1 + E2 with a degeneracy of four. Hopefully these bear some relation to the results obtained by taking the interaction into account. Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission. 3.3 Two particle states and centre of mass p̂2 p̂2 Suppose we have a Hamiltonian Ĥ = 2m1 + 2m2 + V (x̂1 − x̂2 ) defined on H2 . We can separate out the centre of mass motion by letting p̂ = P̂ = p̂1 + p̂2 1 2 1 2 (p̂1 − p̂2 ) x̂ = x̂1 − x̂2 . X̂ = (x̂1 + x̂2 ) i Then X̂i , P̂j = ı~δij , [x̂i , p̂j ] = ı~δij and X̂, P̂ and x̂, p̂ commute respectively. h 2 2 P̂ We can rewrite the Hamiltonian as Ĥ = 2M + ĥ, ĥ = p̂m + V (x̂), where M = 2m and 2 we can decompose H into HCM ⊗ Hint . HCM is acted on by X̂ and P̂ and has wavefunctions φ(X). Hint is acted on by x̂, p̂ and any spin operators. It has wavefunctions ψ(x, s1 , s2 ). We take wavefunctions Ψ(x1 , s1 ; x2 , s2 ) = Φ(X)ψ(x, s1 , s2 ) in H2 . P.X This simplifies the Schrödinger equation, we can just have φ(X) = eı ~ and then 2 P + Eint . We thus need only to solve the one particle equation ĥψ = Eint ψ. E = 2M Under the particle exchange operator Û we have ψ(x, s1 , s2 ) 7→ ψ(−x, s2 , s1 ) = ±ψ(x, s1 , s2 ), with a plus sign for bosons and a minus sign for fermions. In the spinless case then ψ(x) = ψ(−x). If we have a potential V (|x̂|) then we may separate variables to get µ ¶ x R(|x|)χ(s1 , s2 ) ψ(x, s1 , s2 ) = Yl |x| ³ ´ ³ ´ x x with Yl − |x| = (−1)l Yl |x| . For spinless bosons we therefore require l to be even. 3.4 Observation Consider the tensor product of two systems H1 and H2 . A general state |Ψi in H1 ⊗H2 can be written as X |Ψi = aij |ψi i1 |φj i2 i,j with |ψi i1 ∈ H1 and |φj i2 ∈ H2 assumed orthonormal bases for their respective vector spaces. Suppose we make a measurement on the first system leaving the second P system unchanged, and find the first system in a state |ψi i1 . Then 1hψi |Ψi = j aij |φj i2 , which we write as Ai |φi2 , where |φi2 is a normalised state of the second system. We CHAPTER 3. MULTIPARTICLE SYSTEMS 18 2 Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission. interpret |Ai | as the probability of finding system 1 in state |ψi i1 . After measurement system 2 is in a state |φi2 . If aij = λi δij (no summation) then Ai = λi and measurement of system 1 as |ψi i1 determines system 2 to be in state |φi i2 . Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission. Chapter 4 Perturbation Expansions 4.1 Introduction Most problems in quantum mechanics are not exactly solvable and it it necessary to find approximate answers to them. The simplest method is a perturbation expansion. We write Ĥ = Ĥ0 + Ĥ ′ where Ĥ0 describes a solvable system with known eigenvalues and eigenvectors, and Ĥ ′ is in some sense small. We write Ĥ(λ) = Ĥ0 + λĤ ′ and expand the eigenvalues and eigenvectors in powers of λ. Finally we set λ = 1 to get the result. Note that we do not necessarily have to introduce λ; the problem may have some small parameter which we can use. This theory can be applied to the time dependent problem but here we will only discuss the time independent Schrödinger equation. 