Physical Chemistry 2 Quiz: Degeneracy And Perturbation

20 questions · exam conditions

0:00

Degeneracy And PerturbationQuestion 1 of 20

A quantum system has energy levels $E_n = n\hbar\omega$ for $n = 0, 1, 2, ...$ with no degeneracy. A perturbation $H' = V(|n⟩⟨n+1| + |n+1⟩⟨n|)$ is applied, where $V$ is a small real constant. This perturbation couples only nearest-neighbor energy levels. For the ground state $|0⟩$ , what is the second-order energy correction?

E_0^{(2)} = -\frac{V^2}{\hbar\omega}

E_0^{(2)} = -\frac{V^2}{2\hbar\omega}

E_0^{(2)} = 0

because the ground state cannot be coupled to any lower energy states.

E_0^{(2)} = \frac{V^2}{\hbar\omega}

E_0^{(2)} = -\frac{2V^2}{\hbar\omega}

← Back to quizzes

Physical Chemistry 2 Quiz

Physical Chemistry 2 Quiz: Degeneracy And Perturbation

Practice Degeneracy And Perturbation in Physical Chemistry 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Degeneracy And Perturbation, giving you a quick way to practice the rules, question types, and explanations that matter most for Physical Chemistry 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$E_0^{(2)} = -\frac{V^2}{\hbar\omega}$ (correct answer)
$E_0^{(2)} = -\frac{V^2}{2\hbar\omega}$
$E_0^{(2)} = 0$ because the ground state cannot be coupled to any lower energy states.
$E_0^{(2)} = \frac{V^2}{\hbar\omega}$
$E_0^{(2)} = -\frac{2V^2}{\hbar\omega}$

Explanation: When you encounter perturbation theory problems, you need to systematically apply the formulas for energy corrections. The second-order energy correction is given by

E_n^{(2)} = \sum_{k \neq n} \frac{|\langle k|H'|n \rangle|^2}{E_n^{(0)} - E_k^{(0)}}

, where you sum over all states except the one you're analyzing. For the ground state

|0⟩

, you need to find which states the perturbation

H' = V(|n⟩⟨n+1| + |n+1⟩⟨n|)

can couple it to. The matrix element

\langle k|H'|0⟩

is only non-zero when

k = 1

, giving

\langle 1|H'|0⟩ = V

. Therefore, only the

|1⟩

state contributes to the sum. Calculating the second-order correction:

E_0^{(2)} = \frac{|\langle 1|H'|0 \rangle|^2}{E_0^{(0)} - E_1^{(0)}} = \frac{|V|^2}{0 - \hbar\omega} = -\frac{V^2}{\hbar\omega}

Looking at the wrong answers: B)

-\frac{V^2}{2\hbar\omega}

incorrectly includes an extra factor of 2, possibly from confusion about the symmetry of the perturbation. C) misunderstands the physics—while the ground state can't couple to lower states, it can couple to higher ones, which is what matters for second-order corrections. D)

\frac{V^2}{\hbar\omega}

has the wrong sign, likely from incorrectly calculating

E_0^{(0)} - E_1^{(0)}

. Remember: in second-order perturbation theory, always check which states have non-zero matrix elements with your starting state, and be careful with signs when calculating energy differences.

Question 2

Consider a particle in a three-dimensional cubic box with sides of length $L$ . The unperturbed energy levels are $E_{n_x,n_y,n_z} = \frac{\pi^2\hbar^2}{2mL^2}(n_x^2 + n_y^2 + n_z^2)$ . A weak perturbation $H' = αx$ is applied. Which of the following statements about the effect of this perturbation is most accurate?

All degeneracy is immediately lifted because the perturbation breaks the cubic symmetry, and all energy levels become non-degenerate to first order.
The first-order energy corrections are zero for all states, so degeneracy lifting only occurs in second-order perturbation theory. (correct answer)
Only states that differ in the $n_x$ quantum number will have their degeneracy lifted, while degeneracy associated with $n_y$ and $n_z$ permutations remains.
The perturbation couples states with $\Delta n_x = ±1$ and $\Delta n_y = \Delta n_z = 0$ , causing degeneracy lifting through second-order mixing of nearby energy levels.
The degeneracy structure remains completely unchanged because the perturbation commutes with the unperturbed Hamiltonian.

Explanation: When dealing with perturbation theory for degenerate systems, you need to carefully analyze whether the perturbation has non-zero matrix elements between the unperturbed states. The key insight is understanding what happens when you calculate first-order energy corrections. For the perturbation

H' = αx

, the first-order energy correction is

E_n^{(1)} = \langle n|H'|n \rangle = α\langle n|x|n \rangle

. For a particle in a box, the position operator

x

connects the center of the box (where the average position is

L/2

) with the quantum state. However, the crucial point is that

\langle n_x,n_y,n_z|x|n_x,n_y,n_z \rangle = L/2

for all states - this is simply the expectation value of position, which is the same constant for every state regardless of quantum numbers. Since all states get the same energy shift (

αL/2

), no degeneracy is lifted in first order. Option A is incorrect because not all degeneracy is immediately lifted - the first-order correction is identical for all states. Option C misunderstands the mechanism; while the perturbation involves

x

, the first-order diagonal elements don't depend on specific

n_x

values. Option D incorrectly focuses on off-diagonal coupling elements, but the question asks about degeneracy lifting, which first requires examining diagonal corrections. Remember: in degenerate perturbation theory, always check first-order diagonal matrix elements first. If they're equal for degenerate states, degeneracy persists to first order, and you must examine second-order corrections or off-diagonal coupling within the degenerate subspace.

Question 3

Consider the rigid rotor in three dimensions with unperturbed energy levels $E_l = \frac{\hbar^2 l(l+1)}{2I}$ where each level has degeneracy $(2l+1)$ . A perturbation $H' = \beta \cos^2 θ$ is applied, where $θ$ is the polar angle. For the $l = 1$ level, which was initially three-fold degenerate, what happens to the degeneracy structure in first-order perturbation theory?

