Physics 505 Homework No. 8 Solutions S8-1 1. Spinor rotations. Somewhat based on a problem in Schwabl. (a) Suppose n is a unit vector. We are interested in n σ. Show that (n σ)2 = I, where I is the identity matrix. · · Solution 2 We will use the properties of the Pauli matrices that σi = I and σi,σj = 0, provided i = j. { } 6 2 (n σ) =(nxσx + nyσy + nzσz)(nxσx + nyσy + nzσz) · 2 2 2 2 2 2 = nxσx + nyσy + nzσz + nxny(σxσy + σyσx) + nynz(σyσz + σzσy) + nznx(σzσx + σxσz) 2 2 2 =(nx + ny + nz)I = I End Solution The unitary operator which generates a rotation by ϕ about the axis n in spinor space is iϕ n S/¯h U = e · . (b) Show that U = cos ϕ/2 + in σ sin ϕ/2. · Solution First, we note that n S/¯h = n σ/2. Second, we expand the exponential in a Taylor series and note that since· (n σ)2 =· I, even terms in the series will involve the identity matrix and odd terms will involve· n σ to the first power. · iϕ n S/¯h i(ϕ/2)(n σ) U = e · = e · in in−1 = I (ϕ/2)n + i(n σ) (ϕ/2)n n! n! n=0,2,4,... · n=1,3,5,... X X = I cos ϕ/2 + i(n σ) sin ϕ/2 , · End Solution (c) Show that UσU † = n(n σ) n (n σ) cos ϕ +(n σ) sin ϕ. · − × × × You might find it helpful to remember that a product of two Pauli matrices gives the identity matrix or plus or minus i times the third Pauli matrix according to the formula σiσj = δij + iǫijkσk , Copyright c 2012, Edward J. Groth Physics 505 Homework No. 8 Solutions S8-2 where the summation convention is assumed and where we have just used 1 in place of I. Solution UσU † = (cos ϕ/2 + in σ sin ϕ/2)σ(cos ϕ/2 in σ sin ϕ/2) · − · = cos2 ϕ/2 σ + i cos ϕ/2 sin ϕ/2(n σ σ σ n σ) + sin2 ϕ/2 n σ σ n σ · − · · · At this point, we just grind out the ith component of each term. The first term is just 2 cos ϕ/2 σi. For the second term, we need n σ σi σi n σ = njσjσi σinjσj · − · − = nj[σj,σi] =2injǫjikσk = 2iǫijknjσk − = 2i (n σ) , − × i and the middle term is sin ϕ (n σ) . × i For the third term we need n σ σi n σ = njσjσinkσk · · = njnkσjσiσk = njnk(δji + iǫjilσl)σk = ni n σ + injnkǫjil(δlk + iǫlkmσm) · = ni n σ njǫjilǫlkmnkσm · − = ni n σ njǫjil(n σ)l · − × = ni n σ + ǫijlnj(n σ)l · × = ni n σ +(n (n σ)) · × × i The second term looks like what we want, but we still have more work to do on the first and third terms. We combine the two terms and use the identities cos2 ϕ/2 = (1+cos ϕ)/2 and sin2 ϕ/2=(1 cos ϕ)/2 to get − (σ + n(n σ) + n (n σ))/2 + cos ϕ(σ n(n σ) n (n σ))/2) . · × × − · − × × We now use the vector identity A (B C) = B(A C) C(A B). In the first term above, × × · − · (σ + n(n σ) + n (n σ))/2=(σ + n(n σ) + n(n σ) σ(n n))/2 = n(n σ) . · × × · · − · · Copyright c 2012, Edward J. Groth Physics 505 Homework No. 8 Solutions S8-3 In the second term, we have cos ϕ(σ n(n σ) n (n σ))/2) = cos ϕ(σ n(n σ) n(n σ) + σ(n n))/2 − · − × × − · − · · = cos ϕ(σ n(n σ)) − · = cos ϕ(n (n σ)) . − × × Putting everything together, we get the desired result: UσU † = n(n σ) n (n σ) cos ϕ +(n σ) sin ϕ. · − × × × End Solution (d) Consider the special case n = ez and discuss infinitesimal rotations. Solution In this case, 1 + iδϕ/2 0 U 1 + iσzδϕ/2 = , → 0 1 iδϕ/2 − and ′ † σ = UσU σ exσyδϕ + eyσxδϕ. → − The rotation about the z-axis carries a little bit of the x-component of the Pauli operator into the y-direction and a little bit of the y-component into the negative x-direction, just as what happens with an ordinary vector under rotation. End Solution (e) For a spinor α χ = + , α − ′ calculate the transformed spinor χ for n = ez. Solution ′ α+ χ = Uχ = (cos ϕ/2 + iσz sin ϕ/2) α− 1 0 1 0 α = cos ϕ/2 + i sin ϕ/2 + 0 1 0 1 α− − α cos ϕ/2 + iα sin ϕ/2 α e+iϕ/2 = + + = + . α− cos ϕ/2 iα− sin ϕ/2 iϕ/2 − α−e− ! End Solution Copyright c 2012, Edward J. Groth Physics 505 Homework No. 8 Solutions S8-4 (f) What happens to the spinor when ϕ =2π? Solution