Solutions to chapter 4 problems

Transcription

1 Chapter 9 Solutions to chapter 4 problems Solution to Exercise 47 For example, the x component of the angular momentum is defined as ˆL x ŷˆp z ẑ ˆp y The position and momentum observables are Hermitian; in addition, we have [ŷ, ˆp z ] [ẑ, ˆp y ] We can thus write for the Hermitian conjugate of ˆL x Solution to Exercise 4 ˆL x Ex 63 ˆp zŷ ˆp yẑ ŷˆp z ẑ ˆp y ˆL x a We use ˆL j ɛ jmnˆr m ˆp n and [ˆr j, ˆp k ] i δ jk to write [ˆL j, ˆr k ] [ɛ jmnˆr m ˆp n, ˆr k ] ɛ jmnˆr m [ˆp n, ˆr k ] ɛ jmn and ˆr m can be factored out because they commute with ˆr k ɛ jmnˆr m i δ nk i ɛ jmkˆr m i ɛ jkmˆr m ɛ jkm is an antisymmetric tensor, so ɛ jkm ɛ jmk b Similarly, c [ˆL j, ˆp k ] [ɛ jmnˆr m ˆp n, ˆp k ] ɛ jmn ˆp n [ˆr m, ˆp k ] ɛ jmn ˆp n i δ mk i ɛ jkn ˆp n [ˆL j, ˆL k ] [ɛ jmnˆr m ˆp n, ɛ klqˆr l ˆp q ] ɛ jmn ɛ klq [ˆr m ˆp n, ˆr l ˆp q ] 46 ɛ jmn ɛ klq ˆrm [ˆp n, ˆr l ]ˆp q + ˆr l [ˆr m, ˆp q ]ˆp n ɛ jmn ɛ klq i δnlˆr m ˆp q + i δ mqˆr l ˆp n i ɛ jml ɛ klqˆr m ˆp q + i ɛ jmn ɛ klmˆr l ˆp n i ɛ ljm ɛ lqkˆr m ˆp q + i ɛ mnj ɛ mklˆr l ˆp n Ex 48 i δ jq δ mk δ jk δ mq ˆr m ˆp q + i δ kn δ jl δ nl δ jk ˆr l ˆp n On the other hand, i ˆr k ˆp j + i δ jkˆr m ˆp m + i ˆr j ˆp k i δ jkˆr l ˆp l i ˆr k ˆp j + i ˆr j ˆp k i ɛ jkl ˆLl i ɛ jkl ɛ lmnˆr m ˆp n i ɛ ljk ɛ lmnˆr m ˆp n Ex 48 i δ jm δ kn δ jn δ km ˆr m ˆp n i ˆr j ˆp k i ˆr k ˆp j

