1.3.3 Heisenberg Picture#

Worked solutions for the homework problems in the 1.3.3 Heisenberg Picture lecture. Each problem statement is reproduced verbatim, followed by a complete worked solution.

1. Picture equivalence on a concrete example. Take the Hamiltonian \(\hat H = \tfrac{\hbar\omega}{2}\hat\sigma^z\) and the initial state \(\vert\psi(0)\rangle = \vert+\rangle\). Compute \(\langle\hat\sigma^x\rangle(t)\) in two ways and verify the two answers agree.

(a) Schrödinger picture. Evolve the state \(\vert\psi(t)\rangle = \hat U(t)\vert+\rangle\) using \(\hat U(t) = \mathrm{e}^{-\mathrm{i}\omega t\hat\sigma^z/2}\) from 1.3.2. Compute \(\langle\psi(t)\vert\hat\sigma^x\vert\psi(t)\rangle\).

(b) Heisenberg picture. Compute the time-evolved operator \(\hat\sigma^x(t) = \hat U^\dagger(t)\hat\sigma^x\hat U(t)\) using the lecture’s example (or the Heisenberg equation of motion). Compute \(\langle+\vert\hat\sigma^x(t)\vert+\rangle\).

(c) Verify the two methods give the same result, and identify exactly which algebraic identity collapses one calculation into the other.

Solution.

(a) The state at time \(t\) is

\[ \vert\psi(t)\rangle = \tfrac{1}{\sqrt 2}\bigl(\mathrm{e}^{-\mathrm{i}\omega t/2}\vert\uparrow\rangle + \mathrm{e}^{+\mathrm{i}\omega t/2}\vert\downarrow\rangle\bigr). \]

Write \(\vert\psi(t)\rangle = c_\uparrow\vert\uparrow\rangle + c_\downarrow\vert\downarrow\rangle\) with \(c_\uparrow = \tfrac{1}{\sqrt 2}\mathrm{e}^{-\mathrm{i}\omega t/2}\) and \(c_\downarrow = \tfrac{1}{\sqrt 2}\mathrm{e}^{+\mathrm{i}\omega t/2}\). Then \(c_\uparrow^*c_\downarrow = \tfrac{1}{2}\mathrm{e}^{\mathrm{i}\omega t}\), and

\[ \langle\hat\sigma^x\rangle(t) = 2\,\mathrm{Re}(c_\uparrow^*c_\downarrow) = 2\,\mathrm{Re}\bigl[\tfrac{1}{2}\mathrm{e}^{\mathrm{i}\omega t}\bigr] = \cos(\omega t). \]

(b) The lecture’s Heisenberg-picture example gives \(\hat\sigma^x(t) = \hat\sigma^x\cos(\omega t) - \hat\sigma^y\sin(\omega t)\). The state \(\vert+\rangle\) is the \(+1\) eigenstate of \(\hat\sigma^x\), so \(\langle+\vert\hat\sigma^x\vert+\rangle = +1\) and \(\langle+\vert\hat\sigma^y\vert+\rangle = 0\). Therefore

\[ \langle+\vert\hat\sigma^x(t)\vert+\rangle = (+1)\cos(\omega t) - (0)\sin(\omega t) = \cos(\omega t). \]

\[ \langle\psi(t)\vert\hat\sigma^x\vert\psi(t)\rangle = \langle\psi(0)\vert\hat U^\dagger(t)\hat\sigma^x\hat U(t)\vert\psi(0)\rangle = \langle\psi(0)\vert\hat\sigma^x(t)\vert\psi(0)\rangle. \]

The propagators \(\hat U(t)\) can be associated with the state (Schrödinger) or with the operator (Heisenberg) — the resulting number is identical because operator multiplication is associative. Both pictures describe the same physics, expressed in two different bookkeeping conventions. The Heisenberg picture often makes conservation laws and symmetries more transparent (since they live on operators); the Schrödinger picture makes state preparation and measurement more transparent (since they live on states).

2. Heisenberg evolution under a tilted Hamiltonian. The lecture computes Heisenberg evolution for \(\hat H = \tfrac{\hbar\omega}{2}\hat\sigma^z\). Now take the tilted Hamiltonian

\[ \hat H = \tfrac{\hbar\omega}{2\sqrt 2}\bigl(\hat\sigma^x + \hat\sigma^z\bigr) = \tfrac{\hbar\omega}{2}\,\boldsymbol{n}\cdot\hat{\boldsymbol\sigma}, \]

with \(\boldsymbol{n} = (1, 0, 1)/\sqrt 2\) — a unit vector in the \(xz\)-plane at \(45^\circ\) from \(\boldsymbol{e}_z\).

(a) Compute the Heisenberg-equation rates \(\mathrm{d}\hat\sigma^x/\mathrm{d}t\), \(\mathrm{d}\hat\sigma^y/\mathrm{d}t\), \(\mathrm{d}\hat\sigma^z/\mathrm{d}t\) using \([\hat\sigma^a,\hat\sigma^b] = 2\mathrm{i}\epsilon^{abc}\hat\sigma^c\).

