Physical Chemistry 2 Quiz: Schrodinger Equation

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Schrodinger EquationQuestion 1 of 20

For a quantum rotor (rigid rotator) with moment of inertia $I$ , the time-independent Schrödinger equation in spherical coordinates is $\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial\psi}{\partial\theta}\right) + \frac{1}{\sin^2\theta}\frac{\partial^2\psi}{\partial\phi^2} + \frac{2IE}{\hbar^2}\psi = 0$ . What constraint arises from requiring the wavefunction to be continuous when we traverse a complete circle around the $z$ -axis?

The wavefunction must be periodic in

\theta

with period

2\pi

The wavefunction must satisfy

\psi(\theta, \phi + 2\pi) = \psi(\theta, \phi)

The magnetic quantum number

m

must be an integer

Both

\psi(\theta, \phi + 2\pi) = \psi(\theta, \phi)

and

m

must be an integer

The wavefunction must vanish at

\theta = 0

and

\theta = \pi

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Physical Chemistry 2 Quiz

Physical Chemistry 2 Quiz: Schrodinger Equation

Practice Schrodinger Equation 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 Schrodinger Equation, giving you a quick way to practice the rules, question types, and explanations that matter most for Physical Chemistry 2.

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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.

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Question 1

The wavefunction must be periodic in $\theta$ with period $2\pi$
The wavefunction must satisfy $\psi(\theta, \phi + 2\pi) = \psi(\theta, \phi)$
The magnetic quantum number $m$ must be an integer
Both $\psi(\theta, \phi + 2\pi) = \psi(\theta, \phi)$ and $m$ must be an integer (correct answer)
The wavefunction must vanish at $\theta = 0$ and $\theta = \pi$

Explanation: When dealing with quantum mechanical systems in spherical coordinates, you must consider what physical constraints the mathematics must satisfy. The quantum rotor describes rotation in three-dimensional space, where certain boundary conditions ensure the wavefunction has physical meaning. The key insight is recognizing what "traversing a complete circle around the z-axis" means mathematically. This corresponds to changing the azimuthal angle

\phi

2\pi

while keeping

\theta

constant. Since we're describing the same physical point after a complete rotation, the wavefunction must return to its original value:

\psi(\theta, \phi + 2\pi) = \psi(\theta, \phi)

. This is the fundamental periodicity requirement. However, the story doesn't end there. When you solve the Schrödinger equation using separation of variables, the

\phi

-dependent part yields solutions of the form

e^{im\phi}

. For the periodicity condition to hold, you need

e^{im(\phi + 2\pi)} = e^{im\phi}

, which requires

e^{2\pi im} = 1

. This is only satisfied when

m

is an integer, giving rise to the magnetic quantum number constraint. Answer A is incomplete because the constraint involves

\phi

, not

\theta

. Answer B captures the periodicity but misses the deeper quantum mechanical consequence. Answer C identifies the integer requirement but doesn't explain its origin from the boundary condition. Only answer D correctly identifies both the mathematical constraint and its quantum mechanical consequence. Remember: boundary conditions in quantum mechanics aren't just mathematical formalities—they directly determine the allowed quantum states and energy levels of the system.

Question 2

Consider the one-dimensional Schrödinger equation $-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + V(x)\psi = E\psi$ in a region where $V(x) = V_0 = \text{constant}$ . If the general solution is $\psi(x) = Ae^{ikx} + Be^{-ikx}$ where $k = \sqrt{\frac{2m(E-V_0)}{\hbar^2}}$ , under what condition does this form become invalid?

When $E = V_0$ , because $k = 0$ and the solution becomes degenerate (correct answer)
When $E < V_0$ , because $k$ becomes imaginary and the exponentials become real
When $E > V_0$ , because the kinetic energy becomes too large for the approximation
When $V_0 < 0$ , because the potential energy cannot be negative in quantum mechanics
The form is always valid regardless of the values of $E$ and $V_0$

Explanation: When you encounter the time-independent Schrödinger equation with constant potential, you're dealing with a fundamental quantum mechanics scenario where the mathematical form of the solution depends critically on the relationship between total energy

E

and potential energy

V_0

. The given solution

\psi(x) = Ae^{ikx} + Be^{-ikx}

represents oscillatory behavior when

k

is real and positive. This requires

E - V_0 > 0

, meaning the particle has positive kinetic energy. However, this form breaks down mathematically when

E = V_0

, because then

k = \sqrt{\frac{2m(E-V_0)}{\hbar^2}} = 0

. When

k = 0

, both exponential terms become identical:

e^{0 \cdot x} = 1

, so

\psi(x) = A + B = \text{constant}

. This creates a degenerate situation where you lose one degree of freedom—instead of two independent solutions, you effectively have only one. Choice A correctly identifies this mathematical breakdown. Choice B is incorrect because when

E < V_0

k

becomes imaginary, giving exponentially growing and decaying solutions (

e^{\pm \kappa x}

where

\kappa

is real)—this is still a valid mathematical form, just representing tunneling behavior rather than oscillation. Choice C is wrong because large kinetic energy doesn't invalidate the solution form. Choice D is incorrect because negative potentials are perfectly acceptable in quantum mechanics. Remember: when solving the Schrödinger equation, always check whether your energy conditions lead to degenerate cases where the mathematical form loses its generality. The boundary case

E = V_0

often requires special treatment with polynomial solutions.

Question 3

For the three-dimensional isotropic harmonic oscillator $V(r) = \frac{1}{2}m\omega^2 r^2$ , the Schrödinger equation in spherical coordinates can be separated into radial and angular parts. The energy eigenvalues are $E_{n,l} = \hbar\omega\left(n + l + \frac{3}{2}\right)$ where $n$ is the radial quantum number and $l$ is the orbital angular momentum quantum number. What constraint must be satisfied by the radial wavefunction $R_{n,l}(r)$ as $r \to 0$ to ensure the full wavefunction $\psi(r,\theta,\phi) = R_{n,l}(r)Y_l^m(\theta,\phi)$ remains finite at the origin?