The level splits into three non-degenerate levels with different energies corresponding to $m = -1, 0, +1$ .
The level splits into two levels: one non-degenerate level and one two-fold degenerate level. (correct answer)
All three levels remain degenerate because $\cos^2 θ$ has the same symmetry properties as the unperturbed Hamiltonian.
The level splits completely, but the energy differences between the split levels are second-order effects, so they appear degenerate in first order.
The degeneracy is enhanced, resulting in a six-fold degenerate level due to coupling with higher $l$ states.

Explanation: When you encounter degenerate perturbation theory problems, you need to calculate matrix elements of the perturbation between the degenerate states and find which combinations have the same energy. For the

l = 1

level, the three degenerate states are

|1,-1\rangle

|1,0\rangle

, and

|1,+1\rangle

. The perturbation

H' = \beta \cos^2 θ

can be rewritten using

\cos^2 θ = \frac{1}{3}(1 + 2P_2(\cos θ))

, where

P_2

is the second Legendre polynomial. This connects to spherical harmonics since

P_2(\cos θ) \propto Y_2^0(θ,φ)

. The key insight is that only the

Y_2^0

component creates off-diagonal matrix elements. When you calculate

\langle 1,m'|H'|1,m\rangle

, you find that the

m = \pm 1

states mix with each other but not with the

m = 0

state due to selection rules for angular momentum coupling. This gives you a

3 \times 3

matrix that blocks into a

1 \times 1

block (for

m = 0

) and a

2 \times 2

block (for

m = \pm 1

). The

m = 0

state shifts in energy but remains non-degenerate, while the

m = \pm 1

states form two new combinations that happen to have the same energy in first order, maintaining two-fold degeneracy. Answer B correctly describes this splitting pattern. Answer A incorrectly assumes complete splitting into three different energies. Answer C is wrong because the perturbation doesn't have spherical symmetry—it depends on

θ

. Answer D misunderstands the order of the effect; the splitting occurs in first order, not second. Remember: in degenerate perturbation theory, look for symmetry properties that determine which states can mix.

Question 4

A two-level quantum system has degenerate ground states $|1⟩$ and $|2⟩$ with energy $E_0$ . A time-independent perturbation is applied with matrix elements $\langle 1|H'|1⟩ = \Delta$ , $\langle 2|H'|2⟩ = -\Delta$ , and $\langle 1|H'|2⟩ = \langle 2|H'|1⟩ = V$ where $\Delta$ and $V$ are real constants. If $|V| >> |\Delta|$ , which statement best describes the perturbed eigenstates and their energies?

The perturbed energies are approximately $E_0 ± V$ with eigenstates approximately $\frac{1}{\sqrt{2}}(|1⟩ ± |2⟩)$ . (correct answer)
The perturbed energies are approximately $E_0 + \Delta$ and $E_0 - \Delta$ with eigenstates remaining as $|1⟩$ and $|2⟩$ .
The perturbed energies are $E_0 + \sqrt{\Delta^2 + V^2}$ and $E_0 - \sqrt{\Delta^2 + V^2}$ with equal mixing coefficients.
The system becomes non-degenerate with energies $E_0$ and $E_0 + 2\sqrt{\Delta^2 + V^2}$ .
The perturbed energies are approximately $E_0 ± V ± \frac{\Delta^2}{2V}$ with eigenstates that are predominantly $\frac{1}{\sqrt{2}}(|1⟩ ± |2⟩)$ .

Explanation: When you encounter degenerate states with a perturbation that has both diagonal and off-diagonal elements, you need to solve the problem exactly by diagonalizing the perturbation matrix, especially when the off-diagonal coupling is large. The perturbation matrix in the basis

\{|1⟩, |2⟩\}

is:

To find the eigenvalues, you solve

\det(H' - \lambda I) = 0

(\Delta - \lambda)(-\Delta - \lambda) - V^2 = 0

\lambda^2 - \Delta^2 - V^2 = 0

This gives

\lambda = ±\sqrt{\Delta^2 + V^2}

, so the perturbed energies are

E_0 ± \sqrt{\Delta^2 + V^2}

. However, since

|V| >> |\Delta|

, we can approximate:

\sqrt{\Delta^2 + V^2} ≈ \sqrt{V^2} = |V| = V

(assuming

V > 0

). The energies become approximately

E_0 ± V

. The corresponding eigenvectors are

\frac{1}{\sqrt{2}}(|1⟩ ± |2⟩)

when the off-diagonal coupling dominates. Choice A correctly captures this limit. Choice B ignores the off-diagonal coupling entirely, which is wrong when

|V| >> |\Delta|

. Choice C gives the exact result but misses that we can approximate when one parameter dominates. Choice D incorrectly calculates the energy splitting. Remember: when off-diagonal matrix elements dominate in degenerate perturbation theory, the system favors symmetric and antisymmetric combinations of the original states, with energy splitting determined primarily by the coupling strength.

Question 5

Consider a quantum harmonic oscillator in one dimension with an additional quartic perturbation $H' = \lambda x^4$ where $\lambda$ is a small positive constant. The unperturbed energy levels are $E_n = \hbar\omega(n + 1/2)$ . For the first excited state ( $n = 1$ ), what is the most accurate statement about the first-order energy correction?

The first-order correction vanishes because $x^4$ has no matrix elements between states of the same parity.
The first-order correction is $\lambda \langle 1|x^4|1⟩ = \frac{3\lambda\hbar^2}{4m^2\omega^2}$ . (correct answer)
The first-order correction is zero because the harmonic oscillator wavefunctions are orthogonal to $x^4$ .
The first-order correction is $\lambda \langle 1|x^4|1⟩ = \frac{\lambda\hbar^2}{2m^2\omega^2}$ .
The first-order correction cannot be calculated because $x^4$ couples the $n = 1$ state to multiple other states simultaneously.

