Physical Chemistry 2 Quiz: Orthonormality And Inner Products

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Orthonormality And Inner ProductsQuestion 1 of 19

For a two-level quantum system with basis states $|0\rangle$ and $|1\rangle$ , the density matrix $\hat{\rho} = 0.6|0\rangle\langle 0| + 0.4|1\rangle\langle 1|$ represents a mixed state. What is $\text{Tr}(\hat{\rho}^2)$ ?

\text{Tr}(\hat{\rho}^2) = 1.0

\text{Tr}(\hat{\rho}^2) = 0.52

\text{Tr}(\hat{\rho}^2) = 0.48

\text{Tr}(\hat{\rho}^2) = 0.24

\text{Tr}(\hat{\rho}^2) = 0.6

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Physical Chemistry 2 Quiz

Physical Chemistry 2 Quiz: Orthonormality And Inner Products

Practice Orthonormality And Inner Products 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 Orthonormality And Inner Products, 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.

All questions

Question 1

$\text{Tr}(\hat{\rho}^2) = 1.0$
$\text{Tr}(\hat{\rho}^2) = 0.52$ (correct answer)
$\text{Tr}(\hat{\rho}^2) = 0.48$
$\text{Tr}(\hat{\rho}^2) = 0.24$
$\text{Tr}(\hat{\rho}^2) = 0.6$

Explanation: When you encounter density matrices and trace calculations in quantum mechanics, you're dealing with fundamental tools for describing quantum states and measuring their "purity." The trace of

\hat{\rho}^2

tells you whether a state is pure (

\text{Tr}(\hat{\rho}^2) = 1

) or mixed (

\text{Tr}(\hat{\rho}^2) < 1

). To find

\text{Tr}(\hat{\rho}^2)

, you first need to calculate

\hat{\rho}^2

. Since

\hat{\rho} = 0.6|0\rangle\langle 0| + 0.4|1\rangle\langle 1|

, you multiply this by itself:

\hat{\rho}^2 = (0.6|0\rangle\langle 0| + 0.4|1\rangle\langle 1|)^2

Using the orthogonality of basis states (

\langle 0|1\rangle = 0

), the cross terms vanish, giving:

\hat{\rho}^2 = (0.6)^2|0\rangle\langle 0| + (0.4)^2|1\rangle\langle 1| = 0.36|0\rangle\langle 0| + 0.16|1\rangle\langle 1|

The trace is the sum of diagonal elements:

\text{Tr}(\hat{\rho}^2) = 0.36 + 0.16 = 0.52

. Choice A (1.0) would be correct for a pure state, but this mixed state has lower purity. Choice C (0.48) likely comes from incorrectly calculating

2 \times 0.6 \times 0.4

, confusing this with a cross-term calculation. Choice D (0.24) results from multiplying the probabilities together rather than squaring them individually. Remember: for any mixed state described by classical probabilities

p_i

, the purity is

\sum p_i^2

, which always lies between 0 and 1, reaching 1 only for pure states.

Question 2

A set of functions $\{f_n(x)\}$ satisfies $\int_{-1}^{1} f_m(x) f_n(x) dx = \frac{2}{2n+1} \delta_{mn}$ . To construct an orthonormal set $\{g_n(x)\}$ from these functions, what transformation should be applied?

$g_n(x) = f_n(x) \sqrt{\frac{2n+1}{2}}$ (correct answer)
$g_n(x) = f_n(x) \sqrt{\frac{2}{2n+1}}$
$g_n(x) = f_n(x) \sqrt{2n+1}$
$g_n(x) = f_n(x) \frac{1}{\sqrt{2n+1}}$
$g_n(x) = f_n(x) \sqrt{\frac{2n+1}{4}}$

Explanation: When you encounter orthogonality and normalization problems, you're dealing with the fundamental requirement that orthonormal functions must satisfy two conditions: orthogonality (different functions integrate to zero) and normalization (each function integrated with itself equals 1). The given functions already satisfy orthogonality since

\int_{-1}^{1} f_m(x) f_n(x) dx = 0

when

m \neq n

(due to the Kronecker delta

\delta_{mn}

). However, they're not normalized because when

m = n

, the integral equals

\frac{2}{2n+1}

instead of 1. To find the correct transformation, you need the new functions

g_n(x)

to satisfy

\int_{-1}^{1} g_n(x) g_n(x) dx = 1

. If

g_n(x) = C_n f_n(x)

where

C_n

is a normalization constant, then:

\int_{-1}^{1} [C_n f_n(x)]^2 dx = C_n^2 \int_{-1}^{1} f_n(x)^2 dx = C_n^2 \cdot \frac{2}{2n+1} = 1

Solving for

C_n

C_n = \sqrt{\frac{2n+1}{2}}

, making answer A correct. Answer B gives

C_n = \sqrt{\frac{2}{2n+1}}

, which would make the normalization integral equal

\frac{2}{2n+1}

(unchanged). Answer C omits the factor of 2 in the denominator, while answer D would give a normalization integral of

\frac{2}{2n+1} \cdot \frac{1}{2n+1}

, making it too small. Study tip: For normalization problems, always set up the equation

C^2 \times \text{(original norm)} = 1

and solve for the constant

C

Question 3

Two wavefunctions $\psi_A = \sin(\pi x/L)$ and $\psi_B = \sin(2\pi x/L)$ are defined on the interval $[0,L]$ . If a third function $\psi_C = \alpha \psi_A + \beta \psi_B$ is to be orthogonal to $\psi_A$ , what constraint must be satisfied by $\alpha$ and $\beta$ ?

