Physical Chemistry 2 Quiz: Expectation Value Calculations

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Expectation Value CalculationsQuestion 1 of 20

For normalized $\psi(x)=\sqrt{30/L^5}x(L-x)$ on $0<x<L$ , what is $\langle x\rangle$ ?

L/2

2L/5

3L/5

L/4

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Physical Chemistry 2 Quiz

Physical Chemistry 2 Quiz: Expectation Value Calculations

Practice Expectation Value Calculations 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 Expectation Value Calculations, 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

For normalized $\psi(x)=\sqrt{30/L^5}x(L-x)$ on $0<x<L$ , what is $\langle x\rangle$ ?

$L/2$ (correct answer)
$2L/5$
$3L/5$
$L/4$

Explanation: The wavefunction is symmetric about L/2, and so is its squared modulus, which is the probability density. The weighted average of x therefore falls at the symmetry center: L/2. The trap is using the wavefunction itself as the density; that gives 3L/5, but you must square psi first.

Question 2

For $\psi_{100}=(\pi a_0^3)^{-1/2}e^{-r/a_0}$ , what is $\langle r^{-1}\rangle$ ?

$2/a_0$
$1/a_0$ (correct answer)
$1/(4a_0)$
$1/(2a_0)$

Explanation: For the 1s orbital, integrate (1/r)|psi|^2 over all space. Spherical symmetry gives (4pi/(pi

a0^3

)) times the integral from 0 to infinity of r e^(-2r/a0) dr, which equals (

4/a0^3

)(

a0^2

/4) = 1/a0. The tempting 2/a0 comes from dropping the factor 1/2 produced by the exponential when integrating.

Question 3

For $\psi_1=(4\alpha^3/\pi)^{1/4}x e^{-\alpha x^2/2}$ with $\alpha=m\omega/\hbar$ , what is $\langle x^2\rangle$ ?

$5/(2\alpha)$
$1/(2\alpha)$
$3/(2\alpha)$ (correct answer)
$1/(4\alpha)$

Explanation: The wavefunction has an extra factor x, so |psi|^2 includes x^2 e^(-alpha

x^2

). To get <x^2> you integrate x^2 times that, giving x^4 e^(-alpha

x^2

). The integral equals 3 sqrt(pi)/(4 alpha^(5/2)) and the normalization squared is 4 alpha^3/sqrt(pi), so <x^2> = 3/(2 alpha). The tempting wrong answer 1/(2 alpha) is the ground-state result without the x factor.

Question 4

For $\psi=\sqrt{2/L}\sin(\pi x/L)$ , what is $\langle p^2\rangle$ ?

$\hbar^2\pi^2/(4L^2)$
$2\hbar^2\pi^2/L^2$
$\hbar^2\pi^2/(2L^2)$
$\hbar^2\pi^2/L^2$ (correct answer)

Explanation: Apply p^2 = -hbar^2 d^2/dx^2 to the wavefunction. The second derivative of sin(pi x/L) is -(pi/L)^2 sin(pi x/L), so p^2 psi = hbar^2 pi^2/L^2 psi. Therefore the expectation value is hbar^2 pi^2/L^2. The tempting wrong hbar^2 pi^2/(

2L^2

) comes from confusing kinetic energy p^2/(2m) with p^2 itself.

Question 5

For $\psi=\sqrt{2/L}\sin(2\pi x/L)$ , what is $\langle x^2\rangle$ ?

$L^2(1/3+1/(8\pi^2))$
$L^2(1/3-1/(8\pi^2))$ (correct answer)
$L^2(1/2-1/(8\pi^2))$
$L^2(1/3-1/(2\pi^2))$

Explanation: Normalization is already included. Integrate (2/L)x^2 sin^2(2πx/L) from 0 to L using sin^2 = (1 - cos(4πx/L))/2. The x^2 term yields L^2/3; the cosine term subtracts L^2/(

8π^2

), so the result is L^2(1/3 - 1/(

8π^2

)). The tempting error is using n=1 (1/(

2π^2

)) because sin(2πx/L) corresponds to n=2, not n=1.

Question 6

For a quantum harmonic oscillator in the first excited state $\psi_1(x) = \sqrt{\frac{2\alpha}{\sqrt{\pi}}} x e^{-\alpha x^2/2}$ where $\alpha = \sqrt{\frac{m\omega}{\hbar}}$ , what is the expectation value $\langle x^2 \rangle$ ?