Explanation: When you encounter perturbation theory problems, you need to calculate the first-order energy correction using

E_n^{(1)} = \langle n|H'|n \rangle

, where

H'

is the perturbation and

|n\rangle

is the unperturbed state. For the first excited state with perturbation

H' = \lambda x^4

, you need to evaluate

\langle 1|x^4|1 \rangle

. Using the harmonic oscillator position operator in terms of creation and annihilation operators:

x = \sqrt{\frac{\hbar}{2m\omega}}(a + a^\dagger)

, so

x^4 = \left(\frac{\hbar}{2m\omega}\right)^2(a + a^\dagger)^4

. When you expand

(a + a^\dagger)^4

and apply it to

|1\rangle

, the terms that survive the matrix element

\langle 1|...|1\rangle

are those that neither raise nor lower the quantum number. After careful calculation of all terms (including cross terms like

a^\dagger a

), you get

\langle 1|(a + a^\dagger)^4|1\rangle = 3

. Therefore,

E_1^{(1)} = \frac{3\lambda\hbar^2}{4m^2\omega^2}

, confirming answer B. A is wrong because

x^4

is even, so it does have diagonal matrix elements between states of the same parity. C misunderstands orthogonality—you're calculating a diagonal matrix element

\langle 1|x^4|1\rangle

, not between different states. D has the wrong coefficient; it's missing the factor of 3 that comes from properly accounting for all contributing terms in the expansion. Study tip: In harmonic oscillator perturbation problems, always express

x

in terms of ladder operators and carefully track which terms contribute to diagonal matrix elements.

Question 6

A particle moves in a two-dimensional isotropic harmonic oscillator potential. The unperturbed energy levels are $E_{n_x,n_y} = \hbar\omega(n_x + n_y + 1)$ where $n_x, n_y = 0, 1, 2, ...$ A small anisotropic perturbation $H' = \epsilon(x^2 - y^2)$ is applied, where $\epsilon << \hbar\omega$ . How does this perturbation affect the degeneracy of the first excited energy level ( $n_x + n_y = 1$ )?

The two-fold degeneracy is completely lifted, creating two non-degenerate levels with energies differing by $2\epsilon\hbar/(m\omega)$ .
The degeneracy remains unchanged because the perturbation preserves the rotational symmetry of the harmonic oscillator.
The degeneracy is lifted, with the $(1,0)$ state having energy $E_1 + \epsilon\hbar/(m\omega)$ and the $(0,1)$ state having energy $E_1 - \epsilon\hbar/(m\omega)$ . (correct answer)
The perturbation causes the two degenerate states to mix, but their energies remain the same to first order in $\epsilon$ .
The degeneracy is enhanced because the perturbation couples these states to higher energy levels, creating a four-fold degenerate multiplet.

Explanation: When you encounter perturbation theory problems with degenerate states, you need to evaluate how the perturbation affects each degenerate state individually, then determine if the degeneracy is lifted. For the first excited level (

n_x + n_y = 1

), there are two degenerate states:

|1,0\rangle

and

|0,1\rangle

, both with unperturbed energy

E_1 = 2\hbar\omega

. To find the first-order energy corrections, calculate the expectation values of

H' = \epsilon(x^2 - y^2)

for each state. Using the harmonic oscillator position operator

x = \sqrt{\frac{\hbar}{2m\omega}}(a_x + a_x^\dagger)

and similar for

y

, you find

\langle x^2 \rangle = \frac{\hbar}{2m\omega}(2n_x + 1)

and

\langle y^2 \rangle = \frac{\hbar}{2m\omega}(2n_y + 1)

. For state

|1,0\rangle

E_1^{(1)} = \epsilon\langle 1,0|x^2 - y^2|1,0\rangle = \epsilon\frac{\hbar}{m\omega}

For state

|0,1\rangle

E_1^{(1)} = \epsilon\langle 0,1|x^2 - y^2|0,1\rangle = -\epsilon\frac{\hbar}{m\omega}

This confirms answer C is correct. A gives the wrong energy difference (factor of 2 error). B is wrong because the perturbation

x^2 - y^2

explicitly breaks rotational symmetry—it distinguishes between x and y directions. D incorrectly suggests the energies don't shift; while the states don't mix (the perturbation matrix is diagonal in this basis), the energies definitely change. Key strategy: In degenerate perturbation theory, always check if the perturbation matrix is diagonal in your chosen basis. If so, degeneracy lifting is straightforward—just compute diagonal matrix elements.

Question 7

Consider a system where degenerate perturbation theory must be applied to a three-fold degenerate level. The perturbation matrix $W$ in the degenerate subspace has been diagonalized, yielding eigenvalues $w_1 < w_2 < w_3$ . If second-order perturbation theory is required to find the next correction to the energies, which statement about the second-order energy corrections is most accurate?

All three perturbed states have identical second-order corrections because they originated from the same degenerate level.
The second-order corrections depend only on matrix elements between the three degenerate states and are independent of coupling to other energy levels.
Each perturbed state has a different second-order correction that depends on its coupling to non-degenerate levels, with the denominators involving differences $E_0 - E_k$ where $E_k$ are non-degenerate level energies. (correct answer)
The second-order corrections are proportional to $w_i^2$ for each state $i$ , representing the square of the first-order correction.
Second-order corrections cannot be calculated until third-order perturbation theory is applied to resolve any remaining accidental degeneracies.

Explanation: When you encounter degenerate perturbation theory problems, remember that the process occurs in stages: first you diagonalize the perturbation matrix within the degenerate subspace, then apply standard perturbation theory to find higher-order corrections. After diagonalizing the perturbation matrix

W

, you've found the first-order corrections (

w_1, w_2, w_3

) and the correct linear combinations of the original degenerate states. Now each perturbed state has energy

E_0 + w_i

, where

E_0

was the original degenerate energy. For second-order corrections, you apply the standard formula: each state couples to all non-degenerate levels through matrix elements, with denominators

E_0 - E_k

where

E_k

are the energies of non-degenerate states. Since each perturbed state has different matrix elements with these outside states, their second-order corrections will generally be different. Answer A is wrong because the three states, though originating from the same level, now have different wavefunctions after diagonalization, leading to different couplings with external states. Answer B misses the crucial point that second-order corrections come from coupling to states outside the degenerate subspace—the internal couplings were already handled in first-order through diagonalization. Answer D incorrectly suggests the correction is simply

w_i^2

, ignoring the actual second-order perturbation formula involving sums over non-degenerate states. Remember: degenerate perturbation theory first "unmixes" the degenerate states, then treats each resulting state independently for higher-order corrections.

Question 8

A particle in a one-dimensional infinite square well of width $L$ has unperturbed energy levels $E_n = \frac{n^2\pi^2\hbar^2}{2mL^2}$ . A delta-function perturbation $H' = \alpha\delta(x - L/2)$ is placed at the center of the well, where $\alpha$ is a small constant. Which statement about the first-order energy corrections is most accurate?