$\alpha = 0$ (any value of $\beta$ is acceptable)
$\alpha \langle\psi_A|\psi_A\rangle + \beta \langle\psi_A|\psi_B\rangle = 0$ (correct answer)
$\alpha + \beta = 0$
$\alpha^2 + \beta^2 = 1$
$\alpha \langle\psi_A|\psi_A\rangle = \beta \langle\psi_B|\psi_B\rangle$

Explanation: When you encounter orthogonality problems in quantum mechanics, you're dealing with the fundamental requirement that the inner product (overlap integral) between orthogonal functions must equal zero. For two functions to be orthogonal on an interval, we need

\langle\psi_C|\psi_A\rangle = 0

. Let's work through this systematically. The inner product of

\psi_C

with

\psi_A

is:

\langle\psi_C|\psi_A\rangle = \langle\alpha\psi_A + \beta\psi_B|\psi_A\rangle

Using the linearity of the inner product, this expands to:

\alpha\langle\psi_A|\psi_A\rangle + \beta\langle\psi_B|\psi_A\rangle

For orthogonality, this entire expression must equal zero, giving us the constraint in answer B. Answer A (

\alpha = 0

) is incorrect because it's unnecessarily restrictive. Even with

\alpha \neq 0

, orthogonality is possible as long as the full constraint in B is satisfied. Answer C (

\alpha + \beta = 0

) represents a common algebraic misconception. This would only be correct if both inner products

\langle\psi_A|\psi_A\rangle

and

\langle\psi_A|\psi_B\rangle

were equal to 1, which isn't generally true. Answer D (

\alpha^2 + \beta^2 = 1

) confuses orthogonality with normalization. This constraint ensures

\psi_C

has unit norm, but says nothing about its orthogonality to

\psi_A

. Study tip: Always set up the orthogonality condition

\langle f|g\rangle = 0

explicitly and use linearity to expand it. Don't assume shortcuts based on the coefficients alone—the actual overlap integrals determine the constraint.

Question 4

A wavefunction $\Psi(x,t) = c_1 \psi_1(x) e^{-iE_1t/\hbar} + c_2 \psi_2(x) e^{-iE_2t/\hbar}$ is constructed from two orthonormal energy eigenstates. At $t = 0$ , measurements show $|c_1|^2 = 0.7$ and $|c_2|^2 = 0.3$ . What is $\langle\Psi(x,t)|\Psi(x,t)\rangle$ at time $t > 0$ ?

$\langle\Psi(x,t)|\Psi(x,t)\rangle = 1.0$ (correct answer)
$\langle\Psi(x,t)|\Psi(x,t)\rangle = 0.7 + 0.3 \cos[(E_2-E_1)t/\hbar]$
$\langle\Psi(x,t)|\Psi(x,t)\rangle = 0.7 e^{-2iE_1t/\hbar} + 0.3 e^{-2iE_2t/\hbar}$
$\langle\Psi(x,t)|\Psi(x,t)\rangle = 1.0 + 2\sqrt{0.21}\cos[(E_2-E_1)t/\hbar]$
$\langle\Psi(x,t)|\Psi(x,t)\rangle = 0.7 + 0.3 + 2c_1^*c_2 e^{i(E_1-E_2)t/\hbar}$

Explanation: When you encounter a linear combination of energy eigenstates in quantum mechanics, remember that the inner product

\langle\Psi|\Psi\rangle

represents the normalization condition - the total probability of finding the particle somewhere in space, which must always equal 1. To find

\langle\Psi(x,t)|\Psi(x,t)\rangle

, you calculate:

\langle\Psi|\Psi\rangle = \langle c_1 \psi_1 e^{-iE_1t/\hbar} + c_2 \psi_2 e^{-iE_2t/\hbar} | c_1 \psi_1 e^{-iE_1t/\hbar} + c_2 \psi_2 e^{-iE_2t/\hbar}\rangle

Expanding this gives four terms:

|c_1|^2\langle\psi_1|\psi_1\rangle + |c_2|^2\langle\psi_2|\psi_2\rangle + c_1^*c_2 e^{i(E_1-E_2)t/\hbar}\langle\psi_1|\psi_2\rangle + c_2^*c_1 e^{i(E_2-E_1)t/\hbar}\langle\psi_2|\psi_1\rangle

Since

\psi_1

and

\psi_2

are orthonormal,

\langle\psi_1|\psi_1\rangle = \langle\psi_2|\psi_2\rangle = 1

and

\langle\psi_1|\psi_2\rangle = \langle\psi_2|\psi_1\rangle = 0

. The cross terms vanish, leaving only

|c_1|^2 + |c_2|^2 = 0.7 + 0.3 = 1.0

. Answer A is correct. Answer B incorrectly includes a cosine term that would arise from interference between states, but this appears in expectation values of observables, not in the normalization. Answer C shows complex exponentials that don't belong in a probability calculation - the time-dependent phases cancel when you take the modulus squared. Answer D adds an extra interference term that violates orthogonality. Remember: wavefunctions must stay normalized regardless of time evolution. The coefficients

|c_i|^2

represent probabilities that don't change with time in energy eigenstates.

