Physical Chemistry 2 Quiz: Variational Principle

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Variational PrincipleQuestion 1 of 20

Two normalized trial wavefunctions $\psi_1$ and $\psi_2$ for a quantum system yield variational energies $E_1 = 5.2 \text{ eV}$ and $E_2 = 4.8 \text{ eV}$ , respectively. A linear combination $\psi_c = c_1\psi_1 + c_2\psi_2$ with $|c_1|^2 + |c_2|^2 = 1$ is constructed. What can be definitively concluded about the energy $E_c$ obtained using $\psi_c$ ?

E_c

will always be between 4.8 eV and 5.2 eV regardless of the values of

c_1

and

c_2

E_c

will be minimized when

c_1 = c_2 = 1/\sqrt{2}

, giving

E_c = 5.0 \text{ eV}

E_c \geq 4.8 \text{ eV}

with the minimum achieved when

c_2 = 1

and

c_1 = 0

E_c

depends on the overlap integral

\langle\psi_1|\psi_2\rangle

and cannot be determined from the given information

E_c

will always equal

c_1^2 E_1 + c_2^2 E_2

due to the linearity of the Hamiltonian operator

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Physical Chemistry 2 Quiz

Physical Chemistry 2 Quiz: Variational Principle

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

How to use this quiz

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

All questions

Question 1

$E_c$ will always be between 4.8 eV and 5.2 eV regardless of the values of $c_1$ and $c_2$
$E_c$ will be minimized when $c_1 = c_2 = 1/\sqrt{2}$ , giving $E_c = 5.0 \text{ eV}$
$E_c \geq 4.8 \text{ eV}$ with the minimum achieved when $c_2 = 1$ and $c_1 = 0$
$E_c$ depends on the overlap integral $\langle\psi_1|\psi_2\rangle$ and cannot be determined from the given information (correct answer)
$E_c$ will always equal $c_1^2 E_1 + c_2^2 E_2$ due to the linearity of the Hamiltonian operator

Explanation: When you encounter variational method problems involving linear combinations of trial wavefunctions, the key insight is that the energy expression depends critically on how the individual wavefunctions interact - specifically through their overlap integral. For a linear combination

\psi_c = c_1\psi_1 + c_2\psi_2

, the variational energy is:

E_c = \frac{\langle\psi_c|\hat{H}|\psi_c\rangle}{\langle\psi_c|\psi_c\rangle}

When you expand this expression, you get terms involving

E_1

E_2

, the coefficients

c_1

and

c_2

, and crucially, cross terms containing

\langle\psi_1|\hat{H}|\psi_2\rangle

and the overlap integral

S_{12} = \langle\psi_1|\psi_2\rangle

. Without knowing these overlap and interaction terms, you cannot determine

E_c

. Answer D is correct because the overlap integral

\langle\psi_1|\psi_2\rangle

fundamentally affects the energy calculation and isn't provided. Answer A incorrectly assumes

E_c

is simply bounded by the individual energies - this would only be true if the wavefunctions were orthogonal and certain other conditions held. Answer B makes the unjustified assumption that equal mixing coefficients minimize energy and that the result is the arithmetic mean - this ignores the cross terms entirely. Answer C incorrectly assumes you can definitively conclude that

\psi_2

gives the minimum energy. While

E_2 < E_1

, the linear combination could potentially yield an even lower energy depending on the overlap terms. Study tip: In variational problems with linear combinations, always check whether overlap integrals are given. If they're missing, the energy cannot be uniquely determined from the individual variational energies alone.

Question 2

For a variational calculation on the hydrogen atom using the trial wavefunction $\psi_t(r) = Ne^{-\alpha r}$ where $\alpha$ is a variational parameter, the optimal value is found to be $\alpha_{opt} = Z_{eff}/a_0$ . If this approach is applied to the He $^+$ ion, what is the relationship between the variational energy at $\alpha_{opt}$ and the exact ground state energy?

The variational energy equals the exact energy because the trial function has the correct exponential form for hydrogen-like atoms (correct answer)
The variational energy is higher than the exact energy due to the lack of angular dependence in the trial function
The variational energy is lower than the exact energy because the optimized $\alpha$ overcompensates for electron-nuclear attraction
The variational energy approaches the exact energy only in the limit of large nuclear charge $Z$
The variational energy equals exactly half the exact energy due to the spherical symmetry assumption

Explanation: When you encounter variational method problems, focus on whether the trial wavefunction can exactly represent the true wavefunction. The variational principle states that any trial wavefunction will give an energy that's greater than or equal to the exact ground state energy, with equality achieved only when the trial function is identical to the true wavefunction. For hydrogen-like atoms (including He⁺), the exact ground state wavefunction has the form

\psi = Ne^{-Zr/a_0}

, where

Z

is the nuclear charge. Your trial function

\psi_t(r) = Ne^{-\alpha r}

has exactly this exponential form. When you optimize

\alpha

, you get

\alpha_{opt} = Z_{eff}/a_0 = Z/a_0

for He⁺ (since there's no electron shielding in this one-electron ion). This means your optimized trial function becomes identical to the exact wavefunction. Answer A is correct because the trial function, when optimized, perfectly matches the exact hydrogen-like wavefunction, yielding the exact energy. Answer B incorrectly suggests the lack of angular dependence matters, but hydrogen-like ground states are spherically symmetric (s orbitals), so no angular dependence is needed. Answer C wrongly claims the variational energy is lower than exact energy, which violates the variational principle - variational energies can never be below the true ground state. Answer D is incorrect because the relationship holds for any hydrogen-like ion, not just in the large

Z

limit. Study tip: Remember that variational calculations give exact results when the trial function can perfectly represent the true wavefunction - this happens for hydrogen-like atoms with exponential trial functions.

