Foundations of quantum mechanics

Foundations of Quantum Mechanics Dr H Osborn1 Michælmas 1997 A LT EXed by Paul Metcalfe – comments and corrections to pdm23@cam.ac.uk Revision: 2.5 Date: 1999-06-06 14:10:19+01 The following people have maintained these notes – date Paul Metcalfe Contents Introduction 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 8 The Harmonic Oscillator 2.1 Relation to wavefunctions 2.2 More comments 10 11 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 12 fermions 3.3 Two particle states and centre of mass 3.4 Observation 13 13 14 14 15 16 17 17 Perturbation Expansions 4.1 Introduction 4.2 Non-degenerate perturbation theory 4.3 Degeneracy 19 19 19 21 General theory of angular momentum 5.1 Introduction 5.1.1 Spin 12 particles 5.1.2 Spin particles 5.1.3 Electrons 5.2 Addition of angular momentum 5.3 The meaning of quantum mechanics 23 23 24 25 25 26 27 iii iv CONTENTS Introduction These notes are based on the course “Foundations of Quantum Mechanics” given by Dr H Osborn in Cambridge in the Michælmas Term 1997 Recommended books are discussed in the bibliography at the back Other sets of notes are available for different courses At the time of typing these courses were: Probability Analysis Methods Fluid Dynamics Geometry Foundations of QM Methods of Math Phys Waves (etc.) General Relativity Physiological Fluid Dynamics Slow Viscous Flows Acoustics Seismic Waves Discrete Mathematics Further Analysis Quantum Mechanics Quadratic Mathematics Dynamics of D.E.’s Electrodynamics Fluid Dynamics Statistical Physics Dynamical Systems Bifurcations in Nonlinear Convection Turbulence and Self-Similarity Non-Newtonian Fluids They may be downloaded from http://www.istari.ucam.org/maths/ or http://www.cam.ac.uk/CambUniv/Societies/archim/notes.htm or you can email soc-archim-notes@lists.cam.ac.uk to get a copy of the sets you require v Copyright (c) The Archimedeans, Cambridge University All rights reserved Redistribution and use of these notes in electronic or printed form, with or without modification, are permitted provided that the following conditions are met: Redistributions of the electronic files must retain the above copyright notice, this list of conditions and the following disclaimer Redistributions in printed form must reproduce the above copyright notice, this list of conditions and the following disclaimer All materials derived from these notes must display the following acknowledgement: This product includes notes developed by The Archimedeans, Cambridge University and their contributors Neither the name of The Archimedeans nor the names of their contributors may be used to endorse or promote products derived from these notes Neither these notes nor any derived products may be sold on a for-profit basis, although a fee may be required for the physical act of copying You must cause any edited versions to carry prominent notices stating that you edited them and the date of any change THESE NOTES ARE PROVIDED BY THE ARCHIMEDEANS AND CONTRIBUTORS “AS IS” AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE ARE DISCLAIMED IN NO EVENT SHALL THE ARCHIMEDEANS OR CONTRIBUTORS BE LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL DAMAGES HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY OUT OF THE USE OF THESE NOTES, EVEN IF ADVISED OF THE POSSIBILITY OF SUCH DAMAGE Chapter 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 → 