Question 5

Three functions $f_1(x) = 1$ , $f_2(x) = x$ , and $f_3(x) = x^2$ are defined on the interval $[-1,1]$ . Using the Gram-Schmidt process to construct an orthogonal set, what is the second orthogonal function $g_2(x)$ ?

$g_2(x) = x$
$g_2(x) = x - \frac{1}{2}$
$g_2(x) = x - 0$ (correct answer)
$g_2(x) = x - \frac{\langle f_2|g_1\rangle}{\langle g_1|g_1\rangle} g_1$
$g_2(x) = x - \frac{\int_{-1}^1 x dx}{\int_{-1}^1 1 dx}$

Explanation: The Gram-Schmidt process systematically converts a set of linearly independent functions into orthogonal ones by removing projections of each function onto previously constructed orthogonal functions. When you encounter orthogonalization problems, always start with the first function unchanged and work sequentially. The first orthogonal function is simply

g_1(x) = f_1(x) = 1

. To find the second orthogonal function, you apply the Gram-Schmidt formula:

g_2(x) = f_2(x) - \frac{\langle f_2|g_1\rangle}{\langle g_1|g_1\rangle} g_1(x)

. Now calculate the required inner products on the interval

[-1,1]

$\langle f_2|g_1\rangle = \int_{-1}^{1} x \cdot 1 \, dx = \left[\frac{x^2}{2}\right]_{-1}^{1} = \frac{1}{2} - \frac{1}{2} = 0$
$\langle g_1|g_1\rangle = \int_{-1}^{1} 1^2 \, dx = 2$

Since

\langle f_2|g_1\rangle = 0

, the projection term becomes zero, giving

g_2(x) = x - 0 = x

. Answer C is correct because the calculation yields

g_2(x) = x - 0

. Answer A gives the same final result but doesn't show the intermediate step. Answer B incorrectly assumes a non-zero projection, possibly from using the wrong interval or making a calculation error. Answer D shows the correct formula but isn't the simplified final answer. Study tip: When functions have odd symmetry (like

x

) and you're integrating over symmetric intervals, their inner products with even functions (like constants) will always be zero. This makes Gram-Schmidt calculations much simpler—look for this pattern to save time.

Question 6

For hydrogen-like wavefunctions, $\psi_{200}$ and $\psi_{210}$ refer to the $n=2, l=0, m=0$ and $n=2, l=1, m=0$ states respectively. Without explicit calculation, what can be concluded about $\langle\psi_{200}|\psi_{210}\rangle$ ?

$\langle\psi_{200}|\psi_{210}\rangle = 0$ because they have different $l$ values (correct answer)
$\langle\psi_{200}|\psi_{210}\rangle = 0$ because they have different $m$ values
$\langle\psi_{200}|\psi_{210}\rangle \neq 0$ because they have the same $n$ value
$\langle\psi_{200}|\psi_{210}\rangle = 0$ because they are degenerate energy eigenstates
The value cannot be determined without explicit integration

Explanation: When you encounter questions about hydrogen-like wavefunctions and their overlap integrals, you need to think about orthogonality conditions—the fundamental principle that eigenfunctions of Hermitian operators are orthogonal when they correspond to different eigenvalues. Hydrogen-like wavefunctions can be written as products:

\psi_{nlm} = R_{nl}(r)Y_l^m(\theta,\phi)

, where

R_{nl}

is the radial part and

Y_l^m

are spherical harmonics. The overlap integral

\langle\psi_{200}|\psi_{210}\rangle

becomes a product of radial and angular integrals. The key insight is that both states have the same

n

value (so their radial parts aren't orthogonal), but they have different

l

values. The angular momentum operator

\hat{L}^2

has eigenvalues

\hbar^2 l(l+1)

, so

\psi_{200}

and

\psi_{210}

are eigenfunctions with different eigenvalues:

0

versus

2\hbar^2

. Since

\hat{L}^2

is Hermitian, its eigenfunctions with different eigenvalues must be orthogonal, making the overlap integral zero. Choice A correctly identifies this orthogonality due to different

l

values. Choice B is wrong because both states have

m=0

—they don't differ in

m

. Choice C incorrectly suggests that having the same

n

value prevents orthogonality, but different

l

values override this. Choice D mentions degeneracy but misses the point—degenerate states aren't automatically orthogonal to each other. Study tip: Remember that hydrogen wavefunctions are orthogonal whenever any quantum number differs, but the physical reason matters. Different

l

means different angular momentum eigenvalues, which guarantees orthogonality through the properties of Hermitian operators.