All energy levels receive the same first-order correction $\alpha|\psi_n(L/2)|^2$ regardless of the quantum number $n$ .
Even- $n$ states ( $n = 2, 4, 6, ...$ ) receive no first-order correction because their wavefunctions vanish at $x = L/2$ .
Odd- $n$ states receive corrections $\frac{2\alpha}{L}$ , while even- $n$ states receive no correction. (correct answer)
All states receive correction $\frac{\alpha}{L}$ because the delta function has unit normalization.
The first-order corrections alternate in sign: positive for odd $n$ and negative for even $n$ , but with the same magnitude $\frac{2\alpha}{L}$ .

Explanation: When you encounter perturbation theory problems with delta functions, focus on how the delta function evaluates the wavefunction at a specific point. The first-order energy correction is given by

E_n^{(1)} = \langle\psi_n|H'|\psi_n\rangle = \alpha\int_0^L|\psi_n(x)|^2\delta(x-L/2)dx = \alpha|\psi_n(L/2)|^2

. For the infinite square well, the normalized wavefunctions are

\psi_n(x) = \sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right)

. At the center

x = L/2

, we get

\psi_n(L/2) = \sqrt{\frac{2}{L}}\sin\left(\frac{n\pi}{2}\right)

. Here's the key insight:

\sin(n\pi/2)

equals 1 for odd

n

and 0 for even

n

. Therefore, odd-

n

states have

|\psi_n(L/2)|^2 = \frac{2}{L}

, giving corrections

E_n^{(1)} = \frac{2\alpha}{L}

, while even-

n

states receive zero correction. Answer A incorrectly suggests all states receive the same correction—this ignores that

|\psi_n(L/2)|^2

depends on

n

. Answer B correctly identifies that even-

n

states receive no correction but fails to calculate the specific correction for odd states. Answer D claims all states receive

\frac{\alpha}{L}

, which confuses the wavefunction normalization with the actual evaluation at

x = L/2

. The correct answer is C: odd-

n

states receive

\frac{2\alpha}{L}

while even-

n

states receive zero correction. Remember: with delta function perturbations, always evaluate the unperturbed wavefunction at the delta function's location—the symmetry properties often create selection rules that make some corrections vanish.

Question 9

A quantum system has a two-fold degenerate level with states $|a⟩$ and $|b⟩$ . A perturbation $H' = \epsilon(3|a⟩⟨a| - |b⟩⟨b| + 2i|a⟩⟨b| - 2i|b⟩⟨a|)$ is applied, where $\epsilon$ is real and small. What is the correct approach to find the first-order energy corrections?

Diagonalize the matrix $$ \begin{pmatrix} 3\epsilon & 2i\epsilon \ -2i\epsilon & -\epsilon \end{pmatrix} (correct answer)
Use only the diagonal elements $3\epsilon$ and $-\epsilon$ as the first-order corrections, ignoring the off-diagonal terms.
The perturbation is invalid because it is not Hermitian due to the imaginary off-diagonal elements.
Apply standard non-degenerate perturbation theory to each state separately, giving corrections $3\epsilon$ and $-\epsilon$ .
Average the diagonal elements to get a single first-order correction of $\epsilon$ for both states.

Explanation: When you encounter a quantum system with degenerate energy levels and apply a perturbation, you must use degenerate perturbation theory, not the standard non-degenerate approach. The key insight is that degenerate states can mix under perturbation, so you need to account for all matrix elements of the perturbation operator. The correct approach is to construct the perturbation matrix in the basis of degenerate states

\{|a⟩, |b⟩\}

. From the given perturbation

H'

, you get:

The first-order energy corrections are the eigenvalues of this matrix, which you find by diagonalization. This is exactly what option A describes. Option B fails because it ignores the off-diagonal coupling terms, which is only valid for non-degenerate perturbation theory. When states are degenerate, these mixing terms are crucial. Option C is wrong because the matrix is actually Hermitian—note that

⟨b|H'|a⟩ = -2i\epsilon = (2i\epsilon)^* = ⟨a|H'|b⟩^*

, satisfying the Hermitian condition. Option D incorrectly applies non-degenerate perturbation theory to degenerate states, missing the state mixing entirely. Study tip: Whenever you see degenerate energy levels with a perturbation, immediately think "diagonalize the perturbation matrix." The diagonal vs. off-diagonal elements tell you whether degeneracy is broken and how states mix.

Question 10

Consider the hydrogen atom in the $n = 2$ level, which has four degenerate states: $|2,0,0⟩$ , $|2,1,0⟩$ , $|2,1,1⟩$ , and $|2,1,-1⟩$ . A weak electric field $\vec{E} = E_0\hat{z}$ is applied (linear Stark effect), giving perturbation $H' = eE_0z$ . Which statement about the resulting energy level structure is most accurate?

All four levels split into distinct energies because the electric field breaks the spherical symmetry completely.
The level splits into exactly two distinct energies: one three-fold degenerate level and one non-degenerate level.
The level splits into three distinct energies: two non-degenerate levels and one two-fold degenerate level. (correct answer)
No splitting occurs in first order because all diagonal matrix elements $⟨n,l,m|z|n,l,m⟩$ vanish due to parity.
The splitting pattern depends on whether $E_0$ is positive or negative, giving different degeneracy structures for different field directions.

Explanation: When you encounter Stark effect problems, you're dealing with how electric fields lift degeneracy in atomic energy levels. The key is analyzing which matrix elements of the perturbation are non-zero between the degenerate states. For the perturbation

H' = eE_0z

, you need to calculate matrix elements

⟨n',l',m'|z|n,l,m⟩

between all pairs of the four

n=2

states. The operator

z

has specific selection rules:

\Delta l = \pm 1

and

\Delta m = 0

. This means

z

can only connect states with different

l

values. The only non-zero matrix elements are between

|2,0,0⟩

and the

m=0

state from the

l=1

manifold:

⟨2,0,0|z|2,1,0⟩

and

⟨2,1,0|z|2,0,0⟩

. The states

|2,1,1⟩

and

|2,1,-1⟩

remain unperturbed because they can't couple to anything. Diagonalizing the

2×2

subspace involving

|2,0,0⟩

and

|2,1,0⟩

yields two new energy levels with different first-order corrections. Meanwhile,

|2,1,1⟩

and

|2,1,-1⟩

stay degenerate at the original energy. This gives three distinct energy levels total. Answer D is wrong because while diagonal elements do vanish, off-diagonal elements don't, making first-order perturbation theory applicable. Answer A incorrectly assumes all degeneracy is lifted. Answer B has the degeneracies backwards—it should be one doubly-degenerate and two non-degenerate levels. Remember: In Stark effect problems, focus on selection rules for the electric field operator to determine which states can mix.

