1.3.1 Unitary Evolution#

Prompts

A unit complex number satisfies \(z^*z = 1\) and can be written \(z = \mathrm{e}^{\mathrm{i}\theta}\). How does a unitary operator generalize this to matrices? What physical quantity plays the role of \(\theta\)?
If a Hermitian operator \(\hat{G}\) has eigenvalues \(g_n\) and eigenstates \(\vert g_n\rangle\), how do you compute \(\mathrm{e}^{\mathrm{i}\hat{G}\theta}\) using the spectral decomposition? Why is this easier than summing the Taylor series?
Why must quantum time evolution be unitary? What physical principle demands this?
Starting from \(\hat{U}(t) = \mathrm{e}^{-\mathrm{i}\hat{H}t/\hbar}\), how would you derive the Schrödinger equation? How would you extract the Hamiltonian from a given time-evolution operator?
What are energy eigenstates, and why do they evolve only by a global phase?

Lecture Notes#

Overview#

Quantum systems evolve in time. The central postulate of quantum mechanics is that this evolution is unitary: it preserves the norm of any state and the inner product between states. Unitary evolution is not a choice or convenience—it is a fundamental constraint that follows from requiring quantum mechanics to preserve information. This section develops the theory of unitary operators, explains why time evolution must be unitary, and introduces the Hamiltonian as the generator of unitary transformations.

Unitary Operators#

Recall that a complex number \(z\) with \(|z|=1\) satisfies \(z^* z = 1\), and every such number can be written as \(z = \mathrm{e}^{\mathrm{i}\theta}\) with \(\theta \in \mathbb{R}\). The matrix generalization is:

Unitary Operator

A linear operator \(\hat{U}\) is unitary if

\[ \hat{U}^\dagger \hat{U} = \hat{I} \]

equivalently, \(\hat{U}^{-1} = \hat{U}^\dagger\). This is the matrix analog of \(z^*z = 1\) for unit complex numbers.

Why does this definition matter? Suppose we apply \(\hat{U}\) to two states \(\vert\psi\rangle\) and \(\vert\phi\rangle\):

\[ \vert\psi'\rangle = \hat{U} \vert\psi\rangle, \quad \vert\phi'\rangle = \hat{U} \vert\phi\rangle \]

Then the inner product is preserved:

\[ \langle\phi' \vert \psi'\rangle = \langle\phi \vert \hat{U}^\dagger \hat{U} \vert\psi\rangle = \langle\phi \vert \psi\rangle \]

In particular, the norm is preserved: \(\langle\psi'\vert\psi'\rangle = \langle\psi\vert\psi\rangle\), and so are all measurement probabilities.

Why Time Evolution Must Be Unitary#

The requirement of unitarity follows from a single physical principle: information is never lost under time evolution.

What does this mean concretely? Consider two identical, isolated quantum systems. The preservation of quantum information demands:

Distinct states remain distinct. If two systems start in different (orthogonal) states, they must remain in different states at all later times—and also at all earlier times. Otherwise, the dynamics would erase information about the initial preparation.
Identical states follow identical evolution. If two systems start in the same state, they must follow the same trajectory—toward both the future and the past. Otherwise, the dynamics would create information out of nothing.

Deterministic evolution between measurements

Although measurement introduces apparent randomness, the evolution of the quantum state between measurements is fully deterministic. Unitarity governs this deterministic evolution.

