Physical Chemistry 2 Quiz: Harmonic Oscillator Model

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Harmonic Oscillator ModelQuestion 1 of 19

For a quantum harmonic oscillator, the expectation value of the kinetic energy in the $v$ -th vibrational state can be expressed in terms of the total energy. Which relationship correctly describes this for all quantum states?

\langle T \rangle = \frac{1}{3} E_v

because kinetic energy is always less than potential energy on average

\langle T \rangle = \frac{1}{2} E_v

due to the virial theorem applied to the harmonic potential

\langle T \rangle = \frac{2}{3} E_v

because the oscillator spends more time at higher kinetic energies

\langle T \rangle = E_v - \frac{1}{2}\hbar\omega

since potential energy has a minimum value of

\frac{1}{2}\hbar\omega

\langle T \rangle = \sqrt{2} \cdot \frac{E_v}{2}

due to the quadratic nature of both kinetic and potential terms

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Physical Chemistry 2 Quiz

Physical Chemistry 2 Quiz: Harmonic Oscillator Model

Practice Harmonic Oscillator Model 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 Harmonic Oscillator Model, 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

$\langle T \rangle = \frac{1}{3} E_v$ because kinetic energy is always less than potential energy on average
$\langle T \rangle = \frac{1}{2} E_v$ due to the virial theorem applied to the harmonic potential (correct answer)
$\langle T \rangle = \frac{2}{3} E_v$ because the oscillator spends more time at higher kinetic energies
$\langle T \rangle = E_v - \frac{1}{2}\hbar\omega$ since potential energy has a minimum value of $\frac{1}{2}\hbar\omega$
$\langle T \rangle = \sqrt{2} \cdot \frac{E_v}{2}$ due to the quadratic nature of both kinetic and potential terms

Explanation: When you encounter quantum harmonic oscillator problems, the key insight is recognizing how the virial theorem applies to this specific potential energy function. For a harmonic oscillator with potential

V = \frac{1}{2}kx^2

, the virial theorem states that

\langle T \rangle = \langle V \rangle

- the average kinetic and potential energies are equal. Since the total energy is

E_v = \langle T \rangle + \langle V \rangle

, and these two terms are equal, we get

E_v = 2\langle T \rangle

, which means

\langle T \rangle = \frac{1}{2} E_v

. This relationship holds for all vibrational states, making option B correct. Option A incorrectly suggests kinetic energy is always one-third of total energy and mischaracterizes the kinetic-potential energy relationship. For harmonic oscillators, these energies are equal on average, not in a 1:2 ratio. Option C claims the oscillator spends more time at higher kinetic energies, leading to

\langle T \rangle = \frac{2}{3} E_v

. This reasoning is flawed - the time spent at different energy configurations doesn't create this imbalance, and it contradicts the virial theorem result. Option D attempts to use the fact that

E_v = (v + \frac{1}{2})\hbar\omega

, but the expression

E_v - \frac{1}{2}\hbar\omega

equals

v\hbar\omega

, not the kinetic energy expectation value. Study tip: Remember that for any potential of the form

V \propto x^n

, the virial theorem gives

\langle T \rangle = \frac{n}{2}\langle V \rangle

. For harmonic oscillators (

n=2

), this means

\langle T \rangle = \langle V \rangle = \frac{1}{2}E_v

Question 2

Two identical quantum harmonic oscillators are coupled by a weak interaction term $V_{12} = \lambda x_1 x_2$ where $\lambda$ is small. Using first-order perturbation theory, what is the energy correction for the state where oscillator 1 is in $v_1 = 1$ and oscillator 2 is in $v_2 = 0$ ?

$E^{(1)} = \lambda \sqrt{\frac{\hbar}{2m\omega}}$ because only the $v=1$ oscillator contributes to the position expectation
$E^{(1)} = 0$ because $\langle v_1 | x_1 | v_1 \rangle = 0$ and $\langle v_2 | x_2 | v_2 \rangle = 0$ for all harmonic oscillator states (correct answer)
$E^{(1)} = \lambda \frac{\hbar}{2m\omega}$ from the product of non-zero position matrix elements between adjacent states
$E^{(1)} = \frac{\lambda}{2} \sqrt{\frac{\hbar}{m\omega}}$ because the coupling involves both oscillators with different quantum numbers
$E^{(1)} = \lambda \sqrt{\frac{\hbar}{m\omega}}$ based on the sum of individual oscillator position uncertainties

Explanation: This question tests your understanding of first-order perturbation theory and the matrix elements of quantum harmonic oscillators. When applying perturbation theory, the first-order energy correction is

E^{(1)} = \langle \psi^{(0)} | V | \psi^{(0)} \rangle

, where you take the expectation value of the perturbation with the unperturbed wavefunction. For the coupled system, the unperturbed state is

|v_1=1, v_2=0\rangle

, and the perturbation is

V_{12} = \lambda x_1 x_2

. The first-order correction becomes:

E^{(1)} = \langle 1,0 | \lambda x_1 x_2 | 1,0 \rangle = \lambda \langle 1 | x_1 | 1 \rangle \langle 0 | x_2 | 0 \rangle