Question 3

A student applies the variational method to find the ground state of a perturbed harmonic oscillator $\hat{H} = \frac{\hat{p}^2}{2m} + \frac{1}{2}m\omega^2\hat{x}^2 + \lambda\hat{x}^4$ using the trial function $\psi_t(x) = \left(\frac{\beta}{\pi}\right)^{1/4} e^{-\beta x^2/2}$ . After minimizing $\langle E \rangle$ with respect to $\beta$ , the student finds $\beta_{opt} > m\omega/\hbar$ . What does this result indicate about the effect of the $\lambda\hat{x}^4$ perturbation?

The perturbation decreases the effective force constant, leading to a more spread out wavefunction
The perturbation increases the effective force constant, leading to a more localized wavefunction (correct answer)
The perturbation has no effect on the spatial extent because $\langle\hat{x}^4\rangle$ vanishes for Gaussian wavefunctions
The result indicates an error in the calculation since $\beta_{opt}$ should always equal $m\omega/\hbar$ for harmonic systems
The perturbation couples the ground state to excited states, requiring a multi-configuration trial function

Explanation: When you encounter variational method problems, focus on how the trial wavefunction's parameters change in response to the Hamiltonian. The variational parameter

\beta

controls the width of the Gaussian trial function—larger

\beta

means a more localized (narrower) wavefunction. For the unperturbed harmonic oscillator, the optimal

\beta

equals

m\omega/\hbar

. Here, you found

\beta_{opt} > m\omega/\hbar

, meaning the optimized wavefunction is more localized than the original harmonic oscillator ground state. This happens because the

\lambda\hat{x}^4

term adds an additional confining potential that grows rapidly at large

|x|

. The system responds by contracting the wavefunction to minimize this quartic contribution to the energy, effectively increasing the force constant. Choice A is backwards—if the perturbation decreased the effective force constant,

\beta_{opt}

would be smaller than

m\omega/\hbar

, giving a more spread out wavefunction. Choice C contains a fundamental error:

\langle\hat{x}^4\rangle

definitely doesn't vanish for Gaussian functions. In fact, for a Gaussian with parameter

\beta

\langle x^4\rangle = \frac{3}{4\beta^2}

, which is always positive. Choice D misunderstands the variational method—the optimal

\beta

only equals

m\omega/\hbar

for the pure harmonic oscillator; any perturbation will generally change this value. Study tip: Remember that quartic terms (

x^4

) always provide additional confinement since they're always positive, leading to wavefunction contraction. Watch for how perturbations affect the "width" of trial functions in variational problems.

Question 4

In a variational treatment of the helium atom ground state, a student uses the trial wavefunction $\psi_t(\mathbf{r}_1, \mathbf{r}_2) = \psi_{1s}(\mathbf{r}_1; Z_{eff})\psi_{1s}(\mathbf{r}_2; Z_{eff})$ where $\psi_{1s}$ is a hydrogen-like 1s orbital with effective nuclear charge $Z_{eff}$ . After optimization, $Z_{eff} = 1.69$ . What does this result suggest about electron-electron interactions in helium?

Each electron experiences the full nuclear charge $Z = 2$ because $Z_{eff} ≈ 2Z/2.37$
Each electron effectively screens approximately 0.31 units of nuclear charge from the other electron (correct answer)
The electron-electron repulsion is negligible because $Z_{eff} > 1$ , indicating net attraction to the nucleus
The trial wavefunction overestimates electron correlation because $Z_{eff} < Z$
Each electron behaves as if it experiences a nuclear charge reduced by exactly half the other electron's charge

Explanation: When you encounter variational method problems, focus on how the optimized parameters reveal physical insights about the system being studied. The effective nuclear charge tells a story about electron shielding. In helium, each electron experiences both attraction to the nucleus (charge +2) and repulsion from the other electron. The variational principle finds the effective nuclear charge that minimizes the energy. With

Z_{eff} = 1.69

, each electron "feels" a nuclear charge reduced from the full

Z = 2

to 1.69. This reduction of

2.00 - 1.69 = 0.31

units represents how much nuclear charge one electron effectively screens from the other due to electron-electron repulsion. Option A misinterprets the relationship—the ratio

2/2.37

doesn't represent the full nuclear charge experience. Option C contains a fundamental error:

Z_{eff} < Z

(not

Z_{eff} > 1

), and even if

Z_{eff} > 1

, this wouldn't mean repulsion is negligible—it means net attraction exists despite significant repulsion. Option D confuses the physics:

Z_{eff} < Z

correctly reflects electron shielding, and this simple trial wavefunction actually underestimates correlation effects (it can't capture the instantaneous electron-electron interactions that reduce repulsion when electrons avoid each other). The correct answer is B because the screening of 0.31 units quantifies how electron-electron repulsion reduces the effective nuclear attraction each electron experiences. Study tip: In variational problems, optimized parameters always reveal physical effects. When

Z_{eff} < Z_{nuclear}

, calculate the difference—that's the screening effect from electron-electron interactions.

Question 5

A variational calculation for the first excited state of a quantum harmonic oscillator uses the trial function $\psi_t(x) = Nxe^{-\alpha x^2/2}$ . The student finds that minimizing $\langle E \rangle$ gives $\alpha_{opt} = m\omega/\hbar$ , yielding $E_{var} = \frac{3\hbar\omega}{2}$ . What can be concluded about this result?