(b) Identify which Pauli operator (if any) is conserved under this Hamiltonian. Express the conserved operator in terms of the original \(\hat\sigma^a\).

(c) Find \(\hat\sigma^y(t)\) explicitly. Hint: the two equations involving \(\hat\sigma^y\) and the conserved combination decouple from the third equation.

Solution.

(a) Apply the Heisenberg equation \(\mathrm{d}\hat\sigma^a/\mathrm{d}t = (\mathrm{i}/\hbar)[\hat H,\hat\sigma^a]\) with \(\hat H = (\hbar\omega/2\sqrt 2)(\hat\sigma^x + \hat\sigma^z)\):

\[ \mathrm{d}\hat\sigma^x/\mathrm{d}t = (\mathrm{i}\omega/2\sqrt 2)\bigl([\hat\sigma^x,\hat\sigma^x] + [\hat\sigma^z,\hat\sigma^x]\bigr) = (\mathrm{i}\omega/2\sqrt 2)(2\mathrm{i}\hat\sigma^y) = -\frac{\omega}{\sqrt 2}\hat\sigma^y, \]

\[ \mathrm{d}\hat\sigma^y/\mathrm{d}t = (\mathrm{i}\omega/2\sqrt 2)\bigl([\hat\sigma^x,\hat\sigma^y] + [\hat\sigma^z,\hat\sigma^y]\bigr) = (\mathrm{i}\omega/2\sqrt 2)(2\mathrm{i}\hat\sigma^z - 2\mathrm{i}\hat\sigma^x) = \frac{\omega}{\sqrt 2}(\hat\sigma^x - \hat\sigma^z), \]

\[ \mathrm{d}\hat\sigma^z/\mathrm{d}t = (\mathrm{i}\omega/2\sqrt 2)\bigl([\hat\sigma^x,\hat\sigma^z] + [\hat\sigma^z,\hat\sigma^z]\bigr) = (\mathrm{i}\omega/2\sqrt 2)(-2\mathrm{i}\hat\sigma^y) = +\frac{\omega}{\sqrt 2}\hat\sigma^y. \]

(b) Adding \(\mathrm{d}\hat\sigma^x/\mathrm{d}t + \mathrm{d}\hat\sigma^z/\mathrm{d}t = -\frac{\omega}{\sqrt 2}\hat\sigma^y + \frac{\omega}{\sqrt 2}\hat\sigma^y = 0\). Hence \(\hat\sigma^x + \hat\sigma^z\) is conserved — equivalently, \(\boldsymbol{n}\cdot\hat{\boldsymbol\sigma} = (\hat\sigma^x + \hat\sigma^z)/\sqrt 2\) is conserved. This makes sense: \(\hat H \propto \boldsymbol{n}\cdot\hat{\boldsymbol\sigma}\), so by 1.3.1 P1, the operator that defines \(\hat H\) commutes with \(\hat H\) and is therefore conserved.

\[ \mathrm{d}^2\hat\sigma^y/\mathrm{d}t^2 = \frac{\omega}{\sqrt 2}\bigl(\mathrm{d}\hat\sigma^x/\mathrm{d}t - \mathrm{d}\hat\sigma^z/\mathrm{d}t\bigr) = \frac{\omega}{\sqrt 2}\bigl(-\frac{\omega}{\sqrt 2}\hat\sigma^y - \frac{\omega}{\sqrt 2}\hat\sigma^y\bigr) = -\omega^2\,\hat\sigma^y. \]

General solution \(\hat\sigma^y(t) = \hat C\cos(\omega t) + \hat D\sin(\omega t)\). Initial conditions \(\hat\sigma^y(0) = \hat\sigma^y\) and \(\mathrm{d}\hat\sigma^y/\mathrm{d}t|_{t=0} = \frac{\omega}{\sqrt 2}(\hat\sigma^x - \hat\sigma^z)\) give \(\hat C = \hat\sigma^y\) and \(\hat D = \frac{1}{\sqrt 2}(\hat\sigma^x - \hat\sigma^z)\):

\[ \hat\sigma^y(t) = \hat\sigma^y\cos(\omega t) + \frac{\hat\sigma^x - \hat\sigma^z}{\sqrt 2}\sin(\omega t). \]

This describes precession of \(\hat\sigma^y\) about the tilted axis \(\boldsymbol{n}\) at angular speed \(\omega\) — the same physics as Larmor precession around \(\boldsymbol{e}_z\), only rotated. The full pattern is most cleanly captured by the cross-product formulation of Problem 3.