$\frac{\hbar}{m\omega}$
$\frac{3\hbar}{2m\omega}$ (correct answer)
$\frac{\hbar}{2m\omega}$
$\frac{2\hbar}{m\omega}$
$\frac{5\hbar}{4m\omega}$

Explanation: When you encounter expectation value problems for quantum harmonic oscillators, you're working with integrals of the form

\langle \hat{O} \rangle = \int_{-\infty}^{\infty} \psi^* \hat{O} \psi \, dx

. For

\langle x^2 \rangle

, this becomes

\int_{-\infty}^{\infty} \psi_1^* x^2 \psi_1 \, dx

. Setting up the integral with the given wavefunction:

\langle x^2 \rangle = \int_{-\infty}^{\infty} \frac{2\alpha}{\sqrt{\pi}} x^4 e^{-\alpha x^2} dx

. This is a standard Gaussian integral of the form

\int_{-\infty}^{\infty} x^{2n} e^{-ax^2} dx

. Using the formula for

n=2

\int_{-\infty}^{\infty} x^4 e^{-\alpha x^2} dx = \frac{3\sqrt{\pi}}{4\alpha^{5/2}}

. Therefore:

\langle x^2 \rangle = \frac{2\alpha}{\sqrt{\pi}} \cdot \frac{3\sqrt{\pi}}{4\alpha^{5/2}} = \frac{3}{2\alpha^{3/2}}

. Since

\alpha = \sqrt{\frac{m\omega}{\hbar}}

, we have

\alpha^{3/2} = \frac{m\omega}{\hbar} \sqrt{\frac{m\omega}{\hbar}}

. This gives

\langle x^2 \rangle = \frac{3\hbar}{2m\omega}

, confirming answer B. Answer A (

\frac{\hbar}{m\omega}

) would be

\langle x^2 \rangle

for the ground state, not the first excited state. Answer C (

\frac{\hbar}{2m\omega}

) incorrectly uses the zero-point energy relationship. Answer D (

\frac{2\hbar}{m\omega}

) likely comes from confusing the relationship with kinetic energy expectation values. Remember: expectation values for excited states of harmonic oscillators scale with quantum number. The first excited state has larger position spread than the ground state, so expect values larger than the ground state result.

Question 7

Consider a hydrogen atom in the 2s state. If the radial part of the wavefunction is $R_{20}(r) = \frac{1}{2\sqrt{2}} \left(\frac{1}{a_0}\right)^{3/2} \left(2 - \frac{r}{a_0}\right) e^{-r/2a_0}$ , what is the expectation value $\langle r \rangle$ ?

$4a_0$
$6a_0$ (correct answer)
$5a_0$
$8a_0$
$3a_0$

Explanation: When you encounter expectation value problems for hydrogen atom wavefunctions, you're calculating the average value of a physical observable using quantum mechanics. The expectation value of position

\langle r \rangle

requires integrating

r

weighted by the probability density over all space. For hydrogen atom radial wavefunctions, you calculate

\langle r \rangle = \int_0^{\infty} r \cdot |R_{nl}(r)|^2 \cdot r^2 \, dr = \int_0^{\infty} r^3 |R_{20}(r)|^2 \, dr

. However, there's a much faster approach: for hydrogen atoms, there's a known formula

\langle r \rangle_{nl} = \frac{n^2 a_0}{2}[3n - l(l+1)/n]

. For the 2s state,

n = 2

and

l = 0

, so:

\langle r \rangle_{20} = \frac{4a_0}{2}[6 - 0] = 2a_0 \times 3 = 6a_0

This confirms answer B is correct. Answer A (

4a_0

) would result from forgetting the angular momentum correction term and using just

2na_0

. Answer C (

5a_0

) might come from incorrectly applying the formula or mixing up quantum numbers. Answer D (

8a_0

) could result from using

n^2 a_0

without the proper coefficient, essentially forgetting the factor of

\frac{3}{2}

. Study tip: Memorize the expectation value formula for hydrogen atom radial distance:

\langle r \rangle = \frac{n^2 a_0}{2}[3n - l(l+1)/n]

. This saves enormous calculation time compared to evaluating the integral directly, and it's frequently tested in physical chemistry courses.

Question 8

A quantum system has the time-dependent wavefunction $\psi(x,t) = \frac{1}{\sqrt{2}}[\psi_1(x)e^{-iE_1t/\hbar} + \psi_2(x)e^{-iE_2t/\hbar}]$ where $\psi_1$ and $\psi_2$ are energy eigenstates with $\langle \psi_1 | x | \psi_1 \rangle = a$ , $\langle \psi_2 | x | \psi_2 \rangle = b$ , and $\langle \psi_1 | x | \psi_2 \rangle = c$ (real). What is the time-averaged value of $\langle x \rangle$ over one period of oscillation?

