Degeneracy & Perturbation - Physical Chemistry 2

In a cubic crystal field, the $d$ orbitals of a transition metal ion split into $t_{2g}$ (threefold degenerate) and $e_g$ (twofold degenerate) levels. If a small tetragonal distortion is applied along the z-axis, making the crystal field slightly weaker in that direction, what is the correct order of orbital energies from lowest to highest?
Answer: $d_{xy}, d_{xz}=d_{yz}, d_{z^2}, d_{x^2-y^2}$ In cubic symmetry, $t_{2g}$ ( $d_{xy}, d_{xz}, d_{yz}$ ) are lower in energy than $e_g$ ( $d_{z^2}, d_{x^2-y^2}$ ). A weaker field along z-axis means orbitals with z-character are stabilized. The $d_{xy}$ orbital (no z-character) is least affected and lowest. The $d_{xz}$ and $d_{yz}$ orbitals have some z-character and are degenerate at intermediate energy. The $d_{z^2}$ orbital is more stabilized than $d_{x^2-y^2}$ due to its z-character. Choice B incorrectly places $d_{z^2}$ lowest. Choice C maintains cubic degeneracies that should be lifted. Choice D has the wrong relative ordering of the split levels.
Consider a particle in a 2D square box where the unperturbed energies are $E_{n_x,n_y} = \frac{h^2}{8ma^2}(n_x^2 + n_y^2)$ . A small perturbation $H' = \lambda xy$ is applied. Using first-order perturbation theory, which quantum states will have their energies shifted from their unperturbed values?
Answer: No states will have first-order energy shifts because the $xy$ perturbation has odd parity The unperturbed wavefunctions are $\psi_{n_x,n_y} = \frac{2}{a}\sin(\frac{n_x\pi x}{a})\sin(\frac{n_y\pi y}{a})$ . First-order energy shifts are $E^{(1)} = \langle\psi|H'|\psi\rangle = \lambda\langle\psi|xy|\psi\rangle$ . Since the unperturbed states have definite parity and $xy$ is an odd function in both coordinates, all diagonal matrix elements vanish by symmetry. Choice A is wrong because ground state coupling isn't automatic. Choice B ignores the parity argument. Choice D incorrectly assumes only degenerate states can have energy shifts in perturbation theory.
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?
Answer: The perturbed energies are approximately $E_0 ± V$ with eigenstates approximately $\frac{1}{\sqrt{2}}(|1⟩ ± |2⟩)$ . 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: $H' = \begin{pmatrix} \Delta & V \\ V & -\Delta \end{pmatrix}$ 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.
A system has three degenerate energy levels at $E_0$ , with basis states $|1\rangle$ , $|2\rangle$ , and $|3\rangle$ . A perturbation $H'$ has matrix elements: $\langle 1|H'|1\rangle = a$ , $\langle 2|H'|2\rangle = a$ , $\langle 3|H'|3\rangle = -2a$ , $\langle 1|H'|2\rangle = \langle 2|H'|1\rangle = b$ , and all other matrix elements are zero. What condition must be satisfied for the perturbation to completely remove the degeneracy?
Answer: $a \neq 0$ and $b$ can be any value including zero The perturbation matrix is $\begin{pmatrix} a & b & 0 \\ b & a & 0 \\ 0 & 0 & -2a \end{pmatrix}$ . The eigenvalue $E_0 - 2a$ (from state 3) is always distinct if $a \neq 0$ . The $2×2$ block has eigenvalues $E_0 + a ± b$ . These are distinct from each other when $b \neq 0$ , but more importantly, they're distinct from $E_0 - 2a$ whenever $a \neq 0$ . Even if $b = 0$ , we get eigenvalues $E_0 + a$ (doubly degenerate) and $E_0 - 2a$ , partially lifting degeneracy. Complete degeneracy removal requires $a \neq 0$ . Choice A incorrectly prioritizes $b$ . Choice C requires both when only $a$ is essential. Choice D gives an irrelevant magnitude condition.
