Quantum Mechanics Domain Guide

Concise answer

Quantum mechanics on this exam is a small, closed system of ideas. A state is a wavefunction; observables are operators; measurement returns an eigenvalue with probability equal to the squared overlap of the state with that eigenvector; and the state between measurements evolves by the Schrodinger equation. Three exactly solvable systems — the infinite well, the harmonic oscillator, and hydrogen — supply nearly every energy spectrum the test uses, so knowing how each spectrum scales matters far more than being able to solve any differential equation.

Definitions

Wavefunction and probability density: $\Psi(x,t)$ is the state; $|\Psi|^{2}$ is a probability density with units of inverse length, so probability is $\int_a^b |\Psi|^{2}dx$ . Normalisation, $\int_{-\infty}^{\infty}|\Psi|^{2}dx = 1$ , is a requirement, not an option.
Operator and expectation value: An observable is a Hermitian operator: $\hat{x} = x$ , $\hat{p} = -i\hbar\,\partial/\partial x$ , $\hat{H} = \hat{p}^{2}/2m + V$ . Its expectation value in a state is $\langle A\rangle = \int \Psi^{*}\hat{A}\Psi\,dx$ , the mean of many measurements on identically prepared systems — not the result of one.
Eigenstate and measurement: If $\hat{A}\psi_n = a_n\psi_n$ then measuring $A$ on $\psi_n$ certainly returns $a_n$ . For a general state $\Psi = \sum_n c_n\psi_n$ , the probability of $a_n$ is $|c_n|^{2}$ and the state collapses to $\psi_n$ afterwards.
Commutator: $[\hat{A},\hat{B}] = \hat{A}\hat{B}-\hat{B}\hat{A}$ . Commuting observables share eigenstates and can be known simultaneously; $[\hat{x},\hat{p}] = i\hbar$ is precisely why position and momentum cannot.
Stationary state: An energy eigenstate. Its time dependence is the phase $e^{-iEt/\hbar}$ , which cancels in $|\Psi|^{2}$ , so every probability density is time-independent. Only a superposition of different energies produces motion.
Degeneracy: The number of independent states sharing one energy. Hydrogen's level $n$ has $n^{2}$ orbital states ( $2n^{2}$ counting spin), which is where the periodic table's shell capacities come from.

Intuition

Confinement costs energy. Squeezing a particle into a region of size $L$ forces a momentum spread of at least $\hbar/2L$ , so the kinetic energy cannot be zero — that is the whole content of zero-point energy, and it explains in one sentence why the well's ground state is not at the bottom, why the oscillator keeps $\tfrac{1}{2}\hbar\omega$ at absolute zero, and why the hydrogen electron does not fall into the nucleus.

Almost every quantum item is a ratio. The exam rarely wants a number in joules; it wants $E_3/E_1$ , or the wavelength of the photon emitted between two levels, or how the spectrum changes when the box is halved. Once you know $E \propto n^{2}/L^{2}m$ for the well, $E \propto (n+\tfrac{1}{2})\sqrt{k/m}$ for the oscillator, and $E \propto -Z^{2}/n^{2}$ for hydrogen-like atoms, you can answer these in seconds without touching Planck's constant.

Amplitudes interfere; probabilities do not. When two paths lead to the same outcome you add complex amplitudes and then square, which is why the double-slit pattern exists at all and why 'the electron went through one slit or the other' is the wrong picture. This is the same rule that makes a superposition's probability density oscillate at the Bohr frequency $(E_2-E_1)/\hbar$ while its energy probabilities stay perfectly constant.

Concept walkthrough

ETS lists quantum mechanics as a content area of its own, and our learning model keeps it as a single canonical domain because the exam's items draw on one shared formalism rather than on separable sub-topics. Start with the postulates in the compact form the exam uses: the state is $\Psi$ ; observables are Hermitian operators; a measurement of $A$ yields an eigenvalue $a_n$ with probability $|c_n|^{2}$ where $c_n = \langle\psi_n|\Psi\rangle$ ; the state collapses to $\psi_n$ ; and between measurements $i\hbar\,\partial\Psi/\partial t = \hat{H}\Psi$ .

