Quantum Physics Practice

GitHub Changelog - Last modified Aug 24, 2026

Suggest edit

0 of 27 completed

Create worksheet

Title

Selected problems

No problems selected.

Difficulty

Direct

1.1Find Probability in a Subinterval

Exam I | Problem 1.1 |

A particle has wave function

\psi(x) = \frac{1}{2}

for $0 \le x \le 4$, and $\psi(x)=0$ elsewhere.

What is the probability of finding the particle in the interval $[0,2]$?

Solution

The probability is the integral of $|\psi(x)|^2$ over the interval:

P(0 \le x \le 2) = \int_0^2 \left|\frac{1}{2}\right|^2 dx

P(0 \le x \le 2) = \int_0^2 \frac{1}{4}\,dx = \frac{1}{2}

So the probability is $\frac{1}{2}$.

1.2Normalize a Uniform Wave Function

Exam I | Problem 1.2 |

Let

\psi(x) = A

for $0 \le x \le 3$, and $\psi(x)=0$ elsewhere.

Find a normalized value of $A$.

Solution

Use the normalization condition:

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

Here that becomes

\int_0^3 |A|^2\,dx = 1

$$ 3|A|^2 = 1 $$

So a normalized choice is

A = \frac{1}{\sqrt{3}}

up to an overall phase.

1.3Energy of the Second Level in an Infinite Well

Exam I | Problem 1.3 |

For a particle in an infinite square well of width $L$, what is the energy $E_2$ of the $n=2$ state?

Solution

The note gives

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

Substitute $n=2$:

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

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

1.4Identify the Momentum Operator

Exam I | Problem 1.4 |

What is the one-dimensional momentum operator $\hat{p}$?

Solution

The momentum operator in one dimension is

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

This is the operator used for momentum measurements in the position representation.

1.5Use the de Broglie Relation

Exam I | Problem 1.5 |

A particle has de Broglie wavelength $\lambda$.

What is its momentum $p$?

Solution

Use the de Broglie relation

\lambda = \frac{h}{p}

Solve for $p$:

p = \frac{h}{\lambda}

1.6Photoelectric Maximum Kinetic Energy

Exam I | Problem 1.6 |

A metal has work function $\phi = 4\ \text{eV}$.

Light of energy $h\nu = 6\ \text{eV}$ shines on it.

What is the maximum kinetic energy of the emitted electrons?

Solution

Use the photoelectric equation:

K_{\max} = h\nu - \phi

Substitute the values:

K_{\max} = 6 - 4 = 2\ \text{eV}

So the maximum kinetic energy is $2\ \text{eV}$.

1.7Spin-1/2 Outcomes

Exam I | Problem 1.7 |

For an electron, what are the possible measured values of spin projection along an axis?

Solution

An electron has spin $1/2$, so the only allowed measurement outcomes are

\pm \frac{\hbar}{2}

along the chosen axis.

1.8Write the Uncertainty Bound

Exam I | Problem 1.8 |

If the position uncertainty is $\Delta x$, what is the smallest possible momentum uncertainty $\Delta p$?

Solution

The Heisenberg uncertainty principle says

\Delta x\,\Delta p \ge \frac{\hbar}{2}

Solve for $\Delta p$:

\Delta p \ge \frac{\hbar}{2\Delta x}

1.9Write the One-Dimensional Hamiltonian

Exam I | Problem 1.9 |

For a particle of mass $m$ in a potential $V(x)$, write the one-dimensional Hamiltonian operator $\hat{H}$.

Solution

The Hamiltonian is

\hat{H} = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2} + V(x)

This is the energy operator used in the one-dimensional Schrodinger equation.

1.10Check a Two-Level State

Exam I | Problem 1.10 |

Is the state

|\psi\rangle = \frac{3}{5}|0\rangle + \frac{4}{5}|1\rangle

normalized?

Solution

Check the sum of the squared magnitudes of the coefficients:

\left|\frac{3}{5}\right|^2 + \left|\frac{4}{5}\right|^2 = \frac{9}{25} + \frac{16}{25} = 1

So the state is normalized.

Difficulty

Integrated

2.1Normalize and Use the Density

Exam II | Problem 2.1 |

Let

\psi(x) = Ax

for $0 \le x \le 1$, and $\psi(x)=0$ elsewhere.

Find $A$, then compute the probability of finding the particle in $0 \le x \le \frac{1}{2}$.

