# 1.3.1 Unitary Evolution
Worked solutions for the homework problems in the [1.3.1 Unitary Evolution](../ch1_qubit/1-3-1-unitary-evolution) lecture. Each problem statement is reproduced verbatim, followed by a complete worked solution.

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**1. Conservation of observables commuting with H.** Time evolution is generated by a (time-independent) Hermitian Hamiltonian $\hat H$ via $\hat U(t) = \mathrm{e}^{-\mathrm{i}\hat H t/\hbar}$.

(a) Argue (in one sentence) that $\hat H$ commutes with $\hat U(t)$ for every $t$ — and therefore that $\hat H$ commutes with itself under any time-evolution.

(b) Use this to show that $\langle\hat H\rangle$ is constant in time on any state: $\frac{\mathrm{d}}{\mathrm{d}t}\langle\hat H\rangle_{\psi(t)} = 0$.

(c) Generalize: a Hermitian observable $\hat O$ satisfies $[\hat O, \hat H] = 0$. Show that $\hat O$ commutes with $\hat U(t)$, and conclude that $\langle\hat O\rangle$ is constant in time on any initial state. (Hint: write $\langle\hat O\rangle(t) = \langle\psi(0)\vert\hat U^\dagger\hat O\hat U\vert\psi(0)\rangle$ and use $\hat O\hat U = \hat U\hat O$.)

(d) Apply to the Hamiltonian $\hat H = \Delta\,\hat Z$ (energy splitting $2\Delta$). Compute $\langle\hat Z\rangle(t)$ and $\langle\hat X\rangle(t)$ on the initial state $\vert\psi(0)\rangle = \cos(\theta_0/2)\vert 0\rangle + \mathrm{e}^{\mathrm{i}\varphi_0}\sin(\theta_0/2)\vert 1\rangle$. Verify $\langle\hat Z\rangle$ is conserved and $\langle\hat X\rangle$ oscillates; identify the oscillation frequency.

**Solution.**

(a) $\hat U(t) = \mathrm{e}^{-\mathrm{i}\hat H t/\hbar}$ is built from the Taylor series in $\hat H$, so every term commutes with $\hat H$, and therefore so does the whole exponential: $[\hat H, \hat U(t)] = 0$.

(b) Compute $\langle\hat H\rangle$ at time $t$:

$$
\langle\hat H\rangle_{\psi(t)} = \langle\psi(0)\vert\,\hat U^\dagger(t)\hat H\,\hat U(t)\,\vert\psi(0)\rangle.
$$

Use $\hat H\hat U = \hat U\hat H$ from (a), then $\hat U^\dagger\hat U = \hat I$:

$$
\hat U^\dagger\hat H\hat U = \hat U^\dagger\hat U\hat H = \hat I\hat H = \hat H.
$$

Hence $\langle\hat H\rangle_{\psi(t)} = \langle\psi(0)\vert\hat H\vert\psi(0)\rangle = \langle\hat H\rangle_{\psi(0)}$, constant in time. Energy is conserved.

(c) The same argument works for any $\hat O$ with $[\hat O, \hat H] = 0$. Since $\hat U$ is built from $\hat H$, $\hat O$ also commutes with $\hat U(t)$ (operators that commute with $\hat H$ commute with every power of $\hat H$, hence with the exponential). Then

$$
\langle\hat O\rangle(t) = \langle\psi(0)\vert\hat U^\dagger\hat O\hat U\vert\psi(0)\rangle = \langle\psi(0)\vert\hat U^\dagger\hat U\hat O\vert\psi(0)\rangle = \langle\psi(0)\vert\hat O\vert\psi(0)\rangle,
$$

constant in time. **An observable commuting with the Hamiltonian is a conserved quantity** — and this is the operator-algebra origin of conservation laws in quantum mechanics.

