1.3.1 Unitary Evolution

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.

Attention

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:

  1. Norm preservation (same state remains the same):

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

  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

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

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 \(|c_n|^2\)?

Summary#

  • A unitary operator satisfies \(\hat{U}^\dagger \hat{U} = \hat{I}\) and preserves inner products and norms—the matrix generalization of \(z^*z = 1\) for unit complex numbers.

  • Time evolution must be unitary to preserve quantum information and measurement probabilities.

  • Every unitary operator is the exponential of a Hermitian generator: \(\hat{U}(\theta) = \mathrm{e}^{\mathrm{i}\hat{G}\theta}\).

  • Spectral decomposition gives the practical formula: \(\mathrm{e}^{\mathrm{i}\hat{G}\theta} = \sum_n \mathrm{e}^{\mathrm{i}g_n\theta}\vert g_n\rangle\langle g_n\vert\).

  • The generator is extracted from the derivative: \(\hat{G} = -\mathrm{i}\frac{\mathrm{d}\hat{U}}{\mathrm{d}\theta}\big|_{\theta=0}\).

  • The time-evolution operator is \(\hat{U}(t) = \mathrm{e}^{-\mathrm{i}\hat{H}t/\hbar}\), where \(\hat{H}\) is the Hamiltonian.

  • The Schrödinger equation \(\mathrm{i}\hbar \frac{\mathrm{d}}{\mathrm{d}t}\vert\psi\rangle = \hat{H}\vert\psi\rangle\) encodes unitary evolution.

  • Energy eigenstates \(\vert E_n\rangle\) evolve only by a global phase factor and are stationary.

Homework#

1. For a unit complex number \(z = \mathrm{e}^{\mathrm{i}\theta}\), verify that \(z^*z = 1\). Now let \(\hat{U} = \mathrm{e}^{\mathrm{i}\hat{G}\theta}\) with \(\hat{G}\) Hermitian. Show that \(\hat{U}^\dagger \hat{U} = \hat{I}\) using the Taylor series expansion.

2. The Pauli matrix \(\hat{\sigma}^x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\) has eigenvalues \(\pm 1\) with eigenstates \(\vert\pm\rangle = \frac{1}{\sqrt{2}}(\vert 0\rangle \pm \vert 1\rangle)\). Use the spectral decomposition to compute \(\hat{U}(\theta) = \mathrm{e}^{\mathrm{i}\hat{\sigma}^x \theta/2}\) as a \(2 \times 2\) matrix. Verify your result agrees with \(\cos(\theta/2)\hat{I} + \mathrm{i}\sin(\theta/2)\hat{\sigma}^x\).

3. Given the one-parameter family \(\hat{U}(\theta) = \mathrm{e}^{\mathrm{i}\hat{\sigma}^z\theta/2}\), extract the generator by computing \(\hat{G} = -\mathrm{i}\frac{\mathrm{d}\hat{U}}{\mathrm{d}\theta}\big|_{\theta=0}\) and verify you recover \(\hat{G} = \hat{\sigma}^z/2\).

4. Let \(\hat{H} = E_0 \hat{I} + \Delta \hat{\sigma}^z\) (a two-level system with energy offset \(E_0\) and splitting \(\Delta\)). Write down the time-evolution operator \(\hat{U}(t)\) explicitly using the spectral decomposition and show that each energy eigenstate picks up only a phase factor.

5. Starting from the time-evolution operator \(\hat{U}(t) = \mathrm{e}^{-\mathrm{i}\hat{H}t/\hbar}\), derive the time-dependent Schrödinger equation by differentiating \(\vert\psi(t)\rangle = \hat{U}(t) \vert\psi(0)\rangle\) with respect to \(t\).

6. Prove that if \(\vert\psi(0)\rangle\) is normalized, then \(\vert\psi(t)\rangle = \hat{U}(t)\vert\psi(0)\rangle\) is normalized for all \(t\). What role does unitarity play?

7. 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 energies \(E_1\) and \(E_2\). Find \(\vert\psi(t)\rangle\) and interpret the time-dependent phase difference \((E_2 - E_1) t / \hbar\) physically.

8. Show that \(\hat{U}(t_1)\hat{U}(t_2) = \hat{U}(t_1 + t_2)\) (composition property) for the time-evolution operator. Why is this property essential for the consistency of quantum mechanics?

9. 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?