4.2 Non-degenerate perturbation theory Suppose that Ĥ0 |ni = ǫn |ni for n = 0, 1, . . . . We thus assume discrete energy levels and we assume further that the energy levels are non-degenerate. We also require Ĥ ′ to be sufficiently non-singular to make a power series expansion possible. We have the equation Ĥ(λ)|ψn (λ)i = En (λ)|ψn (λ)i. We suppose that En (λ) tends to ǫn as λ → 0 and |ψn (λ)i → |ni as λ → 0. We pose the power series expansions En (λ) = ǫn + λEn(1) + λ2 En(2) + . . . |ψn (λ)i = N |ni + λ|ψn(1) i + . . . , substitute into the Schrödinger equation and require it to be satisfied at each power of λ. The normalisation constant N is easily seen to be 1 + O(λ2 ). The O(1) equation is automatically satisfied and the O(λ) equation is Ĥ0 |ψn(1) i + Ĥ ′ |ni = En(1) |ni + ǫn |ψn(1) i. (1) (1) Note that we can always replace |ψn i with |ψn i + α|ni and leave this equation (1) unchanged. We can therefore impose the condition hn|ψn i = 0. If we apply hn| to 19 CHAPTER 4. PERTURBATION EXPANSIONS 20 (1) this equation we get En = hn|Ĥ ′ |ni — the first order perturbation in energy. If we apply hr| where r 6= n we see that hr|ψn(1) i = − hr|Ĥ ′ |ni ǫr − ǫn Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission. and therefore |ψn(1) i = − X |rihr|Ĥ ′ |ni . ǫr − ǫn r6=n Note that we are justified in these divisions as we have assumed that the eigenvalues are non-degenerate. On doing the same thing to the O(λ2 ) equation we see that En(2) = hn|Ĥ ′ |ψn(1) i ¯2 ¯ ¯ ¯ X ¯hr|Ĥ ′ |ni¯ =− . ǫr − ǫn r6=n This procedure is valid if ǫr − ǫn is not very small when hr|Ĥ ′ |ni = 6 0. d Using these results we can see that dλ En (λ) = hψn (λ)|Ĥ ′ |ψn (λ)i and X ∂ 1 |ψn (λ)i = − |ψr (λ)ihψr (λ)|Ĥ ′ |ψn (λ)i. ∂λ Er (λ) − En (λ) r6=n Also ∂ Ĥ ∂λ = Ĥ ′ and so ∂ ∂2 En (λ) = 2hψn (λ)|Ĥ ′ |ψn (λ)i. ∂λ2 ∂λ Example: harmonic oscillator 2 p̂ Consider Ĥ = 2m + 21 mω 2 x̂2 + λmω 2 x̂2 , which can be viewed as Ĥ0 + Ĥ ′ , where Ĥ0 is the plain vanilla quantum harmonic oscillator Hamiltonian. Calculating the matrix elements hr|x̂2 |ni required is an extended exercise in manipulations of the annihilation and creation operators and is omitted. The results are ¡ ¢ En(1) = ~ω n + 12 ¡ ¢ En(2) = − 21 ~ω n + 21 . We thus get the perturbation expansion for En′ ¢ ¢¡ ¡ En′ = ~ω n + 21 1 + λ − 12 λ2 + O(λ3 ) . ¡ ¢√ 1 + 2λ which This system can also be solved exactly to give En′ = ~ω n + 12 agrees with the perturbation expansion. 4.3. DEGENERACY 21 Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission. 4.3 Degeneracy The method given here breaks down if ǫr = ǫn for r 6= n. Perturbation theory can be extended to the degenerate case, but we will consider only the first order shift in ǫr . We suppose that the states |n, si, s = 1 . . . Nn have the same energy ǫn . Ns is the degeneracy of this energy level. As before we pose a Hamiltonian Ĥ = Ĥ0 + λĤ ′ such that Ĥ0 |n, si = ǫn |n, si and look for states |ψ(λ)i with energy E(λ) → ǫn as λ → 0. The difference with the previous method is that we expand |ψ(λ)i as a power series in λ in the basis of eigenvectors of Ĥ0 . That is X |ψ(λ)i = |n, sias + λ|ψ (1) i. s As the as are arbitrary we can impose the conditions hn, s|ψ (1) i = 0 for each s and n. We thus have to solve Ĥ|ψ(λ)i = E(λ)|ψ(λ)i with E(λ) = ǫn + λE (1) . If we take the O(λ) equation and apply hn, r| to it we get X as hn, r|Ĥ ′ |n, si = ar Er(1) s which is a matrix eigenvalue problem. Thus the first order perturbations in ǫn are the eigenvalues of the matrix hn, r|Ĥ ′ |n, si. If all the eigenvalues are distinct then the perturbation “lifts the degeneracy”. It is convenient for the purpose of calculation to choose a basis for the space spanned by the degenerate eigenvectors in which this matrix is “as diagonal as possible”.1 1 Don’t ask... Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission. 22 CHAPTER 4. PERTURBATION EXPANSIONS Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission. Chapter 5 General theory of angular momentum For a particle with position and momentum operators x̂ and p̂ with the commutation relations [x̂i , pˆj ] =h ı~δij iwe define L̂ = x̂ ∧ p̂. It can be seen that L̂ is Hermitian and it is easy to show L̂i , L̂j = ı~ǫijk L̂k . 5.1 Introduction More generally we define Hermitian angular momentum operators J with the commutation relation [Ji , Jj ] = ı~ǫijk Jk .1 We ask on what space of states can this algebra of operators be realised, or alternatively, what are the representations? We choose one component of J whose eigenvalues label the states; in accordance with convention we choose J3 . Then we have J3 |mi = m~|mi for m ∈ R (discrete) and we also have hm′ |mi = δm′ m . ¤ £ By applying the commutation relation we get (easily) that J3 , J2 = 0, so as J2 is Hermitian we can choose a simultaneous eigenbasis. That is, J2 |mi = λ~2 |mi. We know that λ ≥ 0 since J2 is the sum of the squares of Hermitian operators. At this stage we choose an alternate basis; that is we split J into J+ , J− and J3 , † with J± = J1 ± ıJ2 . Note that J± = J∓ . The commutation relations for Ji give that [J3 , J± ] = ±ı~J± and [J+ , J− ] = 2~J3 . It will be useful later to note that ( J− J+ + J32 + ~J3 J2 = 12 (J+ J− + J− J+ ) + J32 = J+ J− + J32 − ~J3 . Proof of this is immediate. We can now rewrite the [J3 , J± ] commutation relation as J3 J± = J± (J3 ± ~) and so we see that J± |mi is an eigenvector of J3 with ± eigenvalue (m + 1)~ and so J± |mi = ~Nm |m ± 1i, with some normalisation con± stant Nm . By evaluating the norm of J± |mi, and noting that hm|mi = 1 we see that ± 2 ) = λ − m2 ∓ m. (Nm We can now define states |m ± ni for n = 0, 1, 2, . . . . We can pin them down more ± 2 by noting that (Nm ) ≥ 0 for positive norms. However the formulae we have are, 1 Because we are now grown up we will omit the hats if they do not add to clarity. 23 CHAPTER 5. GENERAL THEORY OF ANGULAR MOMENTUM 24 given λ, negative for sufficiently large |m| and so to avoid this we must have mmax = j ¡ ¢2 such that J+ |ji = 0 and so Nj+ = λ − j 2 − j = 0 and so λ = j(j + 1). We can perform a similar trick with J− ; there must exist mmin = −j ′ such that J− | − j ′ i = 0; thus λ = j ′ (j ′ + 1). So j ′ = j and as −j ′ = −j = j − n for some n ∈ {0, 1, 2, . . . } we have j = 0, 12 , 1, 23 , . . . . In summary the states can be labelled by |j mi such that Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission. J2 |j mi = ~2 j(j + 1)|j mi J3 |j mi = ~m|j mi 1 J± |j mi = ~ ((j ∓ m) (j ± m + 1)) 2 |j m + 1i with m ∈ {−j, −j + 1, . . . , j − 1, j} and j ∈ {0, 12 , 1, 23 , . . . }. There are 