(d) Hamiltonian $\hat H = \Delta\hat Z$ has $\hat U(t) = \mathrm{e}^{-\mathrm{i}\Delta\hat Z t/\hbar}$. Applying to $\vert\psi(0)\rangle$:

$$
\vert\psi(t)\rangle = \mathrm{e}^{-\mathrm{i}\Delta t/\hbar}\cos(\theta_0/2)\vert 0\rangle + \mathrm{e}^{+\mathrm{i}\Delta t/\hbar}\mathrm{e}^{\mathrm{i}\varphi_0}\sin(\theta_0/2)\vert 1\rangle.
$$

Pull out the global phase $\mathrm{e}^{-\mathrm{i}\Delta t/\hbar}$:

$$
\vert\psi(t)\rangle = \mathrm{e}^{-\mathrm{i}\Delta t/\hbar}\bigl[\cos(\theta_0/2)\vert 0\rangle + \mathrm{e}^{\mathrm{i}(\varphi_0 + 2\Delta t/\hbar)}\sin(\theta_0/2)\vert 1\rangle\bigr].
$$

So the state at time $t$ has the same polar angle $\theta_0$ but its azimuthal angle has advanced to $\varphi(t) = \varphi_0 + 2\Delta t/\hbar$. From the Bloch-vector formula,

$$
\langle\hat Z\rangle(t) = \cos\theta_0 = \langle\hat Z\rangle(0), \qquad \langle\hat X\rangle(t) = \sin\theta_0\cos\!\bigl(\varphi_0 + 2\Delta t/\hbar\bigr).
$$

$\langle\hat Z\rangle$ is **constant** ($[\hat Z, \hat H] = 0$ since $\hat H \propto \hat Z$), confirming part (c). $\langle\hat X\rangle$ **oscillates** at angular frequency

$$
\omega_{21} = \frac{2\Delta}{\hbar} = \frac{E_+ - E_-}{\hbar},
$$

the transition frequency between the two energy levels ($E_\pm = \pm\Delta$). $\hat X$ does not commute with $\hat H = \Delta\hat Z$, so it is not conserved — its expectation rotates in time, tracing out the Larmor precession of the Bloch vector around $\boldsymbol{e}_z$ (developed further in 1.3.2).

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**2. Exponential of a Pauli matrix.** The Pauli matrix $\hat{X} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ has eigenvalues $\pm 1$ with eigenstates $\vert\pm\rangle = \tfrac{1}{\sqrt{2}}(\vert 0\rangle \pm \vert 1\rangle)$.

(a) Use the **spectral decomposition** of $\hat X$ to compute $\hat{U}(\theta) = \mathrm{e}^{\mathrm{i}\hat{X} \theta/2}$ as an explicit $2 \times 2$ matrix in the $\{\vert 0\rangle, \vert 1\rangle\}$ basis.

(b) Verify that your result agrees with the closed form $\hat U(\theta) = \cos(\theta/2)\hat{I} + \mathrm{i}\sin(\theta/2)\hat{X}$.

(c) Identify the property of Pauli matrices that makes the closed form possible. (Hint: square the spectral form and use $(\hat X)^2 = \hat I$ from 1.1.3 Problem 7.)

**Solution.**

(a) The spectral decomposition is $\hat{X} = (+1)\vert+\rangle\langle+\vert + (-1)\vert-\rangle\langle-\vert$. The exponential acts as a scalar exponential on each eigenspace:

$$
\hat{U}(\theta) = \mathrm{e}^{\mathrm{i}\theta/2}\vert+\rangle\langle+\vert + \mathrm{e}^{-\mathrm{i}\theta/2}\vert-\rangle\langle-\vert.
$$

Substituting the $2\times 2$ projector matrices,

$$
\vert+\rangle\langle+\vert \mapsto \tfrac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix},
\qquad
\vert-\rangle\langle-\vert \mapsto \tfrac{1}{2}\begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix},
$$

we get

$$
\hat U(\theta) \mapsto \tfrac{\mathrm{e}^{\mathrm{i}\theta/2}}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} + \tfrac{\mathrm{e}^{-\mathrm{i}\theta/2}}{2}\begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix} = \begin{pmatrix} \cos(\theta/2) & \mathrm{i}\sin(\theta/2) \\ \mathrm{i}\sin(\theta/2) & \cos(\theta/2) \end{pmatrix},
$$

using $\tfrac{1}{2}(\mathrm{e}^{\mathrm{i}\theta/2} + \mathrm{e}^{-\mathrm{i}\theta/2}) = \cos(\theta/2)$ for the diagonal and $\tfrac{1}{2}(\mathrm{e}^{\mathrm{i}\theta/2} - \mathrm{e}^{-\mathrm{i}\theta/2}) = \mathrm{i}\sin(\theta/2)$ for the off-diagonal.