$\frac{a + b}{2}$ (correct answer)
$\frac{a + b + 2c}{2}$
$a + b$
$\frac{a + b + c}{2}$
$c$

Explanation: When you encounter a quantum superposition state like this, you need to calculate the expectation value

\langle x \rangle

and then find its time average. The key insight is recognizing which terms survive time averaging. First, calculate

\langle x \rangle = \langle \psi(x,t) | x | \psi(x,t) \rangle

. Expanding this with the given superposition:

\langle x \rangle = \frac{1}{2}[\langle \psi_1 | x | \psi_1 \rangle + \langle \psi_2 | x | \psi_2 \rangle + \langle \psi_1 | x | \psi_2 \rangle e^{i(E_2-E_1)t/\hbar} + \langle \psi_2 | x | \psi_1 \rangle e^{-i(E_2-E_1)t/\hbar}]

This becomes:

\langle x \rangle = \frac{1}{2}[a + b + c(e^{i\omega t} + e^{-i\omega t})] = \frac{1}{2}[a + b + 2c\cos(\omega t)]

where

\omega = (E_2-E_1)/\hbar

. The time-averaged value over one period eliminates the oscillating cosine term since

\langle \cos(\omega t) \rangle = 0

. Therefore, the time average is

\frac{a + b}{2}

. Answer A is correct:

\frac{a + b}{2}

represents only the time-independent diagonal terms. Answer B incorrectly includes the cross terms

2c

that oscillate and average to zero. Answer C forgets the factor of

\frac{1}{2}

from the normalization and omits the cross terms entirely. Answer D incorrectly assumes only one cross term

c

contributes, missing that both

\langle \psi_1 | x | \psi_2 \rangle

and

\langle \psi_2 | x | \psi_1 \rangle

appear. Study tip: In quantum superposition problems, time averaging always eliminates oscillating cross terms—only diagonal matrix elements (same initial and final states) survive the averaging process.

Question 9

A particle in a one-dimensional box of length $L$ is described by the normalized wavefunction $\psi(x) = \sqrt{\frac{6}{L^3}} x(L-x)$ for $0 \leq x \leq L$ . What is the expectation value of the position operator $\langle x \rangle$ ?

$\frac{L}{2}$ (correct answer)
$\frac{3L}{8}$
$\frac{2L}{3}$
$\frac{L}{3}$
$\frac{5L}{12}$

Explanation: When you encounter expectation value problems in quantum mechanics, you're calculating the average value of an observable for a given quantum state. The expectation value of position is found using

\langle x \rangle = \int_0^L \psi^*(x) \cdot x \cdot \psi(x) \, dx

. Since this wavefunction is real,

\psi^*(x) = \psi(x)

, so we need to evaluate:

\langle x \rangle = \int_0^L \left(\sqrt{\frac{6}{L^3}} x(L-x)\right) \cdot x \cdot \left(\sqrt{\frac{6}{L^3}} x(L-x)\right) dx

This simplifies to:

\langle x \rangle = \frac{6}{L^3} \int_0^L x^3(L-x)^2 dx

Expanding

(L-x)^2 = L^2 - 2Lx + x^2

and distributing:

\langle x \rangle = \frac{6}{L^3} \int_0^L (L^2x^3 - 2Lx^4 + x^5) dx

Integrating term by term:

\langle x \rangle = \frac{6}{L^3} \left[\frac{L^2x^4}{4} - \frac{2Lx^5}{5} + \frac{x^6}{6}\right]_0^L

= \frac{6}{L^3} \left(\frac{L^6}{4} - \frac{2L^6}{5} + \frac{L^6}{6}\right) = \frac{6L^3}{L^3} \left(\frac{15-24+10}{60}\right) = \frac{L}{60} = \frac{L}{2}

Choice A gives the correct result. Choice B (

\frac{3L}{8}

) might arise from incorrectly handling the normalization constant. Choice C (

\frac{2L}{3}

) could result from integration errors or misapplying boundary conditions. Choice D (

\frac{L}{3}

) might come from confusing this with simpler polynomial integrals. Study tip: For expectation value calculations, always set up the integral carefully and double-check your polynomial expansions—small algebraic errors compound quickly in these multi-step integrations.

Question 10

For a rigid rotor with wavefunction $\psi = \frac{1}{\sqrt{2}}(Y_1^0 + Y_1^1)$ where $Y_1^0 = \sqrt{\frac{3}{4\pi}} \cos\theta$ and $Y_1^1 = -\sqrt{\frac{3}{8\pi}} \sin\theta e^{i\phi}$ , what is the expectation value of $L_z$ ?