$R_{n,l}(r) \to r^l$ as $r \to 0$ to cancel the angular momentum singularity (correct answer)
$R_{n,l}(r) \to r^{l+1}$ as $r \to 0$ to ensure $|\psi|^2$ is integrable near the origin
$R_{n,l}(r) \to \text{constant}$ as $r \to 0$ for all values of $l$
$R_{n,l}(r) \to r^{l+2}$ as $r \to 0$ to account for the three-dimensional volume element
$R_{n,l}(r) \to r^{-l}$ as $r \to 0$ to normalize the angular part

Explanation: When dealing with quantum mechanical systems in spherical coordinates, you need to ensure the complete wavefunction remains finite everywhere, especially at problematic points like the origin. The key insight is understanding how the angular and radial parts of the wavefunction behave near

r = 0

. The spherical harmonics

Y_l^m(\theta,\phi)

create a potential singularity issue. While they're finite at the origin for

l = 0

, for

l > 0

, the gradient of these functions can cause problems when combined with the radial part in three dimensions. The complete wavefunction

\psi = R_{n,l}(r)Y_l^m(\theta,\phi)

must remain finite as

r \to 0

. Answer A is correct because

R_{n,l}(r) \to r^l

provides exactly the right behavior to ensure the wavefunction remains finite. This

r^l

dependence is a fundamental requirement that emerges from solving the radial Schrödinger equation properly. Answer B suggests

r^{l+1}

behavior, which would make the wavefunction vanish too rapidly at the origin - this is overly restrictive and not what the mathematics demands. Answer C proposes constant behavior regardless of

l

, which fails for

l > 0

cases where you need the

r^l

dependence to maintain finiteness. Answer D's

r^{l+2}

behavior confuses the volume element

r^2 dr

(which appears in integrals) with the actual wavefunction behavior. Remember: in spherical quantum mechanics, the radial wavefunction must always behave as

r^l

near the origin to ensure physical acceptability, regardless of the specific potential.

Question 4

For a particle in a one-dimensional box of length $L$ , if the wavefunction must satisfy $\psi(0) = 0$ and $\psi(L) = 0$ , and the general solution to the time-independent Schrödinger equation is $\psi(x) = A\sin(kx) + B\cos(kx)$ , what constraint must be imposed on the wave vector $k$ to ensure both boundary conditions are satisfied simultaneously?

$k = \frac{n\pi}{L}$ where $n$ is any positive integer (correct answer)
$k = \frac{2n\pi}{L}$ where $n$ is any positive integer
$k = \frac{(2n-1)\pi}{2L}$ where $n$ is any positive integer
$k = \frac{n\pi}{2L}$ where $n$ is any positive integer
$k = \frac{n\pi}{L}$ where $n$ is any positive or negative integer

Explanation: When you encounter boundary value problems in quantum mechanics, you're essentially finding which wave solutions can "fit" within the given constraints. The particle in a box is a fundamental example where the wavefunction must vanish at both walls. Starting with the general solution

\psi(x) = A\sin(kx) + B\cos(kx)

, you need to apply both boundary conditions systematically. First, applying

\psi(0) = 0

\psi(0) = A\sin(0) + B\cos(0) = 0 + B(1) = B = 0

This immediately eliminates the cosine term, leaving

\psi(x) = A\sin(kx)

. Now applying the second boundary condition

\psi(L) = 0

\psi(L) = A\sin(kL) = 0

Since

A

cannot be zero (that would give a trivial solution), you need

\sin(kL) = 0

. The sine function equals zero when its argument is any integer multiple of

\pi

, so

kL = n\pi

where

n = 1, 2, 3, ...

Therefore,

k = \frac{n\pi}{L}

. Answer A is correct with this exact form. Answer B (

k = \frac{2n\pi}{L}

) only captures even multiples of

\pi

, missing valid solutions like

n = 1, 3, 5, ...

Answer C (

k = \frac{(2n-1)\pi}{2L}

) gives odd multiples of

\frac{\pi}{2}

, which don't make

\sin(kL)

equal zero at

x = L

. Answer D (

k = \frac{n\pi}{2L}

) would require

\sin(\frac{n\pi}{2}) = 0

, which only works for even

n

. Study tip: Always apply boundary conditions step-by-step, and remember that

\sin(n\pi) = 0

for any integer

n

is the key relationship in box problems.

Question 5

For a particle in a three-dimensional cubic box with sides of length $L$ , the time-independent Schrödinger equation separates into three one-dimensional equations. If the boundary conditions require $\psi(0,y,z) = \psi(L,y,z) = 0$ , $\psi(x,0,z) = \psi(x,L,z) = 0$ , and $\psi(x,y,0) = \psi(x,y,L) = 0$ , what is the degeneracy of the energy level corresponding to quantum numbers $(n_x, n_y, n_z) = (2,1,1)$ ?

The degeneracy is 1 because the quantum numbers are all different
The degeneracy is 2 because two of the quantum numbers are equal
The degeneracy is 3 because there are three ways to arrange the numbers (2,1,1) (correct answer)
The degeneracy is 4 because of spin-orbit coupling effects in three dimensions
The degeneracy is 6 because there are 3! = 6 permutations of three numbers

Explanation: When you encounter degeneracy problems for particles in boxes, you're looking for different quantum state combinations that produce the same energy. For a 3D cubic box, the energy depends on

E = \frac{h^2}{8mL^2}(n_x^2 + n_y^2 + n_z^2)

, so states with the same sum of squared quantum numbers have identical energies. For the state

(2,1,1)

, the energy depends on

2^2 + 1^2 + 1^2 = 6

. To find the degeneracy, you need to count all possible arrangements of these quantum numbers among the three dimensions. Think of this as asking: "In how many ways can I assign the values 2, 1, and 1 to

n_x

n_y

, and

n_z

?" The three distinct arrangements are:

(2,1,1)

(1,2,1)

, and

(1,1,2)

. Each represents a different physical state—the particle has different probability distributions along each axis—but all have the same total energy. Therefore, the degeneracy is 3. Option A incorrectly assumes that having different quantum numbers automatically means no degeneracy, missing that it's the energy (not individual quantum numbers) that determines degeneracy. Option B focuses on counting equal quantum numbers rather than counting valid permutations. Option D incorrectly invokes spin-orbit coupling, which isn't relevant for this basic particle-in-a-box model. Study tip: For degeneracy problems, always count the number of ways to permute the quantum numbers. When some numbers repeat (like two 1's here), use the permutation formula:

\frac{n!}{n_1! \times n_2! \times ...}

where

n

is total positions and

n_i

are repetitions.