Translate these requirements into mathematics. Let \(\hat{U}(t)\) denote the time-evolution map: \(\vert\psi(t)\rangle = \hat{U}(t)\vert\psi(0)\rangle\). The two conditions above imply:

Norm preservation (same state remains the same):

\[ \langle\psi(0)\vert\psi(0)\rangle = 1 \;\Longrightarrow\; \langle\psi(t)\vert\psi(t)\rangle = \langle\psi(0)\vert \hat{U}(t)^\dagger \hat{U}(t) \vert\psi(0)\rangle = 1 \]

Orthogonality preservation (different states remain different):

\[ \langle\phi(0)\vert\psi(0)\rangle = 0 \;\Longrightarrow\; \langle\phi(t)\vert\psi(t)\rangle = \langle\phi(0)\vert \hat{U}(t)^\dagger \hat{U}(t) \vert\psi(0)\rangle = 0 \]

Now consider a complete orthonormal basis \(\{\vert e_n\rangle\}\). These basis states satisfy \(\langle e_m \vert e_n\rangle = \delta_{mn}\). By the two conditions above, the evolved states \(\hat{U}(t)\vert e_n\rangle\) must also form an orthonormal set:

\[ \langle e_m \vert \hat{U}(t)^\dagger \hat{U}(t) \vert e_n\rangle = \delta_{mn} \]

Since this holds for every pair of basis vectors, the operator \(\hat{U}(t)^\dagger \hat{U}(t)\) has matrix elements \(\delta_{mn}\) in every orthonormal basis—meaning it equals the identity:

Unitarity from Information Preservation

\[ \hat{U}(t)^\dagger \hat{U}(t) = \hat{I} \]

Time evolution is unitary because quantum dynamics preserves information: it maps every orthonormal basis to another orthonormal basis, both forward and backward in time.

Hermitian Generators of Unitary Operators#

Just as every unit complex number can be written as \(z = \mathrm{e}^{\mathrm{i}\theta}\) with a real exponent, every unitary operator can be written as the exponential of a Hermitian operator. If \(\hat{G}\) is Hermitian (\(\hat{G}^\dagger = \hat{G}\)), then

\[ \hat{U}(\theta) = \mathrm{e}^{\mathrm{i}\hat{G}\theta} \]

is unitary for any real parameter \(\theta\).

Defining the matrix exponential. The exponential \(\mathrm{e}^{\mathrm{i}\hat{G}\theta}\) is defined via the Taylor series:

\[ \mathrm{e}^{\mathrm{i}\hat{G}\theta} = \sum_{n=0}^{\infty} \frac{(\mathrm{i}\hat{G}\theta)^n}{n!} = \hat{I} + \mathrm{i}\hat{G}\theta + \frac{(\mathrm{i}\hat{G}\theta)^2}{2!} + \cdots \]

Proof: Unitarity of Matrix Exponentials

Using the Taylor series directly:

\[ \hat{U}^\dagger = \left(\sum_n \frac{(\mathrm{i}\hat{G}\theta)^n}{n!}\right)^\dagger = \sum_n \frac{(-\mathrm{i}\hat{G}\theta)^n}{n!} = \mathrm{e}^{-\mathrm{i}\hat{G}\theta} \]

since \(\hat{G}\) is Hermitian and \((\hat{G}^n)^\dagger = (\hat{G}^\dagger)^n = \hat{G}^n\).

Then

\[ \hat{U}^\dagger \hat{U} = \mathrm{e}^{-\mathrm{i}\hat{G}\theta} \mathrm{e}^{\mathrm{i}\hat{G}\theta} = \mathrm{e}^{0} = \hat{I} \]

using the fact that \(\hat{G}\) commutes with itself.

Computing matrix exponentials via spectral decomposition. The Taylor series definition is conceptually clean but impractical for calculation. A far more efficient method uses the spectral decomposition. If \(\hat{G}\) has eigenvalues \(g_n\) and eigenstates \(\vert g_n\rangle\), then \(\hat{G} = \sum_n g_n \vert g_n\rangle\langle g_n\vert\) and

Matrix Exponential via Spectral Decomposition

\[ \mathrm{e}^{\mathrm{i}\hat{G}\theta} = \sum_n \mathrm{e}^{\mathrm{i}g_n\theta} \vert g_n\rangle\langle g_n\vert \]

Each eigenstate picks up a phase \(\mathrm{e}^{\mathrm{i}g_n\theta}\)—the matrix exponential acts as a scalar exponential on each eigenspace separately.