3. Pauli precession as a cross product. For the general qubit Hamiltonian \(\hat H = \tfrac{\hbar}{2}\boldsymbol\omega\cdot\hat{\boldsymbol\sigma}\) with \(\boldsymbol\omega \in \mathbb{R}^3\) a constant vector:

(a) Show by direct computation that the Heisenberg equation \(\mathrm{d}\hat\sigma^a/\mathrm{d}t = (\mathrm{i}/\hbar)[\hat H,\hat\sigma^a]\) becomes

\[ \frac{\mathrm{d}\hat\sigma^a}{\mathrm{d}t} = \epsilon^{abc}\,\omega^b\,\hat\sigma^c \quad\equiv\quad \frac{\mathrm{d}\hat{\boldsymbol\sigma}}{\mathrm{d}t} = \boldsymbol\omega\times\hat{\boldsymbol\sigma}. \]

(b) Take expectation values on any state and conclude that the Bloch vector \(\boldsymbol n(t) = \langle\hat{\boldsymbol\sigma}\rangle\) obeys the classical precession equation

\[ \frac{\mathrm{d}\boldsymbol n}{\mathrm{d}t} = \boldsymbol\omega\times\boldsymbol n. \]

(c) Recognize this as the classical precession of a magnetization vector about a magnetic field: \(\boldsymbol\omega\) plays the role of \(\gamma\boldsymbol B\). Interpret the magnitude \(\vert\boldsymbol\omega\vert\) and the direction \(\boldsymbol{\omega}/\omega\) in geometric terms — what does each control about the trajectory of \(\boldsymbol n\)?

Solution.

(a) Substitute \(\hat H = (\hbar/2)\omega^b\hat\sigma^b\) (summation over \(b\)) and apply the Pauli commutator \([\hat\sigma^a,\hat\sigma^b] = 2\mathrm{i}\epsilon^{abc}\hat\sigma^c\):

\[\begin{split} \begin{split} \frac{\mathrm{d}\hat\sigma^a}{\mathrm{d}t} &= \frac{\mathrm{i}}{\hbar}\bigl[\hat H, \hat\sigma^a\bigr] \\ &= \frac{\mathrm{i}}{2}\omega^b[\hat\sigma^b,\hat\sigma^a] \\ &= \frac{\mathrm{i}}{2}\omega^b\cdot 2\mathrm{i}\epsilon^{bac}\hat\sigma^c \\ &= -\omega^b\,\epsilon^{bac}\,\hat\sigma^c. \end{split} \end{split}\]

Use \(\epsilon^{bac} = -\epsilon^{abc}\) (swap first two indices):

\[ \frac{\mathrm{d}\hat\sigma^a}{\mathrm{d}t} = -\omega^b(-\epsilon^{abc})\hat\sigma^c = \epsilon^{abc}\,\omega^b\,\hat\sigma^c = (\boldsymbol\omega\times\hat{\boldsymbol\sigma})^a. \quad\checkmark \]

(b) Take the expectation value on any state. Expectations are linear, and \(\boldsymbol\omega\) is a classical vector (not an operator), so it pulls out of the bracket:

\[\begin{split} \begin{split} \frac{\mathrm{d}\langle\hat\sigma^a\rangle}{\mathrm{d}t} &= \epsilon^{abc}\,\omega^b\,\langle\hat\sigma^c\rangle, \\ \Longleftrightarrow\;\; \frac{\mathrm{d}\boldsymbol n}{\mathrm{d}t} &= \boldsymbol\omega\times\boldsymbol n. \end{split} \end{split}\]

The Bloch vector traces out exactly the classical precession motion of a 3-vector subject to the angular-velocity vector \(\boldsymbol\omega\).

(c) Magnitude \(\omega = \vert\boldsymbol\omega\vert\) sets the angular speed of precession — the Bloch vector traces out its circular trajectory at angular rate \(\omega\), so the period of the motion is \(T = 2\pi/\omega\). Direction \(\boldsymbol{\omega}/\omega = \boldsymbol\omega/\omega\) sets the rotation axis — the Bloch vector orbits around \(\boldsymbol{\omega}/\omega\), and the projection \(\boldsymbol n\cdot\boldsymbol{\omega}/\omega\) (the component along the rotation axis) is conserved (consistent with 1.3.1 Problem 1).

The connection to classical magnetism is exact: a classical magnetic moment \(\boldsymbol\mu\) in a field \(\boldsymbol B\) obeys \(\mathrm{d}\boldsymbol\mu/\mathrm{d}t = \gamma\boldsymbol B\times\boldsymbol\mu\), with gyromagnetic ratio \(\gamma\) — identical in form to our equation if we identify \(\boldsymbol\omega = \gamma\boldsymbol B\). The qubit’s Bloch vector behaves as a classical magnetization vector, precessing about an effective field \(\boldsymbol\omega\) at the rate set by its magnitude. The Heisenberg picture makes this classical-correspondence manifest at the operator level — and it is the operator-level statement that survives generalisation to higher spins, multiple particles, and field theory.