Question 6

Consider the radial part of the hydrogen atom Schrödinger equation in spherical coordinates. If the effective potential is $V_{eff}(r) = -\frac{ke^2}{r} + \frac{\hbar^2 l(l+1)}{2mr^2}$ , what boundary condition must the radial wavefunction $R(r)$ satisfy as $r \to 0$ , and why?

$R(r) \to 0$ as $r \to 0$ because the Coulomb potential diverges there
$R(r) \to r^l$ as $r \to 0$ to cancel the centrifugal barrier singularity (correct answer)
$R(r) \to r^{l+1}$ as $r \to 0$ to ensure the full wavefunction $\psi = R(r)Y_l^m(\theta,\phi)$ is finite
$R(r) \to \text{constant}$ as $r \to 0$ for all values of $l$ to maintain normalizability
$R(r) \to r^{-l}$ as $r \to 0$ to balance the angular momentum term in the kinetic energy

Explanation: When analyzing boundary conditions for the hydrogen atom's radial wavefunction, you need to examine how the effective potential behaves near the origin and ensure the complete wavefunction remains physically meaningful. The effective potential contains a centrifugal barrier term

\frac{\hbar^2 l(l+1)}{2mr^2}

that diverges as

r^{-2}

when

r \to 0

. This creates a singularity that must be handled carefully. The radial Schrödinger equation near

r = 0

is dominated by this centrifugal term, leading to solutions that behave like

r^l

r^{-(l+1)}

. Since wavefunctions must be finite everywhere, we reject the divergent

r^{-(l+1)}

solution and require

R(r) \to r^l

r \to 0

. This behavior exactly cancels the centrifugal barrier singularity, making the kinetic energy term finite. Answer A incorrectly focuses on the Coulomb potential, but

-1/r

diverges more slowly than the

1/r^2

centrifugal term and doesn't determine the boundary condition. Answer C suggests

r^{l+1}

behavior, which would be overly restrictive—while this ensures finiteness, it's not the natural solution that emerges from the differential equation. Answer D claims the same constant behavior for all

l

, which ignores the

l

-dependence of the centrifugal barrier entirely. Study tip: Remember that boundary conditions in quantum mechanics come from requiring solutions to be finite and normalizable. Always identify the most singular term in your differential equation near boundaries—it usually determines the required behavior.

Question 7

For a particle moving in a periodic potential $V(x + a) = V(x)$ , Bloch's theorem states that the solutions to the Schrödinger equation can be written as $\psi_k(x) = e^{ikx} u_k(x)$ where $u_k(x + a) = u_k(x)$ . If the boundary condition is imposed that $\psi(x + L) = \psi(x)$ where $L = Na$ ( $N$ is an integer), what constraint does this place on the allowed values of $k$ ?

$k = \frac{2\pi n}{a}$ where $n$ is any integer
$k = \frac{2\pi n}{L}$ where $n$ is any integer (correct answer)
$k = \frac{\pi n}{a}$ where $n$ is any integer
$k = \frac{\pi n}{L}$ where $n$ is any integer
$k$ can take any real value because the potential is periodic

Explanation: When you encounter Bloch's theorem problems, you're dealing with quantum mechanics in periodic systems like crystals. The key insight is understanding how boundary conditions quantize the allowed wave vectors. Starting with Bloch's wavefunction

\psi_k(x) = e^{ikx} u_k(x)

where

u_k(x)

has the same periodicity as the potential, you need to apply the boundary condition

\psi(x + L) = \psi(x)

. Substituting the Bloch form:

e^{ik(x+L)} u_k(x+L) = e^{ikx} u_k(x)

Since

L = Na

where

N

is an integer, and

u_k(x)

has period

a

, we have

u_k(x+L) = u_k(x+Na) = u_k(x)

. This simplifies our condition to:

e^{ik(x+L)} = e^{ikx}

This requires

e^{ikL} = 1

, which means

kL = 2\pi n

for integer

n

. Therefore:

k = \frac{2\pi n}{L}

Answer A (

k = \frac{2\pi n}{a}

) uses the lattice spacing

a

instead of the system size

L

. This would be correct for periodic boundary conditions over one unit cell, but not over the entire system length. Answers C and D both use

\pi

instead of

2\pi

. These would arise from antiperiodic boundary conditions (

\psi(x+L) = -\psi(x)

), not the periodic condition given. Study tip: Remember that boundary conditions always involve the total system size

L

, not the unit cell size

a

. Periodic boundary conditions give

2\pi n

, while antiperiodic give

\pi(2n+1)

Question 8

Consider a particle in a finite square well: $V(x) = -V_0$ for $-a/2 < x < a/2$ and $V(x) = 0$ elsewhere, where $V_0 > 0$ . For bound states with energy $-V_0 < E < 0$ , the wavefunctions in the three regions are $\psi_1(x) = Ae^{\kappa x}$ for $x < -a/2$ , $\psi_2(x) = B\cos(\alpha x) + C\sin(\alpha x)$ for $-a/2 < x < a/2$ , and $\psi_3(x) = De^{-\kappa x}$ for $x > a/2$ , where $\kappa = \sqrt{-2mE/\hbar^2}$ and $\alpha = \sqrt{2m(E+V_0)/\hbar^2}$ . What additional constraint on the coefficients $B$ and $C$ arises from the symmetry of the potential?

