Time Evolution

Everything else in this term has been a snapshot: a state at one instant, a measurement, a density matrix. Now we let time run. The Schrödinger equation is the one law of motion for closed quantum systems, and for a constant Hamiltonian its solution is a single clean object — the unitary propagator $U(t) = e^{-iHt}$. This is literally the "Hermitian generates unitary" theorem from 0.1.6 with the eigenvalues of $H$ now read as energies and the parameter as time. Master this and quantum gates stop being axioms: in the next lesson they become time evolution you chose to run for a chosen duration.

Learning Objectives

After this lesson you will be able to:

1. State the time-dependent Schrödinger equation ($\hbar=1$) and solve it for time-independent Hermitian $H$ to obtain $|\psi(t)\rangle = e^{-iHt}|\psi(0)\rangle$.
2. Identify the propagator $U(t)=e^{-iHt}$ as unitary and verify it is a one-parameter group.
3. Expand any state in the energy eigenbasis and read off its evolution as independent phases $e^{-iE_k t}$, recognizing energy eigenstates as stationary states.
4. Derive the Ehrenfest / Heisenberg equation of motion $\frac{d}{dt}\langle A\rangle = i\langle[H,A]\rangle$ for expectation values.
5. Evolve a density matrix via $\rho(t) = U(t)\rho(0)U(t)^\dagger$ and confirm purity is conserved.

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Intuition

A quantum state is a unit vector. As time passes, a closed system's state must stay a unit vector — probabilities always sum to one — so the only thing allowed to happen is a rotation of the state vector inside Hilbert space. Rotations that preserve length are exactly the unitary operators (0.1.6). So time evolution is a continuous family of unitaries $U(t)$ with $U(0)=I$.

The Hamiltonian $H$ is the "rate" of that rotation — its generator. Diagonalize $H$ and the rotation decouples: along each energy eigen-direction $|E_k\rangle$ the state simply spins its phase at angular frequency $E_k$ (faster for higher energy), and nothing else. A state that lives entirely in one eigen-direction therefore never changes except by an unobservable global phase — it is stationary. Interesting dynamics (oscillations, precession) happen only when a state is a superposition of different energies: the relative phase between the components winds at the energy difference, and relative phases are physical.

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Theory

Symbol note. Throughout this lesson $H$ denotes the Hamiltonian (a Hermitian operator), not the Hadamard gate. We set $\hbar = 1$ (Appendix C), so energies have units of angular frequency and time is dimensionless against them.

The time-dependent Schrödinger equation

The dynamical law of closed-system quantum mechanics — the content of the evolution postulate — is the time-dependent Schrödinger equation (TDSE):

$$\boxed{\,i\,\frac{d}{dt}|\psi(t)\rangle = H\,|\psi(t)\rangle\,}$$

with $H = H^\dagger$ the Hamiltonian (the observable for energy). It is a first-order linear ODE on $\mathbb{C}^n$: given the initial state $|\psi(0)\rangle$, the future is determined.

Solving for time-independent $H$: the propagator

Assume $H$ does not depend on time (the case for the rest of Term 1). The TDSE is then a constant- coefficient linear ODE, formally identical to the scalar equation $\dot{x} = -iHx$, whose solution is the matrix exponential:

$$\boxed{\,|\psi(t)\rangle = U(t)\,|\psi(0)\rangle,\qquad U(t) = e^{-iHt}\,}$$

Verification that this solves the TDSE. Differentiate the power series term by term (it converges absolutely for any bounded $H$, so this is legal):

$$\frac{d}{dt}\,e^{-iHt} = \frac{d}{dt}\sum_{n=0}^{\infty}\frac{(-iHt)^n}{n!} = \sum_{n=1}^{\infty}\frac{(-iH)^n t^{n-1}}{(n-1)!} = (-iH)\,e^{-iHt}.$$

(Here $H$ commutes with itself, so the usual scalar rule survives — no Baker–Campbell–Hausdorff correction needed; see Appendix E.) Hence

$$i\frac{d}{dt}|\psi(t)\rangle = i\,(-iH)\,e^{-iHt}|\psi(0)\rangle = H\,|\psi(t)\rangle,$$

and at $t=0$, $U(0)=e^{0}=I$ gives $|\psi(0)\rangle$. So $U(t)=e^{-iHt}$ is the solution. ∎