The result is incorrect because the variational energy should be higher than the exact first excited state energy $\frac{3\hbar\omega}{2}$
The trial function exactly reproduces the first excited state, so the variational method gives the exact energy (correct answer)
The result violates the variational principle because excited state energies cannot be found using simple trial functions
The energy is correct, but $\alpha_{opt}$ should be different from the ground state value $m\omega/\hbar$
The calculation must be wrong because $\langle E \rangle$ depends on both $\alpha$ and the normalization constant $N$

Explanation: The variational principle is a powerful method in quantum mechanics that provides an upper bound to the ground state energy, but it can also give exact results when the trial function matches the true wavefunction. For excited states, the key insight is that if your trial function has the correct functional form, the variational method can yield the exact energy. The trial function

\psi_t(x) = Nxe^{-\alpha x^2/2}

is actually the exact form of the first excited state harmonic oscillator wavefunction. When you minimize

\langle E \rangle

and obtain

\alpha_{opt} = m\omega/\hbar

with

E_{var} = \frac{3\hbar\omega}{2}

, you've recovered both the exact wavefunction parameter and the exact energy of the first excited state. Choice A incorrectly assumes the variational energy should be higher than

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

. While the variational principle typically gives an upper bound, when the trial function is exact, you get the exact energy—no approximation involved. Choice C reflects a common misconception that the variational method can't handle excited states. While it's true that simple variational calculations often target ground states, the method works for excited states when proper trial functions are chosen. Choice D suggests the energy is right but

\alpha_{opt}

is wrong. However,

m\omega/\hbar

is indeed the correct parameter for both ground and first excited states of the harmonic oscillator—they share the same Gaussian width parameter. Study tip: When using variational methods, always check if your trial function matches known exact solutions. If it does, the variational principle will give you the exact answer, not just an approximation.

Question 6

For a variational calculation on a one-dimensional double well potential with minima at $x = ±a$ , two trial functions are considered: $\psi_S(x) = N_S[e^{-\alpha(x-a)^2} + e^{-\alpha(x+a)^2}]$ and $\psi_A(x) = N_A[e^{-\alpha(x-a)^2} - e^{-\alpha(x+a)^2}]$ . If the exact ground and first excited states have energies $E_0$ and $E_1$ respectively, what relationship between the variational energies $E_S$ and $E_A$ is expected?

$E_S < E_A$ because the symmetric function $\psi_S$ better represents the ground state of the symmetric potential
$E_S > E_A$ because the antisymmetric function $\psi_A$ has lower kinetic energy due to destructive interference
$E_S = E_A$ because both functions are linear combinations of the same Gaussian basis functions
$E_S ≈ E_0 < E_1 ≈ E_A$ because the trial functions approximate the ground and first excited states respectively (correct answer)
The relationship depends on the barrier height between the wells and cannot be determined without additional information

Explanation: When you encounter variational calculations on symmetric potentials, focus on how the symmetry properties of trial functions relate to the energy eigenstates. The variational principle states that any trial wavefunction will give an energy greater than or equal to the true ground state energy. For a symmetric double-well potential, the ground state must be symmetric (even parity) while the first excited state is antisymmetric (odd parity). The trial function

\psi_S

is symmetric since it adds the two Gaussians, making

\psi_S(-x) = \psi_S(x)

. Conversely,

\psi_A

is antisymmetric because it subtracts them, giving

\psi_A(-x) = -\psi_A(x)

. Since these trial functions have the correct symmetries and are well-constructed (localized Gaussians at the potential minima), they should approximate their respective exact states quite well, giving

E_S ≈ E_0

and

E_A ≈ E_1

. Option A correctly identifies that

\psi_S

represents the ground state but doesn't recognize that

\psi_A

approximates the first excited state. Option B incorrectly suggests the antisymmetric function has lower energy - this confuses interference effects with energy ordering. The "destructive interference" actually creates a node at

x = 0

, raising the kinetic energy. Option C ignores the crucial symmetry differences between the functions, incorrectly assuming identical energies just because they use the same basis functions. Remember: in symmetric potentials, match your trial function's symmetry to the target eigenstate's symmetry for optimal variational results.

Question 7

A student applies the variational method to a hydrogen atom using $\psi_t(r,\theta,\phi) = N r^n e^{-\beta r} Y_l^m(\theta,\phi)$ where $Y_l^m$ are spherical harmonics. For the ground state ( $l=0, m=0$ ), what constraint on $n$ is necessary to ensure the trial wavefunction gives a finite kinetic energy?

$n ≥ 0$ to ensure the wavefunction vanishes at $r = 0$
$n ≥ 1$ to ensure proper normalization of the radial part
$n > -1/2$ to ensure the kinetic energy integral converges near $r = 0$ (correct answer)
$n = 0$ exactly to match the hydrogen ground state radial dependence
No constraint is needed because the exponential factor $e^{-\beta r}$ ensures convergence for any $n$

Explanation: The variational method requires that all integrals in the energy expression converge, particularly the kinetic energy term. When you encounter variational trial functions, always check whether the mathematical form leads to finite, well-defined integrals. For the kinetic energy, you need to evaluate

\langle \psi_t | -\frac{\hbar^2}{2m}\nabla^2 | \psi_t \rangle

. The critical part is the radial kinetic energy integral, which involves the second derivative of

r^n e^{-\beta r}

. Near

r = 0

, this second derivative behaves like

r^{n-2}

. For the kinetic energy integral

\int_0^{\infty} r^2 |r^{n-2}|^2 dr = \int_0^{\infty} r^{2n-2} dr

to converge near the lower limit, you need

2n - 2 > -1

, which gives

n > -1/2

. Option A incorrectly focuses on boundary conditions at

r = 0

. While

n ≥ 0

does make the wavefunction vanish there, this isn't the constraint that ensures finite kinetic energy. Option B misidentifies the issue as normalization. The normalization integral has different convergence requirements than the kinetic energy integral. Option D is too restrictive—while

n = 0

works and matches the true hydrogen ground state, other values of

n > -1/2

also give finite kinetic energies in variational calculations. Remember: in variational problems, always verify that your trial wavefunction produces convergent integrals for all energy terms, especially kinetic energy which involves derivatives that can worsen convergence behavior near singular points.