Question 11

A particle moves in a three-dimensional spherically symmetric potential. The $l = 1$ energy level is three-fold degenerate with states characterized by $m = -1, 0, +1$ . A perturbation $H' = \beta L_z^2$ is applied, where $L_z$ is the $z$ -component of angular momentum and $\beta$ is a small constant. How does this perturbation affect the degeneracy structure?

The degeneracy is completely lifted, creating three distinct energy levels at $E_l + 0$ , $E_l + \beta\hbar^2$ , and $E_l + \beta\hbar^2$ .
The level splits into two distinct energies: the $m = 0$ state at energy $E_l$ and the $m = ±1$ states at energy $E_l + \beta\hbar^2$ . (correct answer)
The degeneracy is enhanced because $L_z^2$ commutes with $L^2$ and $L_z$ , preserving all symmetries.
All three states receive the same energy shift $\beta⟨L_z^2⟩_{avg}$ , preserving the three-fold degeneracy.
The perturbation has no effect because $L_z^2$ has no matrix elements within a single $l$ manifold.

Explanation: When you encounter perturbation theory problems involving angular momentum, focus on how the perturbation operator acts on the quantum states. The key is calculating the matrix elements of the perturbation in the basis of unperturbed states. For the perturbation

H' = \beta L_z^2

, you need to evaluate

\langle l,m|L_z^2|l,m\rangle

for each

m

value. Since

L_z|l,m\rangle = m\hbar|l,m\rangle

, we have

L_z^2|l,m\rangle = m^2\hbar^2|l,m\rangle

. Therefore:

For $m = 0$ : $\langle 1,0|L_z^2|1,0\rangle = 0^2\hbar^2 = 0$
For $m = ±1$ : $\langle 1,±1|L_z^2|1,±1\rangle = (±1)^2\hbar^2 = \hbar^2$

This gives energy shifts of

0

for the

m=0

state and

\beta\hbar^2

for both

m=±1

states. The degeneracy partially lifts: the

m=0

state remains at energy

E_l

, while the

m=±1

states shift to

E_l + \beta\hbar^2

, creating two distinct energy levels. Option A incorrectly lists the energy shifts and claims complete degeneracy lifting. Option C is wrong because although

L_z^2

commutes with

L^2

and

L_z

, this doesn't preserve degeneracy—it just means these operators share simultaneous eigenstates. Option D incorrectly suggests all states get the same shift, ignoring that the eigenvalues of

L_z^2

depend on

m^2

. Study tip: In perturbation theory, degeneracy is lifted when the perturbation has different eigenvalues for different degenerate states. Always calculate the matrix elements explicitly—don't assume symmetry preserves degeneracy.

Question 12

Consider a system with a four-fold degenerate energy level at $E_0$ . A perturbation is applied such that the $4 × 4$ perturbation matrix $W$ within the degenerate subspace can be block-diagonalized into two $2 × 2$ blocks due to symmetry. If one block has eigenvalues $λ_1, λ_2$ and the other block has eigenvalues $λ_3, λ_4$ , with all eigenvalues distinct, what can be concluded about the selection rules for transitions between the perturbed states?

All transitions between any two of the four perturbed states are allowed because they all originated from the same degenerate manifold.
Transitions are only allowed between states within the same $2 × 2$ block, while transitions between different blocks are forbidden by the symmetry that caused the block structure. (correct answer)
Only transitions between states with the largest energy differences ( $E_0 + λ_1$ to $E_0 + λ_4$ ) are allowed due to energy conservation requirements.
The block structure implies that transitions are allowed only between states with the same eigenvalue magnitude $|λ_i|$ .
No transitions are allowed between any of the perturbed states because the perturbation has lifted all degeneracy and eliminated transition matrix elements.

Explanation: When you encounter degenerate perturbation theory problems, the key insight is that symmetry determines both how the perturbation matrix blocks and which transitions are allowed. The block-diagonal structure isn't arbitrary—it reflects underlying symmetry operations that persist even after the perturbation is applied. The block-diagonal form tells you that the perturbation preserves certain symmetry elements. States within each

2 × 2

block transform according to the same irreducible representation of the symmetry group, while states in different blocks belong to different irreducible representations. This symmetry classification directly determines selection rules: transitions are only allowed between states of compatible symmetry. Since the blocks remain separate due to symmetry, transitions between different blocks are symmetry-forbidden. Only transitions within each

2 × 2

block are allowed, making answer B correct. Answer A incorrectly assumes that sharing the same original degenerate level means all transitions are allowed—this ignores how symmetry creates selection rules. Answer C misapplies energy conservation; energy differences don't determine whether transitions are allowed or forbidden by symmetry. Answer D confuses eigenvalue magnitudes with symmetry classifications—the numerical values of

λ_i

don't determine selection rules. Remember this pattern: whenever a perturbation matrix block-diagonalizes due to symmetry, the block structure directly maps to selection rules. States in different blocks belong to different symmetry species and cannot interconvert via symmetry-allowed transitions. Always connect mathematical structure to physical symmetry principles.

Question 13

A quantum system has three energy levels: a non-degenerate ground state $|0⟩$ with energy $E_0 = 0$ , and two degenerate excited states $|1⟩$ and $|2⟩$ with energy $E_1 = \hbar\omega$ . A perturbation $H' = V(|0⟩⟨1| + |1⟩⟨0| + |1⟩⟨2| + |2⟩⟨1|)$ is applied. If $V << \hbar\omega$ , what is the second-order energy correction to the ground state?

