Quantum mechanics revision guide
Posted on Mon 13 October 2025 in Physics
This page, which will be updated occasionally, summarises important results and useful formulae for the Part IB Quantum Mechanics course.
Vector spaces, inner products and duality
Let $V$ be an inner product space over $\mathbb{C}$.
We use ket notation for vectors: $|\psi\rangle \in V$
The inner product between two states is notated as $\langle\phi|\psi\rangle$. This satisfies conjugate symmetry: $$ \overline{\langle\phi|\psi\rangle} = \langle\psi|\phi\rangle $$
Elements of the dual space $V^*$ use bra notation $\langle\phi|$ and act on $|\psi\rangle \in V$ by $$\langle\phi|: V \rightarrow \mathbb{C} : |\psi\rangle \mapsto \langle\phi|\psi\rangle$$
Note $\langle \mathrm{bra}|\mathrm{ket}\rangle$ form a bra(c)ket.
Conjugate symmetry means that the bra corresponding to $\alpha|\psi\rangle$ is $\overline\alpha\langle\psi|$.
Linear operators and adjoints
Linear operators $L: V \rightarrow V$ apply to kets as $|\psi\rangle \mapsto L |\psi\rangle$.
The adjoint operator $L^\dagger: V^* \rightarrow V^*$ acts on bras by $\langle\phi| \mapsto \langle\phi| L^\dagger$.
By conjugate symmetry, $\overline{\langle\phi|L|\psi\rangle} = \langle\psi|L^\dagger|\phi\rangle$ .
An operator is self-adjoint if $$ \langle\psi|L^\dagger|\phi\rangle = \langle\psi|L|\phi\rangle $$ It is technically illegal to write $L = L^\dagger$ since $L$ is an endomorphism on $V$ but $L^\dagger$ is on $V^*$, but the inner product notation allows us to gloss over this.
Properties of self-adjoint operators:
- eigenvalues are real
- orthogonality of eigenvectors with different eigenvalues: if $L|m\rangle = \lambda_m|m\rangle$ and $L|n\rangle = \lambda_n|n\rangle$, and $\lambda_m \neq \lambda_n$, then $\langle m|n\rangle = 0$
- eigenvectors are complete
This means:
- we can use the eigenvectors as a basis
- we can pick the basis to be orthonormal, even if there is a repeated eigenvalue (Gram–Schmidt)
Orthonormality: $\langle m | n \rangle = \delta_{mn}$
Principles of quantum mechanics
States are elements of an inner product space $V$ over $\mathbb{C}$.
- finite dimensional: spin states
- infinite dimensional: wavefunctions, function spaces
Magnitudes and overall phase $e^{i\varphi}$ always cancel out when calculating physical quantities, so wlog can assume that states are normalised $\langle\psi|\psi\rangle=1$ .
Observable quantities correspond to self-adjoint linear operators:
- position $\mathbf{x}$ $\rightarrow$ position operator $\hat{\mathbf{x}}$
- momentum $\mathbf{p}$ $\rightarrow$ momentum operator $\hat{\mathbf{p}}$
- energy $E$ $\rightarrow$ Hamiltonian $\hat{H}$
%% * angular momentum $L_z$, $\mathbf{L}^2$ $\rightarrow$ $\hat{L_z}$, $\hat{\mathbf{L}}^2$ respectively %%
These can be vector quantities but for 1D systems we can conflate $\mathbf{x}$ and $x$. Often drop the hats to simplify notation.
Eigenbasis: For $L$ self-adjoint we have a complete set of eigenvectors $|n\rangle$ with corresponding eigenvalues $\lambda_n \in\mathbb{R}$ not necessarily distinct. Completeness means we can pick an orthonormal basis and write $$ |\psi\rangle = \sum_n c_n |n\rangle, \qquad c_n = \langle n|\psi\rangle. $$ The corresponding dual is $$ \langle\psi| = \sum_n \overline{c_n} \langle n|. $$
Normalisation implies $$ \langle\psi|\psi\rangle = \sum_n |c_n|^2 = 1. $$
Using the eigenbasis is nice because we can apply $L$ simply by multiplying: $$ L|\psi\rangle = \sum_n c_n \lambda_n |n\rangle $$
Observations are probabilistic.