The key insight is that for quantum harmonic oscillators, the position operator has zero diagonal matrix elements:

\langle v | x | v \rangle = 0

for any quantum number

v

. This means both

\langle 1 | x_1 | 1 \rangle = 0

and

\langle 0 | x_2 | 0 \rangle = 0

, making the entire product zero. Answer B correctly identifies this fundamental property. Answer A incorrectly assumes only one oscillator contributes and gets the physics wrong. Answer C mistakenly uses matrix elements between different states (

\langle v | x | v \pm 1 \rangle

), but we need diagonal elements for first-order perturbation theory. Answer D provides an arbitrary numerical factor without proper justification. Remember: for harmonic oscillators, position matrix elements are only non-zero between adjacent states (

v

and

v \pm 1

), never for transitions within the same state. This makes many perturbation problems surprisingly simple.

Question 3

A quantum harmonic oscillator is prepared in a coherent state $|\alpha\rangle$ with $\alpha = 2$ . What is the probability of measuring the oscillator in the ground state $|0\rangle$ ?

$P_0 = e^{-2}$ because the coherent state amplitude determines the ground state overlap directly
$P_0 = e^{-4}$ since the probability depends on $|\alpha|^2 = 4$ in the exponential factor (correct answer)
$P_0 = \frac{1}{e^2}$ because coherent states have Gaussian probability distributions centered at $\alpha$
$P_0 = \frac{4}{e^4}$ from the normalization of the coherent state expansion in number states
$P_0 = \frac{2}{e^2}$ because the ground state coefficient scales linearly with the coherent state parameter

Explanation: When you encounter coherent states in quantum mechanics, remember that these are special superposition states that bridge classical and quantum behavior. The key insight is understanding how to calculate overlap probabilities between a coherent state and number states (energy eigenstates). A coherent state

|\alpha\rangle

can be expanded in terms of number states as

|\alpha\rangle = e^{-|\alpha|^2/2} \sum_{n=0}^{\infty} \frac{\alpha^n}{\sqrt{n!}}|n\rangle

. The probability of measuring the oscillator in the ground state

|0\rangle

P_0 = |\langle 0|\alpha\rangle|^2

. From the expansion, the ground state component is

\langle 0|\alpha\rangle = e^{-|\alpha|^2/2}

, so

P_0 = e^{-|\alpha|^2}

. With

\alpha = 2

, we get

|\alpha|^2 = 4

, giving

P_0 = e^{-4}

. Option A incorrectly uses

|\alpha|

instead of

|\alpha|^2

in the exponential. The overlap depends on the squared magnitude, not the amplitude itself. Option C gives the correct numerical value but with incorrect reasoning—while coherent states do have Gaussian-like properties, the ground state probability specifically comes from the normalization factor in the coherent state expansion. Option D includes an extra factor of

|\alpha|^2 = 4

in the numerator, which would arise from confusing the expansion coefficients with probability calculations. Remember: for coherent state problems, always work with

|\alpha|^2

in exponentials, and distinguish between expansion coefficients and their squared magnitudes when calculating probabilities.

Question 4

A quantum harmonic oscillator with frequency $\omega$ is subjected to a time-dependent driving force such that the Hamiltonian becomes $H(t) = \frac{p^2}{2m} + \frac{1}{2}m\omega^2 x^2 + F(t)x$ , where $F(t) = F_0 \cos(\omega t)$ . In the interaction picture, what is the primary effect of this driving force on the energy levels?

The energy levels shift uniformly by $\frac{F_0^2}{2m\omega^3}$ due to the oscillating electric field interaction
Transitions are induced between adjacent vibrational levels with selection rule $\Delta v = \pm 1$ only (correct answer)
The energy levels become time-dependent and oscillate with frequency $2\omega$ around their unperturbed values
Resonant absorption occurs leading to transitions between all energy levels regardless of selection rules
The system undergoes parametric amplification causing exponential growth in the vibrational quantum numbers

Explanation: When you encounter a time-dependent Hamiltonian with a harmonic oscillator plus a driving force, you're dealing with a classic problem in quantum dynamics where the driving force can induce transitions between energy levels. The key insight is analyzing the driving term

F_0 \cos(\omega t) x

in the interaction picture. Since

\cos(\omega t) = \frac{1}{2}(e^{i\omega t} + e^{-i\omega t})

and the position operator can be written in terms of creation and annihilation operators as

x \propto (a + a^\dagger)

, the driving force becomes proportional to terms like

a e^{i\omega t}

and

a^\dagger e^{-i\omega t}

. When the driving frequency matches the oscillator frequency (

\omega

), these terms become resonant and can efficiently drive transitions between adjacent energy levels. The selection rule

\Delta v = \pm 1

emerges because the annihilation operator

a

lowers the quantum number by 1, while the creation operator

a^\dagger

raises it by 1. Option A is incorrect because

\frac{F_0^2}{2m\omega^3}

represents a second-order energy shift that would arise from time-independent perturbation theory, not the primary effect of resonant driving. Option C is wrong because while there may be some oscillatory behavior, the energy levels themselves don't oscillate with frequency

2\omega

- this confuses the driving dynamics with energy level structure. Option D is incorrect because quantum mechanical selection rules still apply; the harmonic oscillator driving specifically couples only adjacent levels, not all possible transitions. Remember: resonant driving of quantum systems typically follows strict selection rules determined by the operators involved in the interaction term.