$E_0^{(2)} = -\frac{V^2}{\hbar\omega}$ (correct answer)
$E_0^{(2)} = -\frac{V^2}{2\hbar\omega}$
$E_0^{(2)} = -\frac{2V^2}{\hbar\omega}$
$E_0^{(2)} = 0$ because the ground state is non-degenerate and cannot mix with degenerate levels in second order.
$E_0^{(2)} = \frac{V^2}{\hbar\omega}$

Explanation: When you encounter perturbation theory problems, you need to systematically apply the formulas for energy corrections. The second-order energy correction for a non-degenerate state is given by:

E_n^{(2)} = \sum_{k \neq n} \frac{|\langle k|H'|n \rangle|^2}{E_n^{(0)} - E_k^{(0)}}

For the ground state

|0⟩

, you need matrix elements

\langle k|H'|0⟩

where

k \neq 0

. Looking at the perturbation

H' = V(|0⟩⟨1| + |1⟩⟨0| + |1⟩⟨2| + |2⟩⟨1|)

, only the term

|1⟩⟨0|

contributes:

\langle 1|H'|0⟩ = V

. The terms involving

|2⟩

don't connect to

|0⟩

, so

\langle 2|H'|0⟩ = 0

. Therefore:

E_0^{(2)} = \frac{|V|^2}{E_0^{(0)} - E_1^{(0)}} = \frac{V^2}{0 - \hbar\omega} = -\frac{V^2}{\hbar\omega}

A is correct with this exact result. B incorrectly includes a factor of 2 in the denominator, possibly from mistakenly thinking both degenerate states contribute equally. C has an extra factor of 2 in the numerator, which might come from incorrectly counting the

|0⟩⟨1|

and

|1⟩⟨0|

terms twice. D reflects a fundamental misunderstanding—non-degenerate states absolutely can have second-order corrections, and degeneracy of other levels is irrelevant to this calculation. Key strategy: In second-order perturbation theory, systematically identify which matrix elements are non-zero, then apply the formula term by term. The degeneracy of excited states doesn't prevent ground state corrections.

Question 14

A quantum system has a four-fold degenerate ground state. A small perturbation $H'$ is applied that has matrix elements $\langle ψ_i | H' | ψ_j \rangle$ where $ψ_1, ψ_2, ψ_3, ψ_4$ are the degenerate basis states. If the perturbation matrix in this basis has eigenvalues $λ_1 = 0$ , $λ_2 = λ_3 = ε$ , and $λ_4 = 2ε$ , what is the degeneracy structure of the perturbed system to first order?

One non-degenerate level at energy $E_0$ , one two-fold degenerate level at $E_0 + ε$ , and one non-degenerate level at $E_0 + 2ε$ . (correct answer)
Two two-fold degenerate levels at energies $E_0$ and $E_0 + ε$ , with no states at $E_0 + 2ε$ .
Four non-degenerate levels at energies $E_0$ , $E_0 + ε/2$ , $E_0 + ε$ , and $E_0 + 2ε$ .
One four-fold degenerate level at energy $E_0 + ε/2$ , representing the average perturbation energy.
Two non-degenerate levels at $E_0$ and $E_0 + 2ε$ , and one two-fold degenerate level at $E_0 + ε$ .

Explanation: When you encounter degenerate perturbation theory problems, you need to diagonalize the perturbation matrix within the degenerate subspace to find how the degeneracy is lifted. Starting with a four-fold degenerate ground state at energy

E_0

, the perturbation

H'

lifts this degeneracy. The key insight is that the eigenvalues of the perturbation matrix directly give you the first-order energy corrections. Since the perturbation matrix has eigenvalues

λ_1 = 0

λ_2 = λ_3 = ε

, and

λ_4 = 2ε

, the perturbed energy levels become:

$E_0 + 0 = E_0$ (one state)
$E_0 + ε$ (two states, since $λ_2 = λ_3$ )
$E_0 + 2ε$ (one state)

This gives you one non-degenerate level at

E_0

, one two-fold degenerate level at

E_0 + ε

, and one non-degenerate level at

E_0 + 2ε

, which is answer A. Answer B incorrectly suggests no states exist at

E_0 + 2ε

, ignoring the

λ_4 = 2ε

eigenvalue. Answer C treats all eigenvalues as distinct, missing that

λ_2 = λ_3

means those states remain degenerate. Answer D incorrectly averages the perturbation energies, which isn't how degenerate perturbation theory works—you must diagonalize the perturbation matrix, not average it. Remember: in degenerate perturbation theory, identical eigenvalues of the perturbation matrix correspond to states that remain degenerate after the perturbation is applied. Count the multiplicity of each eigenvalue to determine the new degeneracy structure.

Question 15

A quantum system exhibits accidental degeneracy where two energy levels $E_a$ and $E_b$ from different parts of the spectrum happen to be equal: $E_a = E_b = E_0$ . When a small perturbation $H'$ is applied, the matrix element $⟨a|H'|b⟩ = V ≠ 0$ , while $⟨a|H'|a⟩ = ⟨b|H'|b⟩ = 0$ . As the perturbation strength increases, which statement best describes the behavior of this 'avoided crossing'?

The energy levels cross at the unperturbed value $E_0$ , with the perturbation having no effect on the crossing behavior.
The levels repel each other, with a minimum energy separation of $2|V|$ occurring when the unperturbed levels would have crossed. (correct answer)
The levels approach each other asymptotically but never actually cross, with the minimum separation approaching zero as the perturbation becomes very weak.
One level rises while the other falls, but they cross at an energy different from $E_0$ due to the perturbation.
The crossing behavior depends on the relative phases of the wavefunctions $|a⟩$ and $|b⟩$ .