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'?
Answer: The levels repel each other, with a minimum energy separation of $2|V|$ occurring when the unperturbed levels would have crossed. 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.
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?
Answer: The level splits into two distinct energies: the $m = 0$ state at energy $E_l$ and the $m = ±1$ states at energy $E_l + \beta\hbar^2$ . 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.
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?
Answer: One non-degenerate level at energy $E_0$ , one two-fold degenerate level at $E_0 + ε$ , and one non-degenerate level at $E_0 + 2ε$ . 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.
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?
Answer: 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. 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.
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?
Answer: 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. 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.
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?
Answer: Odd- $n$ states receive corrections $\frac{2\alpha}{L}$ , while even- $n$ states receive no correction. 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.
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?
Answer: All three levels become non-degenerate with energies $E_0 + a - b$ , $E_0 + a + b$ , and $E_0 + c$ . 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 $\begin{pmatrix} a & b \\ b & a \end{pmatrix}$ , and the third row and column are decoupled. 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.
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?
Answer: Diagonalize the matrix $\begin{pmatrix} 3\epsilon & 2i\epsilon \\ -2i\epsilon & -\epsilon \end{pmatrix}$ and use the eigenvalues as the first-order corrections. 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: $\mathbf{W} = \begin{pmatrix} ⟨a|H'|a⟩ & ⟨a|H'|b⟩ \\ ⟨b|H'|a⟩ & ⟨b|H'|b⟩ \end{pmatrix} = \begin{pmatrix} 3\epsilon & 2i\epsilon \\ -2i\epsilon & -\epsilon \end{pmatrix}$ 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.
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?
Answer: $E_1' = 0$ , $E_2' = \epsilon - \delta$ , $E_3' = \epsilon + \delta$ , $E_4' = 2\epsilon$ 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.
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?
Answer: The first-order energy corrections are zero for all states, so degeneracy lifting only occurs in second-order perturbation theory. 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.
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?
Answer: $E_0^{(2)} = -\frac{V^2}{\hbar\omega}$ 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.
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?
Answer: 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 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.
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?
Answer: The first-order correction is $\lambda \langle 1|x^4|1⟩ = \frac{3\lambda\hbar^2}{4m^2\omega^2}$ . 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.
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?
Answer: $E_0^{(2)} = -\frac{V^2}{\hbar\omega}$ 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.
A rigid rotor molecule with rotational constant $B$ has its $J=2$ level (initially 5-fold degenerate) perturbed by a weak external field that breaks the spherical symmetry. The perturbation can be written as $H' = \alpha Y_2^0(\theta,\phi)$ where $Y_2^0$ is a spherical harmonic and $\alpha$ is a small constant. After applying first-order perturbation theory, how many distinct energy levels will the $J=2$ manifold split into?
Answer: Three levels: $m_J = 0$ , $m_J = ±1$ , and $m_J = ±2$ The perturbation $H' = \alpha Y_2^0(\theta,\phi) = \alpha\sqrt{\frac{5}{16\pi}}(3\cos^2\theta - 1)$ preserves axial symmetry around the z-axis, so $m_J$ remains a good quantum number. First-order energy shifts are $E^{(1)}_{J,m_J} = \langle J,m_J|H'|J,m_J\rangle = \alpha\langle Y_2^{m_J}|Y_2^0|Y_2^{m_J}\rangle$ . Using properties of spherical harmonics, this gives different values for $m_J = 0$ , $m_J = ±1$ , and $m_J = ±2$ , but states with $±m_J$ remain degenerate due to the axial symmetry. Choice A incorrectly groups $m_J = ±1, ±2$ together. Choice C gives the wrong count. Choice D assumes complete degeneracy removal, which doesn't occur when axial symmetry is preserved.
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?
Answer: All three occupied MO energies shift, and the degeneracy of the HOMO level is completely removed 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.

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