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Qualifying Exam
Quantum Mechanics – SOLUTIONS
August 2020
Solve one of the two problems in Part A, and one of the two problems in Part B. Each problem is worth 50 points.
Part A
Problem 1. Let us consider an electron whose squared orbital angular momentum L^2 is measured to be 6¯h^2.
1.a) [5 points] For each one of the two bases described above, list all the possible states for the electron which are compatible with this measurement. We have s = 12 and l = 2 (since l(l + 1) = 6). Since l + s ≤ j ≤ |l − s|, the allowed values for j will be j = 52 , j = 32. As a consequence, there will be ten possible states in the basis {|jmj 〉}: j = 52 with mj = 5 2 ,^
3 2 ,^
1 2 ,^ −
1 2 ,^ −
3 2 ,^ −
5 2 (six states) and^ j^ =^
3 2 with^ mj^ =^
3 2 ,^
1 2 ,^ −
1 2 ,^ −
3 2 (four states). In the basis {|mlms〉}, we can have all possible combinations of ms = −^12 , 12 with ml = 2 , 1 , 0 , − 1 , −2, which is again ten states.
1.b) [15 points] Find the expression of the following basis states {|jmj 〉} in terms of the appropriate elements of the basis {|mlms〉}.
1.b.1 : First state on the ladder, here is only one possible way of satisfying mj = ms + ml, therefore |j =
mj =
〉 = |ml = 2 ms =
1.b.2 : We should apply the lowering operator on both sides to obtain
|j =
mj =
|ml = 1 ms =
|ml = 2 ms = −
and then apply the orthogonality condition to get
|j =
mj =
|ml = 1 ms =
|ml = 2 ms = −
1.b.3 : Not a possible state. We cannot have j = 12 if s = 12 and l = 2.
1
August 2020 First Examination Ph.D. Program in Physics - The Graduate Center CUNY
1.b.4 : We can continue to apply the lowering operator until we reach the minimum of mj , or start from the bottom of the ladder (which is faster). The only combination of ms and ml that allows for mj = −^52 is
|j =
mj = −
〉 = |ml = − 2 ms = −
1.c) [15 points] Now assume the electron to be in the state with j = 32 , and mj = (^12) (and the value of l is the same as in the previous questions). If one measures the z-components of the electron orbital angular momentum and spin, what are the possible values and their probabilities? In our problem, there are only two configurations of ms and ml that are compatible with |j = 32 , and mj = 12 , namely the states |ml = 1 ms = −^12 〉 and |ml = 0 ms = 12 〉. To obtain the probabilities, we start from the expression that we obtained in question 1.b.2), and we apply once more the lowering operator, to obtain:
|j =
mj =
|ml = 0 ms =
|ml = 1 ms = −
which allows to read the probabilities 25 and 35 , respectively for the pair of values (ml = 0, ms = 1 2 ) and (ml^ = 1, ms^ =^ − 1 2 ).
1.d) [15 points] Let us now assume that the electron is in the state with ml = 1 and ms = −^12 (again, l did not change). What are the possible values of j and their probabilities? Here we have the “inverse” problem as 1.c). The state |ml = 1ms = −^12 〉 must be a linear combination of the states |j = 32 mj = 12 〉 (which we know already) and |j = 52 mj = 12 〉 (which we do not have yet). Let us first find |j = 52 mj = 12 〉. We can obtain it by applying the lowering operator to |j = 52 mj = 32 〉 or simply by using the orthogonality with |j = 32 mj = 12 〉. In either way, we get: |j =
mj =
|ml = 0 ms =
|ml = 1 ms = −
We can solve now for |ml = 1 ms = −^12 〉, to get
|ml = 1 ms = −
|j =
mj =
|j =
mj =
. Therefor, we will have j = 52 with probability 25 and j = 32 with probability 35.
August 2020 First Examination Ph.D. Program in Physics - The Graduate Center CUNY
. Dividing both sides by k|A|^2 we get:
∣∣^ B
A
2
k 1 k
∣∣^ E
A
2
That allows for the identifications of T = k k^1
∣∣E
A
and R =
∣∣B
A
, as transmission and reflection probabilities, respectively.