(b) Expand the closed form:

$$
\cos(\theta/2)\hat I + \mathrm{i}\sin(\theta/2)\hat X \mapsto \cos(\theta/2)\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + \mathrm{i}\sin(\theta/2)\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} \cos(\theta/2) & \mathrm{i}\sin(\theta/2) \\ \mathrm{i}\sin(\theta/2) & \cos(\theta/2) \end{pmatrix}. \quad\checkmark
$$

(c) The closed form follows from $(\hat X)^2 = \hat I$ (1.1.3 P7). Split the Taylor series of $\mathrm{e}^{\mathrm{i}\hat X\phi}$ into even and odd powers, using $(\hat X)^{2k} = \hat I$ and $(\hat X)^{2k+1} = \hat X$:

$$
\mathrm{e}^{\mathrm{i}\hat X\phi} = \sum_{k} \frac{(\mathrm{i}\phi)^{2k}}{(2k)!}\hat I + \sum_{k} \frac{(\mathrm{i}\phi)^{2k+1}}{(2k+1)!}\hat X = \cos\phi\,\hat I + \mathrm{i}\sin\phi\,\hat X.
$$

Setting $\phi = \theta/2$ recovers the result. The same identity holds for **any** operator with square $\hat I$ — so any $\boldsymbol{n}\cdot\hat{\boldsymbol\sigma}$ along a unit axis, by 1.1.3 Problem 7. This is the foundation of the Bloch-sphere rotation formula generalised in Problem 6.

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**3. Extracting the generator.** Given the one-parameter family $\hat{U}(\theta) = \mathrm{e}^{\mathrm{i}\hat{Z}\theta/2}$, extract the generator by computing $\hat{G} = -\mathrm{i}\,\dfrac{\mathrm{d}\hat{U}}{\mathrm{d}\theta}\bigg|_{\theta=0}$ and verify you recover $\hat{G} = \hat{Z}/2$.

**Solution.**

By the spectral decomposition $\hat{Z} = (+1)\vert 0\rangle\langle 0\vert + (-1)\vert 1\rangle\langle 1\vert$, the family is diagonal:

$$
\hat{U}(\theta) = \mathrm{e}^{\mathrm{i}\theta/2}\vert 0\rangle\langle 0\vert + \mathrm{e}^{-\mathrm{i}\theta/2}\vert 1\rangle\langle 1\vert \mapsto \begin{pmatrix} \mathrm{e}^{\mathrm{i}\theta/2} & 0 \\ 0 & \mathrm{e}^{-\mathrm{i}\theta/2} \end{pmatrix}.
$$

Differentiate entry by entry:

$$
\frac{\mathrm{d}\hat{U}}{\mathrm{d}\theta} \mapsto \begin{pmatrix} \tfrac{\mathrm{i}}{2}\mathrm{e}^{\mathrm{i}\theta/2} & 0 \\ 0 & -\tfrac{\mathrm{i}}{2}\mathrm{e}^{-\mathrm{i}\theta/2} \end{pmatrix}.
$$

Evaluate at $\theta = 0$:

$$
\frac{\mathrm{d}\hat{U}}{\mathrm{d}\theta}\bigg|_{\theta=0} \mapsto \begin{pmatrix} \tfrac{\mathrm{i}}{2} & 0 \\ 0 & -\tfrac{\mathrm{i}}{2} \end{pmatrix} = \frac{\mathrm{i}}{2}\hat{Z}.
$$

Multiply by $-\mathrm{i}$:

$$
\hat{G} = -\mathrm{i}\cdot\frac{\mathrm{i}}{2}\hat{Z} = \frac{1}{2}\hat{Z}. \quad\checkmark
$$

This matches the generator read straight off the exponent. The general identity is $\hat G = -\mathrm{i}\,\hat U'(0)$, which cancels the $\mathrm{i}$ from the chain rule and returns the Hermitian generator itself. Physically, $\hat U(\mathrm{d}\theta) \approx \hat I + \mathrm{i}\hat G\,\mathrm{d}\theta$ — the generator is what the unitary does at first order, and the inverse map "differentiate at the identity" reads it off. This is the algebraic infrastructure behind every continuous symmetry in quantum mechanics: rotations have angular momentum generators, time translations have Hamiltonian generators (Problem 8), spatial translations have momentum generators.

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**4. Two-level Hamiltonian.** A two-level system has $\hat H = E_0\hat I + \Delta\hat Z$, with an energy offset $E_0$ and a splitting $2\Delta$ between the two levels.

