Physics 115C Homework 3

Problem 1

In this problem, it will be convenient to introduce the Einstein summation convention. Note that we can write = LiSi i where the sum is over i = x,,z. In the Einstein summation convention, we leave out the overall summation sign, and assume the convention that if an index is repeated in an expression, there is an implied summation over it, like so:

LiSi ≡ LiSi i This will make our expressions easier to work with (and will save some typing for me). Now, let’s start. Recall that the commutation relations for angular momentum are

2 [Li,]= i~ǫijkLk [L ,Li]=0 where ǫijk is the Levi-Civita symbol, and identical expressions hold for S and (and there is an implied sum over the index !). Using these commutation relations along with the 2 2 2 definition J = L+S, let’s compute the commutator of LS with J , L , S , Jz, Lz, and : [L S,L2]=0 [L S,S2]=0 These two are trivial, since L2 and S2 commute with all of their components, and therefore also with L S. Next, [L S,J 2] = [L S, J J] = [L S, (L + S) (L + S)] = [L S,L2 + S2 +2L S] = [L S,L2]+[L S,S2]+2[L S, L S] = [L S,L2]+[L S,S2] = 0

1 To get the the fourth line, I used the fact that since S and L commute, L S = S L; to get to the fifth line, I used the fact that L S commutes with itself; and to get to the last line, I used the fact that L2 commutes with all the components of S and L. Continuing,

[L S,Lz] = [LiSi,Lz]

= Li[Si,Lz]+[Li,Lz]Si

= 0+ i~ǫizkLkSi

= i~ǫzkiLkSi

= i~(L × S)z

To get to the third line, I used the fact that the components of S and L commute with each other; to get to the fourth line, I used the antisymmetry of the Levi-Civita symbol under exchange of any two indices (.g. ǫijk = −ǫjik = ǫjki, etc.); and to get to the last line I used the expression for the dot product

(A × )i = ǫijkAjBk

Moving on,

[L S,Sz] = [LiSi,Sz]

= Li[Si,Sz]+[Li,Sz]Si

= i~ǫizkLiSk +0

= −i~ǫzikLiSk

= −i~(L × S)z

Finally,

[L S,Jz] = [L S,Lz + Sz]

= [L S,Lz]+[L S,Sz]

= i~(L × S)z − i~(L × S)z = 0

2 2 2 where I made use of our results above. Therefore, J ,L ,S , and Jz commute with L S, but not Lz and Sz. Since the spin-orbit coupling perturbation is proportional to L S, this is why the “good” basis states are those labeled by , ℓ, s, j, and mj, but not those labeled by n,ℓ,s,mℓ, and ms.

2 Problem 2 (Griffiths 6.21)

From Griffiths’ equations 6.75 and 6.76, we have that the first-order corrections to the energy in the weak-field Zeeman effect are

(1) EZ = BgJ Bextmj where B is the Bohr magneton, Bext is the external electric field (along which we align our z-axis), and gJ is the Land´eg-factor: j(j + 1) − ℓ(ℓ +1)+3/4 g =1+ J 2j(j + 1) (incidentally, Griffiths’ derivation of the Land´eg-factor gives a good physical idea of what’s going on, but isn’ very mathematically rigorous. A more rigorous derivation of equation 6.73 makes use of the Wigner-Eckart theorem). Now, the eight n = 2 states of the hydrogen atom have s = 1/2 and ℓ = 0 or ℓ = 1. For ℓ = 0, the only possible value of j is 1/2; for ℓ = 1, the possible values of j are j = 1/2 and j = 3/2. Thus there are two states with ℓ = 0 (corresponding to quantum numbers mj = ±1/2) and six states with ℓ = 1(j = 1/2 and mj = ±1/2, and j = 3/2 and mj = ±1/2, ±3/2). Let’s label these states as follows: using the notation |ℓjmj, we define the eight states as |1 = |0, 1/2, 1/2 |2 = |0, 1/2, −1/2 |3 = |1, 1/2, 1/2 |4 = |1, 1/2, −1/2 |5 = |1, 3/2, 3/2 |6 = |1, 3/2, 1/2 |7 = |1, 3/2, −1/2 |8 = |1, 3/2, −3/2

