Unitary Time Evolution

Unitary time evolution is the reversible, inner-product-preserving dynamics of a closed quantum system. If the state is known at a reference time $t_0$, then

$$\begin{aligned} \lvert\psi(t)\rangle &= U(t,t_0)\lvert\psi(t_0)\rangle, \\ U^\dagger(t,t_0)U(t,t_0) &=I. \end{aligned}$$

The second relation is not merely a normalization trick. It says that evolution preserves the entire geometry of Hilbert space: norms, angles, orthogonality, and transition probabilities. For a well-posed Schrödinger equation, this structure follows from the self-adjointness of the Hamiltonian.

This page develops that core statement. Detailed construction and composition of $U(t,t_0)$ belong to Time-Evolution Operator, while the functional-analytic converse belongs to Stone Theorem.

Closed-System Evolution as a Unitary Map

A linear operator $U$ on a Hilbert space $\mathcal H$ is unitary when

$$U^\dagger U = UU^\dagger = I.$$

Equivalently, $U$ is a surjective linear map that preserves inner products. Its adjoint is its inverse:

$$U^{-1}=U^\dagger.$$

Consequently, a unitary evolution can always be reversed at the level of the closed-system state. For a two-time evolution operator,

$$U(t,t_0)^{-1} = U(t_0,t) = U^\dagger(t,t_0).$$

Linearity also preserves superposition. If

$$\lvert\chi(t_0)\rangle = a\lvert\psi(t_0)\rangle + b\lvert\phi(t_0)\rangle,$$

then

$$\lvert\chi(t)\rangle = a\lvert\psi(t)\rangle + b\lvert\phi(t)\rangle.$$

In finite dimensions, $U^\dagger U=I$ already implies $UU^\dagger=I$. In an infinite-dimensional Hilbert space, the first identity alone says that $U$ is an isometry; surjectivity must also be established. The unilateral shift in Exercise 7 is the standard counterexample.

For the abstract operator theory behind these statements, see Unitary Operators.

What Unitarity Preserves

Let two states evolve under the same $U(t,t_0)$:

$$\begin{aligned} \lvert\psi(t)\rangle &= U(t,t_0)\lvert\psi_0\rangle,\\ \lvert\phi(t)\rangle &= U(t,t_0)\lvert\phi_0\rangle. \end{aligned}$$

Their inner product is unchanged:

$$\begin{aligned} \langle\phi(t)\vert\psi(t)\rangle &= \langle\phi_0\vert U^\dagger(t,t_0)U(t,t_0) \vert\psi_0\rangle\\ &= \langle\phi_0\vert\psi_0\rangle. \end{aligned}$$

Several consequences follow at once.

Normalization and total probability

Setting $\lvert\phi_0\rangle=\lvert\psi_0\rangle$ gives

$$\lVert\psi(t)\rVert^2 = \lVert\psi_0\rVert^2.$$

A normalized state therefore remains normalized. In wave mechanics this is the global statement that the integral of the probability density remains one. Its local refinement is the continuity equation, derived in Unitarity and Conservation of Probability.

Orthogonality and distinguishability

If $\langle\phi_0\vert\psi_0\rangle=0$, then the evolved states remain orthogonal. More generally,

$$\left| \langle\phi(t)\vert\psi(t)\rangle \right|^2 = \left| \langle\phi_0\vert\psi_0\rangle \right|^2.$$

Thus two closed-system states do not become more or less distinguishable merely because both undergo the same unitary evolution. A unitary may rotate their representation, but it cannot change their overlap.

Relative geometry

Unitary maps preserve linear independence, orthonormal bases, and distances:

$$\left\lVert U\lvert\psi\rangle - U\lvert\phi\rangle \right\rVert = \left\lVert \lvert\psi\rangle - \lvert\phi\rangle \right\rVert.$$

They may nevertheless change relative phases between components of a single state. Those phases can alter later interference probabilities even though the state remains normalized.

What need not remain constant

Unitarity does not imply that the probability of every fixed measurement outcome is constant. For a fixed projector $\Pi_a$,

$$p_a(t) = \langle\psi_0\vert U^\dagger(t,t_0)\Pi_a U(t,t_0) \vert\psi_0\rangle,$$

which generally depends on time. The probability is constant for all initial states only when the relevant projector is invariant under the evolution. For a time-independent Hamiltonian, this is closely related to

$$[H,\Pi_a]=0.$$

The general criterion, including explicit time dependence, is developed in Conservation Laws.