2.d) [25 points] Use the relations that you derived in part 2.b) and 2.c) to compute the transmission probability T. Write C and D in terms of E, using the third and fourth equations.
C =
k 0 − k 1 2 k 0 exp [i(k 1 − k 0 )a] E
D =
k 0 + k 1 2 k 0 exp [i(k 1 + k 0 )a] E
The substitute in the first and second one to write A and B as functions of E. This allows to obtain:
2 kA = (k + k 0 )C + (k − k 0 )D
A =
k + k 0 2 k
k 1 − k 0 2 k 0 exp [i(k 1 − k 0 )a] E + k − k 0 2 k
k 1 + k 0 2 k 0 exp [i(k 1 + k 0 )a] E
A
E
exp [ik 1 a] 4 kk 0 {(k + k 0 )(k 1 − k 0 ) exp [−ik 0 a] + (k − k 0 )(k 1 + k 0 ) exp [ik 0 a] }
A
E
exp [ik 1 a] 4 kk 0
2 cos (k 0 a) (kk 0 + k 0 k 1 ) − 2 i sin (k 0 a) (k 02 + kk 1 )
Then we can use
T = k 1 k
E
A
2
4 k k 02 k 1 cos^2 (k 0 a) (kk 0 + k 0 k 1 )^2 + sin^2 (k 0 a) (k 02 + kk 1 )^2
2.e) [10 points] Check your solution for the transmission probability T in the limit V 1 → 0. In this limit, your solution should reproduce the know value of T for the rectangular symmetric barrier, namely T =
[
1 + V^
02 sin^2 (k^0 a) 4 E(E−V 0 )
]− 1
where k 0 =
2 m(E−V 0 ) ¯h. This can be quite easily computed taking the limit k 1 → k in the expression above for T (note that k 0 is unchanged).
August 2020 First Examination Ph.D. Program in Physics - The Graduate Center CUNY
Formulas for Part A
Schr¨odinger Equation:
i¯h ∂ψ ∂t(x,t )= Hψˆ (x, t) , Hˆ = pˆ 2 2 m +^ V^ (x)^ ,^ Hψˆ E^ (x) =^ EψE^ (x)^ , j(x) = Re
{ −i¯h m ψ ∗(x) ∂ψ(x) ∂x
}
Angular Momentum Operators:
[ Jˆx, Jˆy] = i¯h Jˆz (or [ Jˆi, Jˆj ] = i¯hijk Jˆk) , [ Jˆ^2 , Jˆi] = 0 , Jˆ± = Jˆx ± i Jˆy , Jˆ±|j m〉 = ¯h√j(j + 1) − m(m ± 1) |j m ± 1 〉
August 2020 First Examination Ph.D. Program in Physics - The Graduate Center CUNY
Problems for QMII, 2020 CUNY Graduate Center Li Ge
option to start with a simpler ansatz in the next question.
(IV) How about using ψ(x) = A(x)eikφ(x)? If it works, apply the WKB approximation using A′′^ ≈ 0. Note that the result depends on whether Re[k] vanishes.
Answer: ψ′^ = (A′^ + ikAφ′)eikφ, ′′^ = (A′′^ + 2ikA′φ′^ + ikAφ′′^ − A(kφ′)^2 )eikφ^ = −εk^2 Aeikφ By dropping A′′, we find i(2A′φ′^ + Aφ′′) = Ak((φ′)^2 − ε). If k were real, then the derivation that follows would be exactly the same as in the Schr¨odinger equation. The imaginary part of k makes the result slightly different. Denoting again k = kr + iki (kr,i > 0), we separate the real and imaginary part of the equation above: 2 A′φ′^ + Aφ′′^ = Aki((φ′)^2 − ε), 0 = kr ((φ′)^2 − ε). If kr 6 = 0, the second equation tells us that φ′^ = ±√ε = ±n(x), and the first equation indicates Aφ′^ = const. The wave function is then given by ψ(x) ∝ √^1 n(x) eik^
∫ (^) n(x)dx , which is again the same as in the Schr¨odinger equation. However, there exists zeros of the scattering matrix with kr = 0 for some ε(x), and we find (φ′)^2 6 = ε. As a result, the right hand side of 2A′φ′^ +Aφ′′^ = Aki((φ′)^2 −ε) is not zero either. In other words, in this case the phase and amplitude of the wave function are not simply related.
(V) Assume ε(x) = ε(−x), write down the wave function in the WKB approximation as sine and cosine functions. What is the argument of these sinusoidal functions?
Answer: ψ+(x) = √^1 n(x) cos
( k
∫ (^) x 0
n(x)dx
) and ψ−(x) = √^1 n(x) sin
( k
∫ (^) x 0
n(x)dx
) , one for even-parity and the other for odd-parity. Check:
ψ−(−x) = √^1 n(x) sin
( k
∫ (^) −x 0
n(x)dx
) = √^1 n(x) sin
( −k
∫ (^) x 0
n(−x)dx
) = −ψ−(x).
We have used x → −x in the second step and n(x) = n(−x) in the last step. Similarly, we find ψ+(−x) = ψ+(x).
(VI) Using the boundary condition at either x = L/2 or −L/2, derive the analytical expressions satisfied by k = kr + iki (kr,i > 0), one for even-parity and the other for odd-parity. Assuming ki kr , analyze these express to show that they lead to the same form
k ≈ (^) ¯nL^1
[ q + i ln |^2 αq|
] , (2)
when |α| q. Specify the expressions for α, q, and the average refractive index ¯n inside the cavity.
Answer: Using the boundary condition at x = L/2, we find ψ′− ψ−
∣∣ ∣∣ x=L/ 2
=
√ne[k cos(z/2) − n′ 2 n^2 e^ sin(z/2)] √^1 n e sin(z/2)^
= −inek, or tan(z/2) = (^) −αz + iz ,
P.1 continued on next page... Page 2 of 4
August 2020 First Examination Ph.D. Program in Physics - The Graduate Center CUNY
Problems for QMII, 2020 CUNY Graduate Center Li Ge
where z ≡ ¯nkL/2 and α ≡ ¯nLn′(x = L/2)/ 2 n^2 e ∈ R. ¯n is the average of n(x) inside the cavity, i.e., ¯n = ∫ (^) L/ 2 −L/ 2 n(x)dx/L. Similarly, for the parity-even wave functions, we have
tan(z/2) = α^ −z iz.
α is fixed, and at high energy we find δ ≡ |α|/z 1. To the leading order of δ, we then find e−iz^ = ±iδ, where “+” and “−” are for the two parities. Denoting z = q + iκ, we find q ≈ (m + 0.5)π (m ∈ Z) by noting that e−iz^ is approximately imaginary when δ 1. Finally, we derive Eq. (2) by taking the absolution value of e−iz^ and the approximation δ ≈ |α|/q, assuming q κ.
P.
(I) [15 pts] Consider a one-dimensional harmonic oscillator in the ground state | 0 〉 of the unperturbed Hamiltonian at t = −∞. Let a perturbation H 1 (t) = F xe−t^2 /τ^2 be applied between t = −∞ and +∞. Use the first-order time-dependent perturbation theory to calculate the probability that the oscillator is in the state |n〉 at t = +∞.
Answer: Assuming |ψ(t)〉 = ∑ n dn(t)e−iEnt/ℏ|n〉^ where the (shifted) energy levels are^ En^ =^ nℏω, we know
dn(t) = − ℏi F
∫ (^) ∞ −∞
〈n|x| 0 〉e−t^2 /τ^2 einωtdt. (3)
In the midterm I tested the students about the selection rule based on the parity of the eigenstates. Here it is similar and we find only an odd |n〉 may have a nonzero transition probability. Furthermore, using x =
√ ℏ/ 2 mω(a + a†) and the Gaussian integral we find that only d 1 (+∞) is nonzero:
d 1 (+∞) = − ℏi
√ ℏ 2 mω F τ
∫ (^) ∞ −∞
e−t 2 eiωτ tdt = iF τ
( (^) π 2 mℏω
) 1 / 2 e−ω (^2) τ 2 / 4
. (4)
(II) [5 pts] Prove the adiabatic theorem using this example, i.e., if the Hamiltonian H(t) is slowly varying, then a system starts out in |n〉 at t = −∞ will end up in |n〉 again at t = +∞. Show quantitatively how slow the variation needs to be in terms of the energy difference between two neighboring states.