Question 7

A complete orthonormal set $\{|n\rangle\}$ satisfies the closure relation $\sum_n |n\rangle\langle n| = \hat{I}$ . If an arbitrary state $|\psi\rangle = \sum_n c_n |n\rangle$ , what does the condition $\sum_n |c_n|^2 = 1$ represent?

The closure relation for the basis states
The normalization condition for $|\psi\rangle$ (correct answer)
The orthogonality condition between basis states
The completeness condition for the basis set
The unitarity condition for time evolution

Explanation: When you encounter quantum mechanical state vectors and coefficients, you're dealing with fundamental postulates of quantum mechanics that govern how we describe and normalize quantum states. The condition

\sum_n |c_n|^2 = 1

is the normalization requirement for the quantum state

|\psi\rangle

. In quantum mechanics, any physical state must have unit probability when we consider all possible measurement outcomes. Since

|c_n|^2

represents the probability of finding the system in basis state

|n\rangle

upon measurement, the sum of all these probabilities must equal 1. This ensures the state vector has unit length in the Hilbert space, making

\langle\psi|\psi\rangle = 1

. Let's examine why the other options are incorrect. Option A confuses this with the closure relation

\sum_n |n\rangle\langle n| = \hat{I}

, which expresses completeness of the basis set, not normalization of coefficients. Option C misidentifies this as orthogonality, but orthogonality between basis states is expressed as

\langle m|n\rangle = \delta_{mn}

, involving inner products of basis vectors, not coefficient magnitudes. Option D incorrectly associates this with completeness, which again refers to the closure relation ensuring the basis spans the entire Hilbert space. Remember this pattern: whenever you see

\sum |c_n|^2 = 1

in quantum mechanics, think "normalization" - it's ensuring your state vector represents a valid quantum state with total probability 1. This is distinct from basis properties like orthogonality and completeness.

Question 8

The overlap matrix $S_{ij} = \langle\phi_i|\phi_j\rangle$ for three non-orthogonal basis functions has eigenvalues $\lambda_1 = 2$ , $\lambda_2 = 1$ , and $\lambda_3 = 0.5$ . What does the eigenvalue $\lambda_3 = 0.5$ indicate about the basis set?

The basis functions are linearly independent and well-conditioned
The basis functions are nearly linearly dependent (correct answer)
The basis functions are exactly orthogonal
The basis functions are normalized
The basis set is complete

Explanation: When you encounter overlap matrix eigenvalues in quantum chemistry, you're examining how well-behaved your basis set is. The overlap matrix

S_{ij} = \langle\phi_i|\phi_j\rangle

measures how much basis functions "overlap" with each other, and its eigenvalues reveal the linear independence of your basis functions. For a well-conditioned basis set, all eigenvalues should be close to 1. When eigenvalues approach zero, it signals that your basis functions are becoming nearly linearly dependent - meaning one function can almost be written as a combination of the others. The eigenvalue

\lambda_3 = 0.5

is significantly smaller than the ideal value of 1, indicating the basis functions are nearly linearly dependent and the basis set is poorly conditioned. Looking at the wrong answers: (A) is incorrect because a well-conditioned basis would have eigenvalues near 1, not 0.5. (C) is wrong because orthogonal functions would give an overlap matrix that's the identity matrix (eigenvalues all equal to 1), not the mixed values shown here. (D) is incorrect because normalization relates to diagonal elements

S_{ii} = 1

, not the eigenvalue spectrum of the entire matrix. The correct answer is (B) - the small eigenvalue indicates near linear dependence. Study tip: Remember that overlap matrix eigenvalues near zero are red flags for linear dependence problems. In computational chemistry, eigenvalues below 10⁻⁶ typically indicate serious linear dependence issues that can cause numerical instability in calculations.

Question 9

The Hermite polynomials $H_n(x)$ satisfy the orthogonality relation $\int_{-\infty}^{\infty} H_m(x) H_n(x) e^{-x^2} dx = 2^n n! \sqrt{\pi} \delta_{mn}$ . If $\psi_n(x) = N_n H_n(x) e^{-x^2/2}$ where $N_n$ is chosen for normalization, what is $N_n$ ?

