1 Dirac Notation (15 Pts) 2 the Harmonic Oscillator (15 Pts)

Final exam solutions 137A – Spring 2016 I. Siddiqi / E. Dodds

1 Dirac Notation (15 pts)

Solution. (i) hα| = −i h1| − 2 h2| + i h3| and hβ| = −i h1| + 2 h3|. (ii) hα| βi = hβ| αi∗ = 1 + 2i. (iii)  1 0 2i  |αi hβ| =  2i 0 −4  (1) −1 0 −2i is not Hermitian.

2 The Harmonic Oscillator (15 pts)

1 2 2 V0 Suppose a particle is in a potential that looks like V (x) = mω x + x where V0 is a characteristic 2 a0 energy and a0 is a characteristic length. (i) Write down the time-independent Schrödingerequation (SE). It should look a lot like the SE for the harmonic oscillator, but with an extra term. Eliminate the linear term in x by completing the square. (4 pts) (ii) What are the energy levels of the particle in this potential? (Hint: you may want to define a new variable x∗ that is a shifted version of the original position x. Does this affect the corresponding momentum operator?) (4 pts) (iii) What is the expectation value of the position operator in each eigenstate of this potential? (4 pts)

(iv) Describe qualitatively what will happen to ∆x in each eigenstate as the value of V0 changes. (3 pts)

1 Solution. (i) The SE is 2 ∂2 − ~ ψ(x) + [(m/2)ω2x2 + (V /a )x]ψ(x) = Eψ(x). (2) 2m ∂x2 0 0 To eliminate the linear term, we can rewrite the potential as 2 2 m 2 2 2V0 m 2 V0 V0 V (x) = ω x + 2 x = ω x + 2 − 2 2 . (3) 2 a0mω 2 a0mω 2a0mω Then we can rearrange the Schr¨odingerequation: 2 2 2 2 ~ ∂ m 2 V0 V0 − 2 ψ(x) + ω x + 2 ψ(x) = E + 2 2 ψ(x). (4) 2m ∂x 2 a0mω 2a0mω

V0 (ii) Deﬁne y = x + 2 . The corresponding momentum is the same. The Schr¨odingerequation a0mω now reads 2 2 2 ~ ∂ m 2 2 V0 − 2 ψ(y) + ω y ψ(y) = E + 2 2 ψ(y) (5) 2m ∂y 2 2a0mω which is just the usual SE with a harmonic oscillator potential, except that the energies are shifted. This equation, then, has the same solutions, and we can immediately write down the energies: 2 V0 E + 2 2 = ~ω(n + 1/2), n = 0, 1, 2, 3,... (6) 2a0mω so 2 V0 E = − 2 2 + ~ω(n + 1/2), n = 0, 1, 2, 3,... (7) 2a0mω (iii) The expectation value of y in each of these states is zero for the same reasons the expectation value of x is zero for the usual harmonic oscillator. Since the expectation value is linear, the expectation value of x is V0 V0 hxi = hy − 2 i = − 2 (8) a0mω a0mω

(iv) V0 doesn’t aﬀect ∆x. You can show this by explicit calculation, or you can provide an argument based on the wavefunctions being the same as the usual harmonic oscillator wavefunctions but shifted.

3 Spin 1/2 Particle (20 pts)

Consider a particle with spin s = 1/2.

(i) In the basis where the matrix for Sˆz is diagonal, calculate the matrix representations of the Sˆx, Sˆy, Sˆz, Sˆ+, and Sˆ− operators. You MUST show your work...do not just write down the answer. HINT: You should consider the action of each one of the operators on the basis vectors and evaluate using algebraic tools. (5 pts) 3i Consider the spin state χ = A . 4

2 (ii) Determine the normalization constant A. (5 pts)

(iii) Find the expectation values of Sˆx and Sˆy. (5 pts)

(iv) Find ∆Sz. (5 pts)

Solution. (i) We know the action of each of the operators on the basis vectors:

Sˆ |±zi = ~ |∓zi x 2 Sˆ |±zi = ±i~ |∓zi y 2 Sˆ |±zi = ±~ |±zi z 2

so ˆ ˆ h+z| Sx |+zi h+z| Sx |−zi ~ 0 1 Sˆx −→ = (9) h−z| Sˆx |+zi h−z| Sˆx |−zi 2 1 0 and similarly 0 −i 1 0 Sˆ −→ and Sˆ −→ . (10) y i 0 z 0 −1