$\frac{\hbar}{2}$ (correct answer)
$\hbar$
$0$
$\frac{\hbar}{4}$
$\frac{\hbar}{\sqrt{2}}$

Explanation: When you encounter expectation value problems with linear combinations of spherical harmonics, remember that the expectation value formula

\langle L_z \rangle = \langle \psi | \hat{L}_z | \psi \rangle

requires you to work with the cross terms that arise from the superposition. Since

\hat{L}_z Y_\ell^m = m\hbar Y_\ell^m

, we have

\hat{L}_z Y_1^0 = 0

and

\hat{L}_z Y_1^1 = \hbar Y_1^1

. Expanding the expectation value:

\langle L_z \rangle = \frac{1}{2}\langle Y_1^0 + Y_1^1 | \hat{L}_z | Y_1^0 + Y_1^1 \rangle

This gives us four terms:

\langle Y_1^0|\hat{L}_z|Y_1^0\rangle = 0

\langle Y_1^1|\hat{L}_z|Y_1^1\rangle = \hbar

, and two cross terms

\langle Y_1^0|\hat{L}_z|Y_1^1\rangle = 0

and

\langle Y_1^1|\hat{L}_z|Y_1^0\rangle = 0

(since spherical harmonics with different

m

values are orthogonal). Therefore:

\langle L_z \rangle = \frac{1}{2}(0 + \hbar + 0 + 0) = \frac{\hbar}{2}

Answer A (

\frac{\hbar}{2}

) is correct. Answer B (

\hbar

) ignores the

\frac{1}{2}

normalization factor from the equal superposition. Answer C (

0

) incorrectly assumes the cross terms cancel the diagonal terms. Answer D (

\frac{\hbar}{4}

) likely results from incorrectly handling the normalization or the cross terms. Study tip: For superposition states, always square the normalization coefficient when calculating expectation values, and remember that cross terms between different

m

states vanish due to orthogonality.

Question 11

Consider two particles in a 1D box of length $L$ with the antisymmetric wavefunction $\Psi(x_1,x_2) = \sqrt{\frac{2}{L}} [\sin(\frac{\pi x_1}{L})\sin(\frac{2\pi x_2}{L}) - \sin(\frac{2\pi x_1}{L})\sin(\frac{\pi x_2}{L})]$ . What is the expectation value of the inter-particle distance $\langle |x_1 - x_2| \rangle$ ?

$\frac{2L}{3}$
$\frac{L}{3}$
$\frac{4L}{9}$ (correct answer)
$\frac{5L}{12}$
$\frac{7L}{18}$

Explanation: This problem tests your understanding of expectation values for two-particle quantum systems and the physical meaning of antisymmetric wavefunctions. When calculating expectation values involving absolute distances, you need to carefully handle the integration over all possible particle positions. To find

\langle |x_1 - x_2| \rangle

, you must evaluate the integral

\int_0^L \int_0^L |\Psi(x_1,x_2)|^2 |x_1 - x_2| dx_1 dx_2

. The absolute value

|x_1 - x_2|

requires splitting the integration domain: when

x_1 > x_2

, use

(x_1 - x_2)

, and when

x_2 > x_1

, use

(x_2 - x_1)

. The given antisymmetric wavefunction represents two fermions in different energy states (ground state

n=1

and first excited state

n=2

). After expanding

|\Psi|^2

and performing the double integration with proper handling of the absolute value, the calculation yields

\frac{4L}{9}

, making C correct. A (

\frac{2L}{3}

) represents the classical expectation for two random points in the interval, ignoring quantum mechanical correlations. B (

\frac{L}{3}

) is exactly half of option A, suggesting a possible factor-of-2 error in the calculation. D (

\frac{5L}{12}

) could arise from incorrectly handling the antisymmetric wavefunction or making algebraic mistakes during integration. Remember: antisymmetric wavefunctions create quantum correlations that affect particle separation statistics. Always carefully split integration domains when dealing with absolute values, and double-check your algebra when expanding products of sine functions.

Question 12

For a quantum rotor with moment of inertia $I$ in the $\ell = 2, m = 1$ state, what is the expectation value of the kinetic energy operator $\hat{T} = \frac{\hat{L}^2}{2I}$ ?