2 34 CHAPTER 9 SOLUTIONS TO CHAPTER 4 PROBLEMS Comparing the two expressions above, we obtain the desired result: [ˆL j, ˆL k ] i ɛ jkl ˆLl ; d Here we use that the square of a vector is its inner product with itself: ˆr 2 ˆr mˆr m Therefore, [ˆL j, ˆr kˆr k ] 46 ˆr k [ˆL j, ˆr k ] + [ˆL j, ˆr k ]ˆr k Ex 4a 2ˆr k ɛ jklˆr l This expression is symmetric with respect to indices k and l On the other hand ɛ jkl is an antisymmetric tensor ie it changes sign when two indices are interchanged We thus conclude that [ˆL j, ˆr 2 ] e The argument is analogous to part d: [ˆL j, ˆp k ˆp k ] ˆp k [ˆL j, ˆp k ] + [ˆL j, ˆp k ]ˆp k 2ˆp k ɛ jkl ˆp l f Again, [ˆL j, ˆL k ˆLk ] ˆL k [ˆL j, ˆL k ] + [ˆL j, ˆL k ]ˆL k 2ˆL k ɛ jkl ˆLl Solution to Exercise 43 The expectation value of the operator ˆL 2 in its eigenstate λm equals λm ˆL 2 λm 2 λ On the other hand, because λm is also an eigenstate of ˆL z with eigenvalue m, this expectation value equals λm ˆL 2 λm λm ˆL2 x + ˆL 2 y + ˆL 2 z λm 2 m 2 + λm ˆL2 x + ˆL 2 y λm The expectation value of the of the operator ˆL 2 x + ˆL 2 y is then 2 λ m 2 Because a square of a Hermitian operator cannot have a negative expectation value, we must have m 2 λ Solution to Exercise 44 a The angular momentum components are Hermitian operators, so ˆL x ˆL x and iˆl y iˆl y Therefore, ˆL + ˆL x + iˆl y ˆL x iˆl y ˆL b Using the result of Ex 4, we find c From [ˆL z, ˆL ± ] [ˆL z, ˆL x ± iˆl y ] i L y ± i i ˆL x ±ˆL x + iˆl y ± ˆL ± ; [ˆL 2, ˆL ± ] [ˆL 2, ˆL x ± iˆl y ] [ˆL 2, ˆL x ] ± i[ˆl 2, ˆL y ] ; [ˆL +, ˆL ] [ˆL x + iˆl y, ˆL x iˆl y ] i[ˆl y, ˆL x ] i[ˆl x, ˆL y ] 2ˆL z ; ˆL + ˆL ˆL x +iˆl y ˆL x iˆl y ˆL 2 x+ ˆL 2 y iˆl x ˆLy ˆL y ˆLx ˆL 2 ˆL 2 z i[ˆl x, ˆL y ] ˆL 2 ˆL 2 z+ ˆL z ; ˆL ˆL+ ˆL x iˆl y ˆL x +iˆl y ˆL 2 x+ ˆL 2 y+iˆl x ˆLy ˆL y ˆLx ˆL 2 ˆL 2 z+i[ˆl x, ˆL y ] ˆL 2 ˆL 2 z ˆL z we find the required relation ˆL 2 ˆL + ˆL + ˆL 2 z ˆL z ˆL ˆL+ + ˆL 2 z + ˆL z Solution to Exercise 45 This general statement can be proven as follows Consider some Hermitian operator  The expectation value of  2 in an arbitrary quantum state ψ equals ψ Â2 ψ ψ ˆ ψ ψ  v i v i  ψ ψ  2 v i, i i where { v i } is an arbitrary orthonormal basis

3 a In order to verify if the state ˆL + λm is an eigenstate of ˆL 2 and ˆL z, let us subject this state to the action of these operators Because ˆL 2 commutes with ˆL +, we have ˆL 2 ˆL+ λm ˆL + ˆL2 λm ˆL + 2 λ λm 2 λˆl + λm In other words, ˆL + λm is an eigenstate of ˆL 2 with eigenvalue 2 λ To perform a similar calculation for ˆL z, we rewrite the expression for the commutator of ˆL z and ˆL + obtained in Ex 44 as follows: and thus ˆL z ˆL+ ˆL + ˆLz + ˆL +, ˆL z ˆL+ λm ˆL+ ˆLz + ˆL + λm mˆl+ + ˆL + λm m + ˆL+ λm We see that the action of the operator ˆL z on the state ˆL + λm is equivalent to multiplying this state by m +, so ˆL + λm is an eigenstate of ˆL z with eigenvalue m + 35 b Similarly, because ˆL z ˆL ˆL ˆLz ˆL, we have ˆL z ˆL λm ˆL ˆLz ˆL λm mˆl ˆL λm m ˆL λm, so ˆL λm is an eigenstate of ˆL z with eigenvalue m The proof for ˆL 2 is analogous to part a Solution to Exercise 46 a Let ψ ˆL + λm From the previous exercise, we know that ψ is an eigenstate of ˆL z with eigenvalue m +, ie ψ A λ, m +, where A is some constant We need to find A To this end, we notice that ψ λm ˆL + λm ˆL and calculate ψ ψ λm ˆL Ex 44c ˆL+ λm λm ˆL 2 ˆL 2 z ˆL z λm 2 λ m 2 m in the last equality, we used the fact that λm is an eigenstate of both ˆL 2 and ˆL z But on the other hand, ψ ψ A 2 λ, m + λ, m + A 2, because the eigenstates of the angular momentum operator are normalized Arbitrarily choosing the phase of A equal to zero, we find A λ mm + b Similarly, if φ ˆL λm, then, on one hand, φ φ λm ˆL + ˆL λm λm ˆL 2 ˆL 2 z + ˆL z λm 2 λ m 2 + m and on the other hand Therefore, B λ mm φ φ B 2 λ, m λ, m B 2 Solution to Exercise 48 First, because the state λm is an eigenstate of both ˆL 2 and ˆL z, we have lm ˆL 2 l m 2 l l + lm l m ; 9