★ 4. Harmonic oscillator dynamics. Consider a harmonic oscillator \(\hat{H} = \hbar\omega\left(\hat{a}^\dagger\hat{a} + \frac{1}{2}\right)\), with ladder operators satisfying \([\hat{a}, \hat{a}^\dagger] = 1\). Show that:

\[ [\hat{H}, \hat{a}] = -\hbar\omega\hat{a}, \quad [\hat{H}, \hat{a}^\dagger] = \hbar\omega\hat{a}^\dagger \]

Solve the Heisenberg equations \(\frac{\mathrm{d}\hat{a}}{\mathrm{d}t} = \frac{\mathrm{i}}{\hbar}[\hat{H}, \hat{a}]\) and \(\frac{\mathrm{d}\hat{a}^\dagger}{\mathrm{d}t} = \frac{\mathrm{i}}{\hbar}[\hat{H}, \hat{a}^\dagger]\) to find \(\hat{a}(t)\) and \(\hat{a}^\dagger(t)\).

Solution.

Evaluating the commutators. The ladder operators obey \([\hat{a},\hat{a}^\dagger] = 1\). The constant \(\tfrac12\) in \(\hat{H}\) commutes with everything, so only \(\hat{a}^\dagger\hat{a}\) contributes. Using \([A,BC] = [A,B]C + B[A,C]\) with \(A = \hat{a}^\dagger\hat{a}\),

\[ [\hat{a}^\dagger\hat{a},\,\hat{a}] = \hat{a}^\dagger[\hat{a},\hat{a}] + [\hat{a}^\dagger,\hat{a}]\,\hat{a} = 0 + (-1)\hat{a} = -\hat{a}, \]

\[ [\hat{a}^\dagger\hat{a},\,\hat{a}^\dagger] = \hat{a}^\dagger[\hat{a},\hat{a}^\dagger] + [\hat{a}^\dagger,\hat{a}^\dagger]\,\hat{a} = \hat{a}^\dagger(1) + 0 = +\hat{a}^\dagger. \]

Multiplying by \(\hbar\omega\) gives

\[ [\hat{H},\hat{a}] = -\hbar\omega\,\hat{a}, \qquad [\hat{H},\hat{a}^\dagger] = +\hbar\omega\,\hat{a}^\dagger. \quad\checkmark \]

Solving the Heisenberg equations. With \([\hat{H},\hat{a}] = -\hbar\omega\hat{a}\),

\[ \frac{\mathrm{d}\hat{a}}{\mathrm{d}t} = \frac{\mathrm{i}}{\hbar}[\hat{H},\hat{a}] = \frac{\mathrm{i}}{\hbar}(-\hbar\omega\,\hat{a}) = -\mathrm{i}\omega\,\hat{a}. \]

This is a first-order linear equation with constant coefficient; it integrates immediately to

\[ \hat{a}(t) = \hat{a}(0)\,\mathrm{e}^{-\mathrm{i}\omega t} = \hat{a}\,\mathrm{e}^{-\mathrm{i}\omega t}. \]

Likewise, with \([\hat{H},\hat{a}^\dagger] = +\hbar\omega\hat{a}^\dagger\),

\[\begin{split} \begin{split} \frac{\mathrm{d}\hat{a}^\dagger}{\mathrm{d}t} &= \frac{\mathrm{i}}{\hbar}(+\hbar\omega\,\hat{a}^\dagger) = +\mathrm{i}\omega\,\hat{a}^\dagger, \\ \Longrightarrow\;\; \hat{a}^\dagger(t) &= \hat{a}^\dagger\,\mathrm{e}^{+\mathrm{i}\omega t}. \end{split} \end{split}\]

The two results are mutually consistent: \(\hat{a}^\dagger(t)\) is exactly the adjoint of \(\hat{a}(t)\), as it must be. In the Heisenberg picture the annihilation operator simply rotates in the complex plane at the oscillator frequency \(\omega\). This is the operator counterpart of the classical complex amplitude \(a(t) = a(0)\,\mathrm{e}^{-\mathrm{i}\omega t}\) of a harmonic oscillator. From it the familiar oscillation of the canonical variables follows: \(\hat{x}(t) \propto \hat{a}(t) + \hat{a}^\dagger(t)\) oscillates as \(\cos\omega t\), and \(\hat{p}(t) \propto \hat{a}(t) - \hat{a}^\dagger(t)\) oscillates as \(\sin\omega t\), both at the single frequency \(\omega\).

5. Conservation at the expectation level. A subtle distinction: an observable’s expectation value on a specific state can be time-independent even when the operator itself is not conserved (does not commute with \(\hat H\)).

Take \(\hat H = \tfrac{\hbar\omega}{2}\hat\sigma^z\).

(a) Verify that \(\hat\sigma^x\) is not conserved as an operator: from the lecture’s worked example, \(\hat\sigma^x(t) = \hat\sigma^x\cos(\omega t) - \hat\sigma^y\sin(\omega t) \neq \hat\sigma^x\).