Question 8

Consider a variational calculation for a particle in a box from $x = 0$ to $x = L$ using the trial function $\psi_t(x) = Ax(L-x)(L-2x)$ for $0 ≤ x ≤ L$ . This function satisfies the boundary conditions $\psi_t(0) = \psi_t(L) = 0$ . What can be concluded about the energy obtained from this trial function?

It will approximate the ground state energy because the function has no nodes in the interior
It will approximate the first excited state energy because the function has one node at $x = L/2$ (correct answer)
It will be much higher than any low-lying energy level because the function has poor curvature properties
It cannot give a reliable energy estimate because the function is not symmetric about $x = L/2$
The energy will be exactly $\frac{9\pi^2\hbar^2}{8mL^2}$ due to the cubic polynomial form

Explanation: When analyzing variational trial functions for quantum systems, the key insight is that the number and location of nodes (zeros) in the wavefunction directly correspond to which energy eigenstate the function will best approximate. Let's examine the trial function

\psi_t(x) = Ax(L-x)(L-2x)

. To find nodes, set the function equal to zero: the factors give us zeros at

x = 0

x = L

, and

x = L/2

. The first two are boundary conditions, but

x = L/2

is an interior node that divides the box exactly in half. This nodal structure is crucial because quantum mechanical energy eigenstates are characterized by their node count. The ground state (

n=1

) has no interior nodes, the first excited state (

n=2

) has one node at the center, the second excited state (

n=3

) has two interior nodes, and so on. Since our trial function has exactly one interior node at the box's center, it will best approximate the first excited state. Looking at the wrong answers: (A) incorrectly claims no interior nodes exist - we clearly found one at

x = L/2

. (C) suggests poor curvature properties, but this smooth polynomial actually has reasonable curvature behavior for variational calculations. (D) focuses on asymmetry, but the function is actually antisymmetric about

x = L/2

, which is exactly what the first excited state should be. Study tip: Always count interior nodes in trial functions - this immediately tells you which eigenstate the variational method will approximate, since node number directly correlates with quantum state energy level.

Question 9

A variational calculation for an anharmonic oscillator $V(x) = \frac{1}{2}kx^2 + \lambda x^3$ uses the displaced harmonic oscillator trial function $\psi_t(x) = \left(\frac{\alpha}{\pi}\right)^{1/4} \exp\left[-\frac{\alpha(x-x_0)^2}{2}\right]$ where both $\alpha$ and $x_0$ are variational parameters. For small positive $\lambda$ , what is expected for the optimal displacement $x_0$ ?

$x_0 = 0$ because the anharmonic term preserves the symmetry of the harmonic oscillator
$x_0 > 0$ because the $x^3$ term creates an effective bias toward positive displacements
$x_0 < 0$ because the $x^3$ term destabilizes positive displacements, favoring negative ones (correct answer)
$x_0 = \pm\sqrt{\lambda/k}$ based on the classical equilibrium condition $dV/dx = 0$
The sign of $x_0$ alternates depending on whether $\lambda$ is rational or irrational

Explanation: When you encounter variational calculations with anharmonic potentials, the key is understanding how additional terms affect the optimal wavefunction shape and position compared to the harmonic case. The cubic term

\lambda x^3

(with

\lambda > 0

) creates an asymmetric potential. For positive

x

, this term adds extra repulsion on top of the harmonic restoring force, making the potential steeper. For negative

x

, it actually reduces the effective restoring force, creating a "softer" region. This asymmetry means the ground state wavefunction will shift to minimize its energy by spending more time in the energetically favorable (softer) negative region. The variational principle drives the optimal

x_0

to negative values because this displacement allows the trial wavefunction to better sample the region where the anharmonic correction lowers the potential energy. This gives answer C. Answer A incorrectly assumes the

x^3

term preserves symmetry - it explicitly breaks the even symmetry of the harmonic oscillator. Answer B has the sign wrong; while the

x^3

term does create bias, it favors negative displacements where it reduces the effective potential. Answer D attempts to use classical equilibrium (

dV/dx = kx + 3\lambda x^2 = 0

), but this approach is flawed for quantum ground states, and the algebra doesn't even yield the stated result. Remember: in variational problems with asymmetric anharmonic terms, the optimal wavefunction shifts toward regions where the anharmonic correction is energetically favorable, not where classical turning points occur.

Question 10

A variational calculation for a quantum dot modeled as a 2D harmonic oscillator $V(x,y) = \frac{1}{2}m\omega^2(x^2 + y^2)$ uses the trial function $\psi_t(x,y) = N(x^2 + y^2)e^{-\alpha(x^2 + y^2)}$ where $\alpha$ is optimized. What energy level does this trial function best approximate?

The ground state $(n_x, n_y) = (0,0)$ with energy $\hbar\omega$
The first excited state $(n_x, n_y) = (1,0)$ or $(0,1)$ with energy $2\hbar\omega$
The second excited state $(n_x, n_y) = (2,0)$ , $(0,2)$ , or $(1,1)$ with energy $3\hbar\omega$
A superposition of degenerate second excited states with energy $3\hbar\omega$ (correct answer)
The fourth excited state $(n_x, n_y) = (2,2)$ with energy $5\hbar\omega$

Explanation: When you encounter variational method problems, the key insight is recognizing that trial functions with specific functional forms will best approximate states with similar characteristics—particularly the same radial dependence and symmetry. The trial function

\psi_t(x,y) = N(x^2 + y^2)e^{-\alpha(x^2 + y^2)}

has several important features. The factor

(x^2 + y^2)

gives it radial symmetry around the origin, and this

r^2

dependence creates a node at the center. The exponential decay ensures proper normalization. For a 2D harmonic oscillator, the exact wavefunctions are products of 1D harmonic oscillator states:

\psi_{n_x,n_y}(x,y) = \psi_{n_x}(x)\psi_{n_y}(y)