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 ψ ≡ d3 x |ψ| = 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 = 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 CHAPTER BASICS For two wavefunctions φ and ψ we can define a scalar product (φ, ψ) ≡ d3 x φ∗ ψ ∈ C This has various properties which you can investigate at your leisure Interpretative Postulate Given a particle in a state represented by a wavefunction ψ (henceforth “in a state ψ”) then the probability of finding the particle in state φ is P = |(φ, ψ)| and if the wavefunctions are normalised then ≤ P ≤ P = if ψ ∼ φ We wish to define (linear) operators on our vector space — 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 Hamiltonian H = − 2m ∇2 + V (x) Operators may have either a continuous or discrete 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 A ψ , defined to be 2 i λi |ai | = (ψ, Aψ) We define the square deviation ∆A to be (A − A ψ ) ψ which is in general nonzero Time dependence This is governed by the Schrăodinger equation ∂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) = ∂t if H is Hermitian Thus we can impose the condition (ψ, ψ) = for all time (if ψ is normalisable) If we consider eigenfunctions ψi of H with eigenvalues Ei we can expand a general wavefunction as ψ(x, t) = e − ıEi t ψi (x) If ψ is normalised then the probability of finding the system with energy Ei is |ai |2 1.2 THE DIRAC FORMALISM 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 mechanics is based on operators acting on vectors in some vector space A wavefunction ψ corresponds to some abstract vector |ψ , a ket vector |ψ represents the state of some physical system described by the vector space If |ψ1 and |ψ2 are ket vectors then |ψ = a1 |ψ1 + a2 |ψ2 is a possible ket vector describing a state — this is the superposition principle again We define a dual space of bra vectors φ| and a scalar product φ|ψ , a complex number.1 For any |ψ there corresponds a unique ψ| and we require φ|ψ = ψ|φ ∗ We require the scalar product to be linear such that |ψ = a1 |ψ1 + a2 |ψ2 implies φ|ψ = a1 φ|ψ1 + a2 φ|ψ2 We see that ψ|φ = a∗1 ψ1 |φ + a∗2 ψ2 |φ and so ψ| = a∗1 ψ1 | + a∗2 ψ2 | ˆ We introduce linear operators A|ψ = |ψ and we define operators acting on bra ˆ vectors to the left φ|Aˆ = φ | by requiring φ |ψ = φ|A|ψ for all ψ In general, in ˆ , Aˆ can act either to the right or the left We define the adjoint Aˆ† of Aˆ such φ|A|ψ ˆ that if A|ψ = |ψ then ψ|Aˆ† = ψ | Aˆ is said to be Hermitian if Aˆ = Aˆ† If Aˆ = a1 Aˆ1 + a2 Aˆ2 then Aˆ† = a∗1 Aˆ†1 + a∗2 Aˆ†2 , which can be seen by appealing to ˆ Aˆ as follows: the definitions We also find the adjoint of B ˆ † and the result ˆ † = ψ|Aˆ† B ˆ ˆ ˆ Let B A|ψ = B|ψ = |ψ Then ψ | = ψ |B † ˆ ˆ follows Also, if ψ|A = φ | then |φ = A |φ ˆ We have eigenvectors A|ψ = λ|ψ 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 |φ can be expanded in terms of the basis ket ˆ i = λi |ψi and = ψi |φ If |ψ is normalised vectors, |φ = |ψi where A|ψ ˆ , which is real if Aˆ — ψ|ψ = — then the expected value of Aˆ is Aˆ ψ = ψ|A|ψ is Hermitian The completeness relation for eigenvectors of Aˆ can be written as ˆ1 = i |ψi ψi |, which gives (as before) |ψi ψi |ψ |ψ = ˆ 1|ψ = i We can also rewrite Aˆ = i |ψi λi ψi | and if λj = ∀j then we can define −1 −1 ˆ A = i |ψi λi ψi | We now choose an orthonormal basis {|n } with n|m = δnm and the completeness relation ˆ = n |n n| We can thus expand |ψ = n an |n with an = n|ψ ˆ ˆ and then A|ψ ˆ = m am |m , We now consider a linear operator A, = n an A|n ˆ ˆ ˆ with am = m|A|ψ = n an m|A|n Further, putting Amn = m|A|n we get ˆ am = A a and therefore solving A|ψ = λ|ψ is equivalent to solving the n mn n ˆ matrix equation Aa = λa Amn is called the matrix representation of A We also have ψ| = n a∗n n|, with an ∗ = m a∗m A†mn , where A†mn = A∗nm gives the Hermitian conjugate matrix This is the matrix representation of Aˆ† bra ket Who said that mathematicians have no sense of humour? CHAPTER BASICS 1.2.1 Continuum basis In the above we have assumed discrete eigenvalues λi and normalisable eigenvectors |ψi However, in general, in quantum mechanics operators often have continuous ˆ in dimensions x ˆ must have eigenspectrum — for instance the position operator x ˆ |x = x|x for values x for any point x ∈ R3 There exist eigenvectors |x such that x any x ∈ R3 ˆ must be Hermitian we have x|ˆ As x x = x x| We define the vector space required in the Dirac formalism as that spanned by |x For any state |ψ we can define a wavefunction ψ(x) = x|ψ We also need to find some normalisation criterion, which uses the dimensional Dirac delta function to get x|x = δ (x − x ) Completeness gives d3 x|x x| = We can also recover the ket vector from the wavefunction by |ψ = ˆ1|ψ = d3 x|x ψ(x) ˆ on a wavefunction is multipliAlso x|ˆ x|ψ = xψ(x); the action of the operator x cation by x Something else reassuring is ψ|ψ = ψ|ˆ1|ψ = = d3 x ψ|x x|ψ d3 x |ψ(x)| ˆ is also expected to have continuum eigenvalues We The momentum operator p ˆ |p = p|p We can relate x ˆ and p ˆ using can similarly define states |p which satisfy p ˆ is defined by the commutator, which for two operators Aˆ and B ˆ B ˆ = AˆB ˆ −B ˆ A ˆ A, ˆ and p ˆ is [ˆ x, pˆ] = ı The relationship between x xi , pˆj ] = ı δij In one dimension [ˆ We have a useful rule for calculating commutators, that is: ˆ Cˆ ˆ A, ˆ B ˆ Cˆ + B ˆ B ˆ Cˆ = A, A, This can be easily proved simply by expanding the right hand side out We can use this to calculate xˆ, pˆ2 xˆ, pˆ2 = [ˆ x, pˆ] pˆ + pˆ [ˆ x, pˆ] = 2ı pˆ It is easy to show by induction that [ˆ x, pˆn ] = nı pˆn−1 We can define an exponential by e− ıap ˆ ∞ = n! n=0 − ıaˆ p n 3.2 MULTIPARTICLE SYSTEMS 15 ˆ then so is U ˆ |Ψ Clearly U ˆ = ˆ1 and we Note that if |Ψ is an energy eigenstate of H ˆ ˆ require U to be unitary, which implies that U is Hermitian ˆ |Ψ to be the same states (for identical In quantum mechanics we require |Ψ and U ˆ ˆ = ˆ1 gives that particles) This implies that U |Ψ = λ|Ψ and the requirement U ˆ ) = ±Ψ(ˆ ˆ ) If we λ = ±1 In terms of wavefunctions this means that Ψ(ˆ x1 , x x2 , x 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 , 32 , ).1 ˆij interThe generalisation to N identical particles is reasonably obvious Let U −1 ˆ ˆ ˆ ˆ change particles i and j Then Uij H Uij = H for all pairs (i, j) ˆij |Ψ = ±|Ψ for all pairs The same physical requirement as before gives us that U (i, j) If we have bosons (plus sign) then in terms of wavefunctions we must have ˆ N ) = Ψ(ˆ ˆ pN ), Ψ(ˆ x1 , , x xp1 , , x where (p1 , , pN ) is a permutation of (1, , N ) If we have fermions then ˆ N ) = λΨ(ˆ ˆ pN ), xp1 , , x Ψ(ˆ x1 , , x where λ = +1 if we have an even permutation of (1, , N ) and −1 if we have an odd permutation Remark for pure mathematicians and {±1} are the two possible representations of the permutation group in one dimension 3.2.2 Spinless bosons ˆ and p ˆ ) Suppose (Which means that the only variables for a single particle are x ˆ and we have ˆ = H ˆi + H we have two identical non-interacting bosons Then H ˆ |ψr i = Er |ψr i The general space with two particles is H1 ⊗ H2 which has H a basis {|ψr |ψs }, 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 |ψr |ψr ; √12 (|ψr |ψs + |ψs |ψr ) , r = s The corresponding wavefunctions are ψr (x1 )ψr (x2 ) and √1 (ψr (x1 )ψs (x2 ) + ψs (x1 )ψr (x2 )) and the corresponding eigenvalues are 2Er and Er +Es The factor of 2− just ensures normalisation and √1 (1 ψr |2 ψs | + ψs |2 ψr |) √12 (|ψr |ψs evaluates to δrr δss + δrs δr s ˆ = For N spinless bosons with H √1 N! Spin (|ψr1 |ψrN will be studied later in the course N + |ψs |ψr ) ˆ i we get H + permutations thereof) if ri = rj CHAPTER MULTIPARTICLE SYSTEMS 16 3.2.3 Spin 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 |x |s ≡ |x, s and the wavefunctions can be written ψs (x) = x, s|ψ 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 to be natural to take λ = ± 21 χλ (1) We can also think of the vector χλ = , in which case two possible basis χλ (2) vectors are and Note that χ†λ χλ = δλλ The scalar product is defined in the obvious way: φr λ |φrλ = ψr |ψr χλ |χλ , 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 ˆ we must have Ψ(x1 , s1 ; x2 , s2 ) → −Ψ(x2 , s2 ; x1 , s1 ) The two particle operator U states belong to the antisymmetric combination (H1 ⊗ H2 )A For N electrons the obvious thing can be done Basis for symmetric or antisymmetric particle spin states There is only one antisymmetric basis state χA (s1 , s2 ) = √ χ (s1 )χ− (s2 ) − χ− (s1 )χ (s2 ) , 2 2 and three symmetric possibilities:  χ (s )χ (s )    2 χS (s1 , s2 ) = √12 χ (s1 )χ− (s2 ) + χ− (s1 )χ (s2 ) 2 2   χ (s )χ (s ) − 12 − 12 s1 = s2 ˆ = Hˆ1 + Hˆ2 and take We can now examine two non-interacting electrons, with H Hi independent of spin The single particle states are |ψi |χs The two electron states live in (H1 ⊗ H2 )A , which has a basis |ψr |ψr |χA ; √ (|ψr |ψs + |ψs |ψr ) |χA ; r = s √ (|ψr |ψs − |ψs |ψr ) |χS ; r = s, 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 ˆ 21 ˆ2 p 2e2 2e2 ˆ = p H + − − + 2m 2m 4π |ˆ x1 | 4π |ˆ x2 | 4π e2 ˆ2 | x1 − x |ˆ 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) −2e2 has En = 8π , 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 3.3 Two particle states and centre of mass 2 ˆ = pˆ + pˆ + V (ˆ ˆ ) defined on H2 We can Suppose we have a Hamiltonian H x1 − x 2m 2m separate out the centre of mass motion by letting ˆ 2) (ˆ p1 − p ˆ =p ˆ1 + p ˆ2 P ˆ= p ˆ = (ˆ ˆ2 ) x1 + x X ˆ=x ˆ1 − x ˆ2 x ˆ P ˆ and x ˆ i , Pˆj = ı δij , [ˆ ˆ, p ˆ commute respectively xi , pˆj ] = ı δij and X, Then X ˆ2 ˆ2 p ˆ ˆ ˆ = P We can rewrite the Hamiltonian as H x), where M = 2m and 2M + h, h = m + V (ˆ ˆ and P ˆ and has wavewe can decompose H into HCM ⊗ Hint HCM is acted on by X ˆ, p ˆ and any spin operators It has wavefunctions functions φ(X) Hint is acted on by x 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 Schrăodinger equation, we can just have φ(X) = e and then P2 ˆ = Eint ψ E = 2M + Eint We thus need only to solve the one particle equation hψ ˆ we have Under the particle exchange operator U ψ(x, s1 , s2 ) → 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 (|ˆ x|) then we may separate variables to get ψ(x, s1 , s2 ) = Yl x = (−1)l Yl with Yl − |x| x |x| x |x| R(|x|)χ(s1 , s2 ) 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 |Ψ in H1 ⊗H2 can be written as aij |ψi |φj |Ψ = i,j with |ψi spaces ∈ H1 and |φj ∈ H2 assumed orthonormal bases for their respective vector 18 CHAPTER MULTIPARTICLE SYSTEMS Suppose we make a measurement on the first system leaving the second system unchanged, and find the first system in a state |ψi Then ψi |Ψ = j aij |φj , which we write as Ai |φ , where |φ is a normalised state of the second system We interpret |Ai | as the probability of finding system in state |ψi After measurement system is in a state |φ If aij = λi δij (no summation) then Ai = λi and measurement of system as |ψi determines system to be in state |φi Chapter 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 ˆ where H ˆ describes a solvable system with known eigenvalues ˆ =H ˆ0 +H We write H ˆ is in some sense small and eigenvectors, and H ˆ and expand the eigenvalues and eigenvectors in ˆ ˆ + λH We write H(λ) = H powers of λ Finally we set λ = to get the result Note that we 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 Schrăodinger equation 4.2 Non-degenerate perturbation theory ˆ |n = n |n for n = 0, 1, We thus assume discrete energy levels Suppose that H ˆ and we assume further that the energy levels are non-degenerate We also require H to be sufficiently non-singular to make a power series expansion possible ˆ We have the equation H(λ)|ψ n (λ) = En (λ)|ψn (λ) We suppose that En (λ) tends to n as λ → and |ψn (λ) → |n as λ → We pose the power series expansions En (λ) = n + λEn(1) + λ2 En(2) + |ψn (λ) = N |n + |n(1) + , substitute into the Schrăodinger equation and require it to be satisfied at each power of λ The normalisation constant N is easily seen to be + O(λ2 ) The O(1) equation is automatically satisfied and the O(λ) equation is ˆ |ψn(1) + H ˆ |n = En(1) |n + H (1) (1) (1) n |ψn Note that we can always replace |ψn with |ψn + α|n and leave this equation (1) unchanged We can therefore impose the condition n|ψn = If we apply n| to 19 CHAPTER PERTURBATION EXPANSIONS 20 (1) ˆ |n — the first order perturbation in energy If we this equation we get En = n|H apply r| where r = n we see that r|ψn(1) = − ˆ |n r|H r − n and therefore |ψn(1) = − r=n ˆ |n |r r|H r − 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 ˆ |ψn(1) En(2) = n|H ˆ |n r|H =− r=n r − n ˆ |n = This procedure is valid if r − n is not very small when r|H d ˆ |ψn (λ) and Using these results we can see that dλ En (λ) = ψn (λ)|H ∂ |ψn (λ) = − ∂λ Also ˆ ∂H ∂λ r=n ˆ |ψn (λ) |ψr (λ) ψr (λ)|H Er (λ) − En (λ) ˆ and so =H ∂2 ˆ ∂ |ψn (λ) En (λ) = ψn (λ)|H ∂λ2 ∂λ Example: harmonic oscillator ˆ0 + H ˆ , where ˆ = pˆ2 + mω x ˆ2 + λmω x ˆ2 , which can be viewed as H Consider H 2m ˆ H0 is the plain vanilla quantum harmonic oscillator Hamiltonian Calculating the matrix elements r|ˆ x2 |n required is an extended exercise in manipulations of the annihilation and creation operators and is omitted The results are En(1) = ω n + En(2) = − 1 ω n+ 2 We thus get the perturbation expansion for En En = ω n + 1+λ− λ2 + O(λ3 ) This system can also be solved exactly to give En = ω n + agrees with the perturbation expansion √ + 2λ which 4.3 DEGENERACY 21 4.3 Degeneracy The method given here breaks down if r = n for r = n Perturbation theory can be extended to the degenerate case, but we will consider only the first order shift in r We suppose that the states |n, s , s = Nn have the same energy n Nn is the degeneracy of this energy level ˆ =H ˆ + λH ˆ such that H ˆ |n, s = n |n, s As before we pose a Hamiltonian H and look for states |ψ(λ) with energy E(λ) → n as λ → The difference with the previous method is that we expand |ψ(λ) as a power series ˆ That is in λ in the basis of eigenvectors of H |n, s as + λ|ψ (1) |ψ(λ) = s As the as are arbitrary we can impose the conditions n, s|ψ (1) = for each s and ˆ n We thus have to solve H|ψ(λ) = E(λ)|ψ(λ) with E(λ) = n + λE (1) If we take the O(λ) equation and apply n, r| to it we get s ˆ |n, s = ar En(1) as n, r|H which is a matrix eigenvalue problem Thus the first order perturbations in n are ˆ |n, s If all the eigenvalues are distinct then the eigenvalues of the matrix n, r|H 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 Don’t ask 22 CHAPTER PERTURBATION EXPANSIONS Chapter General theory of angular momentum ˆ and p ˆ with the commutation For a particle with position and momentum operators x ˆ =x ˆ is Hermitian and ˆ∧p ˆ It can be seen that L relations [xˆi , pˆj ] = ı δij we define L ˆ ˆ ˆ it is easy to show Li , Lj = ı ijk Lk 5.1 Introduction We want to find out if there are other Hermitian operators J which satisfy this commutation relation.1 We ask on what space of states can this algebra of operators be realised, or alternatively, what are the representations? We want [Ji , Jj ] = ı ijk Jk We will choose one component of J whose eigenvalues label the states In accordance with convention we choose J3 Note that J2 , J3 = 0, so we can simultaneously diagonalise J2 and J3 Denote the normalised eigenbasis by |λ µ , so that J2 |λ µ = λ|λ µ and J3 |λ µ = µ|λ µ We know that λ ≥ since J2 is the sum of the squares of Hermitian operators † Now define J± = J1 ± ıJ2 These are not Hermitian, but J+ = J− It will be useful to note that [J± , J3 ] = ±J± [J+ , J− ] = 2J3 J2 = (J+ J− + J− J+ ) + J32 = J+ J− − J3 + J32 = J− J+ + J3 + J32 The is taken outside - you can put it back in if you want, it is inessential but may or may not appear in exam questions Since we are now grown up we will omit the hats if they not add to clarity 23 CHAPTER GENERAL THEORY OF ANGULAR MOMENTUM 24 Proof of this is immediate Using the [J± , J3 ] relation we have J3 J± = J± J3 ± J3 , so that J3 J± |λ µ = (µ ± 1) J± |λ µ and J± |λ µ is an eigenstate of J3 with eigenvalue µ ± By similar artifice we can see that J± |λ µ is an eigenstate of J2 with eigenvalue λ Now evaluate the norm of J± |λ µ , which is λ µ|J∓ J± |λ µ = λ − µ2 ∓ µ ≥ We can now define states |λ (µ ± n) for n = 0, 1, 2, We can pin them down more by noting that λ − µ2 ∓ 2µ ≥ for positive norms However the formulae we have are, given λ, negative for sufficiently large |µ| and so to avoid this we must have µmax = j such that J+ |j = 0: hence λ − j − j = and so λ = j(j + 1) We can perform a similar trick with J− ; there must exist µmin = −j such that J− | − j = 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, 32 , In summary the states can be labelled by |j m such that J2 |j m = j(j + 1)|j m J3 |j m = m|j m J± |j m = ((j ∓ m) (j ± m + 1)) |j m ± with m ∈ {−j, −j + 1, , j − 1, j} and j ∈ {0, 12 , 1, 32 , } There are 2j + states with different m for the same j |j m is the standard basis of the angular momentum states ˆ∧p ˆ We have obtained a representation of the algebra labelled by j If J = L = x we must have j an integer ˆ Recall that if we have Aˆ we can define a matrix Aλ λ by A|λ = λ |λ Aλ λ Note that (BA)λ λ = µ Bλ µ Aµλ Given j, we have (J3 )m m = m δm m