Derivation: matrix exponential via spectral decomposition

Insert the spectral decomposition \(\hat{G} = \sum_n g_n \vert g_n\rangle\langle g_n\vert\) into the Taylor series. Since the projectors satisfy \(\vert g_m\rangle\langle g_m \vert g_n\rangle\langle g_n\vert = \delta_{mn} \vert g_n\rangle\langle g_n\vert\), the \(k\)-th power is

\[ \hat{G}^k = \sum_n g_n^k \vert g_n\rangle\langle g_n\vert \]

Substituting:

\[ \mathrm{e}^{\mathrm{i}\hat{G}\theta} = \sum_{k=0}^{\infty} \frac{(\mathrm{i}\theta)^k}{k!} \hat{G}^k = \sum_n \left(\sum_{k=0}^{\infty} \frac{(\mathrm{i}g_n\theta)^k}{k!}\right) \vert g_n\rangle\langle g_n\vert = \sum_n \mathrm{e}^{\mathrm{i}g_n\theta} \vert g_n\rangle\langle g_n\vert \]

Example: Exponential of \(\hat{\sigma}^z\)

Problem. Compute \(\mathrm{e}^{\mathrm{i}\hat{\sigma}^z\theta/2}\).

Solution. The spectral decomposition of \(\hat{\sigma}^z\) is \(\hat{\sigma}^z = (+1)\vert\uparrow\rangle\langle\uparrow\vert + (-1)\vert\downarrow\rangle\langle\downarrow\vert\). Applying the formula:

\[\begin{split} \mathrm{e}^{\mathrm{i}\hat{\sigma}^z\theta/2} = \mathrm{e}^{\mathrm{i}\theta/2}\vert\uparrow\rangle\langle\uparrow\vert + \mathrm{e}^{-\mathrm{i}\theta/2}\vert\downarrow\rangle\langle\downarrow\vert = \begin{pmatrix} \mathrm{e}^{\mathrm{i}\theta/2} & 0 \\ 0 & \mathrm{e}^{-\mathrm{i}\theta/2} \end{pmatrix} \end{split}\]

Extracting the generator from the derivative. Given a one-parameter family \(\hat{U}(\theta) = \mathrm{e}^{\mathrm{i}\hat{G}\theta}\), the generator \(\hat{G}\) can be recovered by differentiating at the identity:

\[ \hat{G} = -\mathrm{i}\frac{\mathrm{d}\hat{U}}{\mathrm{d}\theta}\bigg|_{\theta=0} \]

This is the infinitesimal form of the unitary: \(\hat{U}(\mathrm{d}\theta) \approx \hat{I} + \mathrm{i}\hat{G}\,\mathrm{d}\theta\), so the generator describes what the transformation does “to first order.”

From Hermitian Generator to Time Evolution#

In quantum mechanics, time evolution is generated by the Hamiltonian \(\hat{H}\), which is always Hermitian. The time-evolution operator is:

Time-Evolution Operator

\[ \hat{U}(t) = \mathrm{e}^{-\mathrm{i}\hat{H}t/\hbar} \]

where \(\hbar\) is Planck’s constant and the Hamiltonian \(\hat{H} = \hat{H}^\dagger\) is Hermitian. The Hamiltonian is the generator of time translations: \(\hat{H} = \mathrm{i}\hbar\frac{\mathrm{d}\hat{U}}{\mathrm{d}t}\big|_{t=0}\).