Solution

First normalize:

\int_0^1 |Ax|^2\,dx = 1

A^2 \int_0^1 x^2\,dx = 1

A^2 \cdot \frac{1}{3} = 1

A = \sqrt{3}

Now compute the probability on $[0,\frac{1}{2}]$:

P = \int_0^{1/2} |\sqrt{3}x|^2\,dx

P = 3\int_0^{1/2} x^2\,dx = 3\left[\frac{x^3}{3}\right]_0^{1/2} = \frac{1}{8}

2.2Expectation Value on a Finite Interval

Exam II | Problem 2.2 |

A particle is uniformly distributed on the interval $[0,L]$.

What is the expectation value $\langle x \rangle$?

Solution

For a uniform distribution on $[0,L]$, the probability density is

|\psi(x)|^2 = \frac{1}{L}

\langle x \rangle = \int_0^L x \cdot \frac{1}{L}\,dx

\langle x \rangle = \frac{1}{L}\left[\frac{x^2}{2}\right]_0^L = \frac{L}{2}

2.3Scale the Box Width

Exam II | Problem 2.3 |

For an infinite square well, the energy levels satisfy

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

If the width of the well doubles, by what factor does each energy level change?

Solution

Since $E_n \propto \frac{1}{L^2}$, doubling $L$ multiplies the energy by

\frac{1}{2^2} = \frac{1}{4}

So each energy level becomes one-fourth as large.

2.4Why Position and Momentum Cannot Both Be Sharp

Exam II | Problem 2.4 |

Can position and momentum both be known exactly in the same state? Use the note's operator relation to justify your answer.

Solution

No. The note gives the canonical commutation relation

[\hat{x},\hat{p}] = i\hbar

Because the operators do not commute, the observables cannot both be sharply defined at the same time. This is summarized by

\Delta x\,\Delta p \ge \frac{\hbar}{2}

2.5Compare Barrier Widths

Exam II | Problem 2.5 |

For a barrier of height $V_0$ and particle energy $E<V_0$, the note says

T \propto e^{-2\kappa a}

where $a$ is the barrier width.

If one barrier has width $a$ and another has width $3a$, what is the ratio of their transmission coefficients?

Solution

Let $T_a$ be the transmission for width $a$ and $T_{3a}$ for width $3a$.

Then

T_a \propto e^{-2\kappa a}

and

T_{3a} \propto e^{-2\kappa(3a)} = e^{-6\kappa a}

So the ratio is

\frac{T_{3a}}{T_a} = e^{-4\kappa a}

2.6Separate Time and Space

Exam II | Problem 2.6 |

When the potential does not depend on time, the note says we can try a separated solution of the form

\psi(x,t) = \phi(x)T(t).

What equation does the spatial part satisfy?

Solution

For a time-independent potential, separation of variables leads to the time-independent Schrodinger equation:

\hat{H}\phi = E\phi

So the spatial part is an eigenfunction of the Hamiltonian, and $E$ is the corresponding allowed energy.

2.7Hydrogen Energy Scaling

Exam II | Problem 2.7 |

Hydrogen bound-state energies scale like

E_n \propto -\frac{1}{n^2}.

If the ground-state energy is $E_1$, what is $E_4$ in terms of $E_1$?

Solution

Using the inverse-square scaling,

E_4 = \frac{E_1}{4^2} = \frac{E_1}{16}

Because the hydrogen energies are negative, $E_4$ is still negative but closer to zero than $E_1$.

2.8Ground State and Level Spacing of the Harmonic Oscillator

Exam II | Problem 2.8 |

For the harmonic oscillator,

E_n = \left(n+\frac{1}{2}\right)\hbar\omega.

Find $E_0$, $E_1$, and the spacing between adjacent levels.

Solution

Substitute $n=0$:

E_0 = \frac{1}{2}\hbar\omega

Substitute $n=1$:

E_1 = \frac{3}{2}\hbar\omega

The spacing is

E_1 - E_0 = \hbar\omega

So the ground state has nonzero energy, and the levels are evenly spaced.

Difficulty

Applied

3.1Photoelectric Threshold

Final | Problem 3.1 |

A metal has work function $\phi = 2.1\ \text{eV}$.

It is illuminated by photons with energy $h\nu = 3.6\ \text{eV}$.

Will electrons be emitted? If so, what is $K_{\max}$?

Solution

Since

h\nu > \phi

electrons are emitted.

Use

K_{\max} = h\nu - \phi

K_{\max} = 3.6 - 2.1 = 1.5\ \text{eV}

3.2Why Electron Diffraction Is Quantum

Final | Problem 3.2 |

An experiment produces an interference pattern from a beam of electrons.

