Time-Evolution Operator

The time-evolution operator $U(t,t_0)$ propagates every state of a closed quantum system from a reference time $t_0$ to another time $t$:

$$\lvert\psi(t)\rangle = U(t,t_0)\lvert\psi(t_0)\rangle.$$

It packages the solution of the Schrödinger initial-value problem into one linear operator. Once $U(t,t_0)$ is known, it can evolve arbitrary state vectors, density operators, and transition amplitudes without solving the differential equation again for each initial state.

Definition and Initial Condition

In the Schrödinger picture, $U(t,t_0)$ is defined by

$$\lvert\psi(t)\rangle = U(t,t_0)\lvert\psi_0\rangle, \qquad \lvert\psi_0\rangle = \lvert\psi(t_0)\rangle.$$

The same $U(t,t_0)$ acts on every possible initial vector in the Hilbert space. It depends on the Hamiltonian and the two endpoint times, not on the particular state being propagated.

At equal times,

$$U(t_0,t_0)=I.$$

For a closed system the map is unitary:

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

Unitarity preserves inner products and total probability. Individual measurement probabilities can change as the state evolves relative to a fixed measurement basis. Its physical meaning and its boundary with open-system dynamics are developed in Unitary Time Evolution.

Operator Schrödinger Equations

Suppose every state obeys

$$i\hbar\frac{d}{dt}\lvert\psi(t)\rangle = H(t)\lvert\psi(t)\rangle.$$

Substituting $\lvert\psi(t)\rangle=U(t,t_0)\lvert\psi_0\rangle$ and using the arbitrariness of $\lvert\psi_0\rangle$ gives

$$\begin{aligned} i\hbar \frac{\partial U(t,t_0)}{\partial t} &= H(t)U(t,t_0), \\ U(t_0,t_0) &=I. \end{aligned}$$

This is the forward equation: it differentiates the final-time endpoint while holding $t_0$ fixed.

There is also a backward equation for the initial-time endpoint:

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

The Hamiltonian appears on the right in the backward equation. This ordering matters when operators at different times do not commute.

Both equations can be checked immediately for a time-independent Hamiltonian. They are also consequences of the composition law and uniqueness of the Schrödinger initial-value problem.

For unbounded Hamiltonians, these differential equations are understood on suitable domains. The existence of a unitary propagator requires more than manipulating formal symbols; the finite-dimensional formulas below avoid most of those analytic subtleties.

Composition, Inverse, and Adjoint

Evolution through an intermediate time $t_1$ composes as

$$U(t_2,t_0) = U(t_2,t_1)U(t_1,t_0).$$

The rightmost operator acts first. A state follows the chronological chain

$$\lvert\psi(t_0)\rangle \longmapsto \lvert\psi(t_1)\rangle \longmapsto \lvert\psi(t_2)\rangle.$$

For unitary evolution,

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

Thus

$$U(t_0,t)U(t,t_0) = U(t_0,t_0) = I.$$

The pair of endpoint times is essential. Writing only $U(t)$ is harmless when $t_0=0$ is fixed or when time-translation invariance makes the dependence on $t-t_0$ clear; otherwise it can hide needed information.

Time-Independent Hamiltonians

If $H$ is self-adjoint and time independent, define $\tau=t-t_0$. The solution is

$$U(t,t_0) = \exp\!\left( -\frac{i}{\hbar}H\tau \right).$$

The exponential is a function of the operator $H$, not an entry-by-entry exponential in an arbitrary matrix representation. It may be defined by a convergent power series for bounded operators or, more generally, by the spectral theorem. See Functions of Operators.

Discrete spectral form

For a discrete spectral decomposition

$$H=\sum_n E_nP_n,$$

functional calculus gives

$$U(t,t_0) = \sum_n e^{-iE_n\tau/\hbar}P_n.$$

Every energy eigenspace acquires a phase. Degenerate vectors with the same energy receive the same phase because the formula is written in terms of spectral projectors $P_n$, not an arbitrary basis inside each eigenspace.

If

$$\lvert\psi_0\rangle = \sum_n c_n\lvert E_n\rangle,$$

then

$$\lvert\psi(t)\rangle = \sum_n c_n e^{-iE_n\tau/\hbar} \lvert E_n\rangle.$$

The coefficient magnitudes are fixed, while relative phases between different energies evolve.