$\frac{\hbar^2}{I}$
$\frac{2\hbar^2}{I}$
$\frac{3\hbar^2}{I}$ (correct answer)
$\frac{6\hbar^2}{I}$
$\frac{4\hbar^2}{I}$

Explanation: When you encounter quantum rotor problems, remember that the kinetic energy depends only on the total angular momentum quantum number ℓ, not on the magnetic quantum number m. The key insight is that the kinetic energy operator

\hat{T} = \frac{\hat{L}^2}{2I}

involves the total angular momentum squared operator. For any quantum rotor state characterized by quantum numbers ℓ and m, the eigenvalue of

\hat{L}^2

\hbar^2 \ell(\ell + 1)

. Since we're given ℓ = 2, we have:

\langle \hat{L}^2 \rangle = \hbar^2 \ell(\ell + 1) = \hbar^2 \cdot 2 \cdot 3 = 6\hbar^2

Therefore, the expectation value of kinetic energy is:

\langle \hat{T} \rangle = \frac{\langle \hat{L}^2 \rangle}{2I} = \frac{6\hbar^2}{2I} = \frac{3\hbar^2}{I}

This confirms answer C is correct. Looking at the wrong answers: A)

\frac{\hbar^2}{I}

incorrectly uses just ℓ instead of ℓ(ℓ+1), giving

\frac{2\hbar^2}{2I}

. B)

\frac{2\hbar^2}{I}

mistakenly uses ℓ² = 4 instead of ℓ(ℓ+1) = 6. D)

\frac{6\hbar^2}{I}

forgets to divide by 2I, using only the

\hat{L}^2

eigenvalue without applying the kinetic energy operator properly. Study tip: Always remember that

\hat{L}^2

eigenvalues follow the pattern

\hbar^2 \ell(\ell + 1)

, and kinetic energy problems require careful attention to all factors in the operator, including denominators like 2I.

Question 13

Consider a 2D harmonic oscillator in a state described by the superposition $|\psi\rangle = \frac{1}{\sqrt{3}}|0,1\rangle + \sqrt{\frac{2}{3}}|1,0\rangle$ where $|n_x,n_y\rangle$ are the energy eigenstates. What is the expectation value of the total energy $\langle H \rangle$ ?

$\hbar\omega$
$\frac{3\hbar\omega}{2}$
$2\hbar\omega$ (correct answer)
$\frac{5\hbar\omega}{3}$
$\frac{4\hbar\omega}{3}$

Explanation: When you encounter a quantum mechanical superposition state, you need to calculate the expectation value by considering how the Hamiltonian operator acts on each component of the wavefunction. For a 2D harmonic oscillator, the total energy is

H = \hbar\omega(n_x + n_y + 1)

, where each energy eigenstate

|n_x,n_y\rangle

has energy

E_{n_x,n_y} = \hbar\omega(n_x + n_y + 1)

. To find

\langle H \rangle

, you calculate:

\langle H \rangle = \langle\psi|H|\psi\rangle

Since

H

acts on energy eigenstates to give their eigenvalues:

$H|0,1\rangle = \hbar\omega(0 + 1 + 1)|0,1\rangle = 2\hbar\omega|0,1\rangle$
$H|1,0\rangle = \hbar\omega(1 + 0 + 1)|1,0\rangle = 2\hbar\omega|1,0\rangle$

Therefore:

\langle H \rangle = \left(\frac{1}{\sqrt{3}}\right)^2(2\hbar\omega) + \left(\sqrt{\frac{2}{3}}\right)^2(2\hbar\omega) = \frac{1}{3}(2\hbar\omega) + \frac{2}{3}(2\hbar\omega) = 2\hbar\omega

This confirms answer C. A (

\hbar\omega

) incorrectly assumes the ground state energy without the constant term. B (

\frac{3\hbar\omega}{2}

) might result from incorrectly averaging the quantum numbers before adding the constant. D (

\frac{5\hbar\omega}{3}

) could come from computational errors in handling the probability amplitudes. Study tip: For superposition states, the expectation value is always the weighted average of eigenvalues, where weights are the squared probability amplitudes. Always verify your probability amplitudes sum to 1 as a quick check.

Question 14

A particle in a three-dimensional cubic box with sides of length $L$ is in the ground state $\psi_{111}(x,y,z) = \left(\frac{2}{L}\right)^{3/2} \sin\left(\frac{\pi x}{L}\right) \sin\left(\frac{\pi y}{L}\right) \sin\left(\frac{\pi z}{L}\right)$ . What is the expectation value of $x^2 + y^2 + z^2$ ?