The propagator is unitary — and a group

Because $H$ is Hermitian, Hermitian generates unitary (0.1.6) applies verbatim with $t$ as the parameter:

$$U(t)^\dagger = \big(e^{-iHt}\big)^\dagger = e^{+iH^\dagger t} = e^{+iHt},\qquad U(t)^\dagger U(t) = e^{+iHt}e^{-iHt} = e^{0} = I,$$

where the last step is valid because $+iHt$ and $-iHt$ commute. So $U(t)$ is unitary — it preserves norm and inner products, hence total probability is conserved, exactly as physics demands. It is also a one-parameter group:

$$U(t_1)U(t_2) = e^{-iHt_1}e^{-iHt_2} = e^{-iH(t_1+t_2)} = U(t_1+t_2),\qquad U(t)^{-1} = U(-t).$$

Evolution composes additively in time and is reversible (run $U(-t)$ to undo it) — the seed of reversible computation that gates inherit (Term 2; cf. 0.4.3).

Evolution in the energy eigenbasis; stationary states

Diagonalize the Hamiltonian by the spectral theorem (0.1.5):

$$H = \sum_k E_k\,|E_k\rangle\langle E_k|,\qquad H|E_k\rangle = E_k|E_k\rangle,$$

with real energies $E_k$ and orthonormal energy eigenstates $|E_k\rangle$ (the time- independent Schrödinger equation $H|E_k\rangle = E_k|E_k\rangle$ is just this eigenvalue problem). The function-of-an-operator rule (Appendix E) makes the propagator diagonal in this basis:

$$\boxed{\,U(t) = e^{-iHt} = \sum_k e^{-iE_k t}\,|E_k\rangle\langle E_k|\,}$$

Expand an arbitrary initial state in energies, $|\psi(0)\rangle = \sum_k c_k|E_k\rangle$ with $c_k = \langle E_k|\psi(0)\rangle$. Then evolution is componentwise phase rotation:

$$|\psi(t)\rangle = \sum_k c_k\,e^{-iE_k t}\,|E_k\rangle.$$

Each coefficient just acquires a phase $e^{-iE_k t}$; the moduli $|c_k|$ — hence the probabilities of each energy outcome — never change. Two immediate consequences:

• Stationary states. If $|\psi(0)\rangle = |E_j\rangle$, then $|\psi(t)\rangle = e^{-iE_j t}|E_j\rangle$: the state picks up only a global phase, which is physically invisible (Appendix C). Energy eigenstates are frozen — every measurement statistic of every observable is time-independent. This is why they are called stationary states.
• Dynamics needs an energy spread. Genuine change requires at least two different energies in the superposition; the relative phase between $|E_j\rangle$ and $|E_k\rangle$ winds as $e^{-i(E_j-E_k)t}$, oscillating at the Bohr frequency $\omega_{jk} = E_j - E_k$. Only energy differences are dynamically observable (so we may shift $H \to H + cI$ freely; it only adds an unobservable global phase $e^{-ict}$).

Evolution of expectation values (Ehrenfest)

For an observable $A$ (not explicitly time-dependent) in state $|\psi(t)\rangle$, define $\langle A\rangle_t = \langle\psi(t)|A|\psi(t)\rangle$. Differentiate, using the TDSE for the ket and its adjoint $-i\,\frac{d}{dt}\langle\psi| = \langle\psi|H$ for the bra:

$$\frac{d}{dt}\langle A\rangle_t = \Big(\tfrac{d}{dt}\langle\psi|\Big)A|\psi\rangle + \langle\psi|A\Big(\tfrac{d}{dt}|\psi\rangle\Big) = \langle\psi|(iH)A|\psi\rangle + \langle\psi|A(-iH)|\psi\rangle.$$

Collecting terms gives the Ehrenfest theorem (the Heisenberg-picture equation of motion read in the Schrödinger picture):

$$\boxed{\,\frac{d}{dt}\langle A\rangle_t = i\,\langle[H,A]\rangle_t\,}$$

So an observable's expectation is constant in time iff it commutes with $H$ — such an $A$ is a conserved quantity (a symmetry). In particular $[H,H]=0$, so energy is always conserved: $\frac{d}{dt}\langle H\rangle = 0$. We will turn this equation into Bloch-vector precession in the next lesson.