2j + 1 states with different m for the same j. |j mi is the standard basis of the angular momentum states. We have obtained a representation of the algebra labelled by j. If J = L = x̂ ∧ p̂ we must have j an integer. P ′ ′ ′ Recall that if we have P Â we can define a matrix Aλ λ by Â|λi = λ′ |λ iAλ λ . Note that (BA)λ′ λ = Bλ′ µ Aµλ . Given j, we have (J3 )m′ m = ~m δm′ m and µ p (J± )m′ m = ~ (j ∓ m) (j ± m + 1) δm′ ,m±1 , giving us (2j + 1)×(2j + 1) matrices satisfying the three commutation relations [J3 , J± ] = ±~J± and [J+ , J− ] = 2~J3 . If J are angular momentum operators which act on a vector space V and we have |ψi ∈ V such that J3 |ψi = ~k|ψi and J+ |ψi = 0 then ψ is a state with angular n momentum j = k. The other states are given by J− |ψi, 1 ≤ n ≤ 2k. The conditions 2 2 also give J |ψi = ~ k (k + 1) |ψi 5.1.1 Spin 1 2 particles This is the simplest non-trivial case. We have j = 12 and a two dimensional state space with a basis | 21 12 i and | 12 − 21 i. We have the relations J3 | 12 ± 21 i = ± 21 ~| 12 ± 21 i and J+ | 12 1 2i 1 J− | 2 − 21 i J− | 21 1 2i 1 J+ | 2 − 21 i =0 =0 = ~| 21 − 12 i = ~| 21 1 2 i. It is convenient to introduce explicit matrices σ such that X J| 12 mi = | 12 m′ i 21 ~ (σ)m′ m . m′ are The matrices σ are 2 × 2 matrices (called the Pauli spin matrices). Explicitly, they σ+ = σ1 = 1 (σ+ + σ− ) = 2 µ µ 0 2 0 0 0 1 1 0 ¶ ¶ ¶ µ 1 0 σ3 = 0 −1 µ ¶ ı 0 −ı σ2 = − (σ+ − σ− ) = . ı 0 2 σ− = µ 0 2 ¶ 0 0 Note that σ12 = σ22 = σ32 = 1 and σ † = σ. These satisfy the commutation relations [σi , σj ] = 2ıǫijk σk (a slightly modified angular momentum commutation relation) and we also have σ2 σ3 = ıσ1 (and the relations obtained by cyclic permutation), so 5.2. ADDITION OF ANGULAR MOMENTUM 25 2 σi σj + σj σi = 2δij 1. Thus if n̂ is a unit vector we have (σ.n̂) = 1 and we see that σ.n̂ has eigenvalues ±1. 1 3 2 2 We define the angular momentum 2 ~σ and so s = 4 ~ 1. µ ¶ matrices s µ= ¶ 1 0 and χ− 12 = . The basis states are χ 12 = 0 1 Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission. 5.1.2 Spin 1 particles We apply the theory as above to get 1 0 0 S3 = 0 0 0 0 0 −1 and S− = S+† . 5.1.3 0 S+ = 0 0 √ 2 √0 0 2 0 0 Electrons Electrons are particles with intrinsic spin 21 . The angular momentum J = x̂ ∧ p̂ + s, where s are the spin operators for spin 12 . The basic operators for an electron are x̂, p̂ and s. We can represent these operators by their action on two component wavefunctions: X ψ(x) = ψλ (x)χλ . λ=± 21 In this basis x̂ 7→ x, p̂ 7→ −ı~∇ and s 7→ 21 ~σ. All other operators are constructed in terms of these, for instance we may have a Hamiltonian H= p̂2 + V (x) + U (x)σ.L 2m where L = x̂ ∧ p̂. If V and U depend only on |x| then [J, H] = 0. 5.2 Addition of angular momentum Consider two independent angular momentum operators J(1) and J(2) with J(r) acting on some space V (r) and V (r) having spin jr for r = 1, 2. We now define an angular momentum J acting on V (1) ⊗ V (2) by J = J(1) + J(2) . Using the commutation relations for J(r) we can get [Ji , Jj ] = ı~ǫijk Jk . We want to construct states |J M i forming a standard angular momentum basis, that is such that: J3 |J M i = ~M |J M i ± J± |J M i = ~NJ,M |J M ±1i p ± with NJ,M = (J ∓ M ) (J ± M + 1). We look first for states in V which satisfy J+ |J Ji = 0 and J3 |J Ji = ~J|J Ji. The maximum value of J we can get is j1 + j2 ; and |j1 +j2 j1 +j2 i = |j1 j1 i1 |j2 j2 i2 . Then J+ |j1 +j2 j1 +j2 i = 0. Similarly this is CHAPTER 5. GENERAL THEORY OF ANGULAR MOMENTUM 26 an eigenvector of J3 with eigenvalue ~ (j1 + j2 ). We can now apply J− repeatedly to form all the |J M i states. Applying J− we get |J M −1i = α|j1 j1 −1i1 |j2 j2 i2 + β|j1 j1 i1 |j2 j2 −1i2 . − , and we must The coefficents α and β can be determined from the coefficents Na,b 2 2 have α + β = 1. If we choose |ψi a state orthogonal to this Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission. |ψi = −β|j1 j1 −1i1 |j2 j2 i2 + α|j2 j2 i1 |j2 j2 −1i2 . J3 |ψi can be computed and it shows that |ψi is an eigenvector of J3 with eigenvalue ~ (j1 + j2 − 1). Now 0 = hψ|j1 +j2 j1 +j2 −1i ∝ hψ|J− |j1 +j2 j1 +j2 i and so hψ|J− |φi = 0 for all states |φi in V . Thus J+ |ψi = 0 and hence we have |ψi = |j1 +j2 −1 j1 +j2 −1i. We can then construct the states |j1 +j2 −1 M i by repeatedly applying J− . For each J such that |j1 − j2 | ≤ J ≤ j1 + j2 we can construct a state |J Ji. We define the Clebsch-Gordan coefficients hj1 m1 j2 m2 |J M i, and so X hj1 m1 j2 m2 |J M i|j1 m1 i|j2 m2 i. |J M i = m1 ,m2 The Clebsch-Gordan coefficients are nonzero only when M = m1 + m2 . We can check the number of states; jX 1 +j2 J=|j1 −j2 | (2J + 1) = jX 1 +j2 J=|j1 −j2 | o n 2 (J + 1) − J 2 = (2j1 + 1) (2j2 + 1) . Electrons Electrons have spin 21 and we can represent their spin states with χ± 12 (s). Using this notation we see that two electrons can form a symmetric spin 1 triplet 1 )χ 1 (s ) χ 2 (s ³1 2 2 ´ 1 √ 1 (s1 )χ 1 (s2 ) + χ 1 (s1 )χ 1 (s2 ) χm (s1 , s2 ) = χ − − 2 2 2 2 2 χ 1 (s )χ 1 (s ) −2 1 −2 2 and an antisymmetric spin 0 singlet; ´ 1 ³ χ0 (s1 , s2 ) = √ χ 12 (s1 )χ− 21 (s2 ) − χ− 12 (s1 )χ 21 (s2 ) . 2 5.3 The meaning of quantum mechanics Quantum mechanics deals in probabilities, whereas classical mechanics is deterministic if we have complete information. If we have incomplete information classical mechanics is also probabilistic. Inspired by this we ask if there can be “hidden variables” in quantum mechanics such that the theory is deterministic. Assuming that local effects have local causes, this is not possible. 5.3. THE MEANING OF QUANTUM MECHANICS 27 Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission. We will take a spin example to show this. Consider a spin 12 particle, with two spin states |↑i and |↓i which are eigenvectors of S3 = 21 ~σ3 . If we choose to use two component vectors we have µ ¶ µ ¶ 0 1 . χ↓ = χ↑ = 1 0 Suppose n = (sin θ, 0, cos θ) (a unit vector) and let us find the eigenvectors of µ ¶ cos θ sin θ σ.n = . sin θ − cos θ 2 As (σ.n) = 1 we must have eigenvalues ±1 and an inspired guess gives χ↑,n and χ↓,n as χ↑,n = cos θ2 χ↑ + sin θ2 χ↓ and χ↓,n = − sin θ2 χ↑ + cos θ2 χ↓ . Thus (reverting to ket vector notation) if an electron is in a state |↑i then the probability of finding it in a state |↑, ni is cos2 θ2 and the probability of finding it in a state |↓, ni is sin2 θ2 . Now, suppose we have two electrons in a spin 0 singlet state; 1 |Φi = √ {|↑i1 |↓i2 − |↓i1 |↑i2 } . 