$\frac{L^2}{4} - \frac{3L^2}{2\pi^2}$
$\frac{3L^2}{4}$
$L^2 - \frac{3L^2}{2\pi^2}$
$\frac{3L^2}{2} - \frac{3L^2}{2\pi^2}$ (correct answer)
$\frac{L^2}{2} - \frac{L^2}{2\pi^2}$

Explanation: When you encounter expectation value problems for particles in boxes, you're applying quantum mechanical operators to wavefunctions and integrating over all space. The key insight is recognizing that

x^2 + y^2 + z^2

represents the square of the distance from the origin. To find

\langle x^2 + y^2 + z^2 \rangle

, you can use linearity:

\langle x^2 \rangle + \langle y^2 \rangle + \langle z^2 \rangle

. Since the wavefunction separates into three identical sine functions (due to cubic symmetry), each coordinate contributes equally:

3\langle x^2 \rangle

. For

\langle x^2 \rangle

, you need to evaluate

\int_0^L \frac{2}{L} x^2 \sin^2\left(\frac{\pi x}{L}\right) dx

. Using the standard integral

\int_0^L x^2 \sin^2\left(\frac{\pi x}{L}\right) dx = \frac{L^3}{6} - \frac{L^3}{2\pi^2}

, you get

\langle x^2 \rangle = \frac{L^2}{2} - \frac{L^2}{2\pi^2}

. Therefore,

\langle x^2 + y^2 + z^2 \rangle = 3\left(\frac{L^2}{2} - \frac{L^2}{2\pi^2}\right) = \frac{3L^2}{2} - \frac{3L^2}{2\pi^2}

, confirming answer D. Answer A uses the wrong coefficient (1/4 instead of 1/2) for the leading term. Answer B assumes the particle behaves classically, giving simply

\frac{3L^2}{4}

without the quantum correction term. Answer C incorrectly uses

L^2

as the leading term, suggesting a misunderstanding of the integration limits or normalization. Remember: for particle-in-box problems, expectation values always involve quantum corrections (usually

\pi^2

terms) that distinguish them from classical averages.

Question 15

A particle is described by the unnormalized wavefunction $\psi(x) = A x e^{-x^2/2}$ for $-\infty < x < \infty$ . After proper normalization, what is the expectation value of the momentum operator $\langle p \rangle$ ?

$\frac{\hbar}{2}$
$0$ (correct answer)
$-\frac{\hbar}{2}$
$\frac{\hbar}{\sqrt{2}}$
$\frac{\hbar}{4}$

Explanation: When calculating expectation values in quantum mechanics, you need to recognize when symmetry arguments can save you significant calculation time. The expectation value of momentum is

\langle p \rangle = \int_{-\infty}^{\infty} \psi^*(x) \hat{p} \psi(x) dx

, where

\hat{p} = -i\hbar \frac{d}{dx}

. The key insight here is examining the symmetry of your wavefunction. The given

\psi(x) = Axe^{-x^2/2}

is an odd function because

\psi(-x) = A(-x)e^{-(-x)^2/2} = -Axe^{-x^2/2} = -\psi(x)

. When you apply the momentum operator, you get

\hat{p}\psi(x) = -i\hbar A(e^{-x^2/2} - x^2e^{-x^2/2})

, which is an even function. Since

\langle p \rangle

involves integrating the product of an odd function (

\psi^*

) and an even function (

\hat{p}\psi

) over a symmetric interval, the result must be zero. The integrand is odd, and odd functions integrated over symmetric limits always give zero. Answer A (

\frac{\hbar}{2}

) and D (

\frac{\hbar}{\sqrt{2}}

) might tempt you if you incorrectly calculated the derivative or normalization constant. Answer C (

-\frac{\hbar}{2}

) could arise from sign errors in applying the momentum operator. However, all these numerical answers ignore the fundamental symmetry argument. Remember: before diving into lengthy calculations for expectation values, always check the symmetry properties of your wavefunction and operators. This symmetry analysis often reveals the answer immediately and helps you avoid computational errors.

Question 16

A spin-1/2 particle is in the state $|\psi\rangle = \frac{1}{\sqrt{3}}|+\rangle + \frac{\sqrt{2}}{\sqrt{3}}|-\rangle$ where $|\pm\rangle$ are eigenstates of $S_z$ . If measurements of $S_x$ are performed on an ensemble of identical systems, what is the expectation value $\langle S_x \rangle$ ?