(a) Find the energy eigenvalues and eigenstates. Write the time-evolution operator $\hat U(t)$ in spectral form.

(b) As a $2\times 2$ matrix in the $\{\vert 0\rangle, \vert 1\rangle\}$ basis, factor $\hat U(t)$ into an overall scalar phase (from $E_0\hat I$) and a relative-phase matrix (from $\Delta\hat Z$).

(c) Apply $\hat U(t)$ to each energy eigenstate. Show that the offset $E_0$ contributes only a **global** phase shared by both branches, and that the splitting $\Delta$ controls the **relative** phase between them. Conclude that the absolute zero of energy is unobservable — only energy *differences* matter for dynamics.

**Solution.**

(a) Both $\hat I$ and $\hat Z$ are diagonal in $\{\vert 0\rangle,\vert 1\rangle\}$:

$$
\hat H\vert 0\rangle = (E_0 + \Delta)\vert 0\rangle, \qquad \hat H\vert 1\rangle = (E_0 - \Delta)\vert 1\rangle.
$$

So eigenvalues are $E_\pm = E_0 \pm \Delta$ with eigenstates $\vert 0\rangle, \vert 1\rangle$. The time-evolution operator in spectral form:

$$
\hat U(t) = \mathrm{e}^{-\mathrm{i}(E_0 + \Delta)t/\hbar}\vert 0\rangle\langle 0\vert + \mathrm{e}^{-\mathrm{i}(E_0 - \Delta)t/\hbar}\vert 1\rangle\langle 1\vert.
$$

(b) Factor out $\mathrm{e}^{-\mathrm{i}E_0 t/\hbar}$ (which is the same for both branches because $E_0\hat I$ commutes with everything):

$$
\hat U(t) = \mathrm{e}^{-\mathrm{i}E_0 t/\hbar}\bigl[\mathrm{e}^{-\mathrm{i}\Delta t/\hbar}\vert 0\rangle\langle 0\vert + \mathrm{e}^{+\mathrm{i}\Delta t/\hbar}\vert 1\rangle\langle 1\vert\bigr] = \mathrm{e}^{-\mathrm{i}E_0 t/\hbar}\begin{pmatrix}\mathrm{e}^{-\mathrm{i}\Delta t/\hbar} & 0 \\ 0 & \mathrm{e}^{+\mathrm{i}\Delta t/\hbar}\end{pmatrix}.
$$

(c) Apply to each eigenstate:

$$
\hat U(t)\vert 0\rangle = \mathrm{e}^{-\mathrm{i}(E_0 + \Delta)t/\hbar}\vert 0\rangle, \qquad \hat U(t)\vert 1\rangle = \mathrm{e}^{-\mathrm{i}(E_0 - \Delta)t/\hbar}\vert 1\rangle.
$$

Each picks up only a phase — energy eigenstates are **stationary** (the probability distribution of any measurement is constant in time).

For a superposition $\vert\psi(0)\rangle = c_0\vert 0\rangle + c_1\vert 1\rangle$, the offset $E_0$ contributes the *same* phase $\mathrm{e}^{-\mathrm{i}E_0 t/\hbar}$ to **both** branches:

$$
\vert\psi(t)\rangle = \mathrm{e}^{-\mathrm{i}E_0 t/\hbar}\bigl[c_0\mathrm{e}^{-\mathrm{i}\Delta t/\hbar}\vert 0\rangle + c_1\mathrm{e}^{+\mathrm{i}\Delta t/\hbar}\vert 1\rangle\bigr].
$$

The factor $\mathrm{e}^{-\mathrm{i}E_0 t/\hbar}$ is a global phase — unobservable. Only the *relative* phase $\mathrm{e}^{-\mathrm{i}\cdot 2\Delta t/\hbar}$ between the two branches has physical consequences (Problem 7 traces this in the Bloch sphere).

**Lesson: the absolute zero of energy is unobservable.** Shifting $\hat H \to \hat H + c\hat I$ (with $c$ any real constant) multiplies $\hat U(t)$ by $\mathrm{e}^{-\mathrm{i}c t/\hbar}$, an unobservable global phase. Only energy *differences* $E_+ - E_- = 2\Delta$ enter measurable quantities — this is why "the energy of the ground state" can be set to any convenient value without affecting physics.