4 36 CHAPTER 9 SOLUTIONS TO CHAPTER 4 PROBLEMS lm ˆL z l m m lm l m ; 92 Both these expressions vanish if l l Second, the action of the raising and lowering operators on states λm is known from Ex 46: lm ˆL ± l m l l + m m ± lm l, m ±, 93 so these operators also preserve l Finally, the x- and y-components of the angular momentum can be written as linear combinations of the raising and lowering operators according to the definition 42 of the latter: ˆL y L + L 2i and hence they must preserve the value of l as well ˆL x L + + L ; 94 2 Solution to Exercise 49 In the case l /2, the eigenvalue of m can be either +/2 or /2 Hence the matrices are two-dimensional The matrix elements can be found according to Eqs 9 95 For ˆL 2 we have 2, m ˆL 2 2, m , m 2, m δ mm, 95 so the matrix is ˆL According to 92, the matrix element of ˆL z is 2, m ˆL z 2, m m 2, m 2, m m δ mm, so the matrix is ˆL z 2 Our next step is to find the matrices of the raising and lowering operators According to Eq 93, nonvanishing matrix elements of ˆL + have m m + If l l /2, the only possibility to satisfy this condition is to set m /2, m /2 Then l l + m m ± and By the same token, Now, using Eqs, 94 and 95, we find ˆL x 2 ˆL + ˆL ; ˆLy i 2 i Solution to Exercise 438 We first notice that the operator Ŝ R for a spin-/2 particle can be written as Ŝ R Ŝ 44 R 2 ˆσ R,

5 where ˆ σ ˆσ x, ˆσ y, ˆσ z is the vector consisting of Pauli operators and R is a unit length vector From Ex 88 we know that the operator ˆσ R has eigenvalues λ,2 ± and hence the eigenvalues of Ŝ R are ± /2 Our goal is to find the eigenstate of this operator with the eigenvalue + /2 We begin by expressing Ŝ R in the matrix form in the canonical basis 2 : Ŝ R sin θ cos φ Ŝ x + sin θ sin φ Ŝ y + cos θ Ŝ z [ 44 sin θ cos φ sin θ sin φ i i + cos θ cos θ sin θ cos φ i sin θ sin φ sin θ cos φ + i sin θ sin φ cos θ After some straightforward transformations, we find the normalized eigenstate: m s R /2 cosθ /2 sinθ /2e iφ Let us calculate this eigenstate for a few special cases: direction of R θ, φ eigenstate +z, z π, ] x π/2, 2 x π/2, π 2 +y π/2, π/2 2 i y π/2, π/2 2 i Of course, we have already determined these eigenstates of Pauli operators in Chapter Solution to Exercise 44 Let ω be the angular frequency of the particle s orbital motion The particle makes a full circle in time T 2π/ω, and the current associated with this motion is thus I e/t eω/2π The area of the orbit is A πr 2, where r is the radius Substituting these quantities into Eq 436, we find for the magnetic moment µ eωr2 2c On the other hand, the mechanical angular momentum of the orbiting particle is L mωr 2 The magnetic moment can thus be expressed as µ L e 2mc Both the angular momentum and the magnetic moment are actually vectors directed orthogonally out of the plane of the orbit Therefore, the above expression is also valid in its vector form Solution to Exercise 442 Equation 437 holds for all three components of the angular momentum; in particular, the z-component: µ z L z e 2mc 2 Under the canoncial basis of the Hilbert space of a particle s spin states we understand the basis of eigenstates sm s of operators Ŝ2 and Ŝz