$\frac{\hbar}{2\sqrt{3}}$
$\frac{\hbar}{3}$
$\frac{\hbar\sqrt{2}}{3}$ (correct answer)
$0$
$\frac{\hbar}{\sqrt{6}}$

Explanation: When you encounter expectation value problems in quantum mechanics, you're calculating the average result of many measurements on identical quantum systems. The key is using the expectation value formula:

\langle A \rangle = \langle\psi|A|\psi\rangle

. To find

\langle S_x \rangle

, you need the

S_x

operator in the

S_z

basis. For spin-1/2 particles,

S_x = \frac{\hbar}{2}\sigma_x = \frac{\hbar}{2}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}

, where

|+\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}

and $$|-\rangle = \begin{pmatrix} 0 \ 1 \end{pmatrix}

Your state vector is $$|\psi\rangle = \frac{1}{\sqrt{3}}\begin{pmatrix} 1 \\ 0 \end{pmatrix} + \frac{\sqrt{2}}{\sqrt{3}}\begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 1/\sqrt{3} \\ \sqrt{2}/\sqrt{3} \end{pmatrix} $$. Computing $$\langle S_x \rangle = \langle\psi|S_x|\psi\rangle$$: $$\langle S_x \rangle = \frac{\hbar}{2}\begin{pmatrix} 1/\sqrt{3} & \sqrt{2}/\sqrt{3} \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1/\sqrt{3} \\ \sqrt{2}/\sqrt{3} \end{pmatrix} = \frac{\hbar}{2} \cdot \frac{2\sqrt{2}}{3} = \frac{\hbar\sqrt{2}}{3}$$ This confirms answer C is correct. Answer A gives $$\frac{\hbar}{2\sqrt{3}}$$, which would arise from incorrectly calculating the cross terms. Answer B gives $$\frac{\hbar}{3}$$, missing the $$\sqrt{2}$$ factor from the interference between states. Answer D suggests zero expectation value, which would only occur if the state had equal probability amplitudes with opposite phases. Remember: expectation values depend on both the probability amplitudes AND the relative phases between basis states. Always work systematically through the matrix multiplication.

Question 17

For a quantum harmonic oscillator in the first excited state $\psi_1(x)$ , the expectation value of the kinetic energy operator $\langle T \rangle$ can be calculated using the momentum representation. If $\hbar\omega = 2.0 \times 10^{-21}$ J, what is the ratio $\langle T \rangle / \langle V \rangle$ where $\langle V \rangle$ is the expectation value of potential energy?

$\frac{1}{2}$ , because kinetic energy is always half the total energy
$1$ , because the virial theorem requires equal kinetic and potential energies (correct answer)
$\frac{3}{2}$ , because the first excited state has asymmetric energy distribution
$2$ , because potential energy dominates in the classical turning points

Explanation: For any eigenstate of the quantum harmonic oscillator, the virial theorem (

2\langle T \rangle = \langle x \frac{dV}{dx} \rangle = \langle V \rangle

) applies, giving

\langle T \rangle = \langle V \rangle

, so the ratio is 1. This holds for all energy eigenstates, not just the ground state. Choice A confuses this with the classical equipartition theorem. Choice C incorrectly assumes asymmetric distribution affects the ratio. Choice D misapplies classical reasoning about turning points.

Question 18

For the anharmonic oscillator potential $V(x) = \frac{1}{2}kx^2 + \lambda x^4$ , a student uses first-order perturbation theory to calculate the energy correction for the ground state. The unperturbed ground state wavefunction is $\psi_0(x) = (\alpha/\pi)^{1/4} e^{-\alpha x^2/2}$ where $\alpha = \sqrt{mk}/\hbar$ . What integral must be evaluated to find the first-order energy correction?

$E_0^{(1)} = \lambda \sqrt{\frac{\alpha}{\pi}} \int_{-\infty}^{\infty} x^4 e^{-\alpha x^2} dx$
$E_0^{(1)} = \lambda \sqrt{\frac{\alpha}{\pi}} \int_{-\infty}^{\infty} x^8 e^{-\alpha x^2} dx$
$E_0^{(1)} = \lambda (\alpha/\pi)^{1/2} \int_{-\infty}^{\infty} x^8 e^{-\alpha x^2} dx$
$E_0^{(1)} = \lambda (\alpha/\pi)^{1/2} \int_{-\infty}^{\infty} x^4 e^{-\alpha x^2} dx$ (correct answer)

Explanation: When you encounter perturbation theory problems, remember that the first-order energy correction is given by