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**5. Normalization and unitarity.** Use the unitarity condition $\hat U^\dagger\hat U = \hat I$ to prove that if $\vert\psi(0)\rangle$ is normalised, then $\vert\psi(t)\rangle = \hat U(t)\vert\psi(0)\rangle$ is normalised for every $t$. (One-line calculation expected.) Why is this a non-trivial physical requirement — what would go wrong if time evolution were not unitary?

**Solution.**

Compute the norm:

$$
\langle\psi(t)\vert\psi(t)\rangle = \langle\psi(0)\vert\,\hat U^\dagger(t)\hat U(t)\,\vert\psi(0)\rangle = \langle\psi(0)\vert\hat I\vert\psi(0)\rangle = \langle\psi(0)\vert\psi(0)\rangle = 1. \quad\checkmark
$$

Unitarity collapses the sandwich $\hat U^\dagger\hat U$ to $\hat I$ in a single step. Without it, the norm could drift in time.

**Physical interpretation.** The norm $\langle\psi\vert\psi\rangle$ encodes the total probability summed over a complete measurement: by the Born rule and completeness $\sum_m\hat P_m = \hat I$ from 1-2-3,

$$
\sum_m P(m\vert\psi) = \sum_m \langle\psi\vert\hat P_m\vert\psi\rangle = \langle\psi\vert\hat I\vert\psi\rangle = \langle\psi\vert\psi\rangle.
$$

So $\langle\psi\vert\psi\rangle = 1$ is the statement that *some* outcome is found with certainty whenever a measurement is performed. If time evolution were non-unitary, this total probability could exceed or fall below $1$ — quantum mechanics would lose its probability interpretation. Unitarity is therefore not a mathematical nicety but a hard constraint: the dynamics must preserve the probabilistic content of the state.

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**6. Unitary as Bloch-sphere rotation.** Consider the unitary $\hat U(\theta) = \mathrm{e}^{-\mathrm{i}\hat Z\theta/2}$.

(a) Apply $\hat U(\theta)$ to the general qubit state $\vert\psi\rangle = \cos(\theta_0/2)\vert 0\rangle + \mathrm{e}^{\mathrm{i}\varphi_0}\sin(\theta_0/2)\vert 1\rangle$. Find the resulting state.

(b) Pull out the overall global phase. Read off the new Bloch angles $(\theta_0', \varphi_0')$ in standard form. Conclude that $\hat U(\theta)$ **rotates the Bloch vector by angle $\theta$ about the $\boldsymbol{e}_z$ axis** while leaving the polar angle $\theta_0$ unchanged.

(c) Apply $\hat U(2\pi)$ to a generic state $\vert\psi\rangle$. Show that $\hat U(2\pi) = -\hat I$, so the **state vector** acquires a sign — even though the **physical state** (unchanged under global phase) returns to itself. Explain in one sentence the role of the factor of $1/2$ in the exponent: how many full revolutions are needed for $\hat U$ itself to return to the identity?

**Solution.**

(a) Apply the unitary, using the spectral form $\hat U(\theta) = \mathrm{e}^{-\mathrm{i}\theta/2}\vert 0\rangle\langle 0\vert + \mathrm{e}^{+\mathrm{i}\theta/2}\vert 1\rangle\langle 1\vert$:

$$
\hat U(\theta)\vert\psi\rangle = \mathrm{e}^{-\mathrm{i}\theta/2}\cos(\theta_0/2)\vert 0\rangle + \mathrm{e}^{+\mathrm{i}\theta/2}\mathrm{e}^{\mathrm{i}\varphi_0}\sin(\theta_0/2)\vert 1\rangle.
$$

(b) Factor out $\mathrm{e}^{-\mathrm{i}\theta/2}$:

$$
\hat U(\theta)\vert\psi\rangle = \mathrm{e}^{-\mathrm{i}\theta/2}\Bigl[\cos(\theta_0/2)\vert 0\rangle + \mathrm{e}^{\mathrm{i}(\varphi_0 + \theta)}\sin(\theta_0/2)\vert 1\rangle\Bigr].
$$

Apart from the global phase, this is the standard Bloch parametrization with new angles

$$
\theta_0' = \theta_0, \qquad \varphi_0' = \varphi_0 + \theta.
$$

The polar angle is unchanged; the azimuthal angle has advanced by $\theta$. Geometrically, the Bloch vector has been rotated by $\theta$ about the $\boldsymbol{e}_z$ axis — exactly what one expects from a unitary whose generator is $\hat Z/2$.