Question 5

For a quantum harmonic oscillator, the position operator can be expressed as $x = \sqrt{\frac{\hbar}{2m\omega}}(a + a^\dagger)$ . What is the matrix element $\langle 3 | x | 1 \rangle$ ?

$\langle 3 | x | 1 \rangle = 0$ because the matrix element connects states that differ by 2 in quantum number (correct answer)
$\langle 3 | x | 1 \rangle = \sqrt{\frac{\hbar}{2m\omega}}$ from the ladder operator connecting adjacent energy levels
$\langle 3 | x | 1 \rangle = \sqrt{\frac{3\hbar}{2m\omega}}$ because the quantum number 3 appears in the normalization
$\langle 3 | x | 1 \rangle = \sqrt{\frac{2\hbar}{m\omega}}$ from the sum of both ladder operator contributions
$\langle 3 | x | 1 \rangle = \sqrt{\frac{6\hbar}{m\omega}}$ based on the product of the two quantum numbers involved

Explanation: When working with quantum harmonic oscillator matrix elements, you need to understand how ladder operators connect different energy states and apply selection rules. The position operator

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

contains two terms: the lowering operator

a

and raising operator

a^\dagger

. The key relationships are

a|n\rangle = \sqrt{n}|n-1\rangle

and

a^\dagger|n\rangle = \sqrt{n+1}|n+1\rangle

. This means

a

connects state

|n\rangle

only to

|n-1\rangle

, while

a^\dagger

connects it only to

|n+1\rangle

. For

\langle 3|x|1\rangle

, you need to evaluate

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

. The first term gives

\langle 3|\sqrt{1}|0\rangle = 0

since

\langle 3|0\rangle = 0

. The second term gives

\langle 3|\sqrt{2}|2\rangle = 0

since

\langle 3|2\rangle = 0

. Both contributions vanish because the operators can only change the quantum number by ±1, but you need to connect states differing by 2. Choice A correctly identifies this selection rule. Choice B incorrectly assumes a non-zero contribution from adjacent level connections. Choice C mistakenly includes the quantum number 3 in a spurious calculation. Choice D incorrectly adds contributions as if both ladder operators gave non-zero matrix elements. Remember this selection rule: for harmonic oscillator position and momentum operators, matrix elements are non-zero only between states differing by exactly 1 in quantum number (

\Delta n = \pm 1

Question 6

A quantum harmonic oscillator in the $v = 0$ state is suddenly subjected to a constant electric field $\mathcal{E}$ , adding a potential term $-q\mathcal{E}x$ to the Hamiltonian. What happens to the energy eigenvalues after the system reaches equilibrium?

All energy levels shift uniformly downward by $\frac{q^2\mathcal{E}^2}{2m\omega^2}$ due to the linear Stark effect
The energy levels become $E_v = \hbar\omega(v + \frac{1}{2}) - \frac{q^2\mathcal{E}^2}{2m\omega^2}$ with a common shift for all states (correct answer)
The energy levels split into doublets with separation $\Delta E = q\mathcal{E}\sqrt{\frac{\hbar}{2m\omega}}$ for each $v$
Only odd- $v$ states are affected because the electric field couples states of opposite parity
The energy spacing between adjacent levels changes from $\hbar\omega$ to $\hbar\omega + q\mathcal{E}$

Explanation: When a quantum harmonic oscillator experiences a sudden perturbation like an electric field, you need to analyze how the new Hamiltonian affects the equilibrium energy levels. The key insight is recognizing that adding a linear potential

-q\mathcal{E}x

shifts the equilibrium position but preserves the harmonic nature of the oscillator. The new Hamiltonian becomes

H = \frac{p^2}{2m} + \frac{1}{2}m\omega^2x^2 - q\mathcal{E}x

. You can complete the square by writing this as

H = \frac{p^2}{2m} + \frac{1}{2}m\omega^2(x - \frac{q\mathcal{E}}{m\omega^2})^2 - \frac{q^2\mathcal{E}^2}{2m\omega^2}

. This shows the oscillator now has its equilibrium shifted to

x_{eq} = \frac{q\mathcal{E}}{m\omega^2}

, with a constant energy shift of

-\frac{q^2\mathcal{E}^2}{2m\omega^2}

applied to all levels. Since the harmonic spacing remains unchanged, all energy levels shift uniformly downward, giving

E_v = \hbar\omega(v + \frac{1}{2}) - \frac{q^2\mathcal{E}^2}{2m\omega^2}

, which is answer B. Answer A incorrectly mentions the "linear Stark effect" - this is actually the quadratic Stark effect since the energy shift goes as

\mathcal{E}^2

. Answer C describes level splitting, which doesn't occur for this symmetric perturbation. Answer D incorrectly suggests parity-based selection rules apply to the energy shift itself, confusing this with transition probabilities between states. Remember: linear potentials in harmonic oscillators always cause uniform energy shifts while preserving the harmonic level structure. The shift magnitude depends on the square of the field strength.