From the Schrödinger Equation to Unitarity

Suppose $\lvert\psi(t)\rangle$ and $\lvert\phi(t)\rangle$ satisfy the same Schrödinger equation,

$$i\hbar\lvert\dot\psi\rangle = H\lvert\psi\rangle, \qquad i\hbar\lvert\dot\phi\rangle = H\lvert\phi\rangle.$$

Differentiating their inner product gives

$$\begin{aligned} \frac{d}{dt} \langle\phi\vert\psi\rangle &= \langle\dot\phi\vert\psi\rangle + \langle\phi\vert\dot\psi\rangle\\ &= \frac{i}{\hbar} \langle\phi\vert \left(H^\dagger-H\right) \vert\psi\rangle. \end{aligned}$$

If $H=H^\dagger$, the derivative vanishes. The Schrödinger equation therefore preserves every inner product, not only the norm of one chosen state.

The same argument can be written directly for the evolution operator. If

$$i\hbar\frac{\partial U}{\partial t} = H(t)U, \qquad U(t_0,t_0)=I,$$

and $H(t)$ is self-adjoint, then

$$\begin{aligned} \frac{\partial}{\partial t} \left(U^\dagger U\right) &= \frac{i}{\hbar}U^\dagger H U - \frac{i}{\hbar}U^\dagger H U\\ &=0. \end{aligned}$$

Because $U^\dagger U=I$ at $t=t_0$, it remains $I$ throughout the evolution.

In finite-dimensional systems, this calculation is ordinarily sufficient. For unbounded Hamiltonians, symbols such as $H^\dagger=H$ conceal domain questions: the Schrödinger equation must generate a well-defined evolution on the Hilbert space. This is why the precise requirement is self-adjointness, not merely symmetry of matrix elements on an unspecified domain. See Hermitian vs. Self-Adjoint Operators for the distinction.

Why the Generator Must Be Self-Adjoint

The implication also works locally in the other direction. Write an infinitesimal evolution as

$$U(t+\delta t,t) = I - \frac{i}{\hbar}G(t)\,\delta t + O(\delta t^2).$$

Then

$$U^\dagger U = I + \frac{i}{\hbar} \left(G^\dagger-G\right)\delta t + O(\delta t^2).$$

Unitarity to first order requires

$$G^\dagger=G.$$

The generator of differentiable unitary time evolution is therefore self-adjoint. In quantum dynamics that generator is the Hamiltonian.

For a strongly continuous one-parameter unitary group, Stone Theorem makes this correspondence exact: there is a unique self-adjoint generator $H$ such that

$$U(\tau) = \exp\!\left( -\frac{i}{\hbar}H\tau \right).$$

That group form applies directly to time-translation-invariant dynamics. A driven Hamiltonian $H(t)$ instead supplies a self-adjoint instantaneous generator; its two-time propagator need not depend only on $t-t_0$.

Time-Dependent Hamiltonians

Time dependence does not by itself destroy unitarity. If $H(t)$ is self-adjoint and the time-dependent Schrödinger problem is well posed, then $U(t,t_0)$ remains unitary.

What changes is the formula used to construct it. In general,

$$U(t,t_0) \ne \exp\!\left[ -\frac{i}{\hbar} \int_{t_0}^{t}H(s)\,ds \right]$$

when Hamiltonians at different times fail to commute:

$$[H(t_1),H(t_2)]\ne 0.$$

Time ordering is then required. The canonical derivations are in Time-Evolution Operator and Time-Dependent Hamiltonians.

Density Operators and Mixed States

Unitary evolution acts on a density operator by conjugation:

$$\rho(t) = U(t,t_0)\rho(t_0)U^\dagger(t,t_0).$$

This transformation preserves the defining properties of a density operator.

Trace. Cyclicity gives

$$\operatorname{Tr}\rho(t) = \operatorname{Tr} \left( U^\dagger U\rho(t_0) \right) = \operatorname{Tr}\rho(t_0).$$

Positivity. For every $\lvert\chi\rangle$,

$$\langle\chi\vert\rho(t)\vert\chi\rangle = \langle U^\dagger\chi\vert \rho(t_0) \vert U^\dagger\chi\rangle \ge 0.$$

Spectrum. If

$$\rho(t_0)\lvert r_j\rangle = r_j\lvert r_j\rangle,$$

then

$$\rho(t)U\lvert r_j\rangle = r_jU\lvert r_j\rangle.$$

The eigenvalues $\{r_j\}$ are unchanged. It follows that unitary evolution preserves purity and von Neumann entropy:

$$\begin{aligned} \operatorname{Tr}\rho(t)^2 &= \operatorname{Tr}\rho(t_0)^2,\\ S\!\left(\rho(t)\right) &= S\!\left(\rho(t_0)\right). \end{aligned}$$

A pure state therefore cannot become a mixed state under a unitary acting on that system alone. A subsystem can become mixed, however, when a larger closed system evolves unitarily and develops entanglement. Tracing out the environment discards correlations and is not a unitary operation on the subsystem.

Example: Qubit Precession

Consider

$$H = \frac{\hbar\Omega}{2}\sigma_z, \qquad \tau=t-t_0.$$

The unitary evolution is

$$U(\tau) = \begin{pmatrix} e^{-i\Omega\tau/2} & 0\\ 0 & e^{i\Omega\tau/2} \end{pmatrix}.$$

Start from the $\sigma_x$ eigenstate

$$\lvert+\rangle = \frac{ \lvert0\rangle+\lvert1\rangle }{\sqrt2}.$$

At time $t$,

$$\lvert\psi(t)\rangle = \frac{1}{\sqrt2} \left( e^{-i\Omega\tau/2}\lvert0\rangle + e^{i\Omega\tau/2}\lvert1\rangle \right).$$

The probabilities in the $z$ basis remain

$$p_0(t)=p_1(t)=\frac12.$$

In the $x$ basis, however,

$$\begin{aligned} p_+(t) &= \cos^2\!\left(\frac{\Omega\tau}{2}\right),\\ p_-(t) &= \sin^2\!\left(\frac{\Omega\tau}{2}\right). \end{aligned}$$

Nothing nonunitary has occurred. The norm is fixed, but the Hamiltonian creates a changing relative phase, so probabilities in a basis not aligned with $H$ oscillate. On the Bloch sphere, this is a rotation about the $z$ axis.

Closed Systems and Open Subsystems

Unitary evolution is the dynamical rule for a closed system over the interval being modeled. It is not the most general rule for every state change encountered in practice.

Suppose a system $S$ interacts with an environment $E$. The joint state may evolve unitarily under $V_{SE}$, while the reduced state follows

$$\rho_S(t) = \operatorname{Tr}_E \left[ V_{SE} \left( \rho_S(t_0)\otimes\rho_E \right) V_{SE}^\dagger \right].$$

This reduced map is generally a quantum channel, not conjugation by a unitary on $S$. It can decrease purity, increase entropy, and make initially distinct subsystem states less distinguishable. The missing information resides in system–environment correlations.

Three other cases must also be separated from closed-system unitary evolution:

• Selective measurement update conditions on a recorded outcome and is generally nonunitary and nonlinear after normalization.
• Nonselective measurement and noise are described by completely positive trace-preserving maps, of which unitary channels are a special case.
• Effective non-self-adjoint Hamiltonians can describe loss, decay, postselection, or a restricted sector. Their changing norm signals that some part of the physical description has been omitted or conditioned upon.

The canonical entry point for these general maps is Quantum Channels and Noise. Measurement-specific state changes are introduced in Generalized Measurements Overview.

What Unitarity Does and Does Not Say

Statement	Correct conclusion
$U$ is unitary	Inner products and norms are preserved.
The state evolves unitarily	Evolution of the closed-system state is reversible.
A fixed observable is measured later	Its outcome probabilities may change with time.
$[A,H]=0$ for time-independent $A$ and $H$	The statistics of $A$ are conserved.
A global state evolves unitarily	A reduced subsystem need not evolve unitarily.
A state acquires phases	Relative phases may change interference; a common global phase does not.
The Hamiltonian is time dependent	Evolution can still be unitary, but time ordering may be needed.

Common Mistakes

• Checking only norm preservation for one state instead of inner-product preservation for all states.
• Assuming $U^\dagger U=I$ automatically implies $UU^\dagger=I$ in every infinite-dimensional setting.
• Calling a symmetric differential operator self-adjoint without specifying its domain and boundary conditions.
• Interpreting unitary evolution as conservation of every observable.
• Treating the ordinary exponential of $\int H(t)\,dt$ as valid when Hamiltonians at different times do not commute.
• Expecting a reduced subsystem to evolve unitarily merely because the composite system does.
• Describing selective measurement update as ordinary Hamiltonian evolution.
• Confusing a physically irrelevant global phase with a dynamically relevant relative phase.