$N_n = \frac{1}{\sqrt{2^n n! \sqrt{\pi}}}$
$N_n = \frac{1}{(2^n n! \sqrt{\pi})^{1/4}}$
$N_n = \frac{1}{\sqrt{2^{n-1} n! \sqrt{\pi}}}$
$N_n = \frac{1}{\pi^{1/4} \sqrt{2^n n!}}$ (correct answer)
$N_n = \frac{1}{2^{n/2} \sqrt{n! \pi^{1/2}}}$

Explanation: When you encounter normalization problems involving wavefunctions, you need to ensure that

\int_{-\infty}^{\infty} |\psi_n(x)|^2 dx = 1

. This requires using the given orthogonality relation strategically. For

\psi_n(x) = N_n H_n(x) e^{-x^2/2}

, the normalization condition becomes:

\int_{-\infty}^{\infty} |N_n H_n(x) e^{-x^2/2}|^2 dx = 1

This expands to:

N_n^2 \int_{-\infty}^{\infty} H_n^2(x) e^{-x^2} dx = 1

The key insight is recognizing that

e^{-x^2/2} \cdot e^{-x^2/2} = e^{-x^2}

, which matches the weight function in the given orthogonality relation. Setting

m = n

in the orthogonality relation gives us

\int_{-\infty}^{\infty} H_n^2(x) e^{-x^2} dx = 2^n n! \sqrt{\pi}

. Therefore:

N_n^2 \cdot 2^n n! \sqrt{\pi} = 1

, so

N_n = \frac{1}{\sqrt{2^n n! \sqrt{\pi}}}

. Wait—this matches option A, but let's check the

\pi

factor more carefully. Since

\sqrt{\pi}

appears under the square root, we get

N_n = \frac{1}{\sqrt{2^n n!} \cdot \pi^{1/4}}

, which equals option D. Option A incorrectly keeps

\sqrt{\pi}

inside the square root in the denominator. Option B uses a fourth root instead of square root. Option C has an incorrect power

2^{n-1}

instead of

2^n

. Strategy tip: In normalization problems, always square the wavefunction first, then match the resulting integral to any given orthogonality relations. Watch how exponential terms combine—they often create the exact weight functions you need.

Question 10

A linear operator $\hat{A}$ acts on an orthonormal basis $\{|n\rangle\}$ according to $\hat{A}|1\rangle = 2|1\rangle + |2\rangle$ and $\hat{A}|2\rangle = |1\rangle + 3|2\rangle$ . What is $\langle 1|\hat{A}^\dagger|2\rangle$ ?

$\langle 1|\hat{A}^\dagger|2\rangle = 1$ (correct answer)
$\langle 1|\hat{A}^\dagger|2\rangle = 3$
$\langle 1|\hat{A}^\dagger|2\rangle = 2$
$\langle 1|\hat{A}^\dagger|2\rangle = 5$
$\langle 1|\hat{A}^\dagger|2\rangle = 0$

Explanation: When you encounter questions about Hermitian conjugates (adjoints) of linear operators, remember that the adjoint

\hat{A}^\dagger

satisfies the fundamental relation:

\langle \psi | \hat{A} \phi \rangle = \langle \hat{A}^\dagger \psi | \phi \rangle

for any states

|\psi\rangle

and

|\phi\rangle

. To find

\langle 1|\hat{A}^\dagger|2\rangle

, we use this property by rearranging it:

\langle 1|\hat{A}^\dagger|2\rangle = \langle \hat{A}|1\rangle|2\rangle^*

. Since we're working with real coefficients here, the complex conjugate doesn't change anything. From the given information,

\hat{A}|1\rangle = 2|1\rangle + |2\rangle

. Therefore:

\langle 1|\hat{A}^\dagger|2\rangle = \langle 2|1\rangle + |2\rangle\rangle = \langle 2|2|1\rangle + 1|2\rangle\rangle = 2\langle 2|1\rangle + \langle 2|2\rangle

Since

\{|1\rangle, |2\rangle\}

forms an orthonormal basis,

\langle 2|1\rangle = 0

and

\langle 2|2\rangle = 1

. Thus

\langle 1|\hat{A}^\dagger|2\rangle = 1

, which is answer A. The wrong answers come from misapplying the adjoint property. Answer B (3) would result from incorrectly using

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

. Answer C (2) comes from confusing which coefficient to extract. Answer D (5) appears to add coefficients inappropriately, perhaps from

2 + 3

. Study tip: Always write out the adjoint property explicitly:

\langle i|\hat{A}^\dagger|j\rangle = \langle j|\hat{A}|i\rangle^*

. This systematic approach prevents mixing up which state goes where and ensures you apply orthonormality correctly.

Question 11

Two unnormalized wavefunctions $\phi_1 = x e^{-x^2/2}$ and $\phi_2 = (2x^2 - 1)e^{-x^2/2}$ are defined over $(-\infty, \infty)$ . To determine if they can be made orthogonal by adjusting a linear combination, what integral must be evaluated?

$\int_{-\infty}^{\infty} x(2x^2 - 1) e^{-x^2} dx$
$\int_{-\infty}^{\infty} x^2(2x^2 - 1)^2 e^{-x^2} dx$
$\int_{-\infty}^{\infty} [x e^{-x^2/2}][(2x^2 - 1)e^{-x^2/2}] dx$ (correct answer)
$\int_{-\infty}^{\infty} |x|^2 e^{-x^2} dx$
$\int_{-\infty}^{\infty} x(2x^2 - 1) dx$

Explanation: When dealing with orthogonality of wavefunctions, you need to evaluate the overlap integral between the two functions. Two functions are orthogonal if their overlap integral equals zero, which means

\langle \phi_1 | \phi_2 \rangle = \int_{-\infty}^{\infty} \phi_1^* \phi_2 \, dx = 0

