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Time-Evolution Operator

This is the canonical treatment of the time-evolution operator, including its differential equations, composition law, spectral forms, time ordering, density-operator evolution, and worked examples. The postulate-level definition and properties are summarized at Time-Evolution Operator: Core First Encounter.

The time-evolution operator U(t,t0)U(t,t_0) propagates every state of a closed quantum system from a reference time t0t_0 to another time tt:

∣ψ(t)⟩=U(t,t0)∣ψ(t0)⟩.\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,t0)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.

In the Schrödinger picture, U(t,t0)U(t,t_0) is defined by

∣ψ(t)⟩=U(t,t0)∣ψ0⟩,∣ψ0⟩=∣ψ(t0)⟩.\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,t0)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(t0,t0)=I.U(t_0,t_0)=I.

For a closed system the map is unitary:

U†(t,t0)U(t,t0)=I,U(t,t0)U†(t,t0)=I.\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 probabilities. Its physical meaning and its boundary with open-system dynamics are developed in Unitary Time Evolution.

Suppose every state obeys

iℏddt∣ψ(t)⟩=H(t)∣ψ(t)⟩.i\hbar\frac{d}{dt}\lvert\psi(t)\rangle = H(t)\lvert\psi(t)\rangle.

Substituting ∣ψ(t)⟩=U(t,t0)∣ψ0⟩\lvert\psi(t)\rangle=U(t,t_0)\lvert\psi_0\rangle and using the arbitrariness of ∣ψ0⟩\lvert\psi_0\rangle gives

iℏ∂U(t,t0)∂t=H(t)U(t,t0),U(t0,t0)=I.\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 t0t_0 fixed.

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

−iℏ∂U(t,t0)∂t0=U(t,t0)H(t0).-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.

Evolution through an intermediate time t1t_1 composes as

U(t2,t0)=U(t2,t1)U(t1,t0).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

∣ψ(t0)⟩⟼∣ψ(t1)⟩⟼∣ψ(t2)⟩.\lvert\psi(t_0)\rangle \longmapsto \lvert\psi(t_1)\rangle \longmapsto \lvert\psi(t_2)\rangle.

For unitary evolution,

U(t,t0)−1=U(t0,t),U(t0,t)=U†(t,t0).\begin{aligned} U(t,t_0)^{-1} &= U(t_0,t),\\ U(t_0,t) &= U^\dagger(t,t_0). \end{aligned}

Thus

U(t0,t)U(t,t0)=U(t0,t0)=I.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)U(t) is harmless when t0=0t_0=0 is fixed or when time-translation invariance makes the dependence on t−t0t-t_0 clear; otherwise it can hide needed information.

If HH is self-adjoint and time independent, define τ=t−t0\tau=t-t_0. The solution is

U(t,t0)=exp⁡ ⁣(−iℏHτ).U(t,t_0) = \exp\!\left( -\frac{i}{\hbar}H\tau \right).

The exponential is a function of the operator HH, 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.

For a discrete spectral decomposition

H=∑nEnPn,H=\sum_n E_nP_n,

functional calculus gives

U(t,t0)=∑ne−iEnτ/ℏPn.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 PnP_n, not an arbitrary basis inside each eigenspace.

If

∣ψ0⟩=∑ncn∣En⟩,\lvert\psi_0\rangle = \sum_n c_n\lvert E_n\rangle,

then

∣ψ(t)⟩=∑ncne−iEnτ/ℏ∣En⟩.\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.

The spectral-measure form covers continuous and mixed spectra:

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

and

U(t,t0)=∫Re−iEτ/ℏ dP(E).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.

Time-independent evolution depends only on elapsed time:

U(t,t0)=U(τ).U(t,t_0)=U(\tau).

The composition law becomes

U(τ2)U(τ1)=U(τ1+τ2).U(\tau_2)U(\tau_1) = U(\tau_1+\tau_2).

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

Replacing the Hamiltonian by

H′=H+EcIH'=H+E_{\mathrm c}I

changes the propagator to

U′(t,t0)=e−iEcτ/ℏU(t,t0).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.

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

Umn(t,t0)=⟨m∣U(t,t0)∣n⟩U_{mn}(t,t_0) = \langle m\vert U(t,t_0)\vert n\rangle

are transition amplitudes. If the initial state is

∣ψ0⟩=∑ncn(t0)∣n⟩,\lvert\psi_0\rangle = \sum_n c_n(t_0)\lvert n\rangle,

then its final coefficients are

cm(t)=∑nUmn(t,t0)cn(t0).c_m(t) = \sum_n U_{mn}(t,t_0)c_n(t_0).

