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The Lumenary · First-year graduate

Mathematical foundations of quantum mechanics

Name: ______________________

Date: ______________________

You will learn to

State the postulates of quantum mechanics in Hilbert-space language and compute measurement probabilities with the spectral theorem
Explain why observables are modelled by self-adjoint, possibly unbounded, operators, why their domains matter, and why the canonical commutation relation has no bounded solution
Derive unitary time evolution from the Schrödinger equation and state Stone's theorem
Derive the Robertson–Heisenberg uncertainty relation from commutators and identify the states that saturate it
Solve the quantum harmonic oscillator with ladder operators, including its eigenfunctions and their completeness

1.
A qubit is in the state $\psi = \frac35 e_0 + \frac{4i}{5} e_1$ . Measuring $\sigma_z = \operatorname{diag}(1, -1)$ , what is the probability of the outcome $-1$ ?
- A. $\frac{16}{25}$
- B. $\frac45$
- C. $-\frac{16}{25}$
- D. $\frac{9}{25}$
2.
A three-level system has an observable $A$ with orthonormal eigenvectors $e_1, e_2, e_3$ and eigenvalues $1, 2, 6$ . In the state $\psi = (e_1 + e_2 + e_3)/\sqrt 3$ , what is the expectation $\langle \psi, A\psi \rangle$ ?
3.
In the state $e_0$ (spin up along $z$ ), compare $\sigma_{\sigma_x} \, \sigma_{\sigma_y}$ with $\frac12 |\langle [\sigma_x, \sigma_y] \rangle|$ , using $[\sigma_x, \sigma_y] = 2i\sigma_z$ .
- A.Both equal $1$ : the state saturates Robertson's inequality.
- B.The left side is $0$ because $e_0$ is an eigenvector.
- C.The right side is $0$ because the expectation of a commutator always vanishes.
- D. $1 \ge \frac12$ : the inequality holds strictly.
4.
On $L^2(0, 1)$ let $P_0 = -i\,d/dx$ with domain $\{\psi \in H^1(0, 1) : \psi(0) = \psi(1) = 0\}$ . Which statement is true?
- A. $P_0$ is symmetric but not self-adjoint; its adjoint is $-i\,d/dx$ on all of $H^1(0, 1)$ .
- B. $P_0$ is self-adjoint because $-i\,d/dx$ is formally symmetric.
- C. $P_0$ is not even symmetric.
- D. $P_0$ has real spectrum, like every symmetric operator.
5.
The proof of the Hellinger–Toeplitz theorem shows that the graph of an everywhere-defined symmetric operator $A$ is closed. Which theorem then gives that $A$ is bounded?
- A.The closed graph theorem.
- B.The uniform boundedness principle.
- C.The Riesz representation theorem.
- D.The spectral theorem.
6.
With $[a, a^\dagger] = 1$ , $a h_0 = 0$ and $\|h_0\| = 1$ , compute $\|(a^\dagger)^3 h_0\|^2$ .
7.
Which Lean statement correctly formalises no $n \times n$ complex matrices satisfy $XP - PX = c \cdot 1$ with $c \ne 0$ (and is provable)?
- A.∀ {n : ℕ}, 0 < n → ∀ (X P : Matrix (Fin n) (Fin n) ℂ) {c : ℂ}, c ≠ 0 → X * P - P * X ≠ c • 1
- B.∀ {n : ℕ} (X P : Matrix (Fin n) (Fin n) ℂ) {c : ℂ}, c ≠ 0 → X * P - P * X ≠ c • 1
- C.∀ {n : ℕ}, 0 < n → ∀ (X P : Matrix (Fin n) (Fin n) ℂ) (c : ℂ), X * P - P * X ≠ c • 1
- D.∀ {n : ℕ}, 0 < n → ∀ (X P : Matrix (Fin n) (Fin n) ℂ), X * P ≠ P * X
8.
In Wielandt's proof that $XP - PX = cI$ ( $c \ne 0$ ) is impossible for bounded operators, one needs $P^n \ne 0$ for every $n$ . Why does no power of $P$ vanish?
- A.If $P^{n+1} = 0$ , the identity $XP^{n+1} - P^{n+1}X = (n+1)cP^n$ forces $P^n = 0$ ; descending, $P^0 = I = 0$ , impossible on a nonzero space.
- B.Because $P$ is self-adjoint, and self-adjoint operators are never nilpotent.
- C.Because $\|P^n\| = \|P\|^n$ for every bounded operator.
- D.Because the spectrum of $P$ is non-empty.
9.
Which normalised states $\psi \in L^2(\R)$ satisfy $\sigma_X \sigma_P = \hbar/2$ exactly? Select all that apply.
- A.The oscillator ground state $h_0(x) = \pi^{-1/4}e^{-x^2/2}$ (with $\hbar = 1$ ).
- B.A displaced, boosted, squeezed Gaussian $C\exp\big(-\frac{(x - x_0)^2}{2s^2} + \frac{ip_0x}{\hbar}\big)$ .
- C.The excited oscillator states $h_n$ with $n \ge 1$ .
- D.A plane wave $e^{ikx}$ , which has $\sigma_P = 0$ .
10.
Write expectation(A, psi) returning $\langle \psi, A\psi \rangle = \sum_{i,j} \overline{\psi_i} A_{ij} \psi_j$ for a square matrix A (list of rows) and a vector psi (list of complex numbers), and std_dev(A, psi) returning $\sigma_A = \sqrt{\langle A^2 \rangle - \langle A \rangle^2}$ for a normalised psi and Hermitian A. The tests check Robertson's inequality for Pauli matrices.
Computational — Practice