. For real wavefunctions like these, the overlap integral is simply the product of the two functions integrated over their domain. To determine if

\phi_1 = x e^{-x^2/2}

and

\phi_2 = (2x^2 - 1)e^{-x^2/2}

can be made orthogonal, you must evaluate their direct overlap:

\int_{-\infty}^{\infty} [x e^{-x^2/2}][(2x^2 - 1)e^{-x^2/2}] dx

. Option A incorrectly combines only the polynomial parts while changing the exponential from

e^{-x^2/2}

e^{-x^2}

, which fundamentally alters the functions. Option B squares both functions rather than taking their product, which would give you a normalization-type integral, not an overlap integral. Option D focuses only on

\phi_1

and again uses the wrong exponential form, making it irrelevant to the orthogonality question. The correct answer is C because it preserves the exact form of both wavefunctions and evaluates their product integral—the mathematical definition of overlap. Study tip: For orthogonality problems, always set up the overlap integral

\langle \psi_1 | \psi_2 \rangle

using the exact forms given. Don't modify the exponentials or square individual functions—you're looking for the product of the two wavefunctions integrated over their domain.

Question 12

A wavefunction $\psi(x) = A \sin(\pi x/L)$ for $0 \leq x \leq L$ and $\psi(x) = 0$ elsewhere is normalized. Another function $\phi(x) = B x(L-x)$ for $0 \leq x \leq L$ and $\phi(x) = 0$ elsewhere is also normalized. To evaluate $\langle\psi|\phi\rangle$ , which integral expression is correct?

$\langle\psi|\phi\rangle = AB \int_0^L \sin(\pi x/L) \cdot x(L-x) dx$ (correct answer)
$\langle\psi|\phi\rangle = AB \int_{-\infty}^{\infty} \sin(\pi x/L) \cdot x(L-x) dx$
$\langle\psi|\phi\rangle = A^* B \int_0^L \sin(\pi x/L) \cdot x(L-x) dx$
$\langle\psi|\phi\rangle = AB \int_0^L \sin^2(\pi x/L) \cdot x(L-x) dx$
$\langle\psi|\phi\rangle = AB \int_0^L \sin(\pi x/L) \cdot [x(L-x)]^2 dx$

Explanation: When you encounter inner product calculations in quantum mechanics, you're evaluating the overlap between two wavefunctions. The inner product

\langle\psi|\phi\rangle

represents how much one wavefunction "resembles" another, which is fundamental for calculating transition probabilities and expectation values. The correct approach uses the definition of inner products:

\langle\psi|\phi\rangle = \int \psi^*(x) \phi(x) dx

. Since both functions are real (no imaginary components),

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

, so we need

\int \psi(x) \phi(x) dx

. Both functions are zero outside the interval

[0,L]

, so the integral limits are 0 to L. This gives us

AB \int_0^L \sin(\pi x/L) \cdot x(L-x) dx

, which is answer A. Answer B incorrectly uses infinite limits. While this wouldn't change the result since both functions are zero outside

[0,L]

, it's unnecessarily complex and shows poor understanding of the problem setup. Answer C writes

A^*B

instead of

AB

. This would be correct if

A

were complex, but since

\psi(x)

is purely real (sine function),

A^* = A

. Answer D incorrectly squares the sine function, giving

\sin^2(\pi x/L)

. This fundamental error suggests confusion between inner products (

\langle\psi|\phi\rangle

) and expectation values of operators. Study tip: Always write out the inner product definition first:

\langle\psi|\phi\rangle = \int \psi^*(x) \phi(x) dx

. For real wavefunctions, drop the complex conjugate. Then identify where both functions are non-zero to set your integration limits.

Question 13

A quantum state $|\psi\rangle$ can be expanded in two different orthonormal bases: $|\psi\rangle = \sum_n a_n |n\rangle = \sum_m b_m |m'\rangle$ . If the bases are related by $|m'\rangle = \sum_n U_{mn} |n\rangle$ where $U$ is unitary, what is the relationship between coefficients $a_n$ and $b_m$ ?

$b_m = \sum_n U_{mn} a_n$
$b_m = \sum_n U_{mn}^* a_n$ (correct answer)
$a_n = \sum_m U_{mn} b_m$
$b_m = \sum_n U_{nm} a_n$
$b_m = \sum_n (U^{-1})_{mn} a_n$

Explanation: When you encounter problems involving basis transformations in quantum mechanics, you're dealing with how the same quantum state can be expressed using different sets of orthonormal basis vectors. The key insight is understanding how unitary transformations preserve the physical state while changing its mathematical representation. To find the relationship between coefficients, start with the state expansions:

|\psi\rangle = \sum_n a_n |n\rangle = \sum_m b_m |m'\rangle

. Since

|m'\rangle = \sum_n U_{mn} |n\rangle

, substitute this into the second expansion:

|\psi\rangle = \sum_m b_m \sum_n U_{mn} |n\rangle = \sum_n \left(\sum_m b_m U_{mn}\right) |n\rangle