The spinor changes sign! A rotation of 4π is required to bring the spinor back to its original state! This is a weird property of fermions. However, expectation values involve χ 2 so the sign cancels out! | | End Solution 2. Spin precession. (a) The spinors 1 0 = = , |↑i 0 |↓i 1 are the eigenfunctions of σz. They are an orthonormal, complete basis in which σz is diagonal. In this same basis, what are the eigenfunctions (spinors) of σx and σy? Solution The eigenvalues are 1, so the expression for the eigenspinors of σx is ± 0 1 a a = , 1 0 b ± b which gives b = a. In order that the spinors be normalized, we take a = b = 1/√2. ± | | | | Then the eigenspinors of σx are 1 1 1 1 x = x = . |↑ i √2 1 |↓ i √2 1 − For σy, 0 i a a − = , i 0 b ± b which gives b = ia. Again, we must take a = b =1/√2 and the eigenspinors of σy are ± | | | | 1 1 1 1 y = y = . |↑ i √2 i |↓ i √2 i − End Solution Now we consider a spin 1/2 particle in a uniform magnetic field. For definiteness, we suppose it’s an electron, so the Hamiltonian for the spin degree of freedom is ge e H = S B S B . 2mc · ≈ mc · Copyright c 2012, Edward J. Groth Physics 505 Homework No. 8 Solutions S8-5 χ (b) The state of the system is represented by the spinor χ = + . What is the equation χ− of motion for χ? After writing down the equation for the general case, specialize to the case B = Bez. Your results should depend on the frequency ω = eB/mc which will turn out to be the precession frequency. Solution It’s just the Schroedinger equation (also known as the Pauli equation in this case). ∂ χ+ e χ+ χ+ Bz Bx iBy χ+ i¯h = S B = µBσ B = µB − . ∂t χ− mc · χ− · χ− Bx + iBy Bz χ− − When B = Bez, the equation of motion becomes ∂ χ µBB 1 0 χ ω 1 0 χ i + = + = + , ∂t χ− ¯h 0 1 χ− 2 0 1 χ− − − where ω = 2µBB/¯h = eB/(mc) is the frequency (which will turn out to the precession frequency). End Solution (c) In the case with B = Bez, and in the Heisenberg representation, what is the time dependence of the spin operator, S(t)? Solution In the Heisenberg representation, dS i z = [H, S ]=0 , dt ¯h z so Sz(t) = Sz(0). For the other two components, it’s more convenient to work with Sx iSy. ± d i (Sx iSy) = [H, Sx iSy] dt ± ¯h ± ieB = [Sz, Sx iSy] mc¯h ± ieB = i¯h(Sy iSx) mc¯h ∓ ieB = (Sx iSy) . ± mc ± This means iωt Sx(t) iSy(t) = e (Sx(0) iSy(0)) , ± ± ± or Sx(t) = cos ωt Sx(0) sin ωt Sy(0) − Sy(t) = sin ωt Sx(0) + cos ωt Sy(0) . Copyright c 2012, Edward J. Groth Physics 505 Homework No. 8 Solutions S8-6 End Solution (d) In the case with B = Bez, and in the Schroedinger representation, what is the time dependence of the spinor, χ(t)? Solution iHt/¯h χ(t) = e− χ(0) . The argument of the exponential is iHt ieBt iωt − = − Sz = − Sz = iϕn S/¯h, ¯h mc¯h ¯h · provided we identify ϕ = ωt and n = ez. This form is exactly the form for the unitary operator discussed in problem− 1, so the time dependence of χ is χ (t) 1 0 1 0 χ (0) χ(t) = + = cos( ωt/2) + i sin( ωt/2) + χ−(t) − 0 1 − 0 1 χ−(0) − iωt/2 χ (0) = e− 0 + 0 e+iωt/2 χ−(0) iωt/2 e− χ+(0) = +iωt/2 . e χ−(0) ! This makes sense. The up and down states are the stationary states of the Hamiltonian with energyhω/ ¯ 2 for the up state. The spin in the up state is aligned with B, but the magnetic moment is anti-aligned, so the energy is higher than in the down state which has energy ¯hω/2. − End Solution (e) At t = 0, the spin is aligned along the x axis. What is the probability of gettingh/ ¯ 2 in a measurement of Sz at time t. Solution The initial state must be (to within a phase!) 1 1 χ(0) = x = . |↑ i √2 1 From part (e), we find the state at time t and project that onto z to find the amplitude for going to spin up along z, and then square it to get the probability.h↑ | 2 e iωt/2/√2 1 P ( ) = (1 0) − = , z +iωt/2 ↑ e √2 ! 2 Copyright c 2012, Edward J.