Explanation: When you encounter degenerate quantum states with a perturbation that couples them, you're dealing with an "avoided crossing" phenomenon that requires degenerate perturbation theory rather than standard first-order perturbation theory. Since the unperturbed states have identical energies (

E_a = E_b = E_0

) and the perturbation couples them (

⟨a|H'|b⟩ = V ≠ 0

), you must diagonalize the 2×2 perturbation matrix. With

⟨a|H'|a⟩ = ⟨b|H'|b⟩ = 0

, this matrix becomes:

\begin{pmatrix} 0 & V \\ V^* & 0 \end{pmatrix}

The eigenvalues are

±|V|

, giving perturbed energies of

E_0 + |V|

and

E_0 - |V|

. This creates an energy gap of

2|V|

between the levels, which represents the minimum separation when the unperturbed levels would have crossed. Option A is wrong because the perturbation does affect the crossing—the levels repel rather than cross. Option C incorrectly suggests the levels approach asymptotically; instead, they maintain a finite separation of

2|V|

. Option D is wrong because while one level rises and the other falls, they don't cross at any energy—they're separated by the gap

2|V|

. The correct answer is B: the levels repel each other with a minimum separation of

2|V|

at the would-be crossing point. Study tip: Remember that whenever degenerate states are coupled by a perturbation, you get level repulsion, not crossing. The coupling strength directly determines the minimum energy gap: stronger coupling creates wider avoided crossings.

Question 16

A quantum system initially has a three-fold degenerate energy level. When a perturbation is applied, degenerate perturbation theory shows that the $3 × 3$ perturbation matrix $W$ has the form $W = \begin{pmatrix} a & b & 0 \\ b & a & 0 \\ 0 & 0 & c \end{pmatrix}$ where $a$ , $b$ , and $c$ are real constants with $b ≠ 0$ and $c ≠ a$ . What can be concluded about the degeneracy of the perturbed energy levels?

All three levels become non-degenerate with energies $E_0 + a - b$ , $E_0 + a + b$ , and $E_0 + c$ . (correct answer)
One level remains three-fold degenerate at energy $E_0 + (2a + c)/3$ , representing the average of the diagonal elements.
Two levels become non-degenerate at energies $E_0 + a - b$ and $E_0 + a + b$ , while one remains degenerate at $E_0 + c$ .
The system has one non-degenerate level at $E_0 + c$ and one two-fold degenerate level at $E_0 + a$ .
Two levels remain two-fold degenerate at energies $E_0 + a$ and $E_0 + c$ , with no non-degenerate levels.

Explanation: When you encounter degenerate perturbation theory problems, you need to find the eigenvalues of the perturbation matrix to determine how the degeneracy is lifted. The eigenvalues give you the first-order energy corrections that split the original degenerate level. To find the eigenvalues of this

3 \times 3

matrix, you solve the characteristic equation

\det(W - \lambda I) = 0

. Notice that this matrix has a block structure: the first

2 \times 2

block is $$ \begin{pmatrix} a & b \ b & a \end{pmatrix}

For the $$2 \times 2$$ block, the eigenvalues are $$\lambda = a \pm b$$. The third eigenvalue comes directly from the diagonal element: $$\lambda = c$$. Since $$b \neq 0$$ and $$c \neq a$$, all three eigenvalues are distinct: $$a-b$$, $$a+b$$, and $$c$$. The perturbed energy levels are therefore $$E_0 + (a-b)$$, $$E_0 + (a+b)$$, and $$E_0 + c$$, making all three levels non-degenerate. This confirms answer A. Answer B incorrectly suggests the degeneracy remains, ignoring that distinct eigenvalues split the levels. Answer C mistakenly claims one level stays degenerate, but $$c$$ corresponds to a single eigenvalue with multiplicity one. Answer D incorrectly suggests two-fold degeneracy at $$E_0 + a$$, but $$a$$ itself isn't an eigenvalue—the actual eigenvalues are $$a \pm b$$. **Study tip**: Always diagonalize the perturbation matrix completely. The number of distinct eigenvalues tells you how many different energy levels you'll have after the perturbation.

Question 17

Two identical quantum harmonic oscillators are coupled by a weak interaction $H' = \lambda(a_1^\dagger a_2 + a_1 a_2^\dagger)$ , where $a_i$ and $a_i^\dagger$ are the lowering and raising operators for oscillator $i$ , and $\lambda \ll \hbar\omega$ . Initially, the system is in a state where one oscillator has 2 quanta and the other has 0 quanta. To what extent will this state mix with other states under the perturbation?

It will mix only with the $|0\rangle_1|2\rangle_2$ state, creating symmetric and antisymmetric combinations
It will mix with all states in the two-quantum manifold through a complex coupling network
It will mix with both $|0\rangle_1|2\rangle_2$ and $|1\rangle_1|1\rangle_2$ states since all have the same total energy (correct answer)
No mixing occurs because the perturbation conserves the individual quantum numbers of each oscillator

Explanation: When you encounter coupled quantum harmonic oscillators, the key insight is that the coupling operator determines which states can mix. The interaction

H' = \lambda(a_1^\dagger a_2 + a_1 a_2^\dagger)

transfers one quantum between oscillators while conserving the total number of quanta. Starting with

|2\rangle_1|0\rangle_2

, let's see what states this can connect to. The operator

a_1^\dagger a_2

creates a quantum in oscillator 1 and destroys one in oscillator 2, but since oscillator 2 has zero quanta, this term gives zero. The operator

a_1 a_2^\dagger

destroys a quantum in oscillator 1 and creates one in oscillator 2, yielding

|1\rangle_1|1\rangle_2

. From

|1\rangle_1|1\rangle_2

, the coupling can produce both

|2\rangle_1|0\rangle_2

and

|0\rangle_1|2\rangle_2

. All three states have the same total energy (two quanta total) and form a connected subspace. Answer A is incorrect because it ignores the

|1\rangle_1|1\rangle_2

state, which is directly accessible through the coupling. Answer B overstates the complexity – only states with exactly two total quanta can mix. Answer D misses that while individual quantum numbers aren't conserved, the total quantum number is conserved, allowing mixing within the two-quantum manifold. The correct answer is C: the initial state mixes with both

|0\rangle_1|2\rangle_2

and

|1\rangle_1|1\rangle_2

because they share the same total energy and are connected by the coupling operator. Remember: coupling operators that transfer quanta between subsystems create mixing within constant total quantum number subspaces.

Question 18

A quantum dot can be modeled as a 3D isotropic harmonic oscillator with frequency $\omega$ . The energy levels are $E_n = \hbar\omega(n + \frac{3}{2})$ where $n = n_x + n_y + n_z$ . If a small anisotropy is introduced such that $\omega_z = \omega + \delta\omega$ while $\omega_x = \omega_y = \omega$ (where $\delta\omega \ll \omega$ ), what happens to the first excited level ( $n = 1$ ) which was initially 3-fold degenerate?