Answer key — Mathematical foundations of quantum mechanics

1.
The eigenvalue $-1$ of $\sigma_z$ has eigenvector $e_1$ , so $P(-1) = |\langle e_1, \psi \rangle|^2 = |4i/5|^2 = 16/25$ , and $P(+1) = 9/25$ ; they add to $1$ because $\psi$ is a unit vector. The phase $i$ of the second amplitude does not affect $\sigma_z$ statistics, but it does affect those of $\sigma_x$ and $\sigma_y$ .
2.
$\langle A \rangle = \frac13 \cdot 1 + \frac13 \cdot 2 + \frac13 \cdot 6 = 3$ . Directly: $A\psi = (e_1 + 2e_2 + 6e_3)/\sqrt 3$ and $\langle \psi, A\psi \rangle = (1 + 2 + 6)/3 = 3$ . Note that $3$ is not itself a possible outcome: an expectation is an average over many identically prepared systems.
3.
$\sigma_x e_0 = e_1$ and $\sigma_y e_0 = ie_1$ are orthogonal to $e_0$ , so $\langle \sigma_x \rangle = \langle \sigma_y \rangle = 0$ , while $\sigma_x^2 = \sigma_y^2 = I$ gives variances $1$ . The right side is $\frac12 |\langle e_0, 2i\sigma_z e_0 \rangle| = 1$ . So $1 \cdot 1 = 1$ : equality. In the proof, equality in Cauchy–Schwarz holds because $(\sigma_y - 0)e_0 = i e_1 = i(\sigma_x - 0)e_0$ .
4.
(a). Symmetry follows from integration by parts with vanishing boundary terms. For the adjoint, $\langle \phi, P_0\psi \rangle = \langle -i\phi', \psi \rangle$ for every $\phi \in H^1$ and every $\psi \in D(P_0)$ , because $\psi$ kills the boundary terms; so $D(P_0^*) = H^1(0, 1) \supsetneq D(P_0)$ . The self-adjoint extensions $P_\theta$ impose $\psi(1) = e^{i\theta}\psi(0)$ and have spectra $\theta + 2\pi\Z$ .
5.
(a). If $x_n \to x$ and $Ax_n \to y$ , symmetry gives $\langle y, z \rangle = \langle x, Az \rangle = \langle Ax, z \rangle$ for all $z$ , so $y = Ax$ and the graph is closed. The closed graph theorem (valid because $\mathcal H$ is complete) then gives boundedness. Consequence: unbounded observables such as $X$ and $P$ can only be defined on dense subspaces.
6.
If $N\phi = k\phi$ then $\|a^\dagger\phi\|^2 = \langle \phi, (N + 1)\phi \rangle = (k + 1)\|\phi\|^2$ . Starting from $\|h_0\|^2 = 1$ : $\|a^\dagger h_0\|^2 = 1$ , $\|(a^\dagger)^2 h_0\|^2 = 2$ , $\|(a^\dagger)^3 h_0\|^2 = 3 \cdot 2 = 6 = 3!$ . This is why $h_n = (n!)^{-1/2}(a^\dagger)^n h_0$ is normalised.
7.
(a). Taking traces of X * P - P * X = c • 1 gives 0 = c * n (since trace (X * P) = trace (P * X)), which contradicts c ≠ 0 and n ≠ 0. Dropping either hypothesis makes the statement false, as the counterexamples in (b) and (c) show; (d) is a different and false claim.
8.
(a). If $P^{n+1} = 0$ , then $(n+1)cP^n = XP^{n+1} - P^{n+1}X = 0$ , and since $(n+1)c \ne 0$ , $P^n = 0$ . Repeating, $P^0 = I = 0$ , which is impossible on a nonzero space. Then the norm estimate $(n+1)|c| \|P^n\| \le 2\|X\|\|P\|\|P^n\|$ can be divided by $\|P^n\| > 0$ .
9.
(a) and (b). Equality forces $-i\hbar\psi' - p_0\psi = i\lambda(x - x_0)\psi$ , i.e. $\psi'/\psi = \frac{i p_0}{\hbar} - \frac{\lambda}{\hbar}(x - x_0)$ , whose solutions are the Gaussians in (b) (square-integrable exactly when $\lambda > 0$ ); (a) is the case $x_0 = p_0 = 0$ , $s = 1$ . Excited states have product $(n + \frac12)\hbar$ , and plane waves are not states.
10.
```
import math

def expectation(A, psi):
    n = len(psi)
    a_psi = [sum(A[i][j] * psi[j] for j in range(n)) for i in range(n)]
    return sum(complex(psi[i]).conjugate() * a_psi[i] for i in range(n))

def std_dev(A, psi):
    n = len(psi)
    a_psi = [sum(A[i][j] * psi[j] for j in range(n)) for i in range(n)]
    second = sum(abs(v) ** 2 for v in a_psi)  # <A^2> = ||A psi||^2 for Hermitian A
    mean = expectation(A, psi).real
    return math.sqrt(max(second - mean ** 2, 0.0))
```
The complex conjugate on the left factor is essential: without it, $\psi = (1, i)/\sqrt 2$ would give $\langle \sigma_y \rangle = 0$ instead of $1$ . For Hermitian $A$ , $\langle A^2 \rangle = \langle A\psi, A\psi \rangle = \|A\psi\|^2$ , and the variance is $\|(A - \langle A \rangle)\psi\|^2$ , the quantity bounded in Robertson's inequality.