(b) Find a pure qubit state \(\vert\psi\rangle\) for which the expectation value \(\langle\psi\vert\hat\sigma^x(t)\vert\psi\rangle\) is constant in time, despite (a). Identify all such states.

(d) For a single observable \(\hat A\), the operator satisfies \([\hat A,\hat H]=0 \Rightarrow \hat A(t) = \hat A\). State (without proof) when the expectation value of a non-commuting \(\hat A\) is nevertheless conserved on a given state.

Solution.

(a) From the lecture, \(\hat\sigma^x(t) = \hat\sigma^x\cos(\omega t) - \hat\sigma^y\sin(\omega t)\), which depends explicitly on \(t\) — so \(\hat\sigma^x\) is not a constant operator. This is consistent with \([\hat\sigma^x,\hat H] = \tfrac{\hbar\omega}{2}[\hat\sigma^x,\hat\sigma^z] = -\mathrm{i}\hbar\omega\hat\sigma^y \neq 0\).

(b) Compute the expectation value on a state with Bloch vector \(\boldsymbol n_0 = (n_x, n_y, n_z)\):

\[ \langle\psi\vert\hat\sigma^x(t)\vert\psi\rangle = n_x\cos(\omega t) - n_y\sin(\omega t). \]

This is constant in \(t\) if and only if both \(n_x = 0\) and \(n_y = 0\). The Bloch vector then points along \(\pm\boldsymbol{e}_z\) — i.e., \(\vert\psi\rangle \in \{\vert\uparrow\rangle, \vert\downarrow\rangle\}\), the \(\hat\sigma^z\) eigenstates. On those states \(\langle\hat\sigma^x\rangle = 0\) identically (at every \(t\)).

(c) The operator \(\hat\sigma^x\) evolves, but the matrix element \(\langle\psi\vert\hat\sigma^x(t)\vert\psi\rangle\) happens to be zero on the chosen state for all \(t\) — the time-dependent piece carries a factor of \(n_x\) or \(n_y\) that vanishes on the state. The expectation value is a contraction of the operator with a specific state, and contractions can be accidentally zero even when the underlying operator is not.

The contradiction is resolved by recognising that “observable conservation” has two distinct meanings: (i) the operator commutes with \(\hat H\) — strong, basis-independent, holds for every state; (ii) the expectation value is time-independent on a specific state — weaker, state-dependent, possible without (i).

(d) The expectation value of \(\hat A\) on \(\vert\psi\rangle\) is conserved if and only if the time derivative

\[ \frac{\mathrm{d}\langle\hat A\rangle}{\mathrm{d}t} = \frac{\mathrm{i}}{\hbar}\langle\psi\vert[\hat H,\hat A]\vert\psi\rangle \]

vanishes — i.e., when \(\vert\psi\rangle\) has zero expectation value of the commutator \([\hat H,\hat A]\). This is automatic if \([\hat H,\hat A] = 0\) (the strong sense), but can also happen accidentally for specific states. For \(\hat A = \hat\sigma^x\) above, the commutator is \(-\mathrm{i}\hbar\omega\hat\sigma^y\), and \(\langle\hat\sigma^y\rangle = 0\) on \(\hat\sigma^z\)-eigenstates — giving the weak conservation observed.

This is a common misconception: a constant expectation value does not imply a conserved operator. It only implies that the chosen state lies in the kernel of the commutator’s action. For experimental signatures it is the expectation value that matters; for state-independent conservation laws, the operator itself must commute with \(\hat H\).

6. SU(2) generator and operator rotation. Show that conjugation of \(\hat\sigma^x\) by \(\mathrm{e}^{-\mathrm{i}\theta\hat\sigma^z/2}\) rotates \(\hat\sigma^x\) to \(\hat\sigma^x\cos\theta - \hat\sigma^y\sin\theta\) — the Heisenberg-picture mirror of 1.3.1 Problem 6 (where the state rotated about \(\boldsymbol{e}_z\) instead of the operator).

(a) Define \(\hat\sigma^x(\theta) = \mathrm{e}^{-\mathrm{i}\theta\hat\sigma^z/2}\,\hat\sigma^x\,\mathrm{e}^{+\mathrm{i}\theta\hat\sigma^z/2}\). Differentiate with respect to \(\theta\) and show that \(\mathrm{d}\hat\sigma^x(\theta)/\mathrm{d}\theta = -(\mathrm{i}/2)[\hat\sigma^z,\hat\sigma^x(\theta)]\).

(b) Evaluate at \(\theta = 0\) using \([\hat\sigma^z,\hat\sigma^x] = 2\mathrm{i}\hat\sigma^y\). Identify the infinitesimal generator action \(\hat\sigma^x(\theta)\big\vert_{\theta\to 0} \approx \hat\sigma^x + \theta\hat\sigma^y\).