Continuous and mixed spectra

The spectral-measure form covers continuous and mixed spectra:

$$H = \int_{\mathbb R}E\,dP(E),$$

and

$$U(t,t_0) = \int_{\mathbb R} e^{-iE\tau/\hbar}\,dP(E).$$

This compact expression includes sums over bound states and integrals over continuum energies without treating generalized eigenvectors as ordinary normalizable states. The relevant spectral language is introduced in Spectral Decomposition.

One-parameter group

Time-independent evolution depends only on elapsed time:

$$U(t,t_0)=U(\tau).$$

The composition law becomes

$$U(\tau_2)U(\tau_1) = U(\tau_1+\tau_2).$$

Together with $U(0)=I$ and unitarity, this is a one-parameter unitary group. Its exact relation to a self-adjoint generator is summarized by Stone Theorem.

Shifting the energy origin

Replacing the Hamiltonian by

$$H'=H+E_{\mathrm c}I$$

changes the propagator to

$$U'(t,t_0) = e^{-iE_{\mathrm c}\tau/\hbar} U(t,t_0).$$

Every state gains the same global phase. Closed-system probabilities are unchanged, but relative phases between branches governed by different Hamiltonians can make energy offsets observable in interferometric settings. The statement that the energy zero is arbitrary assumes one common Hamiltonian for the compared alternatives.

Transition Amplitudes

Choose orthonormal bases $\{\lvert n\rangle\}$ at the initial description and $\{\lvert m\rangle\}$ at the final description. The matrix elements

$$U_{mn}(t,t_0) = \langle m\vert U(t,t_0)\vert n\rangle$$

are transition amplitudes. If the initial state is

$$\lvert\psi_0\rangle = \sum_n c_n(t_0)\lvert n\rangle,$$

then its final coefficients are

$$c_m(t) = \sum_n U_{mn}(t,t_0)c_n(t_0).$$

In a discrete complete basis, unitarity implies

$$\sum_m U_{mn}^*(t,t_0)U_{mk}(t,t_0) = \delta_{nk}.$$

Composition becomes a sum over intermediate alternatives:

$$U_{mn}(t_2,t_0) = \sum_k U_{mk}(t_2,t_1) U_{kn}(t_1,t_0).$$

This is the operator origin of the familiar rule “sum amplitudes over unobserved intermediate states.” If an actual measurement at $t_1$ records an outcome, the physical process is different because state update and classical conditioning enter.

Time-Dependent Hamiltonians

For a time-dependent Hamiltonian, the forward equation remains

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

Its integral form is

$$U(t,t_0) = I - \frac{i}{\hbar} \int_{t_0}^{t} H(s)U(s,t_0)\,ds.$$

Iterating this equation generates the Dyson series. With

$$U(t,t_0) = I+U^{(1)}+U^{(2)}+\cdots,$$

the first two nontrivial orders are

$$\begin{aligned} U^{(1)} &= -\frac{i}{\hbar} \int_{t_0}^{t}dt_1\,H(t_1),\\ U^{(2)} &= \left(-\frac{i}{\hbar}\right)^2 \int_{t_0}^{t}dt_1 \times \int_{t_0}^{t_1}dt_2\, H(t_1)H(t_2). \end{aligned}$$

The nested limits enforce chronological ordering: later-time Hamiltonians appear to the left. Formally,

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

The symbol $\mathcal T$ is essential when

$$[H(t_1),H(t_2)]\ne0.$$

If all Hamiltonians commute at different times, then time ordering becomes unnecessary:

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

The detailed construction and convergence issues are explained in Time Ordering.

Short-time evolution

For a differentiable finite-dimensional Hamiltonian and a short interval $\delta t$,

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

For an unbounded generator, this $O(\delta t^2)$ expansion is a statement on suitably regular states, not generally an operator-norm expansion. Multiplying short-time factors in chronological order builds the finite-time propagator when the relevant convergence hypotheses hold. This observation underlies numerical time stepping, product formulas, and the time-ordered exponential.

Piecewise-constant evolution

Suppose $H=H_1$ from $t_0$ to $t_1$ and $H=H_2$ from $t_1$ to $t_2$. Then

$$\begin{aligned} U(t_2,t_0) &= U(t_2,t_1)U(t_1,t_0)\\ &= e^{-iH_2(t_2-t_1)/\hbar} e^{-iH_1(t_1-t_0)/\hbar}. \end{aligned}$$

Even though the interval with $H_1$ occurs first, its exponential is the rightmost factor. If $[H_1,H_2]\ne0$, reversing the factors predicts a different final state.