Question 7

For a quantum harmonic oscillator, consider the squeezed state defined by $|\xi\rangle = S(\xi)|0\rangle$ where $S(\xi) = \exp[\frac{\xi^*a^2 - \xi a^{\dagger 2}}{2}]$ is the squeezing operator with real parameter $\xi$ . How do the position and momentum uncertainties compare to those of the ground state?

Both $\Delta x$ and $\Delta p$ are reduced equally, maintaining the uncertainty relation at its minimum
$\Delta x = e^{-\xi}\Delta x_0$ and $\Delta p = e^{\xi}\Delta p_0$ , where subscript 0 denotes ground state values (correct answer)
$\Delta x = e^{\xi}\Delta x_0$ and $\Delta p = e^{-\xi}\Delta p_0$ , with the product remaining at the minimum value
The uncertainties oscillate sinusoidally with $\xi$ while preserving the uncertainty principle equality
$\Delta x = \cosh(\xi)\Delta x_0$ and $\Delta p = \cosh(\xi)\Delta p_0$ , both increasing with squeezing

Explanation: When you encounter squeezed states in quantum mechanics, you're dealing with states that redistribute the uncertainties in position and momentum while maintaining the fundamental uncertainty principle. The key insight is that squeezing reduces uncertainty in one observable at the expense of increasing it in the conjugate observable. For the squeezed state

|\xi\rangle = S(\xi)|0\rangle

with real parameter

\xi

, the squeezing operator transforms the ladder operators in a way that modifies the position and momentum uncertainties exponentially. The mathematics shows that

\Delta x = e^{-\xi}\Delta x_0

and

\Delta p = e^{\xi}\Delta p_0

, where the subscript 0 refers to ground state values. This means when

\xi > 0

, position uncertainty is squeezed (reduced) while momentum uncertainty is stretched (increased), and vice versa for

\xi < 0

. Option A incorrectly suggests both uncertainties are reduced equally, which would violate the uncertainty principle. Option C has the exponential dependencies reversed - it would give

\xi > 0

reducing momentum uncertainty instead of position uncertainty, contradicting the standard convention for squeezed states. Option D incorrectly describes sinusoidal behavior, which would occur with complex

\xi

but not the real parameter specified in this problem. The correct answer is B because it properly captures how squeezing redistributes quantum noise exponentially between conjugate observables. Study tip: Remember that squeezing always involves an exponential trade-off between conjugate uncertainties - you can't reduce both simultaneously, but you can dramatically reduce one at the cost of increasing the other.

Question 8

A quantum harmonic oscillator potential is modified by adding a small quartic term: $V(x) = \frac{1}{2}m\omega^2 x^2 + \lambda x^4$ where $\lambda$ is small. Using first-order perturbation theory, what is the energy correction for the $v$ -th vibrational level?

$E_v^{(1)} = \lambda \langle v | x^4 | v \rangle = \frac{3\lambda\hbar^2}{4m^2\omega^2}(2v^2 + 2v + 1)$ (correct answer)
$E_v^{(1)} = \frac{\lambda\hbar^2}{m^2\omega^2}(v^2 + v)$ because the quartic term scales with the number of quanta
$E_v^{(1)} = \frac{3\lambda\hbar^2}{2m^2\omega^2}(v + \frac{1}{2})^2$ from the square of the harmonic oscillator quantum number
$E_v^{(1)} = 0$ because the quartic perturbation has no first-order effect on energy levels
$E_v^{(1)} = \frac{\lambda\hbar^2}{4m^2\omega^2}(6v^2 + 6v + 3)$ based on the fourth moment of the harmonic oscillator wavefunction

Explanation: When you encounter a quantum harmonic oscillator with a small perturbation, you need to apply perturbation theory to find energy corrections. The key is recognizing that the first-order energy correction equals the expectation value of the perturbation operator in the unperturbed state:

E_v^{(1)} = \langle v | H' | v \rangle = \lambda \langle v | x^4 | v \rangle

. To evaluate

\langle v | x^4 | v \rangle

, you must use the properties of harmonic oscillator wavefunctions. The position operator can be expressed in terms of ladder operators:

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

. When you calculate

x^4

and take the expectation value with state

|v\rangle

, the result is

\langle v | x^4 | v \rangle = \frac{3\hbar^2}{4m^2\omega^2}(2v^2 + 2v + 1)

. This gives the first-order correction in answer A. Answer B incorrectly suggests a simpler scaling with quantum numbers, missing the correct coefficients and the constant term that arises from the quartic calculation. Answer C attempts to relate the correction to

(v + \frac{1}{2})^2

, which would be the square of the harmonic oscillator energy quantum number, but this relationship doesn't emerge from the proper calculation of

\langle v | x^4 | v \rangle

. Answer D is completely wrong because quartic terms definitely contribute to first-order energy corrections—they don't vanish due to symmetry like odd-power terms would. Study tip: Always remember that first-order perturbation corrections require calculating expectation values of the perturbation operator. Practice working with ladder operators to evaluate powers of the position operator efficiently.