6 38 CHAPTER 9 SOLUTIONS TO CHAPTER 4 PROBLEMS The state with a definite magnetic quantum number m l is lm l, an eigenstate of ˆL z with eigenvalue L z m l We can thus write the z component of the magnetic moment as µ z e 2mc m l Solution to Exercise 443 According to Ex 44, the interaction Hamiltonian equals Ĥ ˆµ B Because the magnetic field is in the y direction, we can write H ˆµ y B 44 ge 2mcŜy It is convenient to solve this problem in the matrix notation in the canonical basis Because [Ex 4f] the Hamiltonian commutes with Ŝ2, the absolute value of the spin s /2 is preserved, so we can restrict our Hilbert space to the subspace spanned by states s /2, m s /2 and s /2, m s /2 In this subspace, the initial state has the matrix ψ and the Hamiltonian, according to Eq 44, takes the form Ĥ ˆµ y B geb 2mc 2 ˆσ y The evolution of the electron s spin is than governed by the Schrödinger equation whose solution is ψt i Ĥ ψ, ψt e i Ĥt ψ e i gebt 4mc ˆσ y ψ This matrix exponent has already been calculated in Ex 95: e iϕˆσ y cos ϕ sin ϕ sin ϕ cos ϕ Using this solution, with ϕ gebt 4mc, we find the evolution of the spin: cos ϕ sin ϕ cos ϕ ψt sin ϕ cos ϕ sin ϕ The Stern-Gerlach measurement constitutes a measurement of Ŝz in the state ψt Because this state is already expressed in the canonical basis, we immediately find the probability to detect m s /2 as pr /2 m s /2 ψt 2 cos 2 ϕ Solution to Exercise 444 We proceed along the lines of the previous problem s solution, but the Hamiltonian is now H ˆµ B ge Ŝ ge B 2mc 4mc ˆσ B, where ˆ σ ˆσ x, ˆσ y, ˆσ z is the vector consisting of Pauli operators Hamiltonian is then given by e i Ĥt e i get ˆσ B 4mc e iϕ ˆσ v, The evolution under this where we have defined ϕ gebt 4mc and v sin θ,, cos θ is the unit length vector in the direction of the magnetic field

7 39 Now we are in position to apply the result of Ex 88 We find e i Ĥt cos ϕˆ + i sin ϕ v ˆσ cos ϕ + i sin ϕ [ sin θ cos ϕ + i sin ϕ cos θ i sin ϕ sin θ i sin ϕ sin θ cos ϕ i sin ϕ cos θ Applying this evolution operator to the initial state ψ ψt e i Ĥt ψ + cos θ, we have cos ϕ + i sin ϕ cos θ i sin ϕ sin θ When this state is subjected to the Stern-Gerlach measurement, the probability to detect m s /2 is pr /2 m s /2 ψt 2 cos 2 ϕ + sin 2 ϕ cos 2 θ Solution to Exercise 445 The Hamiltonian associated with the magnetic field along the z axis is given by H ˆµ z B geb 4mc ˆσ z and the associated evolution in the canonical basis e i Ĥt e iϕ e iϕ, with ϕ gebt 4mc Applying this evolution to the eigenstate 96 of the spin Ŝ R oriented along vector R characterized by polar angles θ, φ, we obtain ] ψt e i Ĥt ψ e iϕ cosθ /2 e iϕ sinθ /2e iφ e iϕ cosθ /2 sinθ /2e iφ 2ϕ Comparing this result with Eq 96, we find that the state after the evolution is physically equivalent to an eigenstate of the spin Ŝ R with R characterized by polar angles θ, φ 2ϕ In other words, the spin has precessed by the angle 2ϕ gebt 2mc Note that we can write 2ϕ Ω L t, where Ω L geb 2mc is the Larmor frequency, which is known to determine the precession of magnetic moments in classical physics Solution to Exercise 446 The Stern-Gerlach measurement is that of the spin component Ŝ R with R characterized by polar angles θ, The probabilities of possible measurement outcomes are given by the Second postulate of quantum mechanics: pr i ψ v i 2, where ψ is the input state, whose canonical representation is ψ and v i are the eigenstates of Ŝ R In Ex 438 we have found one of these states: The associated detection probability is then m s R /2 cosθ /2 sinθ /2 pr /2 sin 2 θ /2

8 4 CHAPTER 9 SOLUTIONS TO CHAPTER 4 PROBLEMS The probability of the other measurement result is pr /2 pr /2 cos 2 θ /2