- The possible results of an observation of $L$ are drawn from the eigenvalues $\lambda_n$ of $L$.
- The probability of measuring $\lambda_n$ is given by Born's rule: $p(n) = |c_n|^2$.
- If $\lambda_n$ is a repeated eigenvalue then each of these contributes a chance to measure that value.
Observations collapse the state.
The state changes instantaneously after an observation. If the result of the measurement is $\lambda_n$, then the state collapses onto its projection onto the corresponding eigenspace (up to normalization).
In particular, if $\lambda_n$ is non-degenerate (i.e. eigenspace has dimension 1) then the state after the measurement is $|n\rangle$.
Time-evolution
Assume that time evolution is governed by a linear operator: $$ |\psi(t)\rangle = U(t) |\psi(0)\rangle $$ $U(0) = I$ must be the identity operator.
Conservation of information implies that orthogonal states need to remain orthogonal, and more generally, inner products need to be conserved. This implies that $U(t)$ must be unitary, $U U^\dagger = U^\dagger U = I$.
Taking the limit $t \rightarrow 0$ we can write $$ U(t) = I - \frac{i}{\hbar}t \hat{H} + O(t^2) $$ for some self-adjoint operator $\hat{H}$ called the Hamiltonian.
The quantity $\hbar$ is the reduced Planck constant; it has units of angular momentum and provides a conversion factor.
Some algebra/calculus then gives the time-dependent Schrodinger equation: $$ i \hbar \frac{\mathrm{d}|\psi\rangle }{\mathrm{d}t}= \hat{H} |\psi\rangle $$
The form of the Hamiltonian $\hat{H}$ does not come from first principles but is chosen to match experimental results and to have analogy (inspiration?) with classical mechanics.
For non-relativistic particles in a potential, the Hamiltonian may be constructed by direct analogy with classical mechanics.
Energy states
Look for eigenstates of $\hat H$ by solving the time-independent Schrodinger equation: $$ \hat H |\chi_n\rangle = E_n |\chi_n\rangle $$
The corresponding time-dependent solution is separable: $$|\psi(t)\rangle = A e^{iE_n t/\hbar} |\chi_n\rangle$$ where $A\in\mathbb{C}$ is arbitrary (it cancels when we calculate observables).
Since eigenstates of $\hat H$ form a complete orthogonal basis, the overall solution may be written in the form $$ |\psi(t)\rangle = \sum_n A_n e^{iE_n t/\hbar} |\chi_n\rangle $$
Uncertainty principle and commutators
More background: [[uncertainty-principle-and-spectrograms]], [[uncertainty-principle-obvious-to-musicians]].
The commutator of two operators is: $$ [A, B] = AB - BA $$ Commuting operators have a shared eigenbasis. If $[A,B] = 0$ then it is possible to find an orthonormal basis $|n\rangle$ such that each $|n\rangle$ is an eigenvector of both $A$ and $B$.
Particle in a 1D potential
For a classical particle of mass $m$ in a 1D potential, the energy is $$ E = \frac{p^2}{2m} + V(x, t) $$ where $p$ is the momentum and $V$ is the potential. (In general $V$ may be a function of $t$, but usually is not in problems that we solve.)
$$ \hat H = \frac{\hat{p}^2}{2m} + \frac{1}{2}m\omega^2 \hat{x}^2 $$
Schrodinger equation: $$ E\psi= \frac{-\hbar^2}{2m} \frac{d^2\psi}{dx^2} + \frac{1}{2}m\omega^2 {x}^2 \psi $$