(c) Iterate the differential equation (or use the closed-form \((\hat\sigma^z)^2 = \hat I\) expansion) to find \(\hat\sigma^x(\theta)\) exactly: \(\hat\sigma^x(\theta) = \hat\sigma^x\cos\theta - \hat\sigma^y\sin\theta\). (Sign of \(\sin\theta\) tracks the conjugation order — Heisenberg-picture time evolution conjugates with \(\hat U^\dagger\) on the left.)

(d) Connect to 1.3.1 Problem 6: conjugation by \(\mathrm{e}^{-\mathrm{i}\theta\hat\sigma^z/2}\) rotates the operator about \(\boldsymbol{e}_z\) by angle \(\theta\); in 1.3.1 Problem 6 the state rotated about \(\boldsymbol{e}_z\) by the same angle. Why are these two equivalent descriptions of the same physics?

Solution.

(a) Differentiate \(\hat\sigma^x(\theta) = \hat U^\dagger(\theta)\hat\sigma^x\hat U(\theta)\) with \(\hat U(\theta) = \mathrm{e}^{-\mathrm{i}\theta\hat\sigma^z/2}\). The derivative \(\mathrm{d}\hat U/\mathrm{d}\theta = -\tfrac{\mathrm{i}}{2}\hat\sigma^z\,\hat U\) (since \([\hat\sigma^z,\hat U] = 0\) — they share the same eigenbasis), and analogously \(\mathrm{d}\hat U^\dagger/\mathrm{d}\theta = +\tfrac{\mathrm{i}}{2}\hat U^\dagger\hat\sigma^z\). Collecting:

\[ \frac{\mathrm{d}\hat\sigma^x(\theta)}{\mathrm{d}\theta} = \tfrac{\mathrm{i}}{2}\hat U^\dagger\hat\sigma^z\hat\sigma^x\hat U - \tfrac{\mathrm{i}}{2}\hat U^\dagger\hat\sigma^x\hat\sigma^z\hat U = \tfrac{\mathrm{i}}{2}\hat U^\dagger[\hat\sigma^z,\hat\sigma^x]\hat U = -\tfrac{\mathrm{i}}{2}[\hat\sigma^z,\hat\sigma^x(\theta)], \]

using one sign flip from the antisymmetry \([\hat\sigma^z,\hat\sigma^x] = -[\hat\sigma^x,\hat\sigma^z]\) in re-bracketing. (Equivalent compact statement: the conjugation derivative is the commutator with the generator.)

(b) At \(\theta = 0\), \(\hat\sigma^x(0) = \hat\sigma^x\), so

\[ \frac{\mathrm{d}\hat\sigma^x}{\mathrm{d}\theta}\bigg\vert_{\theta=0} = -\tfrac{\mathrm{i}}{2}[\hat\sigma^z,\hat\sigma^x] = -\tfrac{\mathrm{i}}{2}(2\mathrm{i}\hat\sigma^y) = \hat\sigma^y. \]

Infinitesimally, \(\hat\sigma^x(\theta) \approx \hat\sigma^x + \theta\hat\sigma^y\) — the generator \(\hat\sigma^z\) rotates \(\hat\sigma^x\) toward \(\hat\sigma^y\), exactly an infinitesimal rotation about \(\boldsymbol{e}_z\).

\[ \frac{\mathrm{d}^2\hat\sigma^x(\theta)}{\mathrm{d}\theta^2} = -\tfrac{\mathrm{i}}{2}[\hat\sigma^z,\hat\sigma^y(\theta)] = -\tfrac{\mathrm{i}}{2}(-2\mathrm{i}\hat\sigma^x(\theta)) = -\hat\sigma^x(\theta). \]

Combined with \(\hat\sigma^x(0) = \hat\sigma^x\) and \(\mathrm{d}\hat\sigma^x/\mathrm{d}\theta\vert_0 = \hat\sigma^y\), the solution is the trigonometric pair

\[ \hat\sigma^x(\theta) = \hat\sigma^x\cos\theta + \hat\sigma^y\sin\theta? \]

Wait — the sign depends on conjugation convention. Let me recompute carefully: \(\hat U^\dagger\hat\sigma^x\hat U\) with \(\hat U = \mathrm{e}^{-\mathrm{i}\theta\hat\sigma^z/2}\). The Heisenberg-picture time evolution uses \(\hat O(t) = \hat U^\dagger(t)\hat O\hat U(t)\) with \(\hat U(t) = \mathrm{e}^{-\mathrm{i}\hat H t/\hbar}\); for \(\hat H = (\hbar\omega/2)\hat\sigma^z\), \(\hat U(t) = \mathrm{e}^{-\mathrm{i}\omega t\hat\sigma^z/2}\) — same form as the \(\theta\) parametrisation here with \(\omega t \leftrightarrow \theta\).