. The second excited states are

\psi_{2,0}

\psi_{0,2}

, and

\psi_{1,1}

, all with energy

3\hbar\omega

. When you form the linear combination

\psi_{2,0} + \psi_{0,2}

, it becomes proportional to

(x^2 + y^2)e^{-\text{constant} \cdot (x^2 + y^2)}

—exactly matching your trial function's form. Option A is wrong because the ground state has no nodes and is purely Gaussian. Option B is incorrect since first excited states have linear, not quadratic, spatial dependence. Option C fails because it suggests a specific single state rather than recognizing the superposition nature. The correct answer is D: your trial function approximates a superposition of degenerate second excited states with energy

3\hbar\omega

. Study tip: In variational problems, match the trial function's symmetry and nodal structure to identify which quantum states it resembles—superpositions often arise naturally when the trial function has higher symmetry than individual eigenstates.

Question 11

In a variational study of the lithium atom, a trial wavefunction $\psi_t = \psi_{1s}(1;Z_1)\psi_{1s}(2;Z_1)\psi_{2s}(3;Z_2)$ is used, where electrons 1 and 2 are in 1s orbitals with effective nuclear charge $Z_1$ , and electron 3 is in a 2s orbital with effective nuclear charge $Z_2$ . After optimization, $Z_1 = 2.69$ and $Z_2 = 1.28$ . What do these results indicate about electron shielding in lithium?

The 1s electrons are more effectively shielded than the 2s electron because $Z_1 > Z_2$
The 2s electron is more effectively shielded than the 1s electrons because it experiences $Z_2 = 1.28 < Z_1 = 2.69$ (correct answer)
Both sets of electrons experience similar shielding since $Z_1 + Z_2 ≈ 2Z = 6$
The results are unphysical because $Z_1 + Z_2 = 3.97 ≠ Z = 3$ violates charge conservation
The 2s electron penetrates the 1s shell, experiencing less shielding than expected from simple screening models

Explanation: When analyzing variational calculations with effective nuclear charges, you need to understand that a smaller effective nuclear charge indicates stronger electron shielding. The effective nuclear charge represents how much nuclear attraction an electron actually feels after accounting for repulsion from other electrons. In lithium (Z = 3), the optimized values

Z_1 = 2.69

for the 1s electrons and

Z_2 = 1.28

for the 2s electron reveal important shielding effects. The 1s electrons experience an effective nuclear charge close to the full nuclear charge (2.69 vs. 3), meaning they're only slightly shielded by each other. In contrast, the 2s electron experiences a much reduced effective nuclear charge (1.28), indicating it's heavily shielded by the inner 1s electrons. Since stronger shielding corresponds to lower effective nuclear charge, the 2s electron is more effectively shielded than the 1s electrons. Option A incorrectly interprets

Z_1 > Z_2

as meaning 1s electrons are more shielded, when the opposite is true. Option C misses the physical meaning by focusing on the sum

Z_1 + Z_2

, which has no direct significance in electron shielding. Option D incorrectly assumes that effective nuclear charges must sum to the actual nuclear charge—this constraint doesn't exist because effective charges account for complex electron-electron interactions, not charge conservation. Remember: in variational calculations, lower effective nuclear charge always indicates stronger electron shielding. Inner electrons shield outer electrons more than outer electrons shield inner ones.

Question 12

A student performs a variational calculation for the ground state of a particle in a 1D box using $\psi_t(x) = N\sin(\pi x/L)\cos(\pi x/L)$ for $0 ≤ x ≤ L$ . After computing $\langle E \rangle$ , the result is found to be exactly $\frac{2\pi^2\hbar^2}{mL^2}$ . How does this compare to the exact ground state energy, and what does it reveal about the trial function?

The trial function gives exactly twice the ground state energy and approximates the first excited state
The trial function gives exactly the second excited state energy because it has the wrong symmetry
The result violates the variational principle since $2\pi^2\hbar^2/(mL^2) < \pi^2\hbar^2/(mL^2)$
The trial function exactly reproduces the first excited state, so the energy equals $E_1 = 4\pi^2\hbar^2/(2mL^2)$ (correct answer)
The trial function is a poor approximation because the energy is much higher than the ground state

Explanation: When you encounter variational calculations, remember that the variational principle states that any trial wavefunction will give an energy greater than or equal to the true ground state energy. However, if your trial function happens to be an exact eigenfunction of the system, you'll get the exact energy for that state. Let's analyze the given trial function

\psi_t(x) = N\sin(\pi x/L)\cos(\pi x/L)

. Using the trigonometric identity

\sin(\theta)\cos(\theta) = \frac{1}{2}\sin(2\theta)

, this becomes

\psi_t(x) = \frac{N}{2}\sin(2\pi x/L)

. This is exactly the first excited state wavefunction for a particle in a box! The exact energy levels are

E_n = \frac{n^2\pi^2\hbar^2}{2mL^2}

, so for

n=2

E_2 = \frac{4\pi^2\hbar^2}{2mL^2} = \frac{2\pi^2\hbar^2}{mL^2}

, which matches the calculated result. Option A incorrectly identifies this as twice the ground state energy approximating the first excited state, but it's exactly the second excited state. Option B mentions wrong symmetry but gets the energy assignment wrong. Option C claims a variational principle violation, but

2\pi^2\hbar^2/(mL^2) > \pi^2\hbar^2/(mL^2)

(the ground state energy), so there's no violation. Option D correctly identifies that the trial function exactly reproduces the first excited state (

n=2

). Study tip: Always check if your trial function can be simplified using trig identities—it might be an exact eigenfunction in disguise, giving you the exact energy of a specific state rather than a variational bound.

