Quantum mechanics revision guide

This page, which will be updated occasionally, summarises important results and useful formulae for the Part IB Quantum Mechanics course.

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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 vectors 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 this acts 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.

It can be shown that $V$ and $V^*$ are isomorphic and that the isomorphism is "natural", i.e. does not depend on a choice of basis. Conjugate symmetry means that the bra corresponding to $\alpha|\psi\rangle$ is $\overline\alpha\langle\psi|$.

Linear operators and adjoints

Linear operators $L: V \rightarrow V$ acts on 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 (the two spaces are isomorphic).

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)

Properties of orthonormal bases:

Given an orthonormal basis $|n\rangle$ the dual basis on $V^*$ is $\langle n|$.

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

Projection (generalisation of Fourier coefficient formula): $$ |\psi\rangle = \sum_n c_n|n\rangle \implies c_n = \langle n|\psi\rangle $$ Sometimes easier to work with non-normalised basis, in which case: $$ c_n = \frac{\langle n |\psi\rangle}{\langle n | n \rangle} $$

Conjugate symmetry: $$ \langle\psi|= \sum_n \overline{c_n} \langle n| $$

Inner products: $$ \begin{align} |\phi\rangle &= \sum_n b_n|n\rangle \ |\psi\rangle &= \sum_n c_n|n\rangle \ \implies \langle\phi|\psi\rangle &= \sum_n \overline{b_n} c_n \end{align} $$

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

A normalised state has $\langle\psi|\psi\rangle=1$ . Magnitudes and any overall phase $e^{i\varphi}$ always cancel out when calculating physical quantities. So states $|\psi\rangle$ and $c|\psi\rangle$ are equivalent for $c\in\mathbb{C}$.

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. Write $$ |\psi\rangle = \sum_n c_n |n\rangle, \qquad c_n = \langle n|\psi\rangle. $$ 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 each eigenstate by the corresponding eigenvalue: $$ L|\psi\rangle = \sum_n c_n \lambda_n |n\rangle $$

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.

Often easier to calculate with the non-normalised version of Born's rule: $$ p(n) = \frac{|\langle n|\psi\rangle|^2}{\langle n|n\rangle \langle\psi|\psi\rangle} $$

Note that the probabilistic interpretation leads to serious philosophical problems! So this can't be the full story, although its predictions agree with experiment. Looking Glass Universe has an excellent video on this.

Observations collapse the state.

The state changes instantaneously during 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 $|n\rangle$.

This means that the state $|\psi\rangle$ isn't fully observable – because observations destroy information. (Again, there are serious problems with this interpretation.)

Neither is any overall phase in $e^{i\varphi}|\psi\rangle$, since those phases cancel when you take quantities such as $\langle\psi|\psi\rangle$.

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. 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.

$\hbar$ is the reduced Planck constant; it has units of angular momentum $\mathrm{kg, m^2, s^{-1}} = \mathrm{J \cdot s}$ and provides a conversion factor.

Some algebra gives the time-dependent Schrodinger equation: $$ i \hbar \frac{\mathrm{d}|\psi\rangle }{\mathrm{d}t}= \hat{H} |\psi\rangle $$

The Hamiltonian $\hat{H}$ does not come from first principles but is chosen to match experimental results, and often by analogy to classical mechanics.

Energy states

Look for energy eigenstates 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 $$ |\psi(t)\rangle = \sum_n A_n e^{iE_n t/\hbar} |\chi_n\rangle $$

Commutators

Commutator of two operators: $$ [A, B] = AB - BA $$ 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 $B$. In other words the eigenspaces "line up" with each other.

Important commutator identities:

Linearity $$ [A+B, C] = [A,C] + [B,C] $$

Anticommutativity $$ [B, A] = -[A, B] $$

Products (Leibniz rule) $$ \begin{align} [A, BC] &= ABC - BCA \ &= ABC - BAC + BAC - BCA \ &= [A, B] C - B [A, C] \end{align} $$ Powers $$ [A, B^n] = [A, B] B^{n-1} + B[A, B] B^{n-2} + \dots + B^{n-1}[A, B] $$ If $[A, B]$ commutes with $B$ (e.g. is a constant): $$ [A, B^n] = n B^{n-1} [A, B] $$ e.g. $[\hat{x}, \hat{p}^2] = 2i\hbar \hat{p}$

Jacobi identity: $$ [A, [B, C]] + [B, [C, A]] + [C, [A, B]] = 0 $$

Canonical relations: $$ [\hat{x}, \hat{p}] = i\hbar $$ Angular momentum components (also spin) $$ [L_i, L_j] = i\hbar, \epsilon_{ijk} ,L_k $$

More background: [[uncertainty-principle-and-spectrograms]], [[uncertainty-principle-obvious-to-musicians]].

measurement of $A$ $\implies$ state collapse to an eigenstate of $A$

If $[A, B] \neq 0$ then the resulting eigenstate is not an eigenstate of $B$

Uncertainty $\sigma_A^2 = (\Delta A)^2 = \langle A^2\rangle - \langle A\rangle^2$

Uncertainty principle (general) $\sigma_A \sigma_B \geq \frac{1}{2} \left|\langle [A, B]\rangle\right|$

Position-momentum $\sigma_x \sigma_p \geq \frac{1}{2}\hbar$

Particle in a 1D potential, QHO

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.)

Quantum harmonic oscillator: $$ \hat H = \frac{\hat{p}^2}{2m} + \frac{1}{2}m\omega^2 \hat{x}^2 $$

TISE: $$ E\psi= \frac{-\hbar^2}{2m} \frac{d^2\psi}{dx^2} + \frac{1}{2}m\omega^2 {x}^2 \psi $$ Scales: $$ \begin{align} & E \sim \frac{\hbar^2}{m r^2} \sim m\omega^2 r^2 \ \implies & E \sim \hbar\omega, \quad r \sim \left(\frac{\hbar}{m\omega}\right)^{1/2} \end{align} $$

Nondimensional form: $$ E\psi = \frac{-1}{2}\frac{d^2\psi}{dx^2} + \frac{1}{2}x^2\psi $$ Look for solutions of the form $$ \begin{align} \psi(x) &= f(x) , e^{-x^2/2} \ \psi'(x) &= \left(f'(x) - x f(x)\right) , e^{-x^2/2} \ \psi''(x) &= \left(f''(x) - 2xf(x) + (x^2 - 1) f(x)\right) , e^{-x^2/2} \end{align} $$ Frobenius solution + convergence $\implies$ $$ E = n + \frac{1}{2} \qquad n = 1, 2, 3, \dots $$ with $f$ a polynomial of degree $n$.

Ladder operators

Hydrogen atom

Spherical harmonics $$ Y_l^m(\theta, \varphi) = N e^{im\varphi} , P_l^m(\cos\theta) $$ $$ l = 0, 1, 2,\dots, \quad m =-l, -l+1, \dots, l $$ where $N$ is a normalisation and $P_l^m$ is an associated Legendre polynomial (not actually a polynomial!)

Spherical harmonics are simultaneous eigenfunctions of $\hat{\mathbf L}^2$ and $\hat{L}_z$ obeying $$ \hat{\mathbf L}^2 Y_l^m = \hbar^2 , l (l+1) , Y_l^m \qquad \hat{L}_x Y_l^m = \hbar, m ,Y_l^m $$

These operators also commute with $\hat{H}$ when the potential is spherically symmetric.