Density-Operator Evolution

Once the propagator is known, a density operator evolves as

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

Differentiating gives the von Neumann equation:

$$i\hbar\dot\rho(t) = [H(t),\rho(t)].$$

Conversely, unitary conjugation solves this equation whenever $U$ solves the corresponding operator Schrödinger equation. Trace, positivity, spectrum, purity, and entropy are preserved for the closed system.

The formula does not describe the most general open-system evolution. Reduced dynamics and noise require quantum channels, even if a larger system plus environment evolves unitarily.

Worked Example: Spin Rotation

Let a spin-$1/2$ Hamiltonian be

$$H = \frac{\hbar\Omega}{2} \boldsymbol n\mathbin{\boldsymbol\cdot}\boldsymbol\sigma, \qquad \lVert\boldsymbol n\rVert=1,\qquad\Omega>0.$$

Because

$$\left( \boldsymbol n\mathbin{\boldsymbol\cdot}\boldsymbol\sigma \right)^2 = I,$$

the even and odd terms of the exponential can be summed separately:

$$\begin{aligned} U(\tau) &= \exp\!\left[ -\frac{i\Omega\tau}{2} \boldsymbol n\mathbin{\boldsymbol\cdot}\boldsymbol\sigma \right]\\ &= \cos\!\left(\frac{\Omega\tau}{2}\right)I - i\sin\!\left(\frac{\Omega\tau}{2}\right) \boldsymbol n\mathbin{\boldsymbol\cdot}\boldsymbol\sigma. \end{aligned}$$

For rotation about the $x$ axis, $\boldsymbol n=\hat{\boldsymbol x}$. Starting from $\lvert0\rangle$,

$$\lvert\psi(t)\rangle = \cos\!\left(\frac{\Omega\tau}{2}\right) \lvert0\rangle - i\sin\!\left(\frac{\Omega\tau}{2}\right) \lvert1\rangle.$$

Therefore

$$\begin{aligned} p_0(t) &= \cos^2\!\left(\frac{\Omega\tau}{2}\right),\\ p_1(t) &= \sin^2\!\left(\frac{\Omega\tau}{2}\right). \end{aligned}$$

The operator exponential has turned the Hamiltonian directly into a rotation with angular frequency $\Omega$.

Common Mistakes

• Writing $U(t)=e^{-iHt/\hbar}$ without stating that $t_0=0$ or that $H$ is time independent.
• Exponentiating matrix entries separately instead of computing an operator or matrix exponential.
• Reversing the order in the composition law or in piecewise-constant evolution.
• Omitting time ordering when Hamiltonians at different times do not commute.
• Treating an energy eigenstate’s phase as evidence that the physical state has changed.
• Forgetting that superpositions of different energies develop observable relative phases.
• Calling $U(t,t_0)$ an observable; it propagates states and is not itself generally a measured quantity.
• Applying closed-system unitary evolution to selective measurement update or reduced open-system dynamics.
• Ignoring domains when using exponentials of unbounded Hamiltonians.

References

• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977, vol. 1, ch. 3.
• P. A. M. Dirac, The Principles of Quantum Mechanics, 4th ed., Oxford University Press, 1958, secs. 26–28.
• 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.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020, ch. 2.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994, chs. 4 and 11.

Exercises

1. Verify that

$$U(t,t_0) = \exp\!\left[ -\frac{i}{\hbar}H(t-t_0) \right]$$

satisfies both the forward equation and the equal-time initial condition for time-independent $H$.

Solution

Because $H$ commutes with every function of itself,

$$\frac{\partial U}{\partial t} = -\frac{i}{\hbar}HU.$$

Multiplying by $i\hbar$ gives

$$i\hbar\frac{\partial U}{\partial t} = HU.$$

At $t=t_0$, the exponent vanishes:

$$U(t_0,t_0)=e^0=I.$$

2. Derive the backward equation

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

from the composition law.

Solution

Insert a short step after $t_0$:

$$U(t,t_0) = U(t,t_0+\delta t) U(t_0+\delta t,t_0).$$

Writing $H_0=H(t_0)$, the short-time factor is

$$U(t_0+\delta t,t_0) = I-\frac{i}{\hbar}H_0\delta t +O(\delta t^2).$$

Also write $U_0=U(t,t_0)$ and $U_+=U(t,t_0+\delta t)$. Rearranging the composition law gives

$$U_+ = U_0 \left[ I+\frac{i}{\hbar}H_0\delta t \right] +O(\delta t^2).$$

Therefore

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

which is equivalent to the stated backward equation.

3. Let $H'=H+E_{\mathrm c}I$, with $H$ time independent. Find $U'(t,t_0)$ and show that all Born probabilities for a state evolved under $H'$ agree with those for the same state evolved under $H$.