. Comparing coefficients of

|n\rangle

gives:

a_n = \sum_m b_m U_{mn}

. To solve for

b_m

, multiply both sides by

U_{km}^*

and sum over

n

. Using the unitary property

\sum_n U_{mn}^* U_{kn} = \delta_{mk}

, you get:

b_m = \sum_n U_{mn}^* a_n

. Answer B is correct. Answer A uses

U_{mn}

instead of its complex conjugate, missing that we need the inverse transformation. Answer C gives the reverse relationship (

a_n

in terms of

b_m

) rather than what's asked. Answer D uses the wrong matrix element

U_{nm}

, which would correspond to using

U^T

instead of

U^\dagger

. Remember: unitary transformations require complex conjugation when inverting the relationship. Always check whether you're using

U

U^\dagger

in basis transformation problems.

Question 14

A quantum mechanical system has two degenerate energy eigenfunctions $\phi_1$ and $\phi_2$ that are normalized but not orthogonal, with $\langle \phi_1 | \phi_2 \rangle = S$ where $0 < S < 1$ . After applying the Gram-Schmidt procedure to create orthonormal functions $\psi_1$ and $\psi_2$ , what is the coefficient of $\phi_1$ in the normalized function $\psi_2$ ?

$\frac{-S}{\sqrt{1-S^2}}$ (correct answer)
$\frac{S}{\sqrt{1-S^2}}$
$-S$
$\frac{1}{\sqrt{1-S^2}}$

Explanation: In Gram-Schmidt,

\psi_1 = \phi_1

and

\psi_2 = N(\phi_2 - S\phi_1)

where

N = 1/\sqrt{1-S^2}

is the normalization constant. Therefore,

\psi_2 = \frac{1}{\sqrt{1-S^2}}\phi_2 - \frac{S}{\sqrt{1-S^2}}\phi_1

, giving coefficient

-S/\sqrt{1-S^2}

. Choice B has wrong sign, C ignores normalization, D gives the coefficient of

\phi_2

Question 15

A linear operator $\hat{A}$ acts on an orthonormal basis set $\{|1\rangle, |2\rangle, |3\rangle\}$ according to: $\hat{A}|1\rangle = 2|1\rangle + |2\rangle$ , $\hat{A}|2\rangle = |1\rangle + 3|2\rangle$ , and $\hat{A}|3\rangle = 4|3\rangle$ . What is the matrix element $\langle 2|\hat{A}|1\rangle$ ?

$1$ (correct answer)
$2$
$3$
$0$

Explanation: To find

\langle 2|\hat{A}|1\rangle

, we apply

\hat{A}

|1\rangle

first:

\hat{A}|1\rangle = 2|1\rangle + |2\rangle

. Then

\langle 2|\hat{A}|1\rangle = \langle 2|(2|1\rangle + |2\rangle) = 2\langle 2|1\rangle + \langle 2|2\rangle = 2(0) + 1 = 1

. Choice B gives the coefficient of

|1\rangle

, C gives an eigenvalue, and D would be true only if the operator were diagonal.

Question 16

Consider the completeness relation for an orthonormal basis $\{|n\rangle\}$ : $\sum_n |n\rangle\langle n| = \hat{I}$ . If a non-normalized state $|\psi\rangle$ is expanded as $|\psi\rangle = \sum_n a_n |n\rangle$ where $\sum_n |a_n|^2 = 4$ , and a measurement of an observable $\hat{O}$ with matrix elements $O_{mn} = \langle m|\hat{O}|n\rangle$ is performed, what is the probability of obtaining eigenvalue $\lambda_k$ if $\hat{O}|k\rangle = \lambda_k |k\rangle$ ?

$\frac{|a_k|}{4}$
$|a_k|^2$
$\frac{|a_k|^2}{4}$ (correct answer)
$\frac{4}{|a_k|^2}$

Explanation: When you encounter quantum measurement problems with non-normalized states, remember that probabilities must always sum to 1, which requires proper normalization of the wavefunction. For a quantum measurement, the probability of obtaining eigenvalue

\lambda_k

depends on the overlap between your state and the corresponding eigenstate

|k\rangle

. Since

|\psi\rangle = \sum_n a_n |n\rangle

and the basis states are orthonormal, the coefficient

a_k

represents the amplitude for finding the system in state

|k\rangle

. However, your state isn't normalized—the normalization condition

\sum_n |a_n|^2 = 4

tells us the total "probability" is 4 instead of 1. To find the actual probability, you need the relative weight:

\frac{|a_k|^2}{\sum_n |a_n|^2} = \frac{|a_k|^2}{4}

. This is answer C. Let's examine why the other options fail: Option A (

\frac{|a_k|}{4}

) uses the amplitude instead of its squared magnitude, forgetting that probabilities come from

|\text{amplitude}|^2

. Option B (

|a_k|^2

) ignores the normalization issue entirely—this would only be correct if

\sum_n |a_n|^2 = 1

. Option D (

\frac{4}{|a_k|^2}

) inverts the relationship completely and would give nonsensical results. Study tip: Always check if wavefunctions are normalized in quantum measurement problems. When they're not, divide the squared amplitude by the total normalization to get the true probability. The pattern is: probability =

\frac{|\text{relevant amplitude}|^2}{\text{total normalization}}

Question 17

Three wavefunctions $\psi_A$ , $\psi_B$ , and $\psi_C$ satisfy the following inner product relationships: $\langle \psi_A | \psi_B \rangle = 0$ , $\langle \psi_A | \psi_C \rangle = 0$ , and $\langle \psi_B | \psi_C \rangle = 0.5$ . If all three functions are normalized, which statement about the set $\{\psi_A, \psi_B, \psi_C\}$ is correct?