The time-independent Schrodinger equation, $-\dfrac{\hbar^{2}}{2m}\dfrac{d^{2}\psi}{dx^{2}} + V\psi = E\psi$ , is the workhorse, and its boundary conditions do more work than its solutions. Requiring $\psi$ to be continuous, single-valued, and normalisable is what quantises the spectrum; the differential equation alone quantises nothing. For the infinite square well of width $L$ this gives $\psi_n = \sqrt{2/L}\,\sin(n\pi x/L)$ and $E_n = n^{2}h^{2}/8mL^{2} = n^{2}\pi^{2}\hbar^{2}/2mL^{2}$ with $n = 1,2,3,\ldots$ — note that $n = 0$ is excluded because it would make $\psi$ vanish everywhere, which is not a state.

The other two solvable systems are worth knowing as spectra rather than as wavefunctions. The harmonic oscillator has $E_n = (n+\tfrac{1}{2})\hbar\omega$ with $n = 0,1,2,\ldots$ : evenly spaced levels, a non-zero ground state, and a ladder whose spacing $\hbar\omega$ is independent of $n$ . Hydrogen has $E_n = -13.6\ \mathrm{eV}/n^{2}$ , levels that crowd together as $n$ grows and converge on the ionisation threshold at zero. Three different confinements, three different scalings — and confusing them is the most reliable way to lose an easy mark, because each has a plausible-looking distractor built from the other two.

Superposition is where the formalism becomes a calculation. Write $\Psi = \sum_n c_n\psi_n$ with $\sum|c_n|^{2}=1$ . Then the probability of measuring $E_n$ is $|c_n|^{2}$ , the expectation value is $\langle E\rangle = \sum|c_n|^{2}E_n$ , and the time evolution attaches $e^{-iE_nt/\hbar}$ to each term. Two facts follow immediately and are tested often: the energy probabilities never change with time, because each phase has modulus one; and the position probability density does change, oscillating at the difference frequency $(E_m-E_n)/\hbar$ between any two occupied levels.

Uncertainty and tunnelling supply the estimation questions. $\Delta x\,\Delta p \ge \hbar/2$ gives ground-state energies to within a factor of order one: setting $\Delta x \sim L$ and $\Delta p \sim \hbar/L$ gives $E \sim \hbar^{2}/2mL^{2}$ , which is the well's true ground state up to $\pi^{2}$ . The companion relation $\Delta E\,\Delta t \ge \hbar/2$ says a short-lived state has a broad energy — a linewidth, not a licence to break energy conservation. For a barrier of height $V_0$ and width $a$ with $E < V_0$ , the transmission falls exponentially, $T \approx e^{-2\kappa a}$ with $\kappa = \sqrt{2m(V_0-E)}/\hbar$ : doubling the barrier width squares the already-small probability, which is why tunnelling rates vary over many orders of magnitude for small changes in geometry.

Angular momentum and spin close the formalism. The magnitude of orbital angular momentum is $|\vec{L}| = \sqrt{\ell(\ell+1)}\,\hbar$ with $\ell = 0,1,\ldots,n-1$ , and its projection is $L_z = m_\ell\hbar$ with $m_\ell$ running from $-\ell$ to $+\ell$ in integer steps — so $L_z$ can never equal $|\vec{L}|$ , a geometric statement of the uncertainty between different components. Electrons additionally carry spin $s = \tfrac{1}{2}$ with $m_s = \pm\tfrac{1}{2}$ , and the Pauli exclusion principle forbids two of them from sharing all four quantum numbers, giving hydrogen's level $n$ a total degeneracy of $2n^{2}$ . First-order perturbation theory then handles small corrections in one line: $E_n^{(1)} = \langle\psi_n^{(0)}|\hat{H}'|\psi_n^{(0)}\rangle$ , the expectation value of the perturbation in the unperturbed state.