In a discrete complete basis, unitarity implies

∑mUmn∗(t,t0)Umk(t,t0)=δnk.\sum_m U_{mn}^*(t,t_0)U_{mk}(t,t_0) = \delta_{nk}.

Composition becomes a sum over intermediate alternatives:

Umn(t2,t0)=∑kUmk(t2,t1)Ukn(t1,t0).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 t1t_1 records an outcome, the physical process is different because state update and classical conditioning enter.

For a time-dependent Hamiltonian, the forward equation remains

iℏ∂U(t,t0)∂t=H(t)U(t,t0).i\hbar \frac{\partial U(t,t_0)}{\partial t} = H(t)U(t,t_0).

Its integral form is

U(t,t0)=I−iℏ∫t0tH(s)U(s,t0) ds.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,t0)=I+U(1)+U(2)+⋯ ,U(t,t_0) = I+U^{(1)}+U^{(2)}+\cdots,

the first two nontrivial orders are

U(1)=−iℏ∫t0tdt1 H(t1),U(2)=(−iℏ)2∫t0tdt1×∫t0t1dt2 H(t1)H(t2).\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\\ &\quad\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,t0)=Texp⁡ ⁣[−iℏ∫t0tH(s) ds].U(t,t_0) = \mathcal T \exp\!\left[ -\frac{i}{\hbar} \int_{t_0}^{t}H(s)\,ds \right].

The symbol T\mathcal T is essential when

[H(t1),H(t2)]≠0.[H(t_1),H(t_2)]\ne0.

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

U(t,t0)=exp⁡ ⁣[−iℏ∫t0tH(s) ds].U(t,t_0) = \exp\!\left[ -\frac{i}{\hbar} \int_{t_0}^{t}H(s)\,ds \right].

The detailed construction and convergence issues belong to Time Ordering.

For a sufficiently regular Hamiltonian and a short interval δt\delta t,

U(t+δt,t)=I−iℏH(t)δt+O(δt2).U(t+\delta t,t) = I - \frac{i}{\hbar}H(t)\delta t + O(\delta t^2).

Multiplying many short-time factors in chronological order builds the finite-time propagator. This observation underlies numerical time stepping, product formulas, and the time-ordered exponential.

Suppose H=H1H=H_1 from t0t_0 to t1t_1 and H=H2H=H_2 from t1t_1 to t2t_2. Then

U(t2,t0)=U(t2,t1)U(t1,t0)=e−iH2(t2−t1)/ℏe−iH1(t1−t0)/ℏ.\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 H1H_1 occurs first, its exponential is the rightmost factor. If [H1,H2]≠0[H_1,H_2]\ne0, reversing the factors predicts a different final state.

Once the propagator is known, a density operator evolves as

ρ(t)=U(t,t0)ρ(t0)U†(t,t0).\rho(t) = U(t,t_0)\rho(t_0)U^\dagger(t,t_0).

Differentiating gives the von Neumann equation:

iℏρ˙(t)=[H(t),ρ(t)].i\hbar\dot\rho(t) = [H(t),\rho(t)].

Conversely, unitary conjugation solves this equation whenever UU 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.

Let a spin-1/21/2 Hamiltonian be

H=ℏΩ2n⋅σ,∥n∥=1.H = \frac{\hbar\Omega}{2} \boldsymbol n\mathbin{\boldsymbol\cdot}\boldsymbol\sigma, \qquad \lVert\boldsymbol n\rVert=1.

Because

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

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

U(τ)=exp⁡ ⁣[−iΩτ2n⋅σ]=cos⁡ ⁣(Ωτ2)I−isin⁡ ⁣(Ωτ2)n⋅σ.\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 xx axis, n=x^\boldsymbol n=\hat{\boldsymbol x}. Starting from ∣0⟩\lvert0\rangle,

∣ψ(t)⟩=cos⁡ ⁣(Ωτ2)∣0⟩−isin⁡ ⁣(Ωτ2)∣1⟩.\lvert\psi(t)\rangle = \cos\!\left(\frac{\Omega\tau}{2}\right) \lvert0\rangle - i\sin\!\left(\frac{\Omega\tau}{2}\right) \lvert1\rangle.

