Unitary Dynamics and Pictures of Motion

The Schrödinger, Heisenberg, and interaction pictures are not competing laws of motion. They are time-dependent changes of representation: evolution can be moved between states and operators while every transition amplitude, Born probability, and consistently transformed matrix element remains unchanged. This page derives that equivalence and identifies the assumptions under which a unitary propagator exists.

Required background. Calculus and Linear Algebra supplies inner products, adjoints, exponentials, and basis changes; Ordinary Differential Equations supplies linear initial-value problems; Quantum States, Observables, and Measurement supplies states, effects, traces, and self-adjoint operators. Helpful background. Classical and Variational Mechanics supplies the relation between continuous symmetries and conserved generators.

Propagators as quantum initial-value maps

We keep $\hbar$ explicit. In a finite-dimensional Hilbert space, let $H(t)$ be a continuous family of self-adjoint Hamiltonians. The propagator $U(t,t_0)$ is the unique solution of

$$i\hbar\,\partial_tU(t,t_0