Question 9

A quantum harmonic oscillator is coupled to a heat bath at temperature $T$ . In thermal equilibrium, what is the probability of finding the oscillator in the $v$ -th excited state?

$P_v = \frac{e^{-\beta E_v}}{Z}$ where $\beta = \frac{1}{k_B T}$ and $Z = \sum_{n=0}^{\infty} e^{-\beta E_n}$
$P_v = \frac{v!}{e^{\hbar\omega/k_B T}}$ from the factorial weighting of quantum states at finite temperature
$P_v = (1 - e^{-\hbar\omega/k_B T})e^{-v\hbar\omega/k_B T}$ representing a geometric distribution in quantum number (correct answer)
$P_v = \frac{\sqrt{2\pi k_B T}}{\hbar\omega} \exp\left(-\frac{E_v}{k_B T}\right)$ from the classical limit approximation
$P_v = \frac{e^{-v\hbar\omega/k_B T}}{\sqrt{2\pi v}}$ based on the Stirling approximation for large quantum numbers

Explanation: When analyzing quantum systems in thermal equilibrium, you're dealing with statistical mechanics where the probability of occupying any energy state follows the Boltzmann distribution. For a quantum harmonic oscillator with energy levels

E_v = \hbar\omega(v + \frac{1}{2})

, the key insight is recognizing how the partition function simplifies. The correct approach starts with the Boltzmann distribution

P_v = \frac{e^{-\beta E_v}}{Z}

, but for the harmonic oscillator, the partition function

Z = \sum_{v=0}^{\infty} e^{-\beta\hbar\omega(v+1/2)}

can be evaluated exactly. This geometric series sums to

Z = \frac{e^{-\beta\hbar\omega/2}}{1-e^{-\beta\hbar\omega}}

. When you substitute this back and simplify, you get

P_v = (1 - e^{-\hbar\omega/k_B T})e^{-v\hbar\omega/k_B T}

, which is answer C. Answer A gives the general Boltzmann form but doesn't perform the crucial simplification specific to the harmonic oscillator. Answer B incorrectly introduces factorial weighting, which has no basis in quantum statistical mechanics for this system. Answer D attempts a classical approximation with an incorrect normalization factor that doesn't arise from proper statistical mechanical treatment. The geometric distribution form in C is elegant because it separates the normalization factor

(1 - e^{-\hbar\omega/k_B T})

from the exponential decay in quantum number v. Remember: for harmonic oscillators in thermal equilibrium, always look for this characteristic geometric distribution pattern rather than stopping at the general Boltzmann expression.

Question 10

A quantum harmonic oscillator is in a superposition state $|\psi\rangle = \frac{1}{\sqrt{3}}|0\rangle + \sqrt{\frac{2}{3}}|1\rangle$ . What is the expectation value of the number operator $\hat{n} = a^\dagger a$ ?

$\langle \hat{n} \rangle = \frac{1}{2}$ because the average of the two quantum numbers is $\frac{0+1}{2}$
$\langle \hat{n} \rangle = \frac{2}{3}$ based on the weighted average using the probability amplitudes squared (correct answer)
$\langle \hat{n} \rangle = \frac{\sqrt{2}}{3}$ from the interference between the ground and first excited states
$\langle \hat{n} \rangle = 1$ because the number operator eigenvalue for the first excited state dominates
$\langle \hat{n} \rangle = \frac{1}{\sqrt{3}}$ corresponding to the amplitude coefficient of the ground state contribution

Explanation: When calculating expectation values for quantum superposition states, you need to use the fundamental formula

\langle \hat{O} \rangle = \langle \psi | \hat{O} | \psi \rangle

, where the probabilities are determined by the squared amplitudes of the wave function coefficients. For the number operator

\hat{n}

, we know that

\hat{n}|n\rangle = n|n\rangle

, so

\hat{n}|0\rangle = 0

and

\hat{n}|1\rangle = 1

. The expectation value becomes:

\langle \hat{n} \rangle = \left(\frac{1}{\sqrt{3}}\right)^2 \cdot 0 + \left(\sqrt{\frac{2}{3}}\right)^2 \cdot 1 = 0 + \frac{2}{3} = \frac{2}{3}

This is a weighted average where each eigenvalue is multiplied by its corresponding probability (amplitude squared), not the amplitude itself. Answer A incorrectly treats this as a simple arithmetic average, ignoring the quantum mechanical probabilities entirely. Answer C mistakenly suggests that interference effects between states contribute to the expectation value of the number operator—but interference terms vanish because

\langle 0|1\rangle = 0

(the states are orthogonal). Answer D wrongly claims the first excited state "dominates" when actually the ground state has probability

\frac{1}{3}

and the first excited state has probability

\frac{2}{3}

. Study tip: Always remember that quantum probabilities come from squared amplitudes, and expectation values are weighted averages using these probabilities. For any observable with orthogonal eigenstates, cross-terms disappear, leaving only the diagonal contributions.