Density-matrix evolution

If the state is described by a density matrix $\rho$ (pure or mixed; Course 1.5), evolve every pure component $|\psi_j\rangle \to U(t)|\psi_j\rangle$ in the ensemble $\rho = \sum_j p_j|\psi_j\rangle\langle\psi_j|$. By linearity the conjugation factors out:

$$\boxed{\,\rho(t) = U(t)\,\rho(0)\,U(t)^\dagger = e^{-iHt}\,\rho(0)\,e^{+iHt}\,}$$

Differentiating gives the Liouville–von Neumann equation, the density-matrix form of the TDSE:

$$\frac{d}{dt}\rho(t) = -iH\rho + i\rho H = -i\,[H,\rho].$$

Unitary evolution conserves the spectrum of $\rho$ (it is a similarity transform by a unitary), so it conserves purity $\mathrm{Tr}(\rho^2)$ and von Neumann entropy: closed-system evolution never turns a pure state mixed. (That requires coupling to an environment — open systems, Term 4.1.) Note the sign convention differs from Ehrenfest: states evolve with $-i[H,\,\cdot\,]$, observables with $+i[H,\,\cdot\,]$ — the two pictures are adjoint to each other.

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Worked Examples

Example 1 — $|+\rangle$ precessing under $H = \tfrac{\omega}{2}Z$

Take the Hamiltonian $H = \tfrac{\omega}{2}Z$ (a qubit with energy splitting $\omega$ between $|0\rangle$ and $|1\rangle$) and initial state $|\psi(0)\rangle = |+\rangle = \tfrac{1}{\sqrt2}(|0\rangle+|1\rangle)$.

The energy eigenstates are the computational basis: $H|0\rangle = +\tfrac{\omega}{2}|0\rangle$, $H|1\rangle = -\tfrac{\omega}{2}|1\rangle$, so $E_0 = +\tfrac{\omega}{2}$, $E_1 = -\tfrac{\omega}{2}$. Evolving each component by its phase:

$$|\psi(t)\rangle = \tfrac{1}{\sqrt2}\big(e^{-i\omega t/2}|0\rangle + e^{+i\omega t/2}|1\rangle\big) = e^{-i\omega t/2}\,\tfrac{1}{\sqrt2}\big(|0\rangle + e^{i\omega t}|1\rangle\big).$$

Drop the global phase $e^{-i\omega t/2}$: the physical state is $\tfrac{1}{\sqrt2}(|0\rangle + e^{i\omega t}|1\rangle)$. The relative phase winds at the Bohr frequency $\omega = E_0 - E_1$. On the Bloch sphere this is a unit vector on the equator rotating about the $z$-axis: at $t=0$ it points along $+\hat x$ ($|+\rangle$), at $\omega t=\tfrac{\pi}{2}$ along $+\hat y$ ($|{+}i\rangle$), at $\omega t=\pi$ along $-\hat x$ ($|-\rangle$). Indeed $U(t) = e^{-i(\omega t)Z/2} = R_z(\omega t)$ is exactly the $z$-rotation gate (Appendix E) — Larmor precession. Note also that the populations are stuck: $|\langle 0|\psi(t)\rangle|^2 = |\langle 1|\psi(t)\rangle|^2 = \tfrac12$ for all $t$, because $|0\rangle,|1\rangle$ are the energy eigenstates and only relative phase moves.

Cross-check with Ehrenfest. $\langle X\rangle_t = \langle\psi(t)|X|\psi(t)\rangle = \cos\omega t$, and indeed $\frac{d}{dt}\langle X\rangle = i\langle[\tfrac{\omega}{2}Z, X]\rangle = i\cdot\tfrac{\omega}{2}\langle 2iY\rangle = -\omega\langle Y\rangle = -\omega\sin\omega t$✓ (using$[Z,X]=2iY$, Appendix E).

Example 2 — A two-level oscillation under $H = \tfrac{\Omega}{2}X$

Now choose $H = \tfrac{\Omega}{2}X$ and start in a non-eigenstate of $H$, namely $|\psi(0)\rangle = |0\rangle$. Here $|0\rangle$ is a superposition of $H$'s eigenstates $|\pm\rangle$ (with energies $\pm\tfrac{\Omega}{2}$), so we expect real motion. Two ways to compute:

(a) Closed form of the propagator. Since $X^2 = I$, the exponential closes into trig (Appendix E, §3):

$$U(t) = e^{-i\frac{\Omega}{2}Xt} = \cos\!\tfrac{\Omega t}{2}\,I - i\sin\!\tfrac{\Omega t}{2}\,X = R_x(\Omega t) = \begin{pmatrix}\cos\frac{\Omega t}{2} & -i\sin\frac{\Omega t}{2}\\ -i\sin\frac{\Omega t}{2} & \cos\frac{\Omega t}{2}\end{pmatrix}.$$

Acting on $|0\rangle$:

$$|\psi(t)\rangle = \cos\!\tfrac{\Omega t}{2}\,|0\rangle - i\sin\!\tfrac{\Omega t}{2}\,|1\rangle.$$

(b) Probabilities. The chance of finding the qubit in $|1\rangle$ oscillates:

$$P_1(t) = |\langle 1|\psi(t)\rangle|^2 = \sin^2\!\tfrac{\Omega t}{2} = \tfrac12\big(1-\cos\Omega t\big), \qquad P_0(t) = \cos^2\!\tfrac{\Omega t}{2}.$$

The population sloshes fully from $|0\rangle$ to $|1\rangle$ and back at angular frequency $\Omega$. This is a Rabi oscillation — the subject of the next lesson, where $H=\tfrac{\Omega}{2}X$ is the resonant drive and the $t = \pi/\Omega$ pulse becomes an $X$ gate. The state vector precesses about the $\hat x$-axis of the Bloch sphere ($\langle X\rangle$ is conserved here because $[H,X]=0$, while $\langle Z\rangle = \cos\Omega t$ and $\langle Y\rangle = -\sin\Omega t$ trace the circle).

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Hands-on (Python)

NumPy plus scipy.linalg.expm turn the propagator into one line. We reproduce both worked examples and plot the oscillating probability.

import numpy as np
from scipy.linalg import expm
import matplotlib.pyplot as plt

# Pauli matrices (Appendix E)
I = np.eye(2, dtype=complex)
X = np.array([[0, 1], [1, 0]], dtype=complex)
Z = np.array([[1, 0], [0, -1]], dtype=complex)

def dag(A):                       # conjugate transpose, ⟨·| from |·⟩
    return A.conj().T

def propagator(H, t):
    """U(t) = exp(-i H t): the solution of i d|ψ⟩/dt = H|ψ⟩ for constant H."""
    return expm(-1j * H * t)

# --- Example 1: |+> under H = (ω/2) Z  →  R_z(ωt), pure phase precession ---
omega = 1.0
H1 = 0.5 * omega * Z
plus = np.array([1, 1], dtype=complex) / np.sqrt(2)

t = np.pi / 2 / omega                      # quarter period: |+> -> |+i>
psi_t = propagator(H1, t) @ plus
print("U(t) unitary? ", np.allclose(dag(propagator(H1, t)) @ propagator(H1, t), I))  # True
print("ψ(t) ≈ |+i>:  ", np.round(psi_t, 3))          # (up to global phase) (|0>+i|1>)/√2
# populations are frozen (|0>,|1> are energy eigenstates):
print("P0, P1:       ", np.round(np.abs(psi_t)**2, 3))   # [0.5 0.5] for all t

# --- Example 2: |0> under H = (Ω/2) X  →  Rabi-like population oscillation ---
Omega = 2.0
H2 = 0.5 * Omega * X
ket0 = np.array([1, 0], dtype=complex)
ket1 = np.array([0, 1], dtype=complex)

def P1_numeric(t):
    psi = propagator(H2, t) @ ket0
    return np.abs(ket1 @ psi) ** 2          # Born probability of outcome "1"

ts = np.linspace(0, 4 * np.pi / Omega, 400)
P1_num = np.array([P1_numeric(tt) for tt in ts])
P1_exact = np.sin(Omega * ts / 2) ** 2      # derived closed form sin²(Ωt/2)

print("max |numeric - exact| =", np.max(np.abs(P1_num - P1_exact)))   # ~1e-15

plt.figure(figsize=(7, 3.2))
plt.plot(ts, P1_num, lw=2, label=r"$P_1(t)$ (expm)")
plt.plot(ts, P1_exact, "k--", lw=1, label=r"$\sin^2(\Omega t/2)$")
plt.axhline(1.0, color="gray", ls=":", lw=0.8)
plt.xlabel("time  t"); plt.ylabel(r"$P_1(t)$")
plt.title(r"Population oscillation under $H=\frac{\Omega}{2}X$")
plt.legend(loc="upper right"); plt.tight_layout()
plt.show()