After this page, you should be able to

Normalise a wavefunction and convert between $|\Psi|^{2}$ , a probability over an interval, and an expectation value.
Read off the energy spectrum and its scaling for the infinite well ( $E_n \propto n^{2}$ ), the harmonic oscillator ( $E_n \propto n + \tfrac{1}{2}$ ), and hydrogen ( $E_n \propto -1/n^{2}$ ).
Expand a state in energy eigenstates and give measurement probabilities, the expectation value of the energy, and the oscillation frequency of the resulting probability density.
Use the uncertainty principle to estimate ground-state energies and to reject impossible answers by inspection.
Apply the angular-momentum rules $|\vec{L}| = \sqrt{\ell(\ell+1)}\,\hbar$ and $L_z = m_\ell\hbar$ with $-\ell \le m_\ell \le \ell$ , and count degeneracies.
Convert between energy, wavelength, and momentum for both photons and matter waves using $hc = 1240\ \mathrm{eV\,nm}$ .

Formulas and assumptions

The Schrodinger equation and normalisation

i\hbar\frac{\partial\Psi}{\partial t} = \hat{H}\Psi, \qquad -\frac{\hbar^{2}}{2m}\frac{d^{2}\psi}{dx^{2}} + V\psi = E\psi, \qquad \int_{-\infty}^{\infty}|\Psi|^{2}dx = 1

Variables

Psi: the state; |Psi|^2 is a probability density, not a probability
V: potential energy function
E: energy eigenvalue

Assumptions

Quantisation comes from the boundary conditions (continuity, single-valuedness, normalisability), not from the differential equation alone.
|Psi|^2 has units of inverse length in one dimension, so it must be integrated over an interval before it means anything.

The three solvable spectra and how each one scales

E_n^{\text{well}} = \frac{n^{2}h^{2}}{8mL^{2}}, \qquad E_n^{\text{SHO}} = \left(n+\tfrac{1}{2}\right)\hbar\omega, \qquad E_n^{\text{H}} = -\frac{13.6\,Z^{2}}{n^{2}}\ \mathrm{eV}

Variables

n: the quantum number, whose starting value differs between the three systems
L: width of the well
omega: classical angular frequency sqrt(k/m) of the oscillator

Assumptions

The well ladder starts at n = 1 and the oscillator ladder at n = 0; swapping them is the classic error.
The hydrogen formula applies only to one-electron (hydrogen-like) systems.

Superposition, measurement, and time evolution

\Psi = \sum_n c_n\psi_n, \quad P(E_n) = |c_n|^{2}, \quad \langle E\rangle = \sum_n |c_n|^{2}E_n, \quad \Psi(t) = \sum_n c_n\psi_n e^{-iE_nt/\hbar}

Variables

c_n: complex expansion coefficient
P(E_n): probability of measuring the energy E_n
omega_mn: Bohr angular frequency (E_m - E_n)/hbar

Assumptions

The coefficients must be normalised before any probability is read off them.
Energy probabilities are constant in time; only relative phases evolve, so the position density is what moves.

Uncertainty relations

\Delta x\,\Delta p \ge \frac{\hbar}{2}, \qquad \Delta E\,\Delta t \ge \frac{\hbar}{2}, \qquad E_{\min} \sim \frac{\hbar^{2}}{2mL^{2}}

Variables

Delta x: standard deviation of position
Delta p: standard deviation of momentum
Delta t: lifetime of a state, giving its natural linewidth

Assumptions

These are lower bounds on statistical spreads over many identical measurements, not limits on one instrument's precision.
The energy-time relation describes linewidth and lifetime; it does not permit energy non-conservation.

Barrier tunnelling

T \approx e^{-2\kappa a}, \qquad \kappa = \frac{\sqrt{2m(V_0-E)}}{\hbar}

Variables

V0: barrier height
a: barrier width
kappa: decay constant of the wavefunction inside the barrier

Assumptions

The exponential form assumes a wide, high barrier; it is an estimate, not the exact transmission coefficient.
Transmission falls exponentially with both width and the square root of the energy deficit, so small geometric changes produce enormous rate changes.