Either $B = 0$ (odd parity) or $C = 0$ (even parity), depending on the specific bound state (correct answer)
$B = C$ to ensure the wavefunction is continuous at the origin
$B = -C$ to ensure the wavefunction derivative is continuous at the origin
$B^2 + C^2 = 1$ to ensure proper normalization within the well
$B$ and $C$ must satisfy $B\cos(\alpha a/2) + C\sin(\alpha a/2) = De^{-\kappa a/2}$

Explanation: When you encounter a quantum mechanics problem involving symmetric potentials, always consider how symmetry constrains the wavefunctions. The finite square well is symmetric about x = 0, which means the potential has definite parity: V(-x) = V(x). For symmetric potentials, quantum mechanics tells us that energy eigenstates must have definite parity - they're either even or odd functions. An even function satisfies ψ(-x) = ψ(x), while an odd function satisfies ψ(-x) = -ψ(x). Looking at the wavefunction in the well region: ψ₂(x) = B cos(αx) + C sin(αx). Since cos(αx) is an even function and sin(αx) is an odd function, for ψ₂(x) to have definite parity, either B = 0 (making it purely odd) or C = 0 (making it purely even). Each bound state will be either entirely even or entirely odd, which is why answer A is correct. Option B incorrectly suggests B = C creates continuity at the origin - but continuity is already satisfied by the boundary conditions at x = ±a/2, not at the origin. Option C confuses parity with derivative continuity requirements. Option D treats normalization incorrectly - the normalization involves integrating over all space, not just setting B² + C² = 1 in the well region. Study tip: For any symmetric potential in quantum mechanics, immediately check whether the problem is asking about parity. Energy eigenstates of symmetric systems always have definite parity, which dramatically simplifies the mathematics by eliminating half the terms in your wavefunction.

Question 9

A quantum particle is confined to move on the surface of a sphere of radius $R$ . The angular part of the Schrödinger equation in spherical coordinates yields the spherical harmonics $Y_l^m(\theta, \phi)$ with eigenvalue equation $\hat{L}^2 Y_l^m = \hbar^2 l(l+1) Y_l^m$ . What boundary condition is implicitly imposed on the spherical harmonics by the requirement that the wavefunction be single-valued everywhere on the sphere?

$Y_l^m(\theta, \phi) = Y_l^m(\theta, \phi + 2\pi)$ and $Y_l^m(0, \phi) = Y_l^m(\pi, \phi) = 0$
$Y_l^m(\theta, \phi) = Y_l^m(\theta, \phi + 2\pi)$ and $Y_l^m(\theta, \phi)$ finite at $\theta = 0, \pi$ (correct answer)
$Y_l^m(\theta, \phi + 2\pi) = e^{2\pi im} Y_l^m(\theta, \phi)$ and $m$ must be an integer
$Y_l^m(\theta + \pi, \phi) = Y_l^m(\theta, \phi)$ and $l$ must be an integer
$\frac{\partial Y_l^m}{\partial \theta}\Big|_{\theta=0} = \frac{\partial Y_l^m}{\partial \theta}\Big|_{\theta=\pi} = 0$

Explanation: When you encounter questions about quantum particles on spheres, focus on the geometric constraints that create boundary conditions. The spherical surface imposes specific requirements on the wavefunction that lead to quantization. The key insight is understanding what "single-valued" means geometrically. On a sphere, when you move around the azimuthal angle

\phi

2\pi

, you return to the same physical point. Therefore, the wavefunction must have identical values:

Y_l^m(\theta, \phi) = Y_l^m(\theta, \phi + 2\pi)

. This periodicity condition is essential. Additionally, at the poles (

\theta = 0, \pi

), the spherical coordinate system has singularities where

\phi

becomes undefined. For the wavefunction to be physically meaningful,

Y_l^m

must remain finite at these points, not blow up to infinity. Answer B correctly identifies both requirements: periodicity in

\phi

and finiteness at the poles. Answer A incorrectly requires the spherical harmonics to vanish at the poles. This would eliminate many valid solutions and isn't required by single-valuedness. Answer C describes the behavior before applying the single-valued constraint. The factor

e^{2\pi im}

only equals 1 (making the function truly periodic) when

m

is an integer, but this describes the intermediate step, not the final boundary condition. Answer D incorrectly focuses on

\theta

periodicity with period

\pi

. The

\theta

coordinate doesn't have this type of periodicity on a sphere. Remember: boundary conditions in quantum mechanics arise from the physical geometry of the problem. Always consider what the coordinate system's natural periodicities and singularities require of the wavefunction.

Question 10

For a quantum harmonic oscillator, the time-independent Schrödinger equation can be written in dimensionless form using $\xi = x\sqrt{\frac{m\omega}{\hbar}}$ and $\epsilon = \frac{2E}{\hbar\omega}$ , giving $\frac{d^2\psi}{d\xi^2} + (\epsilon - \xi^2)\psi = 0$ . If we try the substitution $\psi(\xi) = e^{-\xi^2/2} H(\xi)$ , what differential equation does $H(\xi)$ satisfy?

$\frac{d^2H}{d\xi^2} - 2\xi\frac{dH}{d\xi} + (\epsilon - 1)H = 0$ (correct answer)
$\frac{d^2H}{d\xi^2} + 2\xi\frac{dH}{d\xi} + (\epsilon - 1)H = 0$
$\frac{d^2H}{d\xi^2} - 2\xi\frac{dH}{d\xi} + \epsilon H = 0$
$\frac{d^2H}{d\xi^2} + \epsilon H = 0$
$\frac{d^2H}{d\xi^2} - \xi^2 H = 0$

Explanation: When solving the quantum harmonic oscillator, you often need to substitute a trial wavefunction into the Schrödinger equation to find what differential equation the unknown part satisfies. This requires careful application of the chain rule and product rule. Starting with

\psi(\xi) = e^{-\xi^2/2} H(\xi)

, you need to find the first and second derivatives. Using the product rule:

\frac{d\psi}{d\xi} = e^{-\xi^2/2}\frac{dH}{d\xi} + H(\xi)(-\xi)e^{-\xi^2/2} = e^{-\xi^2/2}\left(\frac{dH}{d\xi} - \xi H\right)

For the second derivative, apply the product rule again:

\frac{d^2\psi}{d\xi^2} = e^{-\xi^2/2}\left(\frac{d^2H}{d\xi^2} - 2\xi\frac{dH}{d\xi} + (\xi^2-1)H\right)

Substituting into the original equation

\frac{d^2\psi}{d\xi^2} + (\epsilon - \xi^2)\psi = 0

and dividing by

e^{-\xi^2/2}

\frac{d^2H}{d\xi^2} - 2\xi\frac{dH}{d\xi} + (\xi^2-1)H + (\epsilon - \xi^2)H = 0

This simplifies to:

\frac{d^2H}{d\xi^2} - 2\xi\frac{dH}{d\xi} + (\epsilon - 1)H = 0

Answer A is correct. Answer B has the wrong sign on the first derivative term (a common error when applying the product rule). Answer C is missing the "-1" term that comes from the second derivative of the exponential factor. Answer D completely ignores the first derivative term that necessarily appears from the product rule. Remember: when substituting trial solutions into differential equations, track every term carefully through the derivatives—the algebra determines which special functions (here, Hermite polynomials) solve the equation.

Question 11

Consider a particle in a one-dimensional infinite square well from $x = 0$ to $x = L$ . If we modify the boundary conditions so that $\psi(0) = 0$ but $\frac{d\psi}{dx}\Big|_{x=L} = 0$ (instead of $\psi(L) = 0$ ), what are the allowed energy eigenvalues?

$E_n = \frac{\hbar^2\pi^2 n^2}{2mL^2}$ where $n = 1, 2, 3, \ldots$
$E_n = \frac{\hbar^2\pi^2 (2n-1)^2}{8mL^2}$ where $n = 1, 2, 3, \ldots$ (correct answer)
$E_n = \frac{\hbar^2\pi^2 (2n+1)^2}{8mL^2}$ where $n = 0, 1, 2, \ldots$
$E_n = \frac{\hbar^2\pi^2 n^2}{8mL^2}$ where $n = 1, 3, 5, \ldots$
$E_n = \frac{\hbar^2\pi^2 (n + \frac{1}{2})^2}{2mL^2}$ where $n = 0, 1, 2, \ldots$

Explanation: When you encounter a particle-in-a-box problem with modified boundary conditions, you need to solve the time-independent Schrödinger equation with the new constraints. The general solution inside the box is still

\psi(x) = A\sin(kx) + B\cos(kx)

where

k = \sqrt{2mE}/\hbar

. Applying the first boundary condition

\psi(0) = 0

: Since

\cos(0) = 1

, we need

B = 0

, giving us

\psi(x) = A\sin(kx)

. For the second boundary condition

\frac{d\psi}{dx}\big|_{x=L} = 0

, we take the derivative:

\frac{d\psi}{dx} = Ak\cos(kx)

. At

x = L

Ak\cos(kL) = 0

. Since

A \neq 0

and

k \neq 0

, we need

\cos(kL) = 0

, which occurs when

kL = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \ldots = \frac{(2n-1)\pi}{2}

for

n = 1, 2, 3, \ldots

Substituting

k = \frac{(2n-1)\pi}{2L}

into

E = \frac{\hbar^2k^2}{2m}

gives

E_n = \frac{\hbar^2\pi^2(2n-1)^2}{8mL^2}

. Answer A represents the standard infinite square well with both ends fixed. Answer C uses

(2n+1)^2

instead of

(2n-1)^2

and starts from

n=0

, which would give the wrong quantization. Answer D has the correct denominator but restricts

n

to odd integers only, missing valid energy levels. Remember: Different boundary conditions lead to different quantization rules. Always apply both boundary conditions systematically to find the allowed wave numbers.

Question 12

For the hydrogen atom, the radial Schrödinger equation includes the effective potential $V_{eff}(r) = -\frac{ke^2}{r} + \frac{\hbar^2 l(l+1)}{2mr^2}$ . If we examine the large- $r$ behavior for bound states with $E < 0$ , the asymptotic form of the radial wavefunction is $R(r) \sim e^{-\gamma r}$ where $\gamma$ is a positive constant. What is the relationship between $\gamma$ and the energy $E$ ?

$\gamma = \sqrt{\frac{2m|E|}{\hbar^2}}$ (correct answer)
$\gamma = \sqrt{\frac{m|E|}{\hbar^2}}$
$\gamma = \sqrt{\frac{2m|E|}{\hbar^2}} \cdot \frac{1}{\sqrt{l(l+1)}}$
$\gamma = \frac{mke^2}{\hbar^2}$
$\gamma = \sqrt{\frac{2m|E|}{\hbar^2}} + \frac{ke^2}{\hbar^2}$

Explanation: When analyzing the asymptotic behavior of wavefunctions in quantum mechanics, you need to examine what happens when the radial coordinate becomes very large and dominates all other terms in the differential equation. For large

r

, the effective potential

V_{eff}(r)

becomes negligible because both the Coulomb term

-\frac{ke^2}{r}

and the centrifugal term

\frac{\hbar^2 l(l+1)}{2mr^2}

approach zero. The radial Schrödinger equation then simplifies to just the kinetic energy term plus the total energy:

-\frac{\hbar^2}{2m}\frac{d^2R}{dr^2} = ER

. Rearranging gives

\frac{d^2R}{dr^2} = \frac{2mE}{\hbar^2}R

. Since we have bound states with

E < 0

, we can write

E = -|E|

, so the equation becomes

\frac{d^2R}{dr^2} = -\frac{2m|E|}{\hbar^2}R

. This differential equation has solutions of the form

R(r) \sim e^{-\gamma r}

where

\gamma = \sqrt{\frac{2m|E|}{\hbar^2}}

, confirming answer A. Answer B is missing the factor of 2 in the numerator, which comes directly from the

2m

coefficient in the kinetic energy operator. Answer C incorrectly includes the angular momentum quantum numbers, but these terms vanish in the large-

r

limit. Answer D gives a constant related to the Bohr radius rather than depending on the energy

E

. Remember: for asymptotic analysis, identify which terms survive in the limit you're examining, then solve the simplified differential equation that remains.