# --- Density-matrix evolution conserves purity ---
rho0 = np.outer(ket0, ket0.conj())          # pure state |0><0|
def rho_t(t):
    U = propagator(H2, t)
    return U @ rho0 @ dag(U)                 # ρ(t) = U ρ(0) U†

for tt in [0.0, 0.7, 1.3, 2.5]:
    r = rho_t(tt)
    purity = np.real(np.trace(r @ r))        # Tr(ρ²): 1 iff pure
    print(f"t={tt:>4}:  Tr(ρ)={np.real(np.trace(r)):.3f}  purity={purity:.6f}")
# Tr(ρ)=1 and purity=1 throughout — closed evolution keeps a pure state pure.

Braket note (optional). Braket's LocalSimulator evolves circuits, not raw Hamiltonians, so we will not exponentiate $H$ on it. But the bridge is direct: since $e^{-i\frac{\Omega}{2}Xt}=R_x(\Omega t)$, the same dynamics is one gate — Circuit().rx(0, Omega * t). We build that connection explicitly in the next lesson and develop Braket properly in Term 2.1.

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Exercises

E1 (easy). For $H = \tfrac{\omega}{2}Z$, write $U(t)$ as a $2\times2$ matrix and confirm it equals $R_z(\omega t)$ from Appendix E.

Solution

$Z$ is diagonal, so $U(t) = e^{-i\frac{\omega}{2}Zt} = \mathrm{diag}(e^{-i\omega t/2}, e^{+i\omega t/2})$. Comparing with $R_z(\theta) = \mathrm{diag}(e^{-i\theta/2}, e^{+i\theta/2})$ and reading $\theta = \omega t$ gives $U(t) = R_z(\omega t)$. ∎

E2 (easy). Show that an energy eigenstate $|E_j\rangle$ remains an energy eigenstate under evolution, and that all expectation values are time-independent in such a state.

Solution

$|\psi(t)\rangle = e^{-iE_j t}|E_j\rangle$, still proportional to $|E_j\rangle$. For any observable $A$, $\langle A\rangle_t = \langle E_j|e^{+iE_j t}A\,e^{-iE_j t}|E_j\rangle = \langle E_j|A|E_j\rangle$: the phases are scalars and cancel, so $\langle A\rangle_t$ is constant. Hence "stationary state." ∎

E3 (medium). Starting from the TDSE, derive the Liouville–von Neumann equation $\dot\rho = -i[H,\rho]$ for a pure state $\rho = |\psi\rangle\langle\psi|$.

Solution

$\dot\rho = (\tfrac{d}{dt}|\psi\rangle)\langle\psi| + |\psi\rangle(\tfrac{d}{dt}\langle\psi|)$. The TDSE gives $\tfrac{d}{dt}|\psi\rangle = -iH|\psi\rangle$ and (adjoint) $\tfrac{d}{dt}\langle\psi| = +i\langle\psi|H$. Substituting: $\dot\rho = -iH|\psi\rangle\langle\psi| + i|\psi\rangle\langle\psi|H = -iH\rho + i\rho H = -i[H,\rho]$. ∎ (Linearity extends it to mixed$\rho=\sum_j p_j|\psi_j\rangle\langle\psi_j|$.)

E4 (medium). Let $H = \tfrac{\omega}{2}Z$ and $|\psi(0)\rangle = |+\rangle$. Compute $\langle X\rangle_t$, $\langle Y\rangle_t$, $\langle Z\rangle_t$ in closed form and identify the Bloch-vector motion.

Solution

From Example 1, $|\psi(t)\rangle = \tfrac1{\sqrt2}(|0\rangle + e^{i\omega t}|1\rangle)$ (global phase dropped). Then $\langle X\rangle = 2\,\mathrm{Re}\,\overline{c_0}c_1 = \cos\omega t$, $\langle Y\rangle = 2\,\mathrm{Im}\,\overline{c_0}c_1 = \sin\omega t$, $\langle Z\rangle = |c_0|^2-|c_1|^2 = 0$. The Bloch vector $\vec r = (\cos\omega t, \sin\omega t, 0)$ traces the equator counterclockwise at rate $\omega$ — Larmor precession about $\hat z$. (Cross-check Ehrenfest: $\dot{\langle X\rangle} = -\omega\langle Y\rangle$, $\dot{\langle Y\rangle} = +\omega\langle X\rangle$, the rotation ODE.)