Therefore

p0(t)=cos⁡2 ⁣(Ωτ2),p1(t)=sin⁡2 ⁣(Ωτ2).\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 Ω\Omega.

  1. Specify both endpoints and the picture. Write U(t,t0)U(t,t_0) unless the reference time is unambiguous.
  2. Classify the Hamiltonian. Decide whether it is time independent, commuting at different times, piecewise constant, or genuinely noncommuting and driven.
  3. Exploit spectral structure. For time-independent HH, diagonalize it or use spectral projectors rather than exponentiating arbitrary matrix entries.
  4. Respect chronological order. In a product, the earliest evolution acts on the right.
  5. Apply the operator once. Use the resulting UU for all desired initial states, amplitudes, or density operators.
  6. Check invariants. Verify U(t0,t0)=IU(t_0,t_0)=I, composition, and unitarity. In numerical work, deviations from these identities are useful error diagnostics.
  7. Separate dynamics from measurement. Unitary propagation between interventions and conditional state update at an intervention are different operations.

Worked Example: Diagonal Two-Level Hamiltonian

Section titled “Worked Example: Diagonal Two-Level Hamiltonian”

For

H=ℏω2σz,H=\frac{\hbar\omega}{2}\sigma_z,

the evolution operator is

U(t,0)=e−iωtσz/2=(e−iωt/200eiωt/2).U(t,0) =e^{-i\omega t\sigma_z/2} = \begin{pmatrix} e^{-i\omega t/2} & 0\\ 0 & e^{i\omega t/2} \end{pmatrix}.

Acting on a state, it changes the relative phase of the two basis components.

  • Writing U(t)=e−iHt/ℏU(t)=e^{-iHt/\hbar} without stating that t0=0t_0=0 or that HH 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,t0)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.
  • P. A. M. Dirac, The Principles of Quantum Mechanics, 4th ed., Oxford University Press, 1958, secs. 26–28.
  • C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977, vol. 1, ch. 3.
  • R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994, chs. 4 and 11.
  • J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020, ch. 2.
  • 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.
  1. Verify that
U(t,t0)=exp⁡ ⁣[−iℏH(t−t0)]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 HH.

Solution

Because HH commutes with every function of itself,

∂U∂t=−iℏHU.\frac{\partial U}{\partial t} = -\frac{i}{\hbar}HU.

Multiplying by iℏi\hbar gives

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

At t=t0t=t_0, the exponent vanishes:

U(t0,t0)=e0=I.U(t_0,t_0)=e^0=I.
  1. Derive the backward equation
−iℏ∂U(t,t0)∂t0=U(t,t0)H(t0)-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 t0t_0:

U(t,t0)=U(t,t0+δt)U(t0+δt,t0).U(t,t_0) = U(t,t_0+\delta t) U(t_0+\delta t,t_0).

Writing H0=H(t0)H_0=H(t_0), the short-time factor is

U(t0+δt,t0)=I−iℏH0δt+O(δt2).U(t_0+\delta t,t_0) = I-\frac{i}{\hbar}H_0\delta t +O(\delta t^2).

Also write U0=U(t,t0)U_0=U(t,t_0) and U+=U(t,t0+δt)U_+=U(t,t_0+\delta t). Rearranging the composition law gives

U+=U0[I+iℏH0δt]+O(δt2).U_+ = U_0 \left[ I+\frac{i}{\hbar}H_0\delta t \right] +O(\delta t^2).

Therefore

∂U(t,t0)∂t0=iℏU(t,t0)H(t0),\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.

  1. Let H′=H+EcIH'=H+E_{\mathrm c}I, with HH time independent. Find U′(t,t0)U'(t,t_0) and show that all Born probabilities for a state evolved under H′H' agree with those for the same state evolved under HH.
Solution

Because HH commutes with the identity,

U′(t,t0)=exp⁡ ⁣[−iℏ(H+EcI)τ]=e−iEcτ/ℏU(t,t0).\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 ∣a⟩\lvert a\rangle,

∣⟨a∣U′∣ψ0⟩∣2=∣e−iEcτ/ℏ∣2∣⟨a∣U∣ψ0⟩∣2.\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.

  1. For
H=ℏΩ2σx,H = \frac{\hbar\Omega}{2}\sigma_x,

compute U(τ)U(\tau) and the first time at which an initial ∣0⟩\lvert0\rangle becomes ∣1⟩\lvert1\rangle up to a global phase.