2 Then the probability of finding electron 1 with spin up is 12 , and after making this measurement electron 2 must be spin down. Similarly, if we find electron 1 with spin down (probability 12 again) then electron 2 must have spin up. More generally, suppose we measure electron 1’s spin along direction n. Then we see that the probability for electron 1 to have spin up in direction n (aligned) is 21 and then electron 2 must be in the state |↓, ni2 . If we have two electrons (say electron 1 and electron 2) in a spin 0 state we may physically separate them and consider independent experiments on them. We will consider three directions as sketched. For electron 1 there are three vari(1) (1) ables which we may measure (in separate experiments); Sz = ±1, Sn = ±1 and (1) Sm = ±1. We can also do this for electron 2. (1) (2) We see that if we find electron 1 has Sz = 1 then electron 2 has Sz = −1 (etc.). If there exists an underlying deterministic theory then we could expect some probability distribution p for this set of experiments; ³ ´ (1) (2) 0 ≤ p Sz(1) , Sn(1) , Sm , Sz(2) , Sn(2) , Sm ≤1 (1) (2) which is nonzero only if Sdirn = −Sdirn and X p({s}) = 1. {s} Bell inequality Suppose we have a probability P distribution p(a, b, c) with a, b, c = ±1. We define partial probabilities pbc (b, c) = a p(a, b, c) and similarly for pac (a, c) and pab (a, b). Then 28 CHAPTER 5. GENERAL THEORY OF ANGULAR MOMENTUM pbc (1, −1) = p(1, 1, −1) + p(−1, 1, −1) ≤ p(1, 1, −1) + p(1, 1, 1) + p(−1, 1, −1) + p(−1, −1, −1) ≤ pab (1, 1) + pac (−1, −1). Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission. Applying this to the two electron system we get ³ ´ ³ ´ ³ ´ (2) (2) P Sn(1) = 1, Sm = 1 ≤ P Sz(1) = 1, Sn(2) = −1 + P Sz(1) = −1, Sm =1 . We can calculate these probabilities from quantum mechanics ³ ´ ³ ´ P Sz(1) = 1, Sn(2) = −1 = P Sz(1) = 1, Sn(1) = 1 = cos2 ³ ´ (2) P Sz(1) = −1, Sm = 1 = cos2 θ+φ and 2 ³ ´ (2) P Sn(1) = 1, Sm = 1 = sin2 θ2 . The Bell inequality gives sin2 φ 2 ≤ cos2 θ 2 + cos2 θ+φ 2 θ 2 which is not in general true. Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission. References ◦ P.A.M. Dirac, The Principles of Quantum Mechanics, Fourth ed., OUP, 1958. I enjoyed this book as a good read but it is also an eminently suitable textbook. ◦ Feynman, Leighton, Sands, The Feynman Lectures on Physics Vol. 3, AddisonWesley, 1964. Excellent reading but not a very appropriate textbook for this course. I read it as a companion to the course and enjoyed every chapter. ◦ E. Merzbacher, Quantum Mechanics, Wiley, 1970. This is a recommended textbook for this course (according to the Schedules). I wasn’t particularly impressed but you may like it. There must be a good, modern textbook for this course. If you know of one please send me a brief review and I will include it if I think it is suitable. In any case, with these marvellous notes you don’t need a textbook, do you? Related courses There are courses on Statistical Physics and Applications of Quantum Mechanics in Part 2B and courses on Quantum Physics and Symmetries and Groups in Quantum Physics in Part 2A. 29

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

### Related manuals

Download PDF

advertisement