E5 (hard). Prove $U(t)=e^{-iHt}$ conserves purity: $\mathrm{Tr}\big(\rho(t)^2\big) = \mathrm{Tr}\big(\rho(0)^2\big)$ for all $t$. Then argue that no unitary can turn a pure state into a mixed one.

Solution

$\rho(t)^2 = U\rho_0 U^\dagger U\rho_0 U^\dagger = U\rho_0^2 U^\dagger$ (using $U^\dagger U = I$). By cyclicity of the trace (Appendix E), $\mathrm{Tr}(U\rho_0^2 U^\dagger) = \mathrm{Tr}(U^\dagger U\rho_0^2) = \mathrm{Tr}(\rho_0^2)$. So purity is invariant. A pure state has purity $1$; since unitary evolution preserves purity, it stays $1$, i.e. pure. Changing purity (decoherence) is impossible for closed/unitary dynamics and requires non-unitary channels — open systems, Term 4.1. ∎

E6 (hard). Suppose $H = H_A\otimes I + I\otimes H_B$ on a two-qubit system. Show the propagator factorizes, $U(t) = e^{-iH_A t}\otimes e^{-iH_B t}$, and conclude such a (non-interacting) Hamiltonian never creates entanglement from a product state.

Solution

The two terms commute: $[H_A\otimes I,\, I\otimes H_B] = (H_A\otimes H_B) - (H_A\otimes H_B) = 0$ (mixed-product rule, Appendix E). For commuting terms $e^{A+B}=e^Ae^B$, and $e^{-i(H_A\otimes I)t} = e^{-iH_A t}\otimes I$ (the $I$-factor exponentiates to $I$), so $U(t) = (e^{-iH_A t}\otimes I)(I\otimes e^{-iH_B t}) = e^{-iH_A t}\otimes e^{-iH_B t}$. Acting on a product $|a\rangle\otimes|b\rangle$ yields $(e^{-iH_A t}|a\rangle)\otimes(e^{-iH_B t}|b\rangle)$ — still a product. Entanglement requires an interaction term that is not a sum of single-system operators (the seed of two-qubit entangling gates, Term 2.1). ∎

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Checkpoint

1. Write the TDSE ($\hbar=1$) and its solution for time-independent Hermitian $H$.
2. Why is the propagator $U(t)=e^{-iHt}$ unitary, and what physical fact does that guarantee?
3. What is a stationary state, and why does it not evolve observably?
4. Give the equation of motion for $\langle A\rangle$ and state the condition for $A$ to be conserved.
5. How does a density matrix evolve, and what quantity does that evolution conserve?

Answers

1. $i\,\frac{d}{dt}|\psi\rangle = H|\psi\rangle$; solution $|\psi(t)\rangle = e^{-iHt}|\psi(0)\rangle$.
2. Hermitian $H$ has real eigenvalues, so $e^{-iE_k t}$ has modulus $1$ and $U^\dagger U = e^{+iHt}e^{-iHt}=I$; unitarity preserves the norm, i.e. total probability is conserved.
3. An energy eigenstate $|E_j\rangle$; it evolves only by the global phase $e^{-iE_j t}$, which is physically invisible, so every measurement statistic is time-independent.
4. $\frac{d}{dt}\langle A\rangle = i\langle[H,A]\rangle$; $A$ is conserved iff $[H,A]=0$. (Energy always is, since $[H,H]=0$.)
5. $\rho(t) = U(t)\rho(0)U(t)^\dagger$, obeying $\dot\rho = -i[H,\rho]$; it conserves the spectrum of $\rho$, hence purity $\mathrm{Tr}(\rho^2)$ and entropy — pure stays pure.

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Further Reading

• [NC] Nielsen & Chuang, §2.2.2 — the evolution postulate and the Schrödinger equation; §2.2.3 for the density-operator picture.
• [Sak] Sakurai & Napolitano, §2.1–2.2 — time evolution, the propagator, stationary states, and the Heisenberg/Ehrenfest equations (with $\hbar$ explicit; set $\hbar=1$ to match us).
• [Gri] Griffiths & Schroeter, §2.1 — stationary states and the separation of the time-dependent Schrödinger equation, at a gentler pace.
• [Pre] Preskill, Ph219, Ch. 2–3 — unitary evolution and the Liouville–von Neumann equation.

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