The Evolution Postulate

How does a quantum state change when you don't measure it? Postulate 3 says: by a unitary transformation. In continuous time this is the Schrödinger equation; in the discrete language of quantum computing it is "apply a gate." The two are the same statement — $U(t) = e^{-iHt}$ — and the link is precisely the "Hermitian generates unitary" theorem from Term 0. This postulate is why every quantum gate is unitary and why (un-measured) quantum computation is reversible.

Learning Objectives

After this lesson you will be able to:

1. State Postulate 3 in both discrete ($\lvert\psi'\rangle = U\lvert\psi\rangle$) and continuous (Schrödinger) forms.
2. Explain why evolution must be unitary (preservation of normalization/probability).
3. Solve the time-independent Schrödinger equation to obtain the propagator $U(t) = e^{-iHt}$.
4. Contrast reversible unitary evolution with irreversible measurement.
5. Evolve states numerically and confirm unitarity.

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Intuition

Between measurements, a closed quantum system glides along deterministically — but the "glide" is a rotation of the state vector inside Hilbert space, never a stretch or shrink (that would change total probability). Rotations of complex vectors that preserve lengths and angles are exactly unitary maps. The generator of this continuous rotation is the system's energy operator, the Hamiltonian $H$. In quantum computing we usually skip the continuous story and just specify the net rotation — a gate $U$ — but physically that gate is "let the right Hamiltonian act for the right amount of time" (Course 1.6, and hardware in Term 4.4).

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Theory

Postulate 3 (Unitary evolution)

Postulate 3 (discrete form). The evolution of a closed quantum system from time $t_1$ to $t_2$ is given by a unitary operator $U$ depending only on $t_1,t_2$:

$$\lvert\psi(t_2)\rangle = U\,\lvert\psi(t_1)\rangle, \qquad U^\dagger U = I.$$

Postulate 3 (continuous form). The time evolution of the state of a closed system is governed by the Schrödinger equation

$$\boxed{\, i\,\frac{d}{dt}\lvert\psi(t)\rangle = H\,\lvert\psi(t)\rangle \,} \qquad(\hbar = 1),$$

where $H = H^\dagger$ is a fixed Hermitian operator, the Hamiltonian (the energy observable).

We set $\hbar = 1$ throughout (see Appendix C); restoring units puts $\hbar$ on the left: $i\hbar\,d_t\lvert\psi\rangle = H\lvert\psi\rangle$.

Why unitary? Probability conservation

Unitarity is forced by Postulate 1: a state must stay normalized. If $\lvert\psi'\rangle = U\lvert\psi\rangle$, then $\langle\psi'|\psi'\rangle = \langle\psi|U^\dagger U|\psi\rangle$, which equals $\langle\psi|\psi\rangle = 1$ for all states iff $U^\dagger U = I$. Equivalently (0.1.6), unitaries are precisely the inner-product-preserving maps — they preserve all overlaps, hence all Born probabilities. Any non-unitary "evolution" would leak or create probability.

From the Schrödinger equation to the propagator

For a time-independent $H$, the Schrödinger equation $i\,d_t\lvert\psi\rangle = H\lvert\psi\rangle$ is a linear ODE with constant operator coefficient. Its solution is the operator exponential

$$\boxed{\ \lvert\psi(t)\rangle = e^{-iHt}\,\lvert\psi(0)\rangle \equiv U(t)\,\lvert\psi(0)\rangle\ }.$$

Verification. Differentiate the series $U(t) = e^{-iHt} = \sum_{n\ge0}\frac{(-iHt)^n}{n!}$ term by term: $\frac{d}{dt}U(t) = -iH\,e^{-iHt} = -iH\,U(t)$, so $i\,d_t(U\lvert\psi(0)\rangle) = H\,U\lvert\psi(0)\rangle$, and $U(0) = I$ gives the right initial condition. ∎ And $U(t)$ is unitary because $H$ is Hermitian — the exact theorem proved in 0.1.6:

$$U(t)^\dagger U(t) = e^{+iH^\dagger t}e^{-iHt} = e^{+iHt}e^{-iHt} = I .$$

So the discrete and continuous forms agree: $U = e^{-iHt}$. Conversely, Stone's theorem (and finite dimensionally, just the matrix logarithm) says every unitary is $e^{-iH}$ for some Hermitian $H$ — so "a gate" and "evolve under some Hamiltonian for some time" are the same notion. This is the bridge from physics to circuits, fully exploited in Course 1.6 and Term 2.