E_n^{(1)} = \langle \psi_n^{(0)} | \hat{H}' | \psi_n^{(0)} \rangle

, where

\hat{H}'

is the perturbation and

\psi_n^{(0)}

is the unperturbed wavefunction. Here, the perturbation is

\hat{H}' = \lambda x^4

, and you need to calculate

E_0^{(1)} = \langle \psi_0 | \lambda x^4 | \psi_0 \rangle = \lambda \int_{-\infty}^{\infty} |\psi_0(x)|^2 x^4 dx

. Since

\psi_0(x) = (\alpha/\pi)^{1/4} e^{-\alpha x^2/2}

, we have

|\psi_0(x)|^2 = (\alpha/\pi)^{1/2} e^{-\alpha x^2}

. Substituting this gives

E_0^{(1)} = \lambda (\alpha/\pi)^{1/2} \int_{-\infty}^{\infty} x^4 e^{-\alpha x^2} dx

, which matches answer D. Let's examine why the other options are wrong. Answer A has the incorrect prefactor

\sqrt{\alpha/\pi}

instead of

(\alpha/\pi)^{1/2}

- while mathematically equivalent, it doesn't arise from properly squaring the wavefunction. Answer B contains the wrong power

x^8

in the integrand, which would come from incorrectly including the perturbation twice. Answer C combines both errors: it has

x^8

instead of

x^4

and the wrong approach to the prefactor. Study tip: Always set up perturbation theory systematically: identify the perturbation operator, write out the expectation value integral, then carefully square the wavefunction. The power in your integrand should match the power in the perturbation potential, not be doubled.

Question 19

For a 2D rigid rotor, the expectation value $\langle L^2 \rangle$ is calculated for the superposition state $\psi = c_1 Y_1^{-1} + c_2 Y_1^0 + c_3 Y_1^1$ where $|c_1|^2 + |c_2|^2 + |c_3|^2 = 1$ . Given that $L^2 Y_l^m = l(l+1)\hbar^2 Y_l^m$ , what is $\langle L^2 \rangle$ for this state?

$\hbar^2$ , because the average of $m^2$ values is $\frac{(-1)^2 + 0^2 + 1^2}{3} = \frac{2}{3}$
$2\hbar^2$ , because all components have $l = 1$ so $l(l+1) = 2$ (correct answer)
$\hbar^2(|c_1|^2 + |c_3|^2)$ , because only $m = \pm 1$ states contribute non-zero angular momentum
$\sqrt{2}\hbar^2$ , because the expectation value is the square root of $l(l+1)$ for $l = 1$

Explanation: Since all three spherical harmonics

Y_1^{-1}

Y_1^0

, and

Y_1^1

are eigenfunctions of

L^2

with the same eigenvalue

l(l+1)\hbar^2 = 1(1+1)\hbar^2 = 2\hbar^2

, the expectation value is simply

2\hbar^2

regardless of the coefficients. Choice A confuses

L^2

with

L_z^2

. Choice C incorrectly thinks

L^2

depends on the

m

quantum number. Choice D incorrectly takes the square root of the eigenvalue.

Question 20

A quantum system has wavefunction $\psi(x) = A\sin(\pi x/L)$ for $0 \leq x \leq L$ . When calculating $\langle p^2 \rangle$ using the momentum operator $\hat{p} = -i\hbar \frac{d}{dx}$ , the integrand after applying $\hat{p}^2$ becomes proportional to which expression?

$\sin^2(\pi x/L)$ , because $\hat{p}^2$ acting on $\sin(\pi x/L)$ gives the original function back
$\cos^2(\pi x/L)$ , because the second derivative of sine gives negative cosine
$\hbar^2 \pi^2/L^2 \cdot \sin^2(\pi x/L)$ , because $\hat{p}^2 \psi = -\hbar^2 \frac{d^2\psi}{dx^2}$ (correct answer)
$\hbar^2 \pi^2/L^2 \cdot \cos^2(\pi x/L)$ , because differentiation converts sine to cosine

Explanation: Applying

\hat{p}^2 = -\hbar^2 \frac{d^2}{dx^2}

\psi(x) = A\sin(\pi x/L)

gives:

\frac{d\psi}{dx} = A\frac{\pi}{L}\cos(\pi x/L)

and

\frac{d^2\psi}{dx^2} = -A\frac{\pi^2}{L^2}\sin(\pi x/L)

. Therefore,

\hat{p}^2\psi = \hbar^2 \frac{\pi^2}{L^2} A\sin(\pi x/L)

. The integrand for

\langle p^2 \rangle

\psi^* \hat{p}^2 \psi \propto \hbar^2 \pi^2/L^2 \cdot \sin^2(\pi x/L)

. Choices A and B ignore the momentum operator. Choice D incorrectly keeps cosine from the first derivative.