(c) At $\theta = 2\pi$ the spectral form gives

$$
\hat U(2\pi) = \mathrm{e}^{-\mathrm{i}\pi}\vert 0\rangle\langle 0\vert + \mathrm{e}^{+\mathrm{i}\pi}\vert 1\rangle\langle 1\vert = (-1)\vert 0\rangle\langle 0\vert + (-1)\vert 1\rangle\langle 1\vert = -\hat I.
$$

Applying to any state, $\hat U(2\pi)\vert\psi\rangle = -\vert\psi\rangle$ — the state vector picks up a sign, which is a global phase. The **physical** state is unchanged ($-\vert\psi\rangle$ is the same physical state as $\vert\psi\rangle$, by 1.1.1 Problem 3), so geometrically the Bloch vector has returned to itself after a $2\pi$ "rotation."

**Role of the $1/2$.** The factor $1/2$ in the exponent halves the "speed" of $\hat U$ relative to the geometric rotation angle. A rotation that completes a full revolution on the Bloch sphere ($\theta = 2\pi$) corresponds to *half* of a full cycle in the unitary's exponent ($\theta/2 = \pi$), so $\hat U(2\pi) = -\hat I$ rather than $+\hat I$. The unitary itself returns to $\hat I$ only at $\theta = 4\pi$:

$$
\hat U(4\pi) = \mathrm{e}^{-\mathrm{i}\cdot 2\pi}\vert 0\rangle\langle 0\vert + \mathrm{e}^{+\mathrm{i}\cdot 2\pi}\vert 1\rangle\langle 1\vert = \hat I.
$$

The qubit is a "double cover" of the Bloch sphere's rotations — a foreshadowing of the deeper spin-1/2 representation theory developed in Chapter 2.

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**7. Superposition time evolution.** Consider the superposition $\vert\psi(0)\rangle = \frac{1}{\sqrt{2}}(\vert E_1\rangle + \vert E_2\rangle)$, where $\vert E_1\rangle$ and $\vert E_2\rangle$ are energy eigenstates with eigenvalues $E_1$ and $E_2$. Find $\vert\psi(t)\rangle$.

(a) Apply $\hat U(t) = \mathrm{e}^{-\mathrm{i}\hat H t/\hbar}$ term by term to the superposition.

(b) Pull out the overall global phase and identify the relative-phase factor between the two branches.

(c) Define the **transition frequency** $\omega_{21} = (E_2 - E_1)/\hbar$. Show that the relative phase advances at angular frequency $\omega_{21}$, and that the populations $\vert c_n\vert^2$ are constant in time (consistent with Problem 1 since $\hat H$ is conserved). Identify the period $T$ after which the relative phase has rotated by $2\pi$.

(d) Briefly explain why a system in a single energy eigenstate is "stationary" (no observable time dependence) while a superposition is not.

**Solution.**

(a) Linearity of $\hat U(t)$ acts term by term:

$$
\vert\psi(t)\rangle = \hat U(t)\vert\psi(0)\rangle = \frac{1}{\sqrt 2}\bigl[\mathrm{e}^{-\mathrm{i}E_1 t/\hbar}\vert E_1\rangle + \mathrm{e}^{-\mathrm{i}E_2 t/\hbar}\vert E_2\rangle\bigr].
$$

(b) Factor out $\mathrm{e}^{-\mathrm{i}E_1 t/\hbar}$:

$$
\vert\psi(t)\rangle = \frac{\mathrm{e}^{-\mathrm{i}E_1 t/\hbar}}{\sqrt 2}\bigl[\vert E_1\rangle + \mathrm{e}^{-\mathrm{i}(E_2 - E_1)t/\hbar}\vert E_2\rangle\bigr].
$$

The leading $\mathrm{e}^{-\mathrm{i}E_1 t/\hbar}$ is a **global phase** — unobservable. The factor $\mathrm{e}^{-\mathrm{i}(E_2 - E_1)t/\hbar}$ is the **relative phase**, multiplying only the $\vert E_2\rangle$ branch.

(c) The relative phase advances at the transition frequency

$$
\omega_{21} = \frac{E_2 - E_1}{\hbar},
$$

with period $T = 2\pi/\omega_{21} = 2\pi\hbar/(E_2 - E_1)$. The populations $\vert c_n\vert^2 = 1/2$ remain constant for both branches: only the relative phase rotates. This is consistent with Problem 1 — $\hat H$ commutes with itself, so $\langle\hat H\rangle$ (and indeed any function of $\hat H$) is conserved; populations in energy eigenstates are constants of motion.