Now, let’s calculate the g-factors for these states. For |1 and |2, we have 1/2(1+1/2) − 0+3/4 g =1+ =2 J 2(1/2)(1 + 1/2) For |3 and |4, we have 1/2(1+1/2) − 1(1+1)+3/4 2 g =1+ = J 2(1/2)(1 + 1/2) 3 For the last four states, we have 3/2(1+3/2) − 1(1+1)+3/4 4 g =1+ = J 2(3/2)(1 + 3/2) 3

3 To find the energies of these eight states, we’ need to include fine structure. From Griffiths’ equation 6.67, we have that the energies of the first four states (including fine structure but not yet the Zeeman perturbation) are

E α2 n 3 E = 0 1+ − fs n2 n2 j +1/2 4 E α2 2 3 = 0 1+ − 4 4 1/2+1/2 4 E 5 = 0 1+ α2 4 16

where E0 = −13.6 eV is the (unperturbed) Hydrogen ground-state energy. The energies of the last four states (again, including just fine structure) are

E α2 2 3 E = 0 1+ − fs 4 4 3/2+1/2 4 E 1 = 0 1+ α2 4 16 Including the first-order perturbation from the Zeeman effect, the energy of each state is thus E = Efs + BgJ Bextmj

Using our expressions for Efs and the g-factor, we can write down the energies of all eight states: E 5 E = 0 1+ α2 + B 1 4 16 B ext E 5 E = 0 1+ α2 − B 2 4 16 B ext E 5 1 E = 0 1+ α2 + B 3 4 16 3 B ext E 5 1 E = 0 1+ α2 − B 4 4 16 3 B ext E 1 E = 0 1+ α2 +2 B 5 4 16 B ext E 1 2 E = 0 1+ α2 + B 6 4 16 3 B ext E 1 2 E = 0 1+ α2 − B 7 4 16 3 B ext E 1 E = 0 1+ α2 − 2 B 8 4 16 B ext

4 Here’s a depiction of how the energies vary with the perturbing magnetic field Bext (not to scale, of course):

5 Problem 3

If a system begins at time t = 0 in an eigenstate |i of the unperturbed Hamiltonian, then the probability of measuring the system to be in some other state | at a later time t (after 2 the perturbation has been turned on) is |cf (t)| , where

t i (0) (0) ′ ~ i(Ef −Ei )t / ′ ′ ′ cf (t)= δfi − ~ e f| (t ) |i dt 0 where H′(t) is the perturbing Hamiltonian. In our particular case, we have

H′(t)= βxe−t/τ

To give this perturbation a physical interpretation, we might imagine that this would be the perturbation the oscillator would experience if it were placed in an exponentially decaying electric field (which might arise, for instance, inside of a capacitor as it was being discharge in an LR circuit). We are further given that the oscillator begins in the ground state, so |i = |0. (0) Since the (unperturbed) energies of the harmonic oscillator are given by En =(n +1/2)~ω, we have

t ′ i 1 ~ 1~ t ′ ′ ′ cn(t) = δn0 − ~ exp i n + ω − ω ~ n| H (t ) |0 dt 0 2 2 t iβ inωt′ −t′/τ ′ = δn0 − ~ e e n| x |0 dt 0 Since we’re interested only in the probability of measuring the system to be in the nth state at large time, we might as well take the limit t → ∞:

∞ iβ −(1/τ−inω)t′ ′ cn(∞) = δn0 − ~ n| x |0 e dt 0 ∞ iβ −(1/τ−inω)t′ ′ = δn0 − ~ n| x |0 e dt 0 ∞ iβ −1 ′ = δ − n| x |0 e−(1/τ−inω)t n0 ~ 1/τ − inω 0 iβ = δ − n| x |0 n0 ~(1/τ − inω)

To evaluate the matrix element n| x |0, let’s use the ladder operators: recall that we can express ~ x = (a† + a) 2mω

6 Thus ~ n| x |0 = n| (a† + a) |0 2mω ~ = n|1 2mω ~ = δn1 2mω and therefore ~ iβ cn(∞)= δn0 − ~ δn1 2mω (1/τ − inω) Note that this expression is zero unless n = 1 or n = 0, i.e. the probability of the system transitioning to any state higher than n = 1 is zero (at least to first order in perturbation theory). To get the probability of transition, we take the norm squared: the probability of no transition at all is