It splits into two levels: one non-degenerate level and one doubly degenerate level (correct answer)
It splits into three non-degenerate levels with distinct energies
It remains 3-fold degenerate because the perturbation is too weak to lift the degeneracy
It splits into two doubly degenerate levels due to the cylindrical symmetry of the perturbed system

Explanation: The

n=1

level has states

(1,0,0)

(0,1,0)

, and

(0,0,1)

with unperturbed energy

E_1 = \frac{5\hbar\omega}{2}

. The perturbation

H' = \hbar\delta\omega(a_z^\dagger a_z)

affects only the z-component. The perturbed energies are:

(1,0,0)

E = \hbar\omega + \frac{\hbar\omega}{2} + \frac{\hbar\omega}{2} = \frac{5\hbar\omega}{2}

(0,1,0)

: same as above,

(0,0,1)

E = \frac{\hbar\omega}{2} + \frac{\hbar\omega}{2} + \hbar(\omega + \delta\omega) = \frac{5\hbar\omega}{2} + \hbar\delta\omega

. Thus we get one doubly degenerate level (states with

n_z = 0

) and one non-degenerate level (state with

n_z = 1

). Choice B suggests three distinct energies. Choice C ignores the perturbation effect. Choice D incorrectly describes the degeneracy pattern.

Question 19

In a molecular orbital treatment of benzene, the six $\pi$ electrons occupy three bonding MOs with energies $\alpha + 2\beta$ , $\alpha + \beta$ , and $\alpha + \beta$ (where the latter two are degenerate). If a weak perturbation breaks the hexagonal symmetry by making one C-C bond slightly longer, which statement best describes the immediate effect on the electronic structure?

The highest occupied degenerate level splits, but the energy ordering of all MOs remains unchanged
The degenerate HOMO levels split and may cross with the lowest unoccupied molecular orbital
Only the non-degenerate $\alpha + 2\beta$ level shifts significantly while the degenerate levels remain nearly unchanged
All three occupied MO energies shift, and the degeneracy of the HOMO level is completely removed (correct answer)

Explanation: When analyzing molecular orbital perturbations, you need to consider how symmetry breaking affects all energy levels, not just the most obvious ones. In benzene's

\pi

system, the molecular orbitals are defined by the molecule's six-fold rotational symmetry. When you break this symmetry by elongating one C-C bond, several things happen simultaneously. The perturbation creates a new molecular environment where the original symmetry-derived energy levels no longer apply. Most importantly, all occupied MO energies will shift because each orbital has some electron density distributed around the entire ring, including the perturbed bond region. The lowest energy

\alpha + 2\beta

orbital, while still remaining the most stable, will experience an energy shift because it too has bonding character involving all C-C interactions. The degenerate HOMO levels (

\alpha + \beta

) will definitely split since degeneracy requires perfect symmetry. Breaking the hexagonal symmetry completely removes the mathematical basis for their equal energies. Option A incorrectly assumes the energy ordering stays the same - perturbations can cause level crossings. Option B focuses too narrowly on HOMO-LUMO interactions while ignoring effects on other levels. Option C wrongly suggests the degenerate levels remain unchanged - any symmetry breaking must split degenerate levels. Answer D correctly recognizes that all MO energies shift under perturbation and that degeneracy is completely removed, reflecting the new lower-symmetry environment. Study tip: Remember that in MO theory, symmetry breaking affects the entire electronic structure, not just the frontier orbitals. Always consider how perturbations influence all occupied levels.

Question 20

A quantum system has an unperturbed Hamiltonian with eigenvalues $E_1 = 0$ , $E_2 = E_3 = \epsilon$ , and $E_4 = 2\epsilon$ . A perturbation is applied such that the only non-zero matrix elements of $H'$ are $\langle 2|H'|3\rangle = \langle 3|H'|2\rangle = \delta$ and $\langle 1|H'|4\rangle = \langle 4|H'|1\rangle = \gamma$ , where $\delta, \gamma \ll \epsilon$ . Using first-order perturbation theory, what are the perturbed energy levels?

$E_1' = \gamma$ , $E_2' = E_3' = \epsilon + \delta$ , $E_4' = 2\epsilon + \gamma$
$E_1' = \gamma$ , $E_2' = \epsilon - \delta$ , $E_3' = \epsilon + \delta$ , $E_4' = 2\epsilon$
$E_1' = 0$ , $E_2' = E_3' = \epsilon$ , $E_4' = 2\epsilon$
$E_1' = 0$ , $E_2' = \epsilon - \delta$ , $E_3' = \epsilon + \delta$ , $E_4' = 2\epsilon$ (correct answer)

Explanation: When you encounter perturbation theory problems, the key is distinguishing between degenerate and non-degenerate cases. Here, states 2 and 3 are degenerate (both have energy

\epsilon

), while states 1 and 4 are non-degenerate. For non-degenerate states, the first-order energy correction is simply the diagonal matrix element of the perturbation:

E_n^{(1)} = \langle n|H'|n\rangle

. Since all diagonal elements of

H'

are zero, states 1 and 4 receive no first-order corrections:

E_1' = 0

and

E_4' = 2\epsilon

. For degenerate states 2 and 3, you must diagonalize the perturbation matrix within the degenerate subspace. The

2×2

matrix is:

\begin{pmatrix} 0 & \delta \\ \delta & 0 \end{pmatrix}

The eigenvalues are

\pm\delta

, so the corrected energies are

E_2' = \epsilon - \delta

and

E_3' = \epsilon + \delta

. Option A incorrectly applies diagonal corrections to all states, ignoring that off-diagonal elements don't directly add to energies. Option B mistakenly gives state 1 a correction of

\gamma

, but

\langle 1|H'|1\rangle = 0

. Option C ignores the perturbation entirely, giving no corrections at all. The correct answer is D:

E_1' = 0

E_2' = \epsilon - \delta

E_3' = \epsilon + \delta

E_4' = 2\epsilon

. Study tip: Always check for degeneracies first. Non-degenerate states get diagonal corrections only, while degenerate states require matrix diagonalization within their subspace.