We can ﬁnd the raising and lowering matrices by Sˆ± = Sˆx ± iSˆy, so 0 1 0 0 Sˆ −→ and Sˆ −→ . (11) + ~ 0 0 − ~ 1 0

(ii) 3i 1 = |A|2 −3i 4 = |A|2(9 + 16) = 25|A|2 ⇒ A = 1/5. (12) 4 Any phase factor times 1/5 is also valid. (iii) We can obtain the expectation values by direct calculation: 1 0 1 3i hS i = −3i 4 ~ = ~ (−12i + 12i) = 0 (13) x 25 2 1 0 4 50 1 0 −i 3i 12 hS i = −3i 4 ~ = ~ (−12 − 12) = − . (14) y 25 2 i 0 4 50 25~

2 2 2 (iv) We want to calculate (∆Sz) = hSz i − hSzi . Each term can be calculate as in the previous part or in a number of equivalent methods. The results are 9 16 7 hS i = ~ − ~ = − (15) z 25 2 25 2 50~ and 2 9 16 2 2 hS2i = ~ + = ~ = ~ . (16) z 2 25 25 2 4 12 Putting these together we have ∆Sz = 25 ~.

3 4 Angular Momentum in Hydrogen (15 pts)

q 2 q 1 What is hLxi in the state Ψ = Rn1(r) 3 Y11 − 3 Y10 ?

Solution. We can proceed by direct calculation. I’ll work in Dirac notation for brevity. ! 1 r2 r1 Lˆ |ψi = (Lˆ + Lˆ ) |n11i − |n10i (17) x 2 + − 3 3

ˆ p and we can use L± |nlmi = ~ l(l + 1) − m(m ± 1) |nlm ± 1i to ﬁnd " # r2√ r1 √ √ Lˆ |ψi = ~ 2 |n10i − ( 2 |n11i + 2 |n1 − 1i) . (18) x 2 3 3

The desired quantity is then

r2 r1√ r1r2√ 2 hLˆ |ψi = − ~ 2 − 2~ = − (19) x 3 2 3 3 3 2 3~

5 Wave function Refocusing (15 pts)

A particle of mass m in a harmonic oscillator potential starts out in the state:

r 2 mω − mω x2 Ψ(x, 0) = A 1 − 2 x e 2~ (20) 2~ where A is a constant. (i) What is the expectation value of the energy ? (10 pts) (ii) At a later time T , the wave function is

r 2 mω − mω x2 Ψ(x, T ) = B 1 + 2 x e 2~ (21) 2~ where B is a constant. What is the smallest possible value of T ? (5 pts)

Solution. (i) There are a few ways to calculate the expectation value of the energy. Perhaps the easiest way is in this case is to write the given state as a sum of energy eigenstates and then use the coeﬃcients in that sum. The necessary harmonic oscillator energy eigenstates are given on the front of the exam. Since they all have the same exponential factor, we just need p to worry about the polynomials. Using α = mω/~ for brevity, the polynomial is (1 − 2αx)2 = 1 − 4αx + 4α2x2. (22)

4 2 Starting by accounting for the x term with H2 and working down to the constant term, we can write this as H2(x) − 2H1(x) + 4H0(x) (23) and then use the fact that the normalization factors for the harmonic oscillator energy eigenstates are a constant times p2−n/n! to write √ √ 2 1 Ψ(x, 0) = A˜( 2 2!ψ2(x) − 2 2 1! + 3ψ0(x)). (24)

We can calculate the new nomalization constant A˜ by summing the absolute squares of the coeﬃcients and ﬁnd that A˜ = 1/5 (or a phase factor times that). The expectation value of the energy is then

X 8 8 9 73 hEi = |c |2E = ω(2 + 1/2) + ω(1 + 1/2) + ω(0 + 1/2) = ω. (25) n n 25~ 25~ 25~ 50~ n

(ii) The state at an arbitrary time t > 0 is √ √ 3 2 2 2 2 Ψ(x, t) = ψ e−iωt/2 − ψ e−3iω/2 + ψ e−5iωt/2 (26) 5 0 5 1 5 2 " √ √ # 3 2 2 2 2 = e−iωt/2 ψ − ψ e−iωt + ψ e−2iωt (27) 5 0 5 1 5 2

If you multiply out the polynomial in the state at time T , you see that it’s the same state 2 as at time 0 except the sign in front of ψ1 has ﬂipped: (1 + 2αx) = H2 + 2H1 + 3H0. This means that the relative phase e−2iωt between the 0th and 2nd energy eigenstates is still 1:

1 = e−2iωT =⇒ ωT = mπ (28)

where m is an arbitrary integer. The relative phase betwween the 1st state and the other two is − 1 = e−iωT =⇒ ωT = mπ (29) for some integer m. These two conditions for T are the same, so the problem has a solution. The smallest possible value of T (given that T > 0) is thus T = π/ω.