Angular momentum, spin, and degeneracy

|\vec{L}| = \sqrt{\ell(\ell+1)}\,\hbar, \quad L_z = m_\ell\hbar, \quad -\ell \le m_\ell \le \ell, \quad g_n = 2n^{2}

Variables

l: orbital angular momentum quantum number
m_l: magnetic quantum number
g_n: total degeneracy of hydrogen level n including spin

Assumptions

The magnitude uses sqrt(l(l+1)), never l, so L_z can never equal the full magnitude.
The 2n^2 count assumes the pure Coulomb spectrum, before fine structure or external fields lift the degeneracy.

Photons, matter waves, and the constant that saves the arithmetic

E = hf = \frac{hc}{\lambda}, \qquad \lambda_{\text{dB}} = \frac{h}{p}, \qquad hc = 1240\ \mathrm{eV\,nm}, \qquad \hbar c = 197.3\ \mathrm{eV\,nm}

Variables

f: frequency
lambda: wavelength
p: momentum

Assumptions

hc = 1240 eV nm turns almost every photon-energy question into one division; carrying it is worth more than carrying h in joule-seconds.
For a non-relativistic particle accelerated through V, p = sqrt(2 m q V), so lambda = h / sqrt(2 m q V).

Worked example

A two-level superposition in a box: what is fixed, what moves

An electron in a one-dimensional infinite well of width $L = 0.50\ \mathrm{nm}$ is prepared in the state $\Psi(x,0) = \tfrac{1}{\sqrt{5}}\left[\psi_1(x) + 2\psi_2(x)\right]$ . Find the ground-state energy, the probability of each energy outcome, the expectation value of the energy, and the wavelength of the photon emitted if the electron later drops from $n=2$ to $n=1$ .

1Check the normalisation first, because every probability depends on it: $|c_1|^{2}+|c_2|^{2} = \tfrac{1}{5}(1^{2}+2^{2}) = \tfrac{1}{5}(5) = 1$ . The state is normalised as written.
2Compute the ground-state energy in electronvolts using $E_1 = \dfrac{h^{2}}{8mL^{2}} = \dfrac{(hc)^{2}}{8(mc^{2})L^{2}}$ , which avoids joules entirely. With $hc = 1240\ \mathrm{eV\,nm}$ , $mc^{2} = 5.11\times10^{5}\ \mathrm{eV}$ , and $L = 0.50\ \mathrm{nm}$ : $E_1 = \dfrac{(1240)^{2}}{8(5.11\times10^{5})(0.25)} = \dfrac{1.5376\times10^{6}}{1.022\times10^{6}} = 1.50\ \mathrm{eV}$ .
3Read off the ladder. $E_n = n^{2}E_1$ , so $E_2 = 4E_1 = 6.02\ \mathrm{eV}$ . Note that this scaling is the entire content of the well solution; nothing further needs solving.
4Give the measurement probabilities: $P(E_1) = |c_1|^{2} = \tfrac{1}{5} = 0.20$ and $P(E_2) = |c_2|^{2} = \tfrac{4}{5} = 0.80$ . These are the only two possible outcomes, and they do not change with time — the phases $e^{-iE_nt/\hbar}$ have modulus one.
5Compute the expectation value: $\langle E\rangle = 0.20E_1 + 0.80(4E_1) = (0.20 + 3.20)E_1 = 3.4E_1 = 3.4(1.50) = 5.1\ \mathrm{eV}$ . Note that $5.1\ \mathrm{eV}$ is not an allowed measurement result — it lies between $E_1$ and $E_2$ . An expectation value is an average over many runs, not a prediction for one.
6Find the emitted photon. The transition energy is $E_2 - E_1 = 3E_1 = 4.51\ \mathrm{eV}$ , so $\lambda = \dfrac{hc}{\Delta E} = \dfrac{1240\ \mathrm{eV\,nm}}{4.51\ \mathrm{eV}} = 275\ \mathrm{nm}$ — in the ultraviolet. The same difference sets the frequency at which the position probability density sloshes back and forth while the superposition survives: $\omega = (E_2-E_1)/\hbar$ .