Solution

Because $H$ commutes with the identity,

$$\begin{aligned} U'(t,t_0) &= \exp\!\left[ -\frac{i}{\hbar} \left(H+E_{\mathrm c}I\right)\tau \right]\\ &= e^{-iE_{\mathrm c}\tau/\hbar} U(t,t_0). \end{aligned}$$

For any outcome vector $\lvert a\rangle$,

$$\left| \langle a\vert U'\vert\psi_0\rangle \right|^2 = \left| e^{-iE_{\mathrm c}\tau/\hbar} \right|^2 \left| \langle a\vert U\vert\psi_0\rangle \right|^2.$$

The phase factor has absolute value one, so the probabilities agree.

4. For

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

compute $U(\tau)$ and the first time at which an initial $\lvert0\rangle$ becomes $\lvert1\rangle$ up to a global phase.

Solution

Since $\sigma_x^2=I$,

$$U(\tau) = \cos\!\left(\frac{\Omega\tau}{2}\right)I - i\sin\!\left(\frac{\Omega\tau}{2}\right)\sigma_x.$$

Acting on $\lvert0\rangle$ gives

$$U(\tau)\lvert0\rangle = \cos\!\left(\frac{\Omega\tau}{2}\right)\lvert0\rangle - i\sin\!\left(\frac{\Omega\tau}{2}\right)\lvert1\rangle.$$

The first complete transfer occurs when

$$\frac{\Omega\tau}{2}=\frac{\pi}{2},$$

so

$$\tau=\frac{\pi}{\Omega}.$$

The final state is $-i\lvert1\rangle$, which differs from $\lvert1\rangle$ only by a global phase.

5. A system evolves with $H_1$ for duration $\tau_1$ and then with $H_2$ for duration $\tau_2$. Write the total propagator. Under what condition may the two exponential factors be reversed without changing the result?

Solution

Chronological composition gives

$$U = e^{-iH_2\tau_2/\hbar} e^{-iH_1\tau_1/\hbar}.$$

The first interval acts first and therefore appears on the right. The factors can be reversed when their exponentials commute. A sufficient condition is

$$[H_1,H_2]=0.$$

Without that condition, reversing the order generally describes a different physical protocol.

6. Suppose

$$H(t)=f(t)A+g(t)I,$$

where $A$ is a fixed self-adjoint operator. Find $U(t,t_0)$ without a time-ordering symbol.

Solution

At any two times,

$$[H(t_1),H(t_2)]=0,$$

because every term is a linear combination of $A$ and $I$. Define

$$\begin{aligned} F(t,t_0) &= \int_{t_0}^{t}f(s)\,ds, \\ G(t,t_0) &= \int_{t_0}^{t}g(s)\,ds. \end{aligned}$$

Since $A$ and $I$ commute, the propagator factors as

$$U(t,t_0) = e^{-iG(t,t_0)/\hbar} e^{-iF(t,t_0)A/\hbar}.$$

7. In an orthonormal basis, prove the composition rule

$$U_{mn}(t_2,t_0) = \sum_k U_{mk}(t_2,t_1) U_{kn}(t_1,t_0).$$

Interpret the index $k$.

Solution

Write $U_{ji}=U(t_j,t_i)$ and $(U_{ji})_{ab}=\langle a\vert U_{ji}\vert b\rangle$. Insert the identity at the intermediate time:

$$I=\sum_k\lvert k\rangle\langle k\rvert.$$

Ordinary matrix multiplication then gives

$$\left(U_{20}\right)_{mn} = \sum_k \left(U_{21}\right)_{mk} \left(U_{10}\right)_{kn}.$$

The index $k$ labels a complete set of intermediate alternatives. Because no measurement outcome is selected at $t_1$, the amplitudes are summed before taking an absolute square.

8. Let

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

Differentiate this expression and derive the von Neumann equation for a possibly time-dependent Hamiltonian.

Solution

Use

$$\dot U = -\frac{i}{\hbar}HU, \qquad \dot U^\dagger = \frac{i}{\hbar}U^\dagger H.$$

Then

$$\begin{aligned} \dot\rho &= \dot U\rho(t_0)U^\dagger + U\rho(t_0)\dot U^\dagger\\ &= -\frac{i}{\hbar}H\rho + \frac{i}{\hbar}\rho H\\ &= -\frac{i}{\hbar}[H,\rho]. \end{aligned}$$

Multiplying by $i\hbar$ gives

$$i\hbar\dot\rho=[H,\rho].$$