This form follows from integrating the fundamental equation of motion:

Time-Dependent Schrödinger Equation

Starting with \(\vert\psi(t)\rangle = \hat{U}(t)\vert\psi(0)\rangle\), differentiate:

\[ \frac{\mathrm{d}}{\mathrm{d}t}\vert\psi(t)\rangle = \frac{\mathrm{d}\hat{U}}{\mathrm{d}t}\vert\psi(0)\rangle \]

For \(\hat{U}(t) = \mathrm{e}^{-\mathrm{i}\hat{H}t/\hbar}\), the derivative is

\[ \frac{\mathrm{d}\hat{U}}{\mathrm{d}t} = -\frac{\mathrm{i}}{\hbar}\hat{H} \mathrm{e}^{-\mathrm{i}\hat{H}t/\hbar} = -\frac{\mathrm{i}}{\hbar}\hat{H} \hat{U}(t) \]

Therefore

\[ \frac{\mathrm{d}}{\mathrm{d}t}\vert\psi(t)\rangle = -\frac{\mathrm{i}}{\hbar}\hat{H} \hat{U}(t)\vert\psi(0)\rangle = -\frac{\mathrm{i}}{\hbar}\hat{H}\vert\psi(t)\rangle \]

This is the time-dependent Schrödinger equation:

\[ \mathrm{i}\hbar \frac{\mathrm{d}}{\mathrm{d}t}\vert\psi(t)\rangle = \hat{H}\vert\psi(t)\rangle \]

The Hamiltonian generates infinitesimal unitary transformations: the \(\hbar\) and \(\mathrm{i}\) are bookkeeping factors that ensure the exponential is unitary.

Energy Eigenstates and Stationary States#

If \(\hat{H}\vert E_n\rangle = E_n \vert E_n\rangle\) is an energy eigenstate, then applying the time-evolution operator is simple:

\[ \vert\psi(t)\rangle = \hat{U}(t) \vert E_n\rangle = \mathrm{e}^{-\mathrm{i}E_n t/\hbar} \vert E_n\rangle \]

The state picks up only a global phase factor \(\mathrm{e}^{-\mathrm{i}E_n t/\hbar}\). Since global phases do not affect measurement probabilities, energy eigenstates are stationary: they do not evolve in time (up to an unobservable phase).

Discussion: phase evolution of energy superpositions

If a state \(\vert\psi(0)\rangle\) is a superposition of energy eigenstates, \(\vert\psi(0)\rangle = \sum_n c_n \vert E_n\rangle\), how does the state evolve? What happens to the relative phases between eigenstates? Does the time evolution preserve the squared amplitudes \(|c_n|^2\)?

Poll: Why unitary evolution?

In quantum mechanics, time evolution via \(\hat{U}(t) = \mathrm{e}^{-\mathrm{i}\hat{H}t/\hbar}\) is unitary. Why is this essential?

(A) Unitarity ensures \(\langle\psi(t)\vert\psi(t)\rangle = 1\) (normalization preserved).

(B) Unitarity ensures energy conservation and reversibility.

(D) All of the above.

Summary#

Unitarity preserves quantum mechanics: A unitary operator \(\hat{U}^\dagger\hat{U}=\hat{I}\) preserves norms and inner products; time evolution must be unitary to preserve probabilities and quantum information.
Unitary decomposition: Every unitary is exponential of a Hermitian generator, \(\hat{U}(\theta) = \mathrm{e}^{\mathrm{i}\hat{G}\theta}\); spectral decomposition gives eigenvalue form \(\mathrm{e}^{\mathrm{i}g_n\theta}\) for each eigenstate.
Time-evolution operator: \(\hat{U}(t) = \mathrm{e}^{-\mathrm{i}\hat{H}t/\hbar}\) encodes the Schrödinger equation \(\mathrm{i}\hbar\frac{\mathrm{d}|\psi\rangle}{\mathrm{d}t} = \hat{H}|\psi\rangle\).
Energy eigenstates are stationary: Eigenstates of \(\hat{H}\) acquire only a global phase factor \(\mathrm{e}^{-\mathrm{i}E_n t/\hbar}\) under time evolution—they are stable, non-decaying states.

Homework#

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.

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.)

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 \(\hat{G} = \hat{Z}/2\).

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.

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?

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?

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.

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.)

★ 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?