Composition and reversibility

Unitaries compose: evolving by $U_1$ then $U_2$ is $U_2 U_1$, again unitary (0.1.6 E2) — this is precisely a quantum circuit, a product of gates. And every unitary is invertible, with $U^{-1} = U^\dagger$: un-measured quantum evolution is reversible. Run the gates backwards (apply each $U^\dagger$ in reverse order) and you return to the start. Contrast measurement (Postulate 2), which collapses the state and destroys information — the one irreversible step in quantum theory. The interplay (reversible evolution + irreversible measurement) is what every algorithm orchestrates.

Open systems caveat. Postulate 3 is for closed systems. A system coupled to an environment evolves non-unitarily (decoherence); that is described by quantum channels on density matrices — Course 1.5 sets up the formalism and Term 4 develops noise.

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Worked Examples

Example 1 — Evolution under $H = \tfrac{\omega}{2}Z$

With $H = \tfrac{\omega}{2}Z$, the propagator is (using the spectral form of $Z$, as in 0.1.6 Example 1):

$$U(t) = e^{-i\frac{\omega}{2}Z t} = \begin{pmatrix} e^{-i\omega t/2} & 0 \\ 0 & e^{+i\omega t/2}\end{pmatrix} = R_z(\omega t).$$

Apply to $\lvert+\rangle$: $U(t)\lvert+\rangle = \tfrac1{\sqrt2}(e^{-i\omega t/2}\lvert0\rangle + e^{+i\omega t/2}\lvert1\rangle)$. The relative phase $e^{i\omega t}$ between components grows linearly in time — the qubit's Bloch vector precesses about the $z$-axis at angular frequency $\omega$ (Course 1.6). Note $\langle Z\rangle$ stays $0$ throughout (the relative phase is invisible to a $Z$ measurement) but $\langle X\rangle = \cos\omega t$ oscillates — phase becomes observable in the $X$ basis.

Example 2 — A gate as finite-time evolution

To realize the NOT gate ($X$) we want $U = X$. Pick $H = \tfrac{\pi}{2}X$ and evolve for $t = 1$:

$$e^{-i\frac{\pi}{2}X} = \cos\tfrac{\pi}{2} I - i\sin\tfrac{\pi}{2} X = -iX,$$

which is $X$ up to the global phase $-i$ — physically the NOT gate. (The half-angle identity is from Appendix E.) This is the seed of Rabi oscillations: a resonant drive is a Hamiltonian, and letting it act for a calibrated time produces a gate (1.6.2).

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Hands-on (Python)

import numpy as np
from scipy.linalg import expm

def dag(A): return A.conj().T
def is_unitary(U): return np.allclose(dag(U) @ U, np.eye(U.shape[0]))

Z = np.array([[1, 0], [0, -1]], dtype=complex)
X = np.array([[0, 1], [1, 0]], dtype=complex)
plus = np.array([1, 1], dtype=complex) / np.sqrt(2)

# Propagator U(t) = exp(-iHt) for H = (ω/2) Z:
omega = 2.0
def U(t, H):  return expm(-1j * H * t)

Ut = U(0.5, (omega / 2) * Z)
print(is_unitary(Ut))                       # True — evolution preserves probability
psi_t = Ut @ plus
print(np.round(np.abs(psi_t)**2, 4))        # [0.5 0.5] — populations unchanged under Rz...

# ...but the relative phase shows up in <X>:
expX = lambda s: np.real(s.conj() @ X @ s)
print(round(expX(psi_t), 4), "vs cos(ωt)=", round(np.cos(omega*0.5), 4))   # matches

# A gate is finite-time evolution: exp(-i (π/2) X) = X up to global phase.
gate = expm(-1j * (np.pi/2) * X)
print(np.round(gate, 3))                    # [[0,-i],[-i,0]] = -i X  (physically NOT)
print(np.allclose(np.abs(gate), np.abs(X))) # same operator up to phase

# Reversibility: U† undoes U.
ket0 = np.array([1, 0], dtype=complex)
print(np.allclose(dag(Ut) @ (Ut @ ket0), ket0))   # True

On Braket. Every gate you place in a Circuit is one of these unitaries; the SDK's Gate.RZ(angle) is literally $R_z(\text{angle}) = e^{-i\,\text{angle}\,Z/2}$. Reversibility is why a circuit can be uncomputed by appending the inverse gates — a trick used constantly from Term 2 on.

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Exercises

E1 (easy). Verify that $U = e^{-i\theta X/2} = R_x(\theta)$ is unitary and compute $R_x(\pi)$ up to global phase.

Solution

$R_x(\theta) = \cos\tfrac\theta2 I - i\sin\tfrac\theta2 X$; $R_x^\dagger R_x = I$ since $X$ is Hermitian and $\theta$ real. $R_x(\pi) = -iX$, i.e. $X$ (NOT) up to the global phase $-i$.