The set forms an orthonormal basis since all functions are normalized
The set forms an orthogonal basis but requires Gram-Schmidt orthogonalization
The set is orthogonal but not orthonormal due to the non-zero overlap between $\psi_B$ and $\psi_C$
The set is neither orthogonal nor orthonormal due to $\langle \psi_B | \psi_C \rangle \neq 0$ (correct answer)

Explanation: For a set to be orthogonal, all pairs must have zero inner product. Since

\langle \psi_B | \psi_C \rangle = 0.5 \neq 0

, the set is not orthogonal and therefore cannot be orthonormal. Choice A incorrectly assumes normalization implies orthonormality, B incorrectly calls it orthogonal, and C incorrectly states it's orthogonal when it's not.

Question 18

Consider a set of functions $\{f_n(x)\}$ that are orthogonal with respect to the inner product $\langle f_m | f_n \rangle = \int_0^1 f_m^*(x) f_n(x) dx = \delta_{mn} N_n$ , where $N_n$ is the normalization constant for function $f_n$ . If a function $g(x) = \sum_{n=1}^\infty c_n f_n(x)$ is expanded in this basis, what is the correct expression for the expansion coefficient $c_k$ ?

$c_k = \langle f_k | g \rangle \cdot N_k$
$c_k = \frac{\langle g | f_k \rangle}{N_k}$
$c_k = \frac{\langle f_k | g \rangle}{N_k}$ (correct answer)
$c_k = \frac{\langle f_k | g \rangle}{\sqrt{N_k}}$

Explanation: When you encounter orthogonal function expansions in physical chemistry, you're working with a fundamental tool for solving quantum mechanics problems. The key insight is understanding how orthogonality conditions allow you to extract specific coefficients from an infinite series expansion. To find the expansion coefficient

c_k

, you need to exploit the orthogonality relationship. Start with the expansion

g(x) = \sum_{n=1}^\infty c_n f_n(x)

and take the inner product of both sides with

f_k

\langle f_k | g \rangle = \langle f_k | \sum_{n=1}^\infty c_n f_n \rangle = \sum_{n=1}^\infty c_n \langle f_k | f_n \rangle

The orthogonality condition

\langle f_k | f_n \rangle = \delta_{kn} N_n

causes all terms in the sum to vanish except when

n = k

, leaving:

\langle f_k | g \rangle = c_k N_k

Therefore:

c_k = \frac{\langle f_k | g \rangle}{N_k}

Option C is correct. Option A incorrectly multiplies by

N_k

instead of dividing, which would give coefficients that are too large by a factor of

N_k^2

. Option B uses the wrong order in the inner product—while this gives the same numerical result for real functions, it's conceptually backwards since we project

g

onto the basis function

f_k

. Option D divides by

\sqrt{N_k}

, which would only be correct if the functions were orthonormal (

N_k = 1

). Remember: when extracting coefficients from orthogonal expansions, always take the inner product with the basis function and divide by its normalization constant.

Question 19

Two wavefunctions $\psi_1(x) = \sqrt{\frac{2}{L}}\sin\left(\frac{\pi x}{L}\right)$ and $\psi_2(x) = \sqrt{\frac{2}{L}}\sin\left(\frac{2\pi x}{L}\right)$ represent energy eigenstates of a particle in a 1D box of length $L$ . When evaluating the overlap integral $\langle \psi_1 | \psi_2 \rangle$ using the identity $\sin A \sin B = \frac{1}{2}[\cos(A-B) - \cos(A+B)]$ , what is the key step that ensures orthogonality?

The normalization constants cancel out during the integration process
The cosine terms integrate to zero due to their different periodicities within the box (correct answer)
The sine functions have different amplitudes that create destructive interference
The boundary conditions at $x = 0$ and $x = L$ force the integral to vanish

Explanation: Using the trigonometric identity,

\sin\left(\frac{\pi x}{L}\right)\sin\left(\frac{2\pi x}{L}\right) = \frac{1}{2}[\cos\left(\frac{-\pi x}{L}\right) - \cos\left(\frac{3\pi x}{L}\right)]

. When integrated from 0 to

L

, both cosine terms complete integer numbers of periods and integrate to zero. Choice A is wrong as normalization doesn't affect orthogonality, C misunderstands the mechanism, and D incorrectly focuses on boundary conditions rather than the integration result.