6 Delta Potential (20 pts)

Consider the 1-D potential which has a centered inﬁnite square well with a delta-function barrier in the middle: ( ∞ for |x| > a V (x) = gδ(x)for |x| ≤ a where g is a positive constant.

5 (i) Write down the general form of an energy eigenfunction in the regions a > x > 0 and −a < x < 0. (4 pts) (ii) What are the four boundary conditions that the wavefunction must satisfy in the two regions, including those arising from the δ(x) function at the origin? (4 pts) (iii) Give an analytic expression for the energy of the odd eigenstates and give their wavefunctions. Plot the two lowest energy odd eigenfunctions. (4 pts) (iv) Derive a transcendental equation (contains trigonometric and polynomial terms) whose solution yields the bound state energies of the even wavefunctions. (4 pts) (v) Sketch the lowest energy even wavefunctions for g = 0, g = ∞, and g ∼ 1. What is the energy of this state for g = 0 and g = ∞? Explain this in terms of the solution to an inﬁnite square well problem without the δ(x) function. (4 pts)

Solution. (i) In both regions the potential is zero, so in the ﬁrst we have

ψI (x) = A sin(kx) + B cos(kx) (30) and in the second we have

ψII (x) = C sin(kx) + D cos(kx) (31) √ where in both cases k = 2mE/~. (ii) The wavefunction must be continuous. Since the wavefunction must be zero for x > a where the potential is inﬁnite, we have

ψI (−a) = 0 =⇒ A sin(−ka) + B cos(−ka) = 0 (32)

ψII (a) = 0 =⇒ C sin(ka) + D cos(ka) = 0. (33) Continuity at x = 0 means ψI (0) = ψII (0) =⇒ B = D. (34)

The fourth boundary condition involves the first derivative of the wavefunction at the origin. Recall that the first derivative is continuous wherever the potential is finite. In the special case of a Dirac delta, we can calculate the jump discontinuity in the first derivative by integrating the Schrödingerequation around the origin: Z 2 d2 Z lim − ~ ψ(x) + gδ(x)ψ(x) dx = lim Eψ(x)dx (35) →0 2m dx2 →0 − − 2 =⇒ − lim ~ [ψ0() − ψ0(−)] + gψ(0) = 0 (36) →0 2m 2m =⇒ lim[ψ0() − ψ0(−)] = gψ(0) (37) →0 ~2 In our case this means

0 0 2m ψII (0) − ψI (0) = gψ(0) (38) ~2 2m =⇒ k(−A + C) = gB (39) ~2

6 (iii) An odd wavefunction has ψ(0) = 0, so the delta doesn’t do anything. The energies and wavefunctions are the same as for the ordinary square well. The energies are

2n2π2 E = ~ (40) n 8ma2 and the wavefunctions are 1 ψ(x) = √ sin(nπx/(2a)) (41) a where n is a positive even integer. (iv) We need to combine the boundary conditions to ﬁnd k or equivalently E. Combining the conditions from continuity of the wavefunction gives

A sin(−ka) = C sin(ka) =⇒ A = −C. (42)

Putting this in the condition we got from the delta gives 2m − 2Ak = gB (43) ~2 which along with our ﬁrst boundary condition is m − k = g tan(ka). (44) ~2

(v) See Figure1.

7 1.0

0.5

-1.0 -0.5 0.5 1.0

-0.5

-1.0

(a) The odd solutions are unaﬀected by the Dirac delta because they are zero at the origin.

1.0

0.8

0.6

0.4

0.2

-1.0 -0.5 0.5 1.0 (b) When the strength of the Dirac delta is zero, the ground state is unaffected (blue line). When the strength is infinite the problem is equivalent to two wells, each with the usual ground state (orange line). For finite delta strength g you should sketch something in between these two extremes.

Figure 1: Plots for problem 6.