(d) An **energy eigenstate** picks up only a global phase $\mathrm{e}^{-\mathrm{i}E_n t/\hbar}$ under time evolution. Global phases are unobservable, so no measurement on the state changes with time — the state is *stationary*. A **superposition** of energy eigenstates also picks up a global phase, but the *relative* phase between branches rotates at frequency $\omega_{21}$, which IS observable through any observable $\hat A$ with $\langle E_1\vert\hat A\vert E_2\rangle \neq 0$: $\langle\hat A\rangle(t)$ then oscillates at frequency $\omega_{21}$. Dynamics live in the **relative phases between energy components**, set by energy *differences* — not by the absolute energies themselves (Problem 4).

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**8. Composition and the time-evolution group.** The time-evolution operator is $\hat U(t) = \mathrm{e}^{-\mathrm{i}\hat H t/\hbar}$.

(a) Show that $\hat U(t_1)\hat U(t_2) = \hat U(t_1 + t_2)$ for any times $t_1, t_2$.

(b) Use $\hat U(0) = \hat I$ together with the composition property to identify $\hat U(t)^{-1}$ in two equivalent ways: as $\hat U(-t)$ and as $\hat U(t)^\dagger$. Conclude that the family $\{\hat U(t)\}$ is closed under inverses and composition — a **one-parameter group**.

(c) The composition property allows time evolution to be sliced into arbitrarily many small intervals: $\hat U(t) = \hat U(t/N)^N$ for any positive integer $N$. Use this to argue why the final state cannot depend on whether the experimenter mentally chops the time interval $[0, t]$ into a single span, two halves, or a thousand slivers — a physical consistency requirement.

(d) What changes if $\hat H$ depends explicitly on time? (Briefly identify what fails — do not derive the time-ordered solution; that comes in Chapter 5.)

**Solution.**

(a) Both exponentials are built from the *same* operator $\hat H$, so the exponents commute: $[\hat H t_1, \hat H t_2] = t_1 t_2[\hat H, \hat H] = 0$. The commuting-exponential identity $\mathrm{e}^{\hat A}\mathrm{e}^{\hat B} = \mathrm{e}^{\hat A + \hat B}$ (valid when $[\hat A, \hat B] = 0$) then gives

$$
\hat U(t_1)\hat U(t_2) = \mathrm{e}^{-\mathrm{i}\hat H t_1/\hbar}\mathrm{e}^{-\mathrm{i}\hat H t_2/\hbar} = \mathrm{e}^{-\mathrm{i}\hat H(t_1 + t_2)/\hbar} = \hat U(t_1 + t_2). \quad\checkmark
$$

(b) From (a) with $t_2 = -t_1$,

$$
\hat U(t)\hat U(-t) = \hat U(0) = \hat I \quad\Longrightarrow\quad \hat U(t)^{-1} = \hat U(-t).
$$

But $\hat U^\dagger(t) = \mathrm{e}^{+\mathrm{i}\hat H t/\hbar} = \hat U(-t)$ as well (Hermiticity of $\hat H$ swaps the sign of $\mathrm{i}$). Equating:

$$
\hat U(t)^{-1} = \hat U(t)^\dagger = \hat U(-t).
$$

The family $\{\hat U(t)\}$ contains $\hat I$ (identity), is closed under composition (a), and is closed under inverses ($\hat U(-t) \in $ family). These are the axioms of a **one-parameter group**.

(c) From (a) applied repeatedly,

$$
\hat U(t/N)^N = \underbrace{\hat U(t/N)\,\hat U(t/N)\cdots\hat U(t/N)}_{N\ \text{times}} = \hat U(t/N + t/N + \cdots + t/N) = \hat U(t).
$$

So slicing $t$ into $N$ equal pieces and evolving through each piece in sequence is *exactly* the same operator as evolving once through the full interval. The physical state at time $t$ does not depend on the slicing — only on the total elapsed time and the Hamiltonian. This is a non-trivial consistency requirement: time evolution is **smoothly continuous**, with no preferred bookkeeping checkpoints. (The construction underlies the path-integral formulation developed in Chapter 3, where $N \to \infty$ infinitesimal slices reconstruct the full propagator.)