E2 (easy). If $\lvert\psi(0)\rangle$ is an eigenstate of $H$ with energy $E$, find $\lvert\psi(t)\rangle$. What is observable about its time dependence?

Solution

$\lvert\psi(t)\rangle = e^{-iEt}\lvert\psi(0)\rangle$ — only a global phase accrues. Nothing observable changes: energy eigenstates are stationary states. (Relative phases between different energy eigenstates in a superposition are observable — Course 1.6.)

E3 (medium). Prove that the product of the propagators over $[0,t_1]$ and $[t_1, t_1+t_2]$ equals the propagator over $[0, t_1+t_2]$ for time-independent $H$.

Solution

$U(t_2)U(t_1) = e^{-iHt_2}e^{-iHt_1} = e^{-iH(t_1+t_2)} = U(t_1+t_2)$, valid because $H$ commutes with itself (so the exponents add — Appendix E, BCH). ∎ This semigroup property is just "evolve, then evolve more."

E4 (medium). Show that if $H$ is not Hermitian (say $H = -i\Gamma$ with $\Gamma$ positive), $U(t) = e^{-iHt}$ is not unitary and the norm decays. Why does this violate Postulate 1?

Solution

$U(t) = e^{-i(-i\Gamma)t} = e^{-\Gamma t}$, with $U^\dagger U = e^{-2\Gamma t}\neq I$; norms shrink as $\lVert\psi(t)\rVert^2 = e^{-2\Gamma t}\lVert\psi(0)\rVert^2$. Total probability is no longer conserved, contradicting the requirement that states stay unit vectors. (Such non-Hermitian "effective Hamiltonians" are used as models of open systems, where the missing probability has leaked to an environment — Term 4.)

E5 (hard). Given a unitary $U$ on a qubit, show how to find a Hermitian $H$ with $U = e^{-iH}$, and apply it to $U = H_{\text{adamard}}$. (Use the spectral decomposition.)

Solution

Diagonalize $U = \sum_k e^{i\theta_k}\lvert v_k\rangle\langle v_k|$ (unitary ⇒ unit-modulus eigenvalues, 0.1.6). Define $H = \sum_k(-\theta_k)\lvert v_k\rangle\langle v_k|$ (real eigenvalues ⇒ Hermitian); then $e^{-iH} = \sum_k e^{i\theta_k}\lvert v_k\rangle\langle v_k| = U$. For the Hadamard, eigenvalues are $\pm1 = e^{i\cdot 0}, e^{i\pi}$ with the eigenvectors from 0.1.5 E4; so $H$ has eigenvalues $0$ and $-\pi$ on those eigenvectors, giving $H = \tfrac{\pi}{2}(I - \tfrac{1}{\sqrt2}(X+Z))$ up to an additive multiple of $I$ (which is just a global phase). ∎

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Checkpoint

1. State Postulate 3 in discrete and continuous forms.
2. Why must closed-system evolution be unitary?
3. Solve $i\,d_t\lvert\psi\rangle = H\lvert\psi\rangle$ for time-independent $H$ and verify the solution.
4. In what sense are a "gate" and "evolution under a Hamiltonian" the same thing?
5. Contrast the reversibility of evolution with measurement.

Answers

1. Discrete: $\lvert\psi(t_2)\rangle = U\lvert\psi(t_1)\rangle$, $U$ unitary. Continuous: $i\,d_t\lvert\psi\rangle = H\lvert\psi\rangle$, $H$ Hermitian.
2. To preserve normalization/Born probabilities for all states, which requires $U^\dagger U = I$.
3. $\lvert\psi(t)\rangle = e^{-iHt}\lvert\psi(0)\rangle$; differentiating $e^{-iHt}$ gives $-iH e^{-iHt}$, satisfying the equation with $U(0)=I$.
4. $U(t) = e^{-iHt}$ for Hermitian $H$, and (Stone/matrix log) every unitary equals $e^{-iH}$ for some Hermitian $H$ — gates are net evolutions.
5. Unitary evolution is invertible ($U^{-1}=U^\dagger$), hence reversible; measurement collapses the state and is irreversible.

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Further Reading

• [NC] Nielsen & Chuang, §2.2.2 (Postulate 2, evolution) and §2.2.3.
• [Sak] Sakurai & Napolitano, §2.1–2.2 — time evolution and the Schrödinger equation.
• [Gri] Griffiths & Schroeter, Ch. 1–2 — the Schrödinger equation (with $\hbar$ explicit).

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