852HW1_Solutions

# 852HW1_Solutions - PHYS852 Quantum Mechanics II Spring 2009...

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Unformatted text preview: PHYS852 Quantum Mechanics II, Spring 2009 HOMEWORK ASSIGNMENT 1: Solutions 1. [25 points] This problem shows you how to derive the matrix representations of spin operators from ﬁrst principles. a.) For spin 1/2, use the eigenvalue equation Sz |ms = ms |ms to ﬁnd the components of the two-by-two matrix representation of Sz in the basis of its own eigenstates. b.) From the deﬁnition S± = Sx + iSy , write Sx and Sy in terms of S+ and S− . c.) The relation J± |j, mj = j (j + 1) − mj (mj ± 1)|j, mj ± 1 is valid for any angular momentum operators. Apply this to the spin-1/2 case to ﬁnd the matrix elements of S+ and S− in the basis of eigenstates of Sz . d.) From your previous answers, derive the matrix representations of Sx and Sy for spin-1/2. e.) Show explicitly that S 2 = S · S = Answer: a.) Sz = +|Sz |+ −|Sz |+ +|Sz |− −|Sz |− = 2 2 2 s(s + 1). +|+ −|+ − 2 +|− − 2 −|− = 2 10 0 −1 b.) From S+ = Sx + iSy we get iSy = S+ − Sx . Putting this into S− = Sx − iSy and solving for Sx gives Sx = 1 (S+ + S− ) 2 Alternatively, from S+ = Sx + iSy we can get Sx = S+ − iSy . 1 Putting this into S− = Sx − iSy and solving for Sy gives Sy = 2i (S+ − S− ) c.) For s = 1/2, m = 1/2, we have s(s + 1) − m(m ± 1) = 0, 1 For s = 1/2, m = −1/2, we have s(s + 1) − m(m ± 1) = 1, 0 This gives: S+ = +|S+ |+ −|S+ |+ +|S− |+ −|S− |+ +|S+ |− −|S+ |− +|S− |− −|S− |− 01 00 01 00 = 0 0 +|+ −|+ 0 0 = 01 00 00 10 01 10 0 −i i0 S− = d.) Sx = Sy = = +|− −|− + − 00 10 00 10 = 1 1 (S+ + S− ) = 2 2 = = 2 2 1 −i (S+ − S− ) = 2i 2 1 e.) 2 2 Sx = 2 2 Sy = 4 01 10 0 −i i0 10 0 −1 32 = 4 01 10 0 −i i0 10 0 −1 21 2 = = = 21 4 2 10 01 10 01 10 01 = 2 2 4 = = I 2 4 2 4 2 4 4 I I 2 2 Sz = 4 4 2 2 2 Sx + Sy + Sz = 3 = 22 2 1 +1 2 s(s + 1) 2 2. [15 points] Consider an electron whose position is held ﬁxed, so that it can be described by a simple two-component spinor. The initial state of the particle is spin-up (ms = 1/2) with respect to the z-axis. At time t = 0 a uniform magnetic ﬁeld is applied along the y-axis. What is the state-vector of this system at any arbitrary time t > 0. Answer: Choosing z as our quantization axis, we have |ψ (0) = |+ . The Hamiltonian for our system is H= In matrix form, this gives H= The eigenvectors of σy are found via det −λ −i i −λ =0 eB0 me 0 −i i0 . 2e 2eB0 eB0 B·S = Sy = σy . me me me which gives the characteristic equation λ2 − 1 = 0, so λ = ±1. 1 the corresponding eigenvectors are |λ± = √2 (|+ ± i|− ). Thus the eigenfrrequencies of our system are ω± = ± and the eigenvectors are eB0 me 1 |ω± = √ (|+ ± i|− ) 2 The state at time t is then |ψ (t) = |ω+ e−iω+ t ω+ |+ + |ω− e−iω− t ω− |+ With ω0 = eB0 me , this gives 1 1 |ψ (t) = (|+ + i|− )e−iω0 t + (|+ − i|− )eiω0 t = |+ cos(ω0 t) + |− sin(ω0 t) 2 2 3 3. [15 points] Cohen-Tannoudji problem 9.1, page 990 Answer: From the expressions, we can deduce that 1 |ψ = √ |R 3 r ⊗ √ 3|00 θφ + |10 θφ ⊗ |+ s + (|11 θφ − |10 θφ ) ⊗ |− s , where the subscripts r , θφ, and s indicate the radial, angular, and spin Hilbert spaces, respectively. The angular Hilbert-space vectors are |ℓm states. a.) from ψ |ψ = 1 we ﬁnd 1 R|R [3 + 1 + 1 + 1] = 2 R|R 3 so we need R|R = 1/2 for normalization. 