Solution

Since σx2=I\sigma_x^2=I,

U(τ)=cos⁡ ⁣(Ωτ2)I−isin⁡ ⁣(Ωτ2)σx.U(\tau) = \cos\!\left(\frac{\Omega\tau}{2}\right)I - i\sin\!\left(\frac{\Omega\tau}{2}\right)\sigma_x.

Acting on ∣0⟩\lvert0\rangle gives

U(τ)∣0⟩=cos⁡ ⁣(Ωτ2)∣0⟩−isin⁡ ⁣(Ωτ2)∣1⟩.\begin{aligned} U(\tau)\lvert0\rangle &= \cos\!\left(\frac{\Omega\tau}{2}\right)\lvert0\rangle \\ &\quad- i\sin\!\left(\frac{\Omega\tau}{2}\right)\lvert1\rangle. \end{aligned}

The first complete transfer occurs when

Ωτ2=π2,\frac{\Omega\tau}{2}=\frac{\pi}{2},

so

τ=πΩ.\tau=\frac{\pi}{\Omega}.

The final state is −i∣1⟩-i\lvert1\rangle, which differs from ∣1⟩\lvert1\rangle only by a global phase.

  1. A system evolves with H1H_1 for duration τ1\tau_1 and then with H2H_2 for duration τ2\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−iH2τ2/ℏe−iH1τ1/ℏ.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

[H1,H2]=0.[H_1,H_2]=0.

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

  1. Suppose
H(t)=f(t)A+g(t)I,H(t)=f(t)A+g(t)I,

where AA is a fixed self-adjoint operator. Find U(t,t0)U(t,t_0) without a time-ordering symbol.

Solution

At any two times,

[H(t1),H(t2)]=0,[H(t_1),H(t_2)]=0,

because every term is a linear combination of AA and II. Define

F(t,t0)=∫t0tf(s) ds,G(t,t0)=∫t0tg(s) ds.\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 AA and II commute, the propagator factors as

U(t,t0)=e−iG(t,t0)/ℏe−iF(t,t0)A/ℏ.U(t,t_0) = e^{-iG(t,t_0)/\hbar} e^{-iF(t,t_0)A/\hbar}.
  1. In an orthonormal basis, prove the composition rule
Umn(t2,t0)=∑kUmk(t2,t1)Ukn(t1,t0).U_{mn}(t_2,t_0) = \sum_k U_{mk}(t_2,t_1) U_{kn}(t_1,t_0).

Interpret the index kk.

Solution

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

I=∑k∣k⟩⟨k∣.I=\sum_k\lvert k\rangle\langle k\rvert.

Ordinary matrix multiplication then gives

(U20)mn=∑k(U21)mk(U10)kn.\left(U_{20}\right)_{mn} = \sum_k \left(U_{21}\right)_{mk} \left(U_{10}\right)_{kn}.

The index kk labels a complete set of intermediate alternatives. Because no measurement outcome is selected at t1t_1, the amplitudes are summed before taking an absolute square.

  1. Let
ρ(t)=U(t,t0)ρ(t0)U†(t,t0).\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

U˙=−iℏHU,U˙†=iℏU†H.\dot U = -\frac{i}{\hbar}HU, \qquad \dot U^\dagger = \frac{i}{\hbar}U^\dagger H.

Then

ρ˙=U˙ρ(t0)U†+Uρ(t0)U˙†=−iℏHρ+iℏρH=−iℏ[H,ρ].\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 iℏi\hbar gives

iℏρ˙=[H,ρ].i\hbar\dot\rho=[H,\rho].

Additional exercises retained from the earlier canonical treatment

Section titled “Additional exercises retained from the earlier canonical treatment”
  1. Show that the time-independent expression U(t,t0)=e−iH(t−t0)/ℏU(t,t_0)=e^{-iH(t-t_0)/\hbar} satisfies the operator differential equation.
Solution

Differentiate:

∂U∂t=−iℏHe−iH(t−t0)/ℏ.\frac{\partial U}{\partial t} =-\frac{i}{\hbar}H e^{-iH(t-t_0)/\hbar}.

Multiplying by iℏi\hbar gives

iℏ∂U∂t=HU,i\hbar\frac{\partial U}{\partial t} =H U,

and at t=t0t=t_0, U(t0,t0)=e0=IU(t_0,t_0)=e^0=I.