Question 11

A quantum harmonic oscillator is prepared in a state $|\psi\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |2\rangle)$ . What is the period of oscillation for the expectation value $\langle x(t) \rangle$ ?

The period is $T = \frac{2\pi}{\omega}$ , the same as the classical harmonic oscillator period
The period is $T = \frac{\pi}{\omega}$ because the superposition involves states differing by 2 in quantum number
The period is $T = \frac{4\pi}{\omega}$ due to the interference between ground and second excited states
The expectation value $\langle x(t) \rangle = 0$ for all times because $x$ has no diagonal matrix elements (correct answer)
The oscillation is aperiodic because the energy difference $2\hbar\omega$ is not commensurate with $\omega$

Explanation: When you encounter quantum harmonic oscillator problems involving superposition states, you need to carefully analyze whether the position operator has non-zero matrix elements between the states involved. For the quantum harmonic oscillator, the position operator

\hat{x}

only connects states that differ by exactly one quantum number (

\Delta n = \pm 1

). This comes from the ladder operator representation:

\hat{x} \propto (\hat{a} + \hat{a}^\dagger)

, where

\hat{a}|n\rangle = \sqrt{n}|n-1\rangle

and

\hat{a}^\dagger|n\rangle = \sqrt{n+1}|n+1\rangle

. In your superposition

|\psi\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |2\rangle)

, calculating

\langle x(t)\rangle

requires matrix elements like

\langle 0|\hat{x}|0\rangle

\langle 2|\hat{x}|2\rangle

\langle 0|\hat{x}|2\rangle

, and

\langle 2|\hat{x}|0\rangle

. The diagonal elements (

\langle 0|\hat{x}|0\rangle

and

\langle 2|\hat{x}|2\rangle

) are zero because

\hat{x}

cannot connect a state to itself. The off-diagonal elements (

\langle 0|\hat{x}|2\rangle

and

\langle 2|\hat{x}|0\rangle

) are also zero because states |0⟩ and |2⟩ differ by 2 quantum numbers, not 1. Therefore,

\langle x(t)\rangle = 0

at all times, making D correct. Choice A assumes classical behavior without considering quantum selection rules. Choice B incorrectly assumes there's oscillation with modified period. Choice C also assumes oscillation occurs but with different timing. Key strategy: Always check the selection rules for quantum operators first. For harmonic oscillators, remember that

\hat{x}

and

\hat{p}

only connect adjacent energy levels (

\Delta n = \pm 1

Question 12

For a quantum harmonic oscillator, the commutator $[x^2, p^2]$ can be evaluated using the canonical commutation relation. What is the result?

$[x^2, p^2] = 0$ because both operators represent measurable quantities that commute
$[x^2, p^2] = 2i\hbar(xp + px)$ from applying the commutation relation twice (correct answer)
$[x^2, p^2] = -4\hbar^2$ because the commutator involves products of non-commuting operators
$[x^2, p^2] = 4i\hbar H$ where $H$ is the harmonic oscillator Hamiltonian
$[x^2, p^2] = i\hbar(x^2 + p^2)$ from the symmetric combination of position and momentum

Explanation: When evaluating commutators involving products of operators, you need to systematically apply the canonical commutation relation

[x,p] = i\hbar

using the product rule for commutators. To find

[x^2, p^2]

, use the identity

[AB, CD] = A[B,C]D + AC[B,D] + [A,C]BD + C[A,D]B

. For our case with

A=C=x

and

B=D=p

[x^2, p^2] = x[x,p]p + xp[x,p] + [x,p]px + p[x,p]x

Since

[x,p] = i\hbar

[x^2, p^2] = x(i\hbar)p + xp(i\hbar) + (i\hbar)px + p(i\hbar)x

= i\hbar(xp + xp + px + px) = 2i\hbar(xp + px)

This confirms answer B is correct. Answer A is wrong because

x

and

p

don't commute, so their squares won't either. The fact that both represent measurable quantities is irrelevant to their commutation properties. Answer C gives a numerical result, but commutators of position and momentum operators typically yield expressions involving the operators themselves, not just constants. The reasoning about "products of non-commuting operators" is vague and doesn't lead to this specific value. Answer D incorrectly relates the result to the Hamiltonian. While

xp + px

appears in some quantum mechanical expressions, it doesn't equal the harmonic oscillator Hamiltonian

H = \frac{p^2}{2m} + \frac{1}{2}m\omega^2x^2

. Study tip: Master the product rule for commutators and always work systematically through each term—commutator problems reward careful algebraic manipulation over conceptual shortcuts.

Question 13

A quantum harmonic oscillator system transitions from the $v = 2$ state to the $v = 1$ state by emitting a photon. If the oscillator frequency is $\omega_0 = 3.0 \times 10^{13}$ rad/s, and we account for the recoil energy of the oscillator, what is the frequency of the emitted photon?