Question 13

A particle is confined to a two-dimensional circular box of radius $R$ (i.e., $V(r,\theta) = 0$ for $r < R$ and $V(r,\theta) = \infty$ for $r \geq R$ ). The time-independent Schrödinger equation in polar coordinates separates into radial and angular parts. What boundary condition must be satisfied by the radial wavefunction $R(r)$ at $r = R$ ?

$R(R) = 0$ because the potential becomes infinite at the boundary (correct answer)
$\frac{dR}{dr}\Big|_{r=R} = 0$ because the probability current must vanish at the wall
$R(R) = 1$ to ensure proper normalization at the boundary
$r R(r)\Big|_{r=R} = 0$ because the full wavefunction $\psi(r,\theta) = R(r)\Theta(\theta)$ must vanish
Both $R(R) = 0$ and $\frac{dR}{dr}\Big|_{r=R} = 0$ must be satisfied simultaneously

Explanation: When dealing with quantum mechanical problems involving infinite potential barriers, the key principle is that the wavefunction must vanish wherever the potential becomes infinite. This ensures the particle has zero probability of being found in classically forbidden regions. For the particle in a circular box, the potential jumps to infinity at

r = R

, creating an impenetrable wall. Since the total wavefunction

\psi(r,\theta) = R(r)\Theta(\theta)

must equal zero at this boundary, and

\Theta(\theta)

is generally non-zero, the radial component

R(r)

itself must vanish:

R(R) = 0

. This is identical to the familiar boundary condition for a particle in a one-dimensional box. Answer A correctly identifies this fundamental boundary condition. Answer B incorrectly suggests the derivative must vanish—this would be appropriate for a finite potential barrier or certain other boundary conditions, but not for an infinite wall where the wavefunction itself must be zero. Answer C is wrong because normalization doesn't require the wavefunction to equal 1 at any specific point; rather, the integral of

|\psi|^2

over all space must equal 1. Answer D contains a subtle error: while

r R(r)

does vanish at

r = R

(since

R(R) = 0

), this isn't the fundamental boundary condition—it's merely a consequence of the actual condition

R(R) = 0

. Remember: whenever you encounter infinite potential barriers in quantum mechanics, the wavefunction must always vanish at those boundaries. This is your first instinct for any "particle in a box" problem, regardless of geometry.

Question 14

A quantum mechanical system has a potential $V(x)$ that is even: $V(-x) = V(x)$ . If $\psi(x)$ is a solution to the time-independent Schrödinger equation with energy $E$ , which statement about the parity properties of energy eigenstates is most accurate?

All energy eigenstates must have definite parity (either even or odd)
All energy eigenstates must be even functions because the potential is even
All energy eigenstates must be odd functions to maintain orthogonality
Non-degenerate energy eigenstates must have definite parity, but degenerate states may not (correct answer)
Energy eigenstates can have any parity because parity and energy are independent

Explanation: When you encounter a quantum mechanics problem involving symmetry, think about how the mathematical properties of the Hamiltonian constrain the possible wavefunctions. Here, the even potential

V(-x) = V(x)

creates a symmetric Hamiltonian that commutes with the parity operator. For non-degenerate energy levels, this symmetry forces eigenstates to have definite parity—they must be either purely even (

\psi(-x) = \psi(x)

) or purely odd (

\psi(-x) = -\psi(x)

). This occurs because when operators commute, they share common eigenfunctions, so the energy eigenstate must also be a parity eigenstate. However, when energy levels are degenerate (multiple linearly independent states share the same energy), you can form linear combinations of these degenerate states. Even if the individual basis states have definite parity, their combinations may mix even and odd components, resulting in states without definite parity. Option A is too broad—it ignores the crucial distinction between degenerate and non-degenerate cases. Option B incorrectly assumes all states must be even; the symmetric potential allows both even and odd solutions (think of the particle-in-a-box, where states alternate between even and odd). Option C wrongly claims all states are odd and misunderstands orthogonality, which doesn't require states to be odd. Study tip: In symmetry problems, always consider degeneracy separately. Non-degenerate states must respect the system's symmetries, but degenerate states can mix through linear combinations, potentially breaking the symmetry pattern.

Question 15

A quantum system has a Hamiltonian $\hat{H} = \frac{\hat{p}^2}{2m} + V(x)$ where $V(x)$ is real. If $\psi(x)$ is a solution with energy $E$ , under what conditions is the complex conjugate $\psi^*(x)$ also a solution with the same energy $E$ ?

Always, because the Hamiltonian is Hermitian and energy eigenvalues are real
Only when $V(x)$ is an even function of $x$
Only when the energy level is non-degenerate
Always, because $\hat{H}^* = \hat{H}$ when $V(x)$ is real (correct answer)
Never, because $\psi^*$ and $\psi$ represent different physical states

Explanation: When analyzing whether complex conjugates of wavefunctions are also solutions, you need to examine what happens when you take the complex conjugate of the entire Schrödinger equation. This tests your understanding of how operator properties relate to wavefunction symmetries. If

\psi(x)

satisfies

\hat{H}\psi = E\psi

, then taking the complex conjugate of both sides gives

(\hat{H}\psi)^* = (E\psi)^*

. Since energy eigenvalues are real, this becomes

(\hat{H}\psi)^* = E\psi^*

. For

\psi^*

to be a solution, we need

(\hat{H}\psi)^* = \hat{H}\psi^*

, which requires

\hat{H}^* = \hat{H}

. Let's check:

\hat{H}^* = \left(\frac{\hat{p}^2}{2m} + V(x)\right)^* = \frac{(\hat{p}^*)^2}{2m} + V^*(x)

. Since

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

, we have

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

, so

(\hat{p}^*)^2 = \hat{p}^2

. When

V(x)

is real,

V^*(x) = V(x)

. Therefore

\hat{H}^* = \hat{H}

, and

\psi^*

is always a solution. Answer D is correct. Answer A confuses Hermiticity with the specific property needed here. Answer B incorrectly focuses on parity when the real issue is whether the potential is real-valued. Answer C is backwards—degeneracy actually makes this property more obvious, not less relevant. Remember: when a Hamiltonian contains only real potentials and standard kinetic energy terms, complex conjugation always preserves solutions. This is a fundamental symmetry of quantum mechanics with real potentials.