$E_1 = 1.50\ \mathrm{eV}$ , $P(E_1)=0.20$ , $P(E_2)=0.80$ , $\langle E\rangle = 3.4E_1 = 5.1\ \mathrm{eV}$ , and the $2\to1$ photon has $\lambda \approx 275\ \mathrm{nm}$ . Halving the well width to $0.25\ \mathrm{nm}$ would multiply every energy by four and divide the photon wavelength by four.

Common traps

Treating $|\Psi|^{2}$ as a probability. It is a density with units of inverse length; a probability only appears after integrating over an interval.
Skipping normalisation and reading the raw coefficients as probabilities. In $\tfrac{1}{\sqrt5}(\psi_1+2\psi_2)$ the probabilities are $1/5$ and $4/5$ , not $1$ and $2$ .
Starting the infinite-well ladder at $n=0$ , or the harmonic-oscillator ladder at $n=1$ . The well excludes $n=0$ because $\psi$ would vanish; the oscillator's ground state is $n=0$ with energy $\tfrac{1}{2}\hbar\omega$ .
Applying $E_n \propto n^{2}$ to the harmonic oscillator or $E_n \propto n$ to the box. Each of the three solvable systems has its own scaling, and the distractors are built from the other two.
Reporting an expectation value as a possible measurement outcome. Only eigenvalues can be measured; $\langle E\rangle$ generally is not one of them.
Reading $\Delta E\,\Delta t \ge \hbar/2$ as permission to violate energy conservation. It relates a state's lifetime to its natural linewidth.
Writing $|\vec{L}| = \ell\hbar$ instead of $\sqrt{\ell(\ell+1)}\,\hbar$ , then concluding that $L_z$ can equal the full magnitude.
Expecting tunnelling probability to fall linearly with barrier width. It falls exponentially, so a modest widening can suppress the rate by orders of magnitude.
Adding probabilities for interfering alternatives. Amplitudes add and are then squared, which is the only reason interference exists.
Assuming a measurement leaves the state alone. It projects the state onto the measured eigenstate, so an immediate repeat measurement returns the same value with certainty.

Infinite Square Well Transitions Photoelectric Effect Quantum Mechanics Foundations GRE Physics overview Atomic Physics Domain Guide Photons, Work Functions, and Matter Waves GRE Physics reference sheet Sign in for current practice

Question depth and domain coverage vary by exam. Practice answers are checked after submission.

Sources

GRE Subject Test Content and Structure — ETS. Accessed 2026-07-06. Use as a cited source for exam facts; do not imply affiliation or reproduce protected test material.
University Physics Volume 3, Section 7.1: Wave Functions — OpenStax. Accessed 2026-08-02. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.
University Physics Volume 3, Section 7.4: The Quantum Particle in a Box — OpenStax. Accessed 2026-08-02. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.
University Physics Volume 3, Section 7.2: The Heisenberg Uncertainty Principle — OpenStax. Accessed 2026-08-02. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.
University Physics Volume 3, Section 6.5: De Broglie's Matter Waves — OpenStax. Accessed 2026-08-02. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.
University Physics Volume 3, Section 6.2: Photoelectric Effect — OpenStax. Accessed 2026-08-02. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.

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Quantum Mechanics Domain Guide

Concise answer

Definitions

Intuition

Concept walkthrough

After this page, you should be able to

Formulas and assumptions

The Schrodinger equation and normalisation

The three solvable spectra and how each one scales

Superposition, measurement, and time evolution

Uncertainty relations

Barrier tunnelling

Angular momentum, spin, and degeneracy

Photons, matter waves, and the constant that saves the arithmetic

Worked example

A two-level superposition in a box: what is fixed, what moves

Common traps

Related pages and practice

Sources

Sources and corrections