$\omega_{photon} = \omega_0$ because the energy difference between adjacent levels is always $\hbar\omega_0$
$\omega_{photon} = \omega_0 - \frac{\hbar\omega_0^2}{2Mc^2}$ where $M$ is the oscillator mass, accounting for recoil (correct answer)
$\omega_{photon} = \omega_0 + \frac{\hbar\omega_0^2}{2Mc^2}$ because recoil energy adds to the photon energy
$\omega_{photon} = \omega_0\sqrt{1 - \frac{\hbar\omega_0}{Mc^2}}$ from relativistic energy-momentum conservation
$\omega_{photon} = \omega_0\left(1 - \frac{v^2}{2c^2}\right)$ where $v$ is the recoil velocity of the oscillator

Explanation: When a quantum harmonic oscillator emits a photon, you must consider conservation of both energy and momentum. While the energy difference between vibrational levels is indeed

\hbar\omega_0

, not all of this energy goes into the photon—some is "lost" to recoil of the oscillator itself. Here's the physics: when the oscillator emits a photon with momentum

p = \hbar\omega_{photon}/c

, conservation of momentum requires the oscillator to recoil with equal and opposite momentum. This recoil gives the oscillator kinetic energy

E_{recoil} = p^2/2M = \hbar^2\omega_{photon}^2/2Mc^2

. Since the total available energy is

\hbar\omega_0

, we have:

\hbar\omega_0 = \hbar\omega_{photon} + \frac{\hbar^2\omega_{photon}^2}{2Mc^2}

Solving for

\omega_{photon}

and using the approximation that the recoil term is small gives

\omega_{photon} = \omega_0 - \frac{\hbar\omega_0^2}{2Mc^2}

, confirming answer B. Answer A ignores recoil entirely—a common oversight when treating the oscillator as infinitely massive. Answer C has the wrong sign, incorrectly suggesting recoil adds energy to the photon rather than subtracting from it. Answer D applies relativistic formulas inappropriately; this is a non-relativistic recoil problem where

\hbar\omega_0 \ll Mc^2

. Study tip: Whenever you see photon emission problems mentioning "recoil," remember that momentum conservation always reduces the photon energy below the naive energy difference. The recoil correction is typically small but conceptually crucial.

Question 14

Two quantum harmonic oscillators have force constants $k_1 = 500$ N/m and $k_2 = 125$ N/m, with the same reduced mass. If oscillator 1 has 4 quanta of vibrational energy above its zero-point energy, how many quanta above zero-point energy must oscillator 2 have to possess the same total vibrational energy as oscillator 1?

2 quanta (accounting for the frequency difference)
4 quanta (same number despite frequency difference)
8 quanta (double due to lower frequency) (correct answer)
16 quanta (quadruple due to frequency scaling)

Explanation: The vibrational frequency scales as

\omega \propto \sqrt{k}

, so

\omega_1/\omega_2 = \sqrt{500/125} = 2

. Oscillator 1 has total energy

E_1 = \hbar\omega_1(4 + 1/2) = 4.5\hbar\omega_1

. For oscillator 2 to have the same energy:

\hbar\omega_2(v_2 + 1/2) = 4.5\hbar\omega_1 = 4.5\hbar(2\omega_2) = 9\hbar\omega_2

. Therefore

v_2 + 1/2 = 9

, giving

v_2 = 8.5

, which rounds to 8 quanta. Choice A uses incorrect scaling. Choice B ignores frequency dependence. Choice D applies incorrect quartic scaling.

Question 15

A diatomic molecule modeled as a quantum harmonic oscillator has a vibrational frequency of $2.0 \times 10^{13}$ Hz. At what temperature will the population of the $v = 1$ state be exactly 10% of the population of the $v = 0$ state?

96 K (using direct Boltzmann factor calculation) (correct answer)
144 K (including zero-point energy correction)
192 K (using classical equipartition approximation)
288 K (using doubled frequency for energy gap)

Explanation: The population ratio follows

N_1/N_0 = e^{-\hbar\omega/k_BT}

. Setting this equal to 0.1:

e^{-\hbar\omega/k_BT} = 0.1

, so

\hbar\omega/k_BT = \ln(10) = 2.303

. With

\hbar\omega = (1.055 \times 10^{-34})(2.0 \times 10^{13}) = 2.11 \times 10^{-21}

J, we get

T = 2.11 \times 10^{-21}/(2.303 \times 1.381 \times 10^{-23}) = 96

K. Choice B incorrectly includes zero-point energies which cancel in the ratio. Choice C uses classical approximation. Choice D uses wrong energy gap.

Question 16

Consider two identical quantum harmonic oscillators that are weakly coupled such that they can exchange energy quanta. If the system starts with oscillator A in the $v = 3$ state and oscillator B in the $v = 1$ state, which final distribution represents a statistically accessible microstate after thermal equilibration?

A: $v = 0$ , B: $v = 4$ (complete energy transfer)
A: $v = 2$ , B: $v = 2$ (equal energy distribution) (correct answer)
A: $v = 1$ , B: $v = 3$ (energy exchange only)
A: $v = 4$ , B: $v = 0$ (reverse initial condition)

Explanation: Energy conservation requires the total energy to remain constant. Initial total energy corresponds to 4 quanta above zero-point. All given options conserve energy, but thermal equilibration favors the distribution that maximizes entropy. For identical oscillators, the most probable distribution is equal sharing when possible. Since we have 4 quanta to distribute, the (2,2) distribution is most probable. Choices A, C, and D are all energetically allowed but statistically less probable than equal distribution.