(d) When $\hat H$ depends on time, $[\hat H(t_1), \hat H(t_2)] \neq 0$ in general, so the commuting-exponential identity used in (a) **fails**: $\mathrm{e}^{-\mathrm{i}\int\hat H\,\mathrm{d}t/\hbar}$ is not the correct propagator. The composition property still holds in a more general form (the evolution remains unitary and forms a *two*-parameter family $\hat U(t_2, t_1)$), but the simple exponential form breaks down. The full machinery — time-ordered exponentials, Dyson series — is developed in Chapter 5.

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★ **9. Observable evolution in time.** A Hermitian operator $\hat{A}$ has eigenvalues $a_1 = 2$ and $a_2 = -3$ with orthonormal eigenstates $\vert a_1\rangle$ and $\vert a_2\rangle$. Compute $\mathrm{e}^{\mathrm{i}\hat{A}\pi}$ as an explicit operator using the spectral decomposition formula. What happens if you use the Taylor series instead?

**Solution.**

*Spectral method.* The spectral decomposition is $\hat{A} = 2\vert a_1\rangle\langle a_1\vert + (-3)\vert a_2\rangle\langle a_2\vert$. The exponential acts as a scalar exponential on each eigenspace:

$$
\mathrm{e}^{\mathrm{i}\hat{A}\pi} = \mathrm{e}^{\mathrm{i}(2)\pi}\vert a_1\rangle\langle a_1\vert + \mathrm{e}^{\mathrm{i}(-3)\pi}\vert a_2\rangle\langle a_2\vert.
$$

Evaluate the two phases: $\mathrm{e}^{\mathrm{i}2\pi} = 1$ and $\mathrm{e}^{-\mathrm{i}3\pi} = \cos(3\pi) - \mathrm{i}\sin(3\pi) = -1$. Hence

$$
\mathrm{e}^{\mathrm{i}\hat{A}\pi} = (+1)\vert a_1\rangle\langle a_1\vert + (-1)\vert a_2\rangle\langle a_2\vert = \vert a_1\rangle\langle a_1\vert - \vert a_2\rangle\langle a_2\vert.
$$

This operator leaves $\vert a_1\rangle$ untouched and flips the sign of $\vert a_2\rangle$ — a reflection in the eigenbasis. (It can also be written purely in terms of $\hat{A}$: using $\vert a_1\rangle\langle a_1\vert = (\hat A + 3\hat I)/5$ and $\vert a_2\rangle\langle a_2\vert = (2\hat I - \hat A)/5$, one finds $\mathrm{e}^{\mathrm{i}\hat A\pi} = (2\hat A + \hat I)/5$, which indeed gives $+1$ on $\vert a_1\rangle$ and $-1$ on $\vert a_2\rangle$.)

*Taylor series.* The series definition gives the same operator, but the computation is longer. Start from

$$
\mathrm{e}^{\mathrm{i}\hat{A}\pi} = \sum_{n=0}^{\infty}\frac{(\mathrm{i}\pi)^{n}}{n!}\,\hat{A}^{n}.
$$

Because $\hat{A}$ is diagonal in its eigenbasis, every power inherits the same structure:

$$
\hat{A}^{n} = 2^{n}\vert a_1\rangle\langle a_1\vert + (-3)^{n}\vert a_2\rangle\langle a_2\vert.
$$

Substituting and interchanging the sums,

$$
\mathrm{e}^{\mathrm{i}\hat{A}\pi}
= \vert a_1\rangle\langle a_1\vert\sum_{n=0}^{\infty}\frac{(2\mathrm{i}\pi)^{n}}{n!}
+ \vert a_2\rangle\langle a_2\vert\sum_{n=0}^{\infty}\frac{(-3\mathrm{i}\pi)^{n}}{n!}
= \mathrm{e}^{\mathrm{i}2\pi}\vert a_1\rangle\langle a_1\vert + \mathrm{e}^{-\mathrm{i}3\pi}\vert a_2\rangle\langle a_2\vert,
$$

— exactly the spectral result. The Taylor series **works**; it does not break, contradict the spectral answer, or produce a different operator. It is simply *more work*: one must compute $\hat{A}^{n}$ in closed form (already requires noticing the projector structure), then carry out two infinite scalar sums that the spectral formula does in one line via $f(\hat{A}) = \sum_a f(a)\vert a\rangle\langle a\vert$. The spectral method *is* the Taylor sum, resummed eigenspace by eigenspace once and for all.