# 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](../../term-0-foundations/c01-linear-algebra/06-special-operators.md) 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.
---
## 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](../../term-0-foundations/c01-linear-algebra/06-special-operators.md)). 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.
---
## Theory
> **Symbol note.** Throughout this lesson $H$ denotes the **Hamiltonian** (a Hermitian operator), not
> the Hadamard gate. We set $\hbar = 1$ ([Appendix C](../../appendix/C-notation-glossary.md)), 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](../c01-postulates/03-evolution-postulate.md) — 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](../../appendix/E-math-identities.md).) 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](../../term-0-foundations/c01-linear-algebra/06-special-operators.md)) 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](../../term-0-foundations/c04-computation-complexity/03-reversible-computation.md)).
### Evolution in the energy eigenbasis; stationary states
Diagonalize the Hamiltonian by the spectral theorem
([0.1.5](../../term-0-foundations/c01-linear-algebra/05-eigen-spectral-theorem.md)):
$$
H = \sum_k E_k\,|E_k\rangle\langle E_k|,\qquad H|E_k\rangle = E_k|E_k\rangle,
$$
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](../../appendix/E-math-identities.md)) 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](../../appendix/C-notation-glossary.md)). 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.
---
## 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](../../appendix/E-math-identities.md)) — **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](../../appendix/E-math-identities.md)).
### 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](../../appendix/E-math-identities.md), §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).
---
## Hands-on (Python)
NumPy plus `scipy.linalg.expm` turn the propagator into one line. We reproduce both worked examples and
plot the oscillating probability.
```python
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
```
```python
# --- 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()
```
```python
# --- 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](../../term-2-qc-core/c01-gates-circuits/01-single-qubit-gates.md).
---
## 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](../../appendix/E-math-identities.md).
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 $U(t) = R_z(\omega t)$. ∎
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." ∎
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 $\rho=\sum_j p_j|\psi_j\rangle\langle\psi_j|$.)
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.)
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](../../appendix/E-math-identities.md)),
$\mathrm{Tr}(U\rho_0^2 U^\dagger) = \mathrm{Tr}(U^\dagger U\rho_0^2) = \mathrm{Tr}(\rho_0^2)$. 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](../../term-4-noise-hardware/c01-open-systems/01-open-systems-decoherence.md). ∎
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](../../appendix/E-math-identities.md)). 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). ∎
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.