Ansswer: We just need to show that d 1 (∞) → 0, which requires ωτ = τ (E 1 − E 0 )/ℏ 1 for a given ω.
(III) [15 pts] Now consider another case. If we start with a quantum bouncing ball in a box potential in one dimension and slowly modulate its length L with period τ , show that the requirement for the adiabatic theorem to hold is the same as in (II). Here we just need the order of magnitude estimation, and you may want to consider the wave-particle duality of the ball.
Answer: A reasonale condition for the adiabatic theorem to hold is the following: the time it takes for the ball to complete a round trip should be much shorther than τ. The energy of the nth level in a box potential is given by n^2 ℏ^2 π^2 / 2 mL^2 , and hence we estimate its momentum to be p ∼ ℏ/L. The round-trip time is then T = L/v = mL/p ∼ mL^2 /ℏ. Note that E 1 − E 0 here is also on the order of ℏ^2 /mL^2 , so T /τ = mL^2 /ℏτ = ℏ/(E 1 − E 0 )τ 1 , (5) which is the same condition as in (II).
P.2 continued on next page... Page 3 of 4
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January 2021 First Examination Ph.D. Program in Physics - The Graduate Center CUNY
January 2021 First Examination Ph.D. Program in Physics - The Graduate Center CUNY
Qualifying Exam
Quantum Mechanics
June 2021
Problem 1. Consider two observables Aˆ and Bˆ in a three-dimensional Hilbert space. In the basis:
| 1 〉 =
the observables Aˆ and Bˆ are represented, respectively, by the matrices
A → a
(^) , B → b
where b << a.
a) [5 points] Show that the observables Aˆ and Bˆ are not compatible. Check the commutator [ A,ˆ Bˆ]. Since Aˆ and Bˆ do not commute, the observables are NOT compatible. in particular, Aˆ and Bˆ do not admit a common set of eigenstates. b) [10 points] Compute the possible outcomes and the corresponding probabilities of separate (independent) measurements of Aˆ and Bˆ in the state |χ〉 =
1 2 (|^2 〉 − |^3 〉). Aˆ: in general, we have three possible outcomes in the eigenvalues a 1 = 1a, a 2 = 2a, a 3 = 6a, which correspond to the three eigenstates | 1 〉, | 2 〉, and | 3 〉. The probabilities are given by Pi = |〈χ|i〉|^2 , where i = 1, 2 , 3. Since |χ〉 =
1 2 (|^2 〉 − |^3 〉), we will measure^ a^2 = 2a^ with a probability of 50% and^ a^3 = 6a with a probability of 50%. Bˆ: the possible outcomes are given by the eigenvalues. Let us firts compute them together with the corresponding eigenstates |B 1 〉, |B 2 〉, and |B 3 〉. The probabilities will be given by Pi = |〈χ|Bi〉|^2 , where i = 1, 2 , 3. The first eigenvalue is b 1 = − 1 b, and |B 1 〉 = | 1 〉. By diagonalizing the sub-matrix
, we find that the additional eigenvalues are b 2 = − 2 b and b 3 = +2b, with respective eigenvectors |B 2 〉 =
1 2 (|^2 〉 − |^3 〉) and^ |B^3 〉
1 2 (|^2 〉^ +^ |^3 〉). Since |χ〉 = |B 2 〉, we will measure b 2 = − 2 b with a probability of 100%. c) [10 points] Compute the possible outcomes and the corresponding probabilities of a mea- surements of Bˆ that follows a measurement of Aˆ, if the system is initially in the state |χ〉. What about a measurement of Aˆ that follows a measurements of Bˆ? After the measurement of Aˆ, the state of the system will be in | 2 〉 or | 3 〉 with a probability of 50% each. Noting that | 2 〉 =
1 2 (|B^2 〉^ +^ |B^3 〉) and^ |^3 〉^ =
1 2 (|B^2 〉 − |B^3 〉), we can conclude
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Quantum Mechanics - Solutions First Examination - June 2021 Ph.D. Program in Physics The Graduate Center CUNY