Cross-Links

• Schrödinger Equation
• Hamiltonians
• Time-Evolution Operator
• Conservation Laws
• Pictures of Motion Overview
• Unitary Operators
• Stone Theorem
• Unitarity and Conservation of Probability
• Reduced Density Matrices
• Quantum Channels and Noise

References

• J. von Neumann, Mathematical Foundations of Quantum Mechanics, Princeton University Press, 1955.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020, ch. 2.
• L. E. Ballentine, Quantum Mechanics: A Modern Development, World Scientific, 1998, chs. 3–4.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013, chs. 5–6.
• M. Reed and B. Simon, Methods of Modern Mathematical Physics, Vol. I: Functional Analysis, revised ed., Academic Press, 1980.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press, 2010, ch. 8.

Exercises

1. Let $U$ act on a finite-dimensional Hilbert space and satisfy $U^\dagger U=I$. Prove that $U$ is unitary and preserves every inner product.

Solution

For any $\lvert\psi\rangle$,

$$\lVert U\lvert\psi\rangle\rVert^2 = \langle\psi\vert U^\dagger U\vert\psi\rangle = \lVert\psi\rVert^2.$$

Thus the kernel of $U$ is trivial, so $U$ is injective. In a finite-dimensional vector space, an injective linear map from the space to itself is also surjective. Hence $U^{-1}$ exists. Multiplying $U^\dagger U=I$ on the right by $U^{-1}$ gives $U^\dagger=U^{-1}$ and therefore $UU^\dagger=I$.

For arbitrary $\lvert\phi\rangle$ and $\lvert\psi\rangle$,

$$\langle U\phi\vert U\psi\rangle = \langle\phi\vert U^\dagger U\vert\psi\rangle = \langle\phi\vert\psi\rangle.$$

2. Two normalized states have transition probability

$$P_{\phi\leftarrow\psi} = \left| \langle\phi\vert\psi\rangle \right|^2.$$

Show that applying the same unitary $U$ to both states leaves this probability unchanged. Why does this not imply that the probability of a fixed detector outcome is time independent?

Solution

The evolved overlap is

$$\langle U\phi\vert U\psi\rangle = \langle\phi\vert U^\dagger U\vert\psi\rangle = \langle\phi\vert\psi\rangle,$$

so its absolute square is unchanged.

A fixed detector outcome corresponds to a fixed projector $\Pi$. Its probability is

$$p(t) = \langle\psi_0\vert U^\dagger(t,t_0)\Pi U(t,t_0) \vert\psi_0\rangle,$$

which can vary because the detector state is not being evolved alongside $\lvert\psi(t)\rangle$. Only the overlap of two states subjected to the same unitary is guaranteed to be constant.

3. Suppose an infinitesimal evolution has the form

$$U(\delta t) = I-\frac{i}{\hbar}G\,\delta t +O(\delta t^2).$$

Use unitarity to show that $G$ must be self-adjoint.

Solution

Taking the adjoint gives

$$U^\dagger(\delta t) = I+\frac{i}{\hbar}G^\dagger\delta t +O(\delta t^2).$$

Therefore

$$U^\dagger U = I + \frac{i}{\hbar} \left(G^\dagger-G\right)\delta t + O(\delta t^2).$$

The coefficient of $\delta t$ must vanish if $U^\dagger U=I$, so $G^\dagger=G$.

4. For the qubit Hamiltonian

$$H=\frac{\hbar\Omega}{2}\sigma_z,$$

start from $\lvert0\rangle$. Show that all $\sigma_z$ probabilities are constant. Then start from $\lvert+\rangle$ and find the probability of obtaining $+1$ in a $\sigma_x$ measurement after time $\tau$.

Solution

Because $\lvert0\rangle$ is an energy eigenstate,

$$U(\tau)\lvert0\rangle = e^{-i\Omega\tau/2}\lvert0\rangle.$$

Only a global phase changes, so $p_0=1$ and $p_1=0$ at every time.