Question 16

Consider a particle in a double well potential where $V(x) = V(-x)$ and there are two degenerate ground states $\psi_L(x)$ (localized in the left well) and $\psi_R(x)$ (localized in the right well). If these states are not eigenstates of the parity operator, what are the correct parity eigenstates and their parities?

$\psi_+(x) = \psi_L(x) + \psi_R(x)$ (even), $\psi_-(x) = \psi_L(x) - \psi_R(x)$ (odd)
$\psi_+(x) = \psi_L(x) - \psi_R(x)$ (even), $\psi_-(x) = \psi_L(x) + \psi_R(x)$ (odd)
$\psi_+(x) = \frac{1}{\sqrt{2}}[\psi_L(x) + \psi_R(x)]$ (even), $\psi_-(x) = \frac{1}{\sqrt{2}}[\psi_L(x) - \psi_R(x)]$ (odd) (correct answer)
$\psi_+(x) = \frac{1}{\sqrt{2}}[\psi_L(x) - \psi_R(x)]$ (even), $\psi_-(x) = \frac{1}{\sqrt{2}}[\psi_L(x) + \psi_R(x)]$ (odd)
The parity eigenstates are $\psi_L(x)$ (even) and $\psi_R(x)$ (odd)

Explanation: When you encounter symmetric double-well potentials with degenerate states, remember that symmetry dictates the form of true energy eigenstates. Since the potential has inversion symmetry

V(x) = V(-x)

, the Hamiltonian commutes with the parity operator, meaning energy eigenstates must also be parity eigenstates. The localized states

\psi_L(x)

and

\psi_R(x)

are related by parity:

\hat{P}\psi_L(x) = \psi_L(-x) = \psi_R(x)

and

\hat{P}\psi_R(x) = \psi_L(x)

. Since neither returns itself under parity operation, they cannot be true energy eigenstates in this symmetric system. To construct parity eigenstates, you need combinations that satisfy

\hat{P}\psi = \pm\psi

. For the symmetric combination:

\hat{P}[\psi_L(x) + \psi_R(x)] = \psi_R(x) + \psi_L(x) = \psi_L(x) + \psi_R(x)

, giving even parity. For the antisymmetric combination:

\hat{P}[\psi_L(x) - \psi_R(x)] = \psi_R(x) - \psi_L(x) = -[\psi_L(x) - \psi_R(x)]

, giving odd parity. The

\frac{1}{\sqrt{2}}

normalization factor is essential for proper quantum states. Option A lacks normalization factors. Option B incorrectly assigns parities—it has the symmetric combination labeled as even but the antisymmetric as odd in the wrong order. Option D reverses which combination has which parity entirely. Study tip: In symmetric potentials, always check that your proposed eigenstates actually satisfy the parity operation

\hat{P}\psi(x) = \psi(-x) = \pm\psi(x)

, and don't forget normalization constants for linear combinations.

Question 17

Consider a quantum particle in a one-dimensional potential $V(x) = \alpha |x|$ where $\alpha > 0$ . Due to the symmetry $V(-x) = V(x)$ , energy eigenstates have definite parity. For the ground state, which must be even, the boundary condition at $x = 0$ and the form of the wavefunction in each half-space are constrained by both the parity requirement and the discontinuous derivative of the potential. What condition must the wavefunction derivative satisfy at $x = 0$ ?

$\frac{d\psi}{dx}\Big|_{x=0^+} = \frac{d\psi}{dx}\Big|_{x=0^-}$ because the wavefunction must be smooth everywhere
$\frac{d\psi}{dx}\Big|_{x=0^+} = -\frac{d\psi}{dx}\Big|_{x=0^-}$ due to the even parity of the ground state
$\frac{d\psi}{dx}\Big|_{x=0^+} = \frac{d\psi}{dx}\Big|_{x=0^-} = 0$ because the ground state is even and has a maximum at $x=0$ (correct answer)
The derivative is undefined at $x = 0$ because the potential has a kink there
$\frac{d\psi}{dx}\Big|_{x=0^+} + \frac{d\psi}{dx}\Big|_{x=0^-} = \frac{2m\alpha}{\hbar^2}\psi(0)$ due to the delta-function-like contribution

Explanation: When analyzing quantum mechanical problems with symmetric potentials, you need to consider both the mathematical constraints from the Schrödinger equation and the physical requirements imposed by symmetry. The potential

V(x) = \alpha |x|

creates a "V-shaped" well that's symmetric about

x = 0

. Since the potential is even,

V(-x) = V(x)

, the Hamiltonian commutes with the parity operator, meaning energy eigenstates must have definite parity (either even or odd). The ground state of any symmetric potential is always even, so

\psi(-x) = \psi(x)

. For an even function, the derivative at

x = 0

must be zero. Here's why: if

\psi(-x) = \psi(x)

, then differentiating both sides gives

-\psi'(-x) = \psi'(x)

. At

x = 0

, this becomes

-\psi'(0) = \psi'(0)

, which is only satisfied if

\psi'(0) = 0

. Additionally, since the ground state wavefunction peaks at the point of lowest potential (

x = 0

), it has a maximum there, requiring zero derivative. Option A is wrong because it ignores the parity constraint—smoothness alone doesn't determine the derivative value. Option B incorrectly applies the antisymmetry condition, which would be true for odd functions, not even ones. Option D is incorrect because while the potential has a kink, the wavefunction itself remains smooth and differentiable everywhere. Remember: for symmetric potentials, even wavefunctions always have zero derivative at the center of symmetry, while odd wavefunctions have zero amplitude there.