Question 13

For a variational treatment of the hydrogen atom using the trial function $\psi_t(r) = Ne^{-\alpha r^2}$ , the optimized parameter gives $\alpha_{opt} = 0.28 \text{ a}_0^{-2}$ where $a_0$ is the Bohr radius. If the variational energy is $E_{var} = -8.9 \text{ eV}$ , what can be concluded about the accuracy of this Gaussian approximation?

The approximation is excellent since $E_{var}$ is within 1% of the exact energy $-13.6 \text{ eV}$
The approximation significantly underestimates the binding energy, missing about 35% of the exact binding energy (correct answer)
The result violates the variational principle since $E_{var} > E_{exact}$ but the trial function should give lower energy
The Gaussian form is inappropriate for hydrogen because it lacks the correct $1/r$ Coulomb singularity
The optimization is incorrect since $\alpha_{opt}$ should equal $1/a_0^2$ for the hydrogen atom

Explanation: The variational principle is fundamental to quantum mechanical approximations: any trial wavefunction will give an energy that is greater than or equal to the true ground state energy. This makes it a powerful tool for finding upper bounds to exact energies. To evaluate this Gaussian approximation's accuracy, you need to compare the variational energy to hydrogen's exact ground state energy of

-13.6 \text{ eV}

. The trial function gives

E_{var} = -8.9 \text{ eV}

, which is indeed higher than the exact value (as required by the variational principle), but the difference is significant:

\frac{13.6 - 8.9}{13.6} \times 100\% = 34.6\%

. This means the approximation captures only about 65% of the true binding energy, representing a substantial underestimation. Answer A is wrong because

8.9/13.6 = 0.65

, nowhere near 99% accuracy. Answer C misunderstands the variational principle—having

E_{var} > E_{exact}

is exactly what the principle predicts, not a violation. Answer D, while noting a real limitation of Gaussian functions (they don't have the correct cusp behavior at the nucleus), doesn't address the question about accuracy of the energy result. The correct answer is B: the approximation significantly underestimates the binding energy by about 35%. Study tip: When evaluating variational calculations, always calculate the percentage error relative to the exact result. The variational principle guarantees

E_{var} \geq E_{exact}

, but a large energy difference indicates poor wavefunction overlap with the true ground state.

Question 14

A variational calculation is performed for a particle in a finite square well of depth $V_0$ and width $a$ using two different trial functions: $\psi_1(x) = A\cos(\pi x/a)$ for $|x| \leq a/2$ and $\psi_2(x) = Be^{-\beta|x|}$ for all $x$ . Both functions are properly normalized. If the exact ground state is known to be bound with energy $E_0 = -2.5 \text{ eV}$ , which comparison of the variational energies $E_1$ and $E_2$ is most likely?

$E_1 < E_2 < E_0$ because the cosine function better matches the bound state character inside the well
$E_0 < E_1 < E_2$ because the exponential function has lower kinetic energy due to its smooth curvature
$E_0 < E_2 < E_1$ because the exponential function correctly describes the evanescent behavior outside the well (correct answer)
$E_1 = E_2 > E_0$ because both trial functions are symmetric and normalized to unity
The comparison depends on the specific values of $V_0$ , $a$ , and $\beta$ , so no general statement can be made

Explanation: When you encounter variational method problems, remember that the variational principle states that any trial wavefunction will give an energy that is greater than or equal to the true ground state energy:

E_{trial} \geq E_0

. This immediately tells you that both

E_1

and

E_2

must be above the exact energy

E_0 = -2.5

eV. The key to comparing

E_1

and

E_2

lies in how well each trial function captures the essential physics of a bound state in a finite well. The exact solution has specific characteristics: it oscillates inside the well and decays exponentially outside the well, with the wavefunction and its derivative being continuous at the boundaries. The correct answer is C because

\psi_2(x) = Be^{-\beta|x|}

naturally incorporates the exponential decay that must occur outside the well for a bound state. This function is smooth everywhere and automatically satisfies the boundary conditions. In contrast,

\psi_1(x) = A\cos(\pi x/a)

only exists inside the well and completely ignores the quantum mechanical tunneling into the classically forbidden region outside. Option A incorrectly suggests

E_1 < E_0

, violating the variational principle. Option B also violates the variational principle with

E_0 < E_1

, and incorrectly assumes lower kinetic energy for the exponential (the smoothness helps, but this isn't the main factor). Option D wrongly claims the energies are equal and misses that symmetry and normalization don't determine variational accuracy. Study tip: For finite well problems, always consider whether your trial function captures both the oscillatory behavior inside and the exponential decay outside the well.

Question 15

Consider applying the variational principle to estimate the binding energy of a hydrogen atom using the trial wavefunction $\psi_t(\mathbf{r}) = Ne^{-\alpha(x^2 + y^2 + z^2)}$ where $\alpha$ is a variational parameter. How does the variational energy $E_{var}$ obtained from this Gaussian trial function compare to the exact ground state energy $E_{exact} = -13.6 \text{ eV}$ ?

$E_{var} < E_{exact}$ because the Gaussian function is more localized than the exponential, providing better electron-nuclear binding
$E_{var} = E_{exact}$ because both trial and exact functions are spherically symmetric and normalized
$E_{var} > E_{exact}$ with the difference arising primarily from the incorrect asymptotic behavior of the Gaussian function (correct answer)
$E_{var}$ approaches $E_{exact}$ in the limit $\alpha \rightarrow \infty$ due to maximum localization near the nucleus
$E_{var}$ cannot be compared to $E_{exact}$ without specifying the normalization constant $N$