For $\lvert+\rangle$,

$$\begin{aligned} \langle+\vert U(\tau)\vert+\rangle &= \frac12 \left( e^{-i\Omega\tau/2} + e^{i\Omega\tau/2} \right)\\ &= \cos\!\left(\frac{\Omega\tau}{2}\right). \end{aligned}$$

Hence

$$p_+(\tau) = \cos^2\!\left(\frac{\Omega\tau}{2}\right).$$

The evolution is unitary even though this fixed-basis probability oscillates.

5. Let $\rho'=U\rho U^\dagger$. Prove that $\rho'$ has the same eigenvalues as $\rho$. Deduce that purity and von Neumann entropy are invariant.

Solution

If $\rho\lvert r_j\rangle=r_j\lvert r_j\rangle$, then

$$\begin{aligned} \rho' U\lvert r_j\rangle &= U\rho U^\dagger U\lvert r_j\rangle\\ &= r_jU\lvert r_j\rangle. \end{aligned}$$

Thus each eigenvalue $r_j$ is preserved, including multiplicity. Both

$$\operatorname{Tr}\rho^2 = \sum_j r_j^2$$

and

$$S(\rho) = -\sum_j r_j\ln r_j$$

depend only on these eigenvalues, so they are unchanged.

6. Let $U(t,t_0)$ satisfy

$$i\hbar\frac{\partial U}{\partial t} = H(t)U$$

with $H(t)=H^\dagger(t)$. Derive directly that $U^\dagger(t,t_0)U(t,t_0)=I$.

Solution

The evolution equation and its adjoint imply

$$\begin{aligned} \frac{\partial U}{\partial t} &= -\frac{i}{\hbar}H(t)U, \\ \frac{\partial U^\dagger}{\partial t} &= \frac{i}{\hbar}U^\dagger H(t). \end{aligned}$$

Therefore

$$\begin{aligned} \frac{\partial}{\partial t} \left(U^\dagger U\right) &= \frac{\partial U^\dagger}{\partial t}U + U^\dagger\frac{\partial U}{\partial t}\\ &= \frac{i}{\hbar}U^\dagger H U - \frac{i}{\hbar}U^\dagger H U\\ &=0. \end{aligned}$$

Since $U(t_0,t_0)=I$, the constant product is $U^\dagger U=I$.

7. On the Hilbert space $\ell^2(\mathbb N_0)$ with orthonormal basis $\{\lvert n\rangle\}_{n=0}^{\infty}$, define the unilateral shift by

$$S\lvert n\rangle = \lvert n+1\rangle.$$

Show that $S^\dagger S=I$ but $SS^\dagger\ne I$. What lesson does this give about infinite-dimensional isometries?

Solution

The adjoint acts as

$$\begin{aligned} S^\dagger\lvert0\rangle &=0,\\ S^\dagger\lvert n\rangle &= \lvert n-1\rangle \quad(n\ge1). \end{aligned}$$

Hence

$$S^\dagger S\lvert n\rangle = \lvert n\rangle$$

for every $n$, so $S^\dagger S=I$. But

$$SS^\dagger\lvert0\rangle=0 \ne \lvert0\rangle.$$

In fact,

$$SS^\dagger = I-\lvert0\rangle\langle0\rvert.$$

The shift preserves all norms but is not surjective, so it is an isometry rather than a unitary operator. In infinite dimensions, the two identities in the unitary definition are not automatically equivalent.

8. A joint unitary on two qubits maps $\lvert00\rangle$ to the Bell state

$$\lvert\Phi^+\rangle = \frac{ \lvert00\rangle+\lvert11\rangle }{\sqrt2}.$$

Find the reduced state of the first qubit before and after the evolution. Explain why its reduced evolution cannot be unitary.

Solution

Initially,

$$\rho_A(t_0) = \lvert0\rangle\langle0\rvert,$$

which is pure. After the joint evolution,

$$\begin{aligned} \rho_A(t) &= \operatorname{Tr}_B \left( \lvert\Phi^+\rangle\langle\Phi^+\rvert \right)\\ &= \frac12 \left( \lvert0\rangle\langle0\rvert + \lvert1\rangle\langle1\rvert \right)\\ &= \frac{I}{2}. \end{aligned}$$

The purity changes from $1$ to $1/2$. Conjugation by a unitary on the first qubit would preserve the eigenvalues and purity of its density operator, so no unitary acting on that qubit alone can produce this change. The composite evolution is unitary; the reduced evolution is not.