1= b.) If Sz is measured, we would obtain /2 with probability p+ = ψ |+ +|ψ = probability p− = ψ |− −|ψ = 1 . 3 c.) If L2 is measured and 0 is obtained, then the state collapses to |ψ ′ = 1 1 |00 00|ψ = |R ⊗ |+ N N 1 √, 2 2 3 and − /2 with from ψ ′ |ψ ′ = 1 we ﬁnd N 2 = 1 R|R +|+ = 1/2, so N = |ψ ′ = if instead, 2 2 which gives √ 2|R ⊗ |+ is obtained, then we have |ψ ′ = 1 1 |R ⊗ (|+ − |− ) (|10 10| + |11 11|) |ψ = √ N 3N 1 √, 3 from ψ ′ |ψ ′ = 1 we ﬁnd 3N 2 = 2 R|R = 1, so N = so that |ψ ′ = |R ⊗ (|+ − |− ) 4 4. [15 points] Cohen-Tonnoudji problem 9.2, page 990 Answer: †† †† †† a.) We have A = S · P = Sx Px + Sy Py + Sz Pz , so A† = Px Sx + Py Sy + Py Sy . Since all the components of S and P are hermitian, and because components of S commute with components of P (since they operate in diﬀerent Hilbert spaces), it is clear that A† = A. b.) The commutator of A with any component of P is zero, because momentum components all commute with each other and with spin operators. Thus A, Px , Py , and Pz form a set of four compatible observables, so that simultaneous eigenstates do exist. If we consider P to be constant, then we have A = px Sx + py Sy + pz Sz . In matrix form this gives A= 2 px 01 10 + py 0 −i i0 + pz 10 0 −1 = 2 pz px − ipy px + ipy −pz c.) Let px = p sin θ cos φ, py = p sin θ sin φ, and pz = p cos θ , then we have A= p 2 cos θ sin θe−iφ sin θeiφ − cos θ we can recognize this as the component of the spin along the θ , φ direction, which is parallel to p. Thus we know the eigenstates are the eigenstates of Sθφ , given by |+θφ = − cos(θ/2)e−iφ/2 |+z + sin(θ/2)eiφ/2 |−z |−θφ = − sin(θ/2)e−iφ/2 |+z + cos(θ/2)eiφ/2 |−z We also know then that the eigenvalues of A must then be p/2 and − p/2. There is a continuum of degenerate eigenstates, corresponding to all p with the same amplitude p. A system of common eigenvectors would be a set with spin up or down along the p direction, |p, ± , which would satisfy |p| A|p, ± = ± 2 Px |p, ± = px |p, ± Py |p, ± = py |p, ± Py |p, ± = py |p, ± The eigenvectors |p, ± break down into coordinate and spin Hilbert space vectors as |p, ± = |p ⊗ |±θφ 5 5. [20 points] Cohen-Tannoudji problem 9.3, page 991 Answer: Starting from H= we can use the identity (B-12) (σ · A)(σ · B ) = A · B + iσ · (A × B ) with A = B = P − q A(R). Keeping in mind that the cross-pruduct of a vector-operator with itself need not be zero, we ﬁnd H= Now P − qa(R) × P − qa(R) = P × P − qa(R) × P − q P × a(R) + q 2 a(R) × a(R) = −qex (ay Pz − az Py + Py az − Pz ay ) − qey (az Px − ax Pz + Pz ax − Px az ) = −qex ([ay , Pz ] − [az , Py ]) + . . . now x|[f (X ), Px ]|ψ = x|f (X )Px |ψ − x|Px f (X )|ψ ∂ x|f (X )|ψ = f (x) x|Px |ψ + i ∂x ∂ − −i f (x)ψ ′ (x) + i f (x)ψ (x) ∂x = −i f ψ ′ + i f ′ ψ + i f ψ ′ −qez (ax Py − ay Px + Px ay − Py ax ) = −qa(R) × P − q P × a(R) 1 P − qa(R) 2m 2 1 σ · P − qa(R) 2m 2 + qu(R, t) + i σ· 2m P − qa(R) × P − qa(R) + qu(R, t) = i f ′ψ so that = i x|f ′ (X )|ψ [f (X ), Px ] = i f ′ (X ) So we have P − qa(R) × P − qa(R) = −i q ex = i q B (R) ∂ ay ∂az − ∂z ∂y + ... = i q ∇ × a(R) Thus we have H= 1 P − qa(R) 2m 2 − q σ · B (R) + qu(R, t) 2m 6 ...
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## This note was uploaded on 10/25/2010 for the course PHYSICS PHYS 851 taught by Professor Michaelmoore during the Fall '08 term at Michigan State University.

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