Explanation: The variational principle is a powerful quantum mechanical method that provides an upper bound to the true ground state energy - no trial wavefunction can give you an energy lower than the exact ground state energy. When you apply this Gaussian trial function to hydrogen, the variational energy will indeed be higher than the exact energy of -13.6 eV. The fundamental issue lies in how the Gaussian function behaves at large distances from the nucleus. The exact hydrogen wavefunction has an exponential decay (

e^{-r}

), which falls off relatively slowly. Your Gaussian trial function decays as

e^{-\alpha r^2}

, which drops to zero much more rapidly at large distances. This creates two competing effects: near the nucleus, the Gaussian can be optimized to capture some binding, but it fails to properly describe the electron's probability distribution in the important intermediate and long-range regions. Looking at the wrong answers: (A) incorrectly suggests that more localization automatically means better binding energy - this ignores the kinetic energy penalty and incorrect wavefunction shape. (B) is wrong because symmetry and normalization alone don't determine energy; the specific functional form matters critically. (D) is false because as

\alpha \rightarrow \infty

, you're forcing the electron closer to the nucleus, which actually increases kinetic energy dramatically and worsens the variational estimate. Remember: when evaluating trial wavefunctions, always consider both the kinetic energy contribution and how well the function captures the correct physical behavior, especially the asymptotic form at large distances.

Question 16

For a variational treatment of the hydrogen molecule ion H $_2^+$ at the equilibrium bond distance, a student obtains energies $E_g = -16.3 \text{ eV}$ for the bonding orbital and $E_u = -9.1 \text{ eV}$ for the antibonding orbital using the LCAO method. Given that the separated atom limit gives $E_{\text{separated}} = -13.6 \text{ eV}$ , what can be concluded about these results?

Both results are unphysical because they violate the variational principle by being lower than the separated atom energy
The bonding energy is reasonable, but the antibonding energy should be higher than the separated atom limit (correct answer)
Both energies are correct because molecular orbitals always have energies bracketing the atomic orbital energies
The antibonding energy is too low due to insufficient basis set size in the LCAO expansion
The results indicate that the bond distance used was not the true equilibrium distance for H $_2^+$

Explanation: When analyzing molecular orbital energies from variational calculations, you need to understand how the variational principle constrains your results and what physically reasonable outcomes should look like. The variational principle states that any trial wavefunction will give an energy that's either equal to or higher than the true ground state energy. For H₂⁺, the separated atom limit (

-13.6

eV) represents the energy when the electron is localized on one hydrogen atom at infinite separation. When atoms approach to form bonds, the bonding orbital should stabilize (become more negative) relative to this separated limit, while the antibonding orbital should destabilize (become less negative, closer to zero). The bonding energy of

-16.3

eV is reasonable because it's more stable than the separated atom limit, which is exactly what you expect from orbital overlap and electron delocalization in the bonding region. However, the antibonding energy of

-9.1

eV is problematic—it's significantly more stable than the separated atom limit when it should be less stable due to the node between nuclei that destabilizes the electron. Choice A is wrong because the bonding orbital can legitimately be more stable than separated atoms. Choice C incorrectly suggests both energies are always acceptable when bracketing atomic energies. Choice D misidentifies the source of error—the issue isn't basis set size but rather the fundamental expectation that antibonding orbitals should be destabilized relative to separated atoms. Remember: bonding orbitals stabilize below the separated atom limit, while antibonding orbitals should destabilize above it. This energy splitting pattern is fundamental to molecular orbital theory.

Question 17

In a variational study of molecular H $_2^+$ , a student uses the trial wavefunction $\psi_t = c_A \phi_A + c_B \phi_B$ where $\phi_A$ and $\phi_B$ are hydrogen 1s orbitals centered on nuclei A and B, respectively. The overlap integral $S = \langle\phi_A|\phi_B\rangle = 0.6$ and the resonance integral $\beta = \langle\phi_A|\hat{H}|\phi_B\rangle = -1.8 \text{ eV}$ . If $\langle\phi_A|\hat{H}|\phi_A\rangle = \langle\phi_B|\hat{H}|\phi_B\rangle = -13.6 \text{ eV}$ , what is the energy of the bonding molecular orbital?

$\frac{-13.6 - 1.8}{1 + 0.6} = -9.6 \text{ eV}$ (correct answer)
$\frac{-13.6 + 1.8}{1 - 0.6} = -29.5 \text{ eV}$
$-13.6 - 1.8 = -15.4 \text{ eV}$
$\frac{-13.6 - 1.8}{1 - 0.6} = -38.5 \text{ eV}$
$-13.6 + \frac{1.8}{0.6} = -10.6 \text{ eV}$

Explanation: When you encounter molecular orbital problems using the linear combination of atomic orbitals (LCAO) method, you're dealing with the variational principle applied to molecular systems. The key is recognizing that you need to solve a secular equation to find the energy eigenvalues. For the trial wavefunction

\psi_t = c_A \phi_A + c_B \phi_B

, the secular equation becomes:

\frac{H_{AA} - E}{H_{AB} - ES} = \frac{H_{AB} - ES}{H_{BB} - E}

Since the system has symmetry (

H_{AA} = H_{BB} = -13.6

eV), this simplifies to:

(-13.6 - E)^2 = (-1.8 - 0.6E)^2

Taking the square root gives two solutions:

E = \frac{-13.6 \pm 1.8}{1 \pm 0.6}

The bonding orbital corresponds to the lower energy solution (negative sign):

E = \frac{-13.6 - 1.8}{1 + 0.6} = \frac{-15.4}{1.6} = -9.6

eV. Answer A correctly applies this formula. Answer B uses the wrong sign combination (

+1.8

in numerator,

-0.6

in denominator), giving the antibonding energy with incorrect overlap treatment. Answer C ignores the overlap integral entirely, simply adding the coulomb and resonance integrals. Answer D uses the correct numerator but wrong denominator sign (

1-0.6

instead of

1+0.6

). Study tip: Remember that bonding orbitals always have the form

\frac{\alpha + \beta}{1 + S}

while antibonding orbitals use

\frac{\alpha - \beta}{1 - S}

, where

\alpha

is the coulomb integral and

\beta

is the (typically negative) resonance integral.