and (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− ] = 2J3 If J are angular momentum operators which act on a vector space V and we have |ψ ∈ V such that J3 |ψ = k|ψ and J+ |ψ = then ψ is a state with angular n |ψ , ≤ n ≤ 2k The conditions momentum j = k The other states are given by J− also give J |ψ = k (k + 1) |ψ 5.1.1 Spin particles This is the simplest non-trivial case We have j = 12 and a two dimensional state space with a basis | 12 12 and | 12 − 12 We have the relations J3 | 12 ± 12 = ± 21 | 12 ± 12 and J+ | 12 J+ | 12 − 2 J− | 12 =0 = | 12 J− | 12 − 2 = | 12 − = 5.1 INTRODUCTION 25 It is convenient to introduce explicit matrices σ such that J| 12 m = m | 12 m (σ)m m The matrices σ are × matrices (called the Pauli spin matrices) Explicitly, they are σ1 = σ+ = 0 (σ+ + σ− ) = 1 σ− = σ2 = −ı (σ+ − σ− ) = 0 σ3 = 0 −1 −ı ı Note that σ12 = σ22 = σ32 = and σ † = σ These satisfy the commutation relations [σi , σj ] = 2ı ijk σk (a slightly modified angular momentum commutation relation) and we also have σ2 σ3 = ıσ1 (and the relations obtained by cyclic permutation), so ˆ is a unit vector we have (σ.ˆ n)2 = and we see that σi σj + σj σi = 2δij Thus if n σ.ˆ n has eigenvalues ±1 We define the angular momentum matrices s = 12 σ and so s2 = 34 1 and χ− 12 = The basis states are χ 12 = 5.1.2 Spin particles We apply the theory as above to get  S+ =  0 √  √0 2 0  S3 =  0 0  0 −1 † and S− = S+ 5.1.3 Electrons ˆ∧p ˆ + s, s Electrons are particles with intrinsic spin 12 The angular momentum J = x are the spin operators for spin 12 ˆ, p ˆ and s We can represent these operators The basic operators for an electron are x by their action on two component wavefunctions: ψ(x) = ψλ (x)χλ λ=± 12 ˆ → x, p ˆ → −ı ∇ and s → σ All other operators are constructed In this basis x in terms of these, for instance we may have a Hamiltonian H= ˆ2 p + V (x) + U (x)σ.L 2m ˆ∧p ˆ where L = x If V and U depend only on |x| then [J, H] = CHAPTER GENERAL THEORY OF ANGULAR MOMENTUM 26 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, 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 ijk Jk We want to construct states |J M forming a standard angular momentum basis, that is such that: J3 |J M = M |J M ± |J M ± J± |J M = NJ,M ± with NJ,M = (J ∓ M ) (J ± M + 1) We look for states in V which satisfy J+ |J J = and J3 |J J = J|J J The maximum value of J we can get is j1 + j2 ; and |j1 + j2 j1 + j2 = |j1 j1 |j2 j2 Then J+ |j1 + j2 j1 + j2 = Similarly this is an eigenvector of J3 with eigenvalue (j1 + j2 ) We can now apply J− repeatedly to form all the |J M states Applying J− we get |J M − = α|j1 j1 − 1 |j2 j2 + β|j1 j1 |j2 j2 − − The coefficents α and β can be determined from the coefficents Na,b , and we must 2 have α + β = If we choose |ψ a state orthogonal to this; |ψ = −β|j1 j1 − 1 |j2 j2 + α|j1 j1 |j2 j2 − J3 |ψ can be computed and it shows that |ψ is an eigenvector of J3 with eigenvalue (j1 + j2 − 1) Now = ψ|j1 + j2 j1 + j2 − ∝ ψ|J− |j1 + j2 j1 + j2 and so ψ|J− |φ = for all states |φ in V Thus J+ |ψ = and |ψ = |j1 + j2 − j1 + j2 − We can then construct the states |j1 + j2 − M by repeatedly applying J− For each J such that |j1 − j2 | ≤ J ≤ j1 + j2 we can construct a state |J J We define the Clebsch-Gordan coefficients j1 m1 j2 m2 |J M , and so j1 m1 j2 m2 |J M |j1 m1 |j2 m2 |J M = m1 ,m2 The Clebsch-Gordan coefficients are nonzero only when M = m1 + m2 We can check the number of states; j1 +j2 j1 +j2 (J + 1)2 − J = (2j1 + 1) (2j2 + 1) (2J + 1) = J=|j1 −j2 | J=|j1 −j2 | 5.3 THE MEANING OF QUANTUM MECHANICS 27 Electrons Electrons have spin 12 and we can represent their spin states with χ± 12 (s) Using this notation we see that two electrons can form a symmetric spin triplet   1  χ (s1 )χ (s2 ) χm (s1 , s2 ) = √2 χ 12 (s1 )χ− 12 (s2 ) + χ− 12 (s1 )χ 12 (s2 )   χ (s )χ (s ) −2 −2 and an antisymmetric spin singlet; χ0 (s1 , s2 ) = √ χ 12 (s1 )χ− 12 (s2 ) − χ− 12 (s1 )χ 12 (s2 ) 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 We will take a spin example to show this Consider a spin 12 particle, with two spin states |↑ and |↓ which are eigenvectors of S3 = σ3 If we choose to use two component vectors we have χ↑ = χ↓ = Suppose n = (sin θ, 0, cos θ) (a unit vector) and let us find the eigenvectors of σ.n = cos θ sin θ sin θ − cos θ As (σ.n) = we must have eigenvalues ±1 and an inspired guess gives χ↑,n and χ↓,n as χ↑,n = cos θ2 χ↑ + sin θ2 χ↓ χ↓,n = − sin θ2 χ↑ + cos θ2 χ↓ and Thus (reverting to ket vector notation) if an electron is in a state |↑ then the probability of finding it in a state |↑, n is cos2 θ2 and the probability of finding it in a state |↓, n is sin2 θ2 Now, suppose we have two electrons in a spin singlet state; |Φ = √ {|↑ |↓ 2 − |↓ |↑ } Then the probability of finding electron with spin up is 12 , and after making this measurement electron must be spin down Similarly, if we find electron with spin down (probability 12 again) then electron must have spin up More generally, suppose we measure electron 1’s spin along direction n Then we see that the probability for CHAPTER GENERAL THEORY OF ANGULAR MOMENTUM 28 electron to have spin up in direction n (aligned) is 12 and then electron must be in the state |↓, n If we have two electrons (say electron and electron 2) in a spin state we may physically separate them and consider independent experiments on them We will consider three directions as sketched For electron 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 this for electron (1) (2) We see that if we find electron has Sz = then electron has Sz = −1 (etc.) If there exists an underlying deterministic theory then we could expect some probability distribution p for this set of experiments; (1) (2) ≤1 ≤ p Sz(1) , Sn(1) , Sm , Sz(2) , Sn(2) , Sm (1) (2) which is nonzero only if Sdirn = −Sdirn and p({s}) = {s} Bell inequality Suppose we have a probability distribution p(a, b, c) with a, b, c = ±1 We define partial probabilities pbc = a p(a, b, c) and similarly for pac and pab Then 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) Applying this to the two electron system we get (2) (2) P Sn(1) = 1, Sm = ≤ P Sz(1) = 1, Sn(2) = −1 + P Sz(1) = −1, Sm =1 We can calculate these probabilities from quantum mechanics P Sz(1) = 1, Sn(2) = −1 = P Sz(1) = 1, Sn(1) = = cos2 (2) P Sz(1) = −1, Sm = = cos2 θ+φ θ and (2) P Sn(1) = 1, Sm = = sin2 θ2 The Bell inequality gives sin2 φ ≤ cos2 θ + cos2 θ+φ which is not in general true 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, 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 ... POSSIBILITY OF SUCH DAMAGE Chapter 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... 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... the course Foundations of Quantum Mechanics given by Dr H Osborn in Cambridge in the Michælmas Term 1997 Recommended books are discussed in the bibliography at the back Other sets of notes are