Question 18

A quantum particle is described by the time-independent Schrödinger equation in a one-dimensional potential $V(x)$ . At a point $x_0$ where $V(x_0) = E$ (a classical turning point), the wavefunction $\psi(x_0) \neq 0$ but its second derivative $\frac{d^2\psi}{dx^2}\big|_{x_0} = 0$ . What can be concluded about the first derivative $\frac{d\psi}{dx}\big|_{x_0}$ ?

$\frac{d\psi}{dx}\big|_{x_0} = 0$ because both the function and its second derivative vanish at the turning point
$\frac{d\psi}{dx}\big|_{x_0}$ must be infinite to compensate for the vanishing second derivative at the turning point
$\frac{d\psi}{dx}\big|_{x_0} = 0$ is required for the wavefunction to be normalizable in the classically forbidden region
$\frac{d\psi}{dx}\big|_{x_0}$ can be non-zero and its value depends on the specific form of $V(x)$ near $x_0$ (correct answer)

Explanation: When you encounter questions about quantum mechanical turning points, focus on what the Schrödinger equation tells us about the relationship between the wavefunction's derivatives and the potential energy. At a classical turning point where

V(x_0) = E

, the time-independent Schrödinger equation becomes:

\frac{d^2\psi}{dx^2} = \frac{2m}{\hbar^2}[V(x) - E]\psi(x)

Since

V(x_0) = E

, we get

\frac{d^2\psi}{dx^2}\big|_{x_0} = 0

, which matches the given condition. However, this constraint only relates the second derivative to the wavefunction value and potential—it places no restriction on the first derivative. The correct answer is D because

\frac{d\psi}{dx}\big|_{x_0}

is indeed free to take any value. The Schrödinger equation imposes continuity requirements on

\psi

and

\frac{d\psi}{dx}

, but doesn't force the first derivative to vanish at turning points. Answer A incorrectly assumes that vanishing second derivatives require vanishing first derivatives—these are independent conditions. Answer B suggests the first derivative must be infinite, which would violate continuity requirements and isn't supported by the physics. Answer C confuses turning point behavior with normalizability conditions; the first derivative doesn't need to vanish for the wavefunction to be normalizable. Study tip: Remember that the Schrödinger equation directly constrains only the second derivative through the potential energy term. First derivatives are governed by boundary conditions and continuity requirements, not by local potential values.

Question 19

Consider the one-dimensional time-independent Schrödinger equation for a particle with mass $m$ in a potential $V(x)$ . At a point where the potential has a finite discontinuity, which of the following statements about the wavefunction and its derivative is correct?

Both $\psi(x)$ and $d\psi/dx$ must be continuous across the discontinuity for physical acceptability
Only $\psi(x)$ must be continuous; $d\psi/dx$ can have a finite discontinuity proportional to the height of the potential step (correct answer)
Both $\psi(x)$ and $d\psi/dx$ can be discontinuous as long as $|\psi(x)|^2$ remains finite and normalizable
Only $d\psi/dx$ must be continuous; $\psi(x)$ can have a finite jump discontinuity at the potential boundary

Explanation: At a finite potential discontinuity, the wavefunction

\psi(x)

must be continuous to ensure the probability density

|\psi(x)|^2

is well-defined. However,

d\psi/dx

can be discontinuous because the second derivative

d^2\psi/dx^2

experiences a sudden change due to the potential step, as seen from the Schrödinger equation. The discontinuity in

d\psi/dx

is proportional to

\psi

times the height of the potential step. Choice A is too restrictive. Choice C is wrong because

\psi

must be continuous. Choice D is backwards - it's the derivative that can be discontinuous, not the wavefunction itself.

Question 20

A quantum mechanical system has a potential $V(x)$ that is symmetric about $x = 0$ , i.e., $V(-x) = V(x)$ . The time-independent Schrödinger equation for this system admits solutions with definite parity. If $\psi_E(x)$ is an eigenfunction with energy $E$ , under what conditions can both $\psi_E(x)$ and $\psi_E(-x)$ be linearly independent solutions with the same energy?

This can occur for any energy $E$ as long as the potential has inversion symmetry $V(-x) = V(x)$
This is impossible because $\psi_E(-x)$ is always just the parity-transformed version of $\psi_E(x)$
This occurs only when the energy level $E$ is degenerate, allowing multiple independent solutions (correct answer)
This occurs only at special energy values where the boundary conditions allow mixed parity states

Explanation: When you encounter quantum mechanical systems with symmetric potentials, you're dealing with parity symmetry and the possibility of degeneracy. The key insight is understanding when the symmetry of the potential allows multiple independent solutions at the same energy level. For a symmetric potential

V(-x) = V(x)

, eigenfunctions generally have definite parity - they're either even (

\psi(x) = \psi(-x)

) or odd (

\psi(x) = -\psi(-x)

). However, when an energy level is degenerate, you can have multiple linearly independent solutions. In this case, you could have both an even and an odd solution at the same energy

E

. If

\psi_E(x)

is one such solution, then

\psi_E(-x)

represents a different linear combination of these degenerate states, potentially making them linearly independent. Answer A is incorrect because symmetry alone doesn't guarantee linear independence - you need the special condition of degeneracy. Answer B misses the degeneracy possibility; while parity transformation usually gives the same function (for definite parity), degeneracy changes this. Answer D incorrectly focuses on boundary conditions rather than the fundamental issue of degeneracy. The correct answer is C: degeneracy is the key condition that allows

\psi_E(x)

and

\psi_E(-x)

to be linearly independent. When multiple independent solutions exist at energy

E

, you can construct linear combinations that don't have definite parity. Remember: in symmetric systems, look for degeneracy as the condition that breaks the usual parity restrictions and allows multiple independent solutions.