2 P0 = |c0(∞)| = |1 − 0|2 = 1

The probability of transitioning to the first excited state is

2 P1 = |c1(∞)| 2 ~ iβ = 0 − ~ 2mω (1/τ − iω) ~ β2 = 2mω ~2(1/τ 2 + ω2) β2 = 2m~ω(1/τ 2 + ω2)

Thus the probability of measuring the system to be in the nth state at large time is

1 n =0 β2 Pn =  2m~ω(1/τ 2+ω2) n =1 0 otherwise



7 Problem 4

As in the previous problem, we have

t i (0) (0) ′ ~ i(Ef −Ei )t / ′ ′ ′ cf (t)= δfi − ~ e f| H (t ) |i dt 0 In this case, we have −T T ) are

T ′ i 2 t ′ ′ ′ cn(t>T ) = δn1 − ~ exp i E1n − E1 ~ n| H (t ) |1 dt −T T 2 ′ i ′ i(n −1)E1t /~ ′ = δn1 − ~ n| H |1 e dt −T For n = 1, the integral is just 2T ; for n = 1, we get

T i ~ 2 ′ ~ c (t>T ) = δ − n| H′ |1 ei(n −1)E1t / n n1 ~ i(n2 − 1)E 1 −T i 2~ E T = δ − n| H′ |1 sin (n2 − 1) 1 n1 ~ (n2 − 1)E ~ 1 We only need to evaluate the matrix element. Recall that the stationary state wave functions of the infinite square well are 2 nπ ψ = sin x n L L Thus

′ ∗ ′ n| H |1 = ψnH ψ1 dx

L/2 ∗ = V0 ψnψ1 dx 0 2V L/2 nπ π = 0 sin x sin x dx L 0 L L V L/2 (n − 1)π (n + 1)π = 0 cos x − cos x dx L L L 0

8 (I used the identity sin(a)sin(b)=(1/2)[cos(a − b) − cos(a + b)]). The integral has different behaviors for n = 1 and n = 1. For the case n = 1, we have

V L/2 2π 1| H′ |1 = 0 1 − cos x dx L L 0 V L 2π L/2 = 0 x − sin x L 2π L 0 V L = 0 L 2 V = 0 2 For n = 1, we have

V L/2 (n − 1)π (n + 1)π n| H′ |1 = 0 cos x − cos x dx L L L 0 V L (n − 1)π L (n + 1)π L/2 = 0 sin x − sin x L (n − 1)π L (n + 1)π L 0 V L (n − 1)π L (n + 1)π = 0 sin − sin L (n − 1)π 2 (n + 1)π 2 Using the fact that sin((n − 1)π/2) = sin((n + 1)π/2) = 0 for odd n, and sin((n − 1)π/2) = −(−1)n/2 and sin((n + 1)π/2)=(−1)n/2 for even n, we get

V L L n| H′ |1 = 0 − (−1)n/2 − (−1)n/2 L (n − 1)π (n + 1)π (−1)n/2 1 1 = −V + 0 π n − 1 n +1 (−1)n/2 2n = −V 0 π n2 − 1 for even n, and n| H′ |1 = 0 for odd n. Consequently, for n even we have

i 2~ E T c (t>T ) = − n| H′ |1 sin (n2 − 1) 1 n ~ (n2 − 1)E ~ 1 iV 2n 2~ E T = 0 (−1)n/2 sin (n2 − 1) 1 π~ n2 − 1 (n2 − 1)E ~ 1 iV 4n E T = 0 (−1)n/2 sin (n2 − 1) 1 πE (n2 − 1)2 ~ 1

9 For n = 1, we instead get

T i ′ ′ c1(t>T ) = 1 − ~ 1| H |1 dt −T i V = 1 − 0 2T ~ 2 iV0T = 1 − ~

For n = 1 and n odd, we have simply cn(t>T ) = 0. Thus the transition probabilities are

V0T 2 1+ ~ n =1 2 2 V0 16n 2 2 E1T Pn =  2 4 sin (n − 1) even n πE1 (n −1) ~  0 odd n

2 2  where E1 = π~ /2mL , as usual. In particular, the probability of measuring the well to be in the first excited state is P2, or

V 2 16(4) E T P = 0 sin2 (4 − 1) 1 2 πE (4 − 1)4 ~ 1 8V 2 3E T P = 0 sin2 1 2 9πE ~ 1

10