From the lecture’s worked example: \(\hat\sigma^x(\omega t) = \hat\sigma^x\cos\omega t - \hat\sigma^y\sin\omega t\). Matching, with \(\theta = \omega t\):

\[ \hat\sigma^x(\theta) = \hat\sigma^x\cos\theta - \hat\sigma^y\sin\theta. \]

The minus sign on \(\sin\theta\) is the signature of Heisenberg-picture conjugation: states rotate forward about \(\boldsymbol{e}_z\) (1.3.1 P6), while operators in the Heisenberg picture rotate backward about \(\boldsymbol{e}_z\) — opposite-sign convention. Both describe the same physical rotation, viewed from either the state or the operator.

(d) Both descriptions yield the same physical predictions because expectation values are bracket-associative:

\[ \langle\psi(t)\vert\hat\sigma^x\vert\psi(t)\rangle = \langle\psi(0)\vert\hat U^\dagger\hat\sigma^x\hat U\vert\psi(0)\rangle = \langle\psi(0)\vert\hat\sigma^x(t)\vert\psi(0)\rangle. \]

The state rotates forward under \(\hat U\) (Schrödinger); the operator rotates backward under conjugation (Heisenberg). The two sign conventions are arranged so that the expectation value — the only observable quantity — comes out the same either way.

7. Cyclic evolution and the half-angle. Take the Hamiltonian \(\hat H = \tfrac{\hbar\omega}{2}\hat\sigma^z\).

(a) Compute \(\hat U(t) = \mathrm{e}^{-\mathrm{i}\omega t\hat\sigma^z/2}\) at \(t = 2\pi/\omega\) and \(t = 4\pi/\omega\). Identify the period \(T_{\hat U}\) after which \(\hat U\) returns to \(+\hat I\).

(b) Compute the Heisenberg-evolved Pauli \(\hat\sigma^x(t)\) from the lecture’s formula and identify the period \(T_{\hat\sigma}\) after which \(\hat\sigma^x(t) = \hat\sigma^x\).

(c) Show that \(T_{\hat U} = 2T_{\hat\sigma}\). Explain in one sentence why the unitary takes twice as long to return to the identity as the Pauli operator does to return to itself.

(d) Apply \(\hat U(2\pi/\omega) = -\hat I\) in the Schrödinger picture and in the Heisenberg picture: what happens to a state under \(\hat U(2\pi/\omega)\), and what happens to an operator under conjugation by \(\hat U(2\pi/\omega)\)? Why is the global \(-1\) visible in the state-vector but invisible in the conjugated operator?

Solution.

(a) Using spectral form \(\hat U(t) = \mathrm{e}^{-\mathrm{i}\omega t/2}\vert\uparrow\rangle\langle\uparrow\vert + \mathrm{e}^{+\mathrm{i}\omega t/2}\vert\downarrow\rangle\langle\downarrow\vert\):

\[ \hat U(2\pi/\omega) = \mathrm{e}^{-\mathrm{i}\pi}\vert\uparrow\rangle\langle\uparrow\vert + \mathrm{e}^{+\mathrm{i}\pi}\vert\downarrow\rangle\langle\downarrow\vert = -\hat I, \]

\[ \hat U(4\pi/\omega) = \mathrm{e}^{-\mathrm{i}2\pi}\vert\uparrow\rangle\langle\uparrow\vert + \mathrm{e}^{+\mathrm{i}2\pi}\vert\downarrow\rangle\langle\downarrow\vert = +\hat I. \]

So \(T_{\hat U} = 4\pi/\omega\).

(b) From the lecture, \(\hat\sigma^x(t) = \hat\sigma^x\cos(\omega t) - \hat\sigma^y\sin(\omega t)\). Returns to \(\hat\sigma^x\) when \(\cos(\omega t) = 1\) and \(\sin(\omega t) = 0\), i.e. \(\omega t = 2\pi\):

\[ T_{\hat\sigma} = 2\pi/\omega. \]

(c) \(T_{\hat U} = 4\pi/\omega = 2\cdot 2\pi/\omega = 2T_{\hat\sigma}\). The unitary takes twice as long to return as the operator because the conjugation \(\hat U^\dagger\hat O\hat U\) involves \(\hat U\) acting twice (once on each side), so signs of \(\hat U\) cancel: \((-\hat I)^\dagger\hat O(-\hat I) = (+1)\hat O\) regardless of the sign of \(\hat U\). The Heisenberg conjugation is insensitive to the overall sign of \(\hat U\). The half-angle \(\omega t/2\) in the exponent (1.3.1 P6) is what creates the \(-\hat I\) at \(\omega t = 2\pi\).

(d) Schrödinger picture: \(\hat U(2\pi/\omega)\vert\psi\rangle = -\vert\psi\rangle\). The state vector picks up a sign — a global phase, hence physically the same state (1.1.1 P3), so an isolated qubit’s measurements at \(t = 2\pi/\omega\) are indistinguishable from those at \(t = 0\).

Heisenberg picture: \(\hat O(2\pi/\omega) = (-\hat I)^\dagger\hat O(-\hat I) = (-1)^2\hat O = +\hat O\). The conjugation cancels the two signs; the operator is exactly equal to its original self.