Question 17

A quantum harmonic oscillator in its ground state has a total energy of $3.5 \times 10^{-21}$ J. If this oscillator is excited to the $v = 3$ vibrational state, what is the ratio of the kinetic energy at the classical turning points to the kinetic energy at the equilibrium position?

0 (kinetic energy is zero at turning points) (correct answer)
0.25 (one-fourth of the equilibrium value)
0.75 (three-fourths of the equilibrium value)
1.0 (equal to the equilibrium kinetic energy)

Explanation: At the classical turning points of a harmonic oscillator, all energy is potential energy and kinetic energy is zero, regardless of the quantum state. This is true both classically and quantum mechanically - at the turning points, the particle momentarily stops before reversing direction. Choice B incorrectly applies a quantum correction factor. Choice C assumes some residual kinetic energy remains. Choice D incorrectly assumes energy equipartition applies at turning points.

Question 18

Two quantum harmonic oscillators with the same frequency $\omega$ are placed in thermal contact. Initially, oscillator 1 has average energy $\langle E_1 \rangle = 5\hbar\omega$ and oscillator 2 has $\langle E_2 \rangle = 3\hbar\omega$ . After reaching thermal equilibrium, what is the most likely energy difference $|\langle E_1 \rangle - \langle E_2 \rangle|$ between the oscillators?

$4\hbar\omega$ (minimal energy transfer occurs)
$\hbar\omega$ (small residual difference remains)
$2\hbar\omega$ (partial equalization to original difference)
$0$ (perfect energy equalization occurs) (correct answer)

Explanation: When you encounter thermal equilibrium problems with quantum systems, remember that equilibrium means both systems reach the same temperature, which directly determines their average energies. For quantum harmonic oscillators in thermal equilibrium, the average energy follows

\langle E \rangle = \frac{\hbar\omega}{e^{\hbar\omega/k_BT} - 1}

. Since both oscillators have identical frequencies

\omega

and reach the same temperature

T

, they must have identical average energies at equilibrium. The total energy is conserved:

E_{total} = 5\hbar\omega + 3\hbar\omega = 8\hbar\omega

. This energy redistributes equally between the two identical oscillators, giving each

4\hbar\omega

. Therefore,

|\langle E_1 \rangle - \langle E_2 \rangle| = |4\hbar\omega - 4\hbar\omega| = 0

. Option A incorrectly assumes minimal energy transfer, suggesting the oscillators somehow resist equilibration. Option B reflects a common misconception that some residual difference persists—this would violate thermal equilibrium since different energies would imply different temperatures. Option C suggests only partial equalization, perhaps confusing this with a non-equilibrium intermediate state during the equilibration process. The key insight is that thermal equilibrium is absolute for identical systems. When you see problems involving identical quantum systems reaching thermal equilibrium, always remember that equal temperatures mean equal average energies, regardless of the initial energy distribution. Focus on conservation of total energy and the requirement that equilibrium temperatures must be identical.

Question 19

A quantum harmonic oscillator undergoes a transition from $v = 2$ to $v = 0$ by emitting two photons simultaneously. If the oscillator frequency is $\omega_0$ , and the two photons have equal energy, what is the frequency of each emitted photon?

$\omega_0/4$ (one-quarter the oscillator frequency)
$\omega_0/2$ (half the oscillator frequency)
$2\omega_0$ (twice the oscillator frequency)
$\omega_0$ (same as oscillator fundamental frequency) (correct answer)

Explanation: This question tests your understanding of quantum harmonic oscillator energy levels and photon emission. When a quantum harmonic oscillator emits photons, you need to apply conservation of energy between the initial and final quantum states. The energy levels of a quantum harmonic oscillator are given by

E_v = \hbar\omega_0(v + \frac{1}{2})

, where

v

is the vibrational quantum number. For the transition from

v = 2

v = 0

, the total energy difference is:

\Delta E = E_2 - E_0 = \hbar\omega_0(2 + \frac{1}{2}) - \hbar\omega_0(0 + \frac{1}{2}) = 2\hbar\omega_0

Since two photons of equal energy are emitted simultaneously, each photon carries half the total energy difference:

E_{photon} = \frac{\Delta E}{2} = \frac{2\hbar\omega_0}{2} = \hbar\omega_0

. Using

E = \hbar\omega

for photons, each photon has frequency

\omega_0

, making (D) correct. (A)

\omega_0/4

would give total energy

\hbar\omega_0/2

, far too small for this transition. (B)

\omega_0/2

would give total energy

\hbar\omega_0

, which equals only a single quantum of oscillator energy, not the two quanta needed. (C)

2\omega_0

would give total energy

4\hbar\omega_0

, twice what's available from this transition. Study tip: For multi-photon processes, always calculate the total energy difference between quantum states first, then divide by the number of photons. Remember that energy spacing between adjacent harmonic oscillator levels is constant at

\hbar\omega_0