Question 18

A trial wavefunction $\psi_t(x) = N(x - a)(b - x)$ for $a \leq x \leq b$ and zero elsewhere is used to approximate the ground state of a particle in an infinite square well from $x = 0$ to $x = L$ . If $a = 0.1L$ and $b = 0.9L$ , which statement best describes the variational energy compared to the exact ground state energy?

The variational energy will be significantly lower than the exact energy because the trial function has nodes at the boundaries
The variational energy will be approximately equal to the exact energy because both functions are symmetric about $x = L/2$
The variational energy will be higher than the exact energy, with the difference primarily due to the trial function being zero near the boundaries (correct answer)
The variational energy will be lower than the exact energy because the trial function has fewer oscillations than the exact wavefunction
The variational energy cannot be determined without knowing the normalization constant $N$

Explanation: When you encounter variational method problems, remember that the variational principle guarantees that any trial wavefunction will yield an energy greater than or equal to the true ground state energy. The key is analyzing how the trial function differs from the exact solution and understanding the physical consequences. The exact ground state wavefunction for a particle in a box from 0 to L is

\psi_0(x) = \sqrt{2/L}\sin(\pi x/L)

, which has maximum amplitude at the center and goes to zero only at the boundaries. Your trial function

\psi_t(x) = N(x-0.1L)(0.9L-x)

is also symmetric and zero at its boundaries, but it's zero over the ranges 0 to 0.1L and 0.9L to L where the exact wavefunction has significant amplitude. This "squeezing" of the wavefunction away from regions where it should have non-zero values forces higher curvature in the allowed region, which increases the kinetic energy. The variational energy will definitely be higher than the exact energy, with the elevation primarily caused by excluding the particle from 20% of the box near the walls. Answer A is wrong because having nodes at boundaries is required, not problematic. Answer B incorrectly assumes symmetry alone ensures accuracy - the shape matters crucially. Answer D violates the variational principle by claiming the trial energy could be lower than the true ground state. Study tip: In variational problems, always ask whether the trial function artificially restricts the particle's motion compared to the exact solution. Confinement typically raises kinetic energy and thus the total energy.

Question 19

A student applies the variational principle to a particle in a box using the trial function $\psi_t(x) = Nx(L-x)$ for $0 \leq x \leq L$ . After calculating the variational energy, they find $E_t = \frac{5\hbar^2\pi^2}{2mL^2}$ . The exact ground state energy is $E_0 = \frac{\hbar^2\pi^2}{2mL^2}$ . What is the primary source of error in this calculation?

The trial function violates boundary conditions at $x = 0$ and $x = L$ , making the calculation invalid
The normalization constant $N$ was calculated incorrectly, leading to an overestimate of the kinetic energy
The trial function lacks the proper oscillatory character of the true ground state wavefunction (correct answer)
The variational principle has been misapplied since $E_t$ should equal $E_0$ for any valid trial function

Explanation: The trial function

\psi_t(x) = Nx(L-x)

satisfies boundary conditions (choice A is wrong) and can be properly normalized. The key issue is that this function is purely parabolic and lacks the oscillatory character of the true ground state

\sin(\pi x/L)

. This leads to a poor approximation of the kinetic energy. Choice B is incorrect - normalization errors don't systematically overestimate kinetic energy. Choice D misunderstands the variational principle -

E_t \geq E_0

always holds.

Question 20

Consider applying the linear variational method to construct molecular orbitals for $\text{H}_2^+$ using the basis $\{\phi_A, \phi_B\}$ where $\phi_A$ and $\phi_B$ are 1s orbitals centered on nuclei A and B. The secular determinant becomes $\begin{vmatrix} H_{AA} - E & H_{AB} - ES_{AB} \\ H_{BA} - ES_{BA} & H_{BB} - E \end{vmatrix} = 0$ . If the bond length increases significantly, which matrix element changes have the greatest impact on the resulting molecular orbital energies?

The diagonal elements $H_{AA}$ and $H_{BB}$ decrease substantially, leading to higher molecular orbital energies overall
The overlap integral $S_{AB}$ increases exponentially, causing numerical instabilities in the secular equation solution
All matrix elements scale proportionally with bond length, maintaining constant relative energy differences between orbitals
The off-diagonal elements $H_{AB}$ and $S_{AB}$ approach zero, reducing the bonding-antibonding energy splitting (correct answer)

Explanation: When you encounter linear variational method problems for molecular orbitals, focus on how atomic orbital overlap changes with internuclear distance. The key insight is understanding which matrix elements are most sensitive to bond length changes. As the bond length in H₂⁺ increases significantly, the atomic orbitals on centers A and B become increasingly isolated from each other. The overlap integral

S_{AB}

decreases exponentially with distance because it depends on the spatial overlap between the 1s orbitals. Similarly, the off-diagonal Hamiltonian elements

H_{AB}

also decrease toward zero since they represent the interaction energy between orbitals on different centers. This dramatic reduction in

H_{AB}

and

S_{AB}

causes the bonding and antibonding molecular orbital energies to converge toward the isolated atomic orbital energy. The energy splitting between bonding and antibonding orbitals, which creates the chemical bond, becomes negligible. Answer D correctly identifies this fundamental behavior. Answer A is wrong because diagonal elements

H_{AA}

and

H_{BB}

represent energies of isolated atomic orbitals, which don't change significantly with bond length. Answer B incorrectly states that

S_{AB}

increases—it actually decreases exponentially. Answer C is false because matrix elements don't scale proportionally; off-diagonal elements are much more sensitive to distance than diagonal elements. Remember this pattern: in molecular orbital theory, bonding strength and orbital energy splitting are controlled by overlap and interaction between atomic orbitals, both of which decay exponentially with distance.