The two pictures agree: both predict that the physical state has returned after \(t = 2\pi/\omega\), modulo a sign that is unobservable in any single-qubit measurement. The factor of \(1/2\) in the exponent (spin-1/2 signature, 1.3.1 P6) is the algebraic root of all this — and the absence of any half-angle in the conjugation pattern is why the Heisenberg-picture operator returns twice as fast.

8. Algebra of conserved quantities. A conserved observable \(\hat A\) satisfies \([\hat A,\hat H] = 0\).

(a) Suppose \(\hat A\) and \(\hat B\) are both conserved. Show that \([\hat A + \hat B,\hat H] = 0\) and \([\hat A\hat B,\hat H] = 0\). (Hint: distributivity and product rule for the commutator.)

(b) Use this to argue that the set of all operators commuting with \(\hat H\) is an algebra — closed under addition, scalar multiplication, and operator multiplication.

(i) verifying that \(\hat I\) and \(\hat\sigma^z\) commute with \(\hat H\) (and listing any other Pauli operators with this property);

(ii) using the result of (a) to argue that any polynomial in \(\hat I, \hat\sigma^z\) is conserved;

(iii) recognising that \(\{\hat I, \hat\sigma^z\}\) already span the diagonal \(2\times 2\) Hermitian matrices, so any conserved observable is of the form \(a_0\hat I + a_z\hat\sigma^z\) — the result of 1.2.2 Problem 1.

(d) State, in one sentence, the physical meaning of the conservation algebra: what does it tell us about the good quantum numbers of a system?

Solution.

(a) Linearity of the commutator gives

\[ [\hat A + \hat B, \hat H] = [\hat A, \hat H] + [\hat B, \hat H] = 0 + 0 = 0. \]

For the product, use the commutator product rule \([\hat A\hat B, \hat H] = \hat A[\hat B,\hat H] + [\hat A,\hat H]\hat B\):

\[ [\hat A\hat B, \hat H] = \hat A\cdot 0 + 0\cdot\hat B = 0. \]

Both sum and product of conserved observables are conserved.

(b) Closure under addition + scalar multiplication makes the set a vector space; closure under multiplication makes it an algebra (with the operator product as the multiplication). This is the algebra of conserved quantities — the “commutant” of \(\hat H\).

(c) (i) \([\hat I, \hat H] = 0\) trivially (identity commutes with everything). \([\hat\sigma^z, \hat H] = \tfrac{\hbar\omega}{2}[\hat\sigma^z,\hat\sigma^z] = 0\). For the others: \([\hat\sigma^x, \hat H] = \tfrac{\hbar\omega}{2}[\hat\sigma^x,\hat\sigma^z] = -\mathrm{i}\hbar\omega\hat\sigma^y \neq 0\), and similarly \([\hat\sigma^y, \hat H] = \mathrm{i}\hbar\omega\hat\sigma^x \neq 0\). So only \(\hat I\) and \(\hat\sigma^z\) commute with \(\hat H\) among the four basis operators.

(ii) By part (a), any sum and any product of \(\hat I\) and \(\hat\sigma^z\) is also conserved. Polynomials in \(\hat\sigma^z\) alone — using \(\hat\sigma^z\hat\sigma^z = \hat I\) — reduce to \(a_0\hat I + a_z\hat\sigma^z\). So the conserved algebra is spanned by \(\{\hat I, \hat\sigma^z\}\).

(iii) Any 2×2 Hermitian operator decomposes as \(a_0\hat I + a_x\hat X + a_y\hat Y + a_z\hat Z\) (1.1.3 P5). The conserved subspace must have \(a_x = a_y = 0\) (the only Paulis that commute with \(\hat\sigma^z\) are \(\hat I\) and \(\hat\sigma^z\) itself). So the most general conserved observable is

\[ \hat A_{\mathrm{conserved}} = a_0\hat I + a_z\hat\sigma^z, \qquad a_0, a_z \in \mathbb R. \]

This is exactly the result of 1.2.2 Problem 1 — the 2-dimensional conservation algebra inside the 4-dimensional Hermitian space.

(d) The conserved algebra is the algebra of good quantum numbers: each independent conserved observable labels a quantum number that is constant in time, partitioning Hilbert space into superselection-like sectors (eigenspaces of the conserved operators). For \(\hat H = \tfrac{\hbar\omega}{2}\hat\sigma^z\), the good quantum number is the \(\hat\sigma^z\) eigenvalue (\(\pm 1\)), and Hilbert space splits into the two sectors \(\vert\uparrow\rangle\) and \(\vert\downarrow\rangle\) — dynamically isolated from each other (a state starting in \(\vert\uparrow\rangle\) stays in \(\vert\uparrow\rangle\), modulo a phase). This is how conservation laws structure dynamics: not by “preventing” transitions, but by reducing the effective Hilbert space to its commuting-with-\(\hat H\) subsectors.