What quantum idea from the note explains this, and when does the effect become noticeable?

Solution

The pattern shows that electrons have wave behavior, captured by the de Broglie relation

\lambda = \frac{h}{p}

The effect becomes noticeable when the wavelength is comparable to the size of the slit spacing, crystal spacing, or other relevant length scale. That is why pure classical trajectories are not enough.

3.3Choose the Right Semiclassical Approximation

Final | Problem 3.3 |

A particle moves through a region where its de Broglie wavelength changes slowly compared with the length scale of the potential.

Which approximation idea from the note is appropriate, and what is its basic intuition?

Solution

The appropriate idea is the WKB approximation.

Its intuition is semiclassical: connect classical motion with quantum phase accumulation when the wavelength changes slowly, rather than solving the full problem exactly.

3.4Stern-Gerlach Measurements on Different Axes

Final | Problem 3.4 |

An electron is measured to be spin-up along the $z$-axis.

It is then measured along the $x$-axis.

What can you say about the second measurement?

Solution

The result along $x$ is not predetermined by the first measurement along $z$.

Measurements along different axes generally do not commute, so the first measurement does not fix the second one with certainty. The second outcome is probabilistic.

3.5Recognize Perturbation Theory

Final | Problem 3.5 |

A Hamiltonian is written as

\hat{H} = \hat{H}_0 + \lambda \hat{V},

where $\lambda$ is small and the solutions of $\hat{H}_0$ are already known.

Which approximation method should you use?

Solution

Use perturbation theory.

The idea is to start from the known eigenstates and energies of $\hat{H}_0$ and compute small corrections from the additional term $\lambda \hat{V}$.

Difficulty

Challenge

4.1Same Probabilities, Different Phase

Final | Problem 4.1 |

Consider the two states

|\psi_+\rangle = \frac{|0\rangle + |1\rangle}{\sqrt{2}}

and

|\psi_-\rangle = \frac{|0\rangle - |1\rangle}{\sqrt{2}}.

In the $|0\rangle,|1\rangle$ basis, they give the same measurement probabilities. Why are they still different states?

Solution

The probabilities of measuring $|0\rangle$ or $|1\rangle$ are the same because the coefficient magnitudes are the same.

However, the relative phase is different. That phase matters when states interfere, so the two states can behave differently under other measurements or later evolution even though the basis probabilities match.

4.2Normalize an Even Wave Function and Find Its Mean Position

Final | Problem 4.2 |

Let

\psi(x) = Ae^{-|x|}

for all real $x$.

Find $A$, then compute $\langle x \rangle$.

Solution

Normalize the wave function:

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

A^2 \int_{-\infty}^{\infty} e^{-2|x|}\,dx = 1

Use symmetry:

\int_{-\infty}^{\infty} e^{-2|x|}\,dx = 2\int_0^\infty e^{-2x}\,dx = 2\cdot \frac{1}{2} = 1

Thus

$$ A^2 = 1 $$

and we may take

$$ A = 1 $$

Now compute the expectation value:

\langle x \rangle = \int_{-\infty}^{\infty} x|\psi(x)|^2\,dx

Since $|\psi(x)|^2$ is even, the integrand $x|\psi(x)|^2$ is odd, so the integral is zero:

\langle x \rangle = 0

4.3Use a Trial Wave Function

Final | Problem 4.3 |

You choose a trial wave function with an adjustable parameter $\alpha$ and compute

E[\psi] = \frac{\langle \psi|\hat{H}|\psi\rangle}{\langle \psi|\psi\rangle}.

What do you do with $\alpha$, and what does the result tell you about the ground-state energy?

Solution

You vary $\alpha$ to minimize the expected energy.

That is the variational method. The minimized value is an estimate of the ground-state energy, and the note says it gives an upper bound to that energy.

4.4Why the Harmonic Oscillator Cannot Have Zero Energy

Final | Problem 4.4 |

A student says the harmonic oscillator should have zero ground-state energy because the particle could sit at $x=0$ with $p=0$.

Use the note to explain why this is wrong.

Solution

The uncertainty principle prevents both position and momentum from being exactly known at the same time:

\Delta x\,\Delta p \ge \frac{\hbar}{2}

So the ground state cannot have both $x=0$ and $p=0$ with zero spread.

The oscillator energies are

E_n = \left(n+\frac{1}{2}\right)\hbar\omega

so the ground-state energy is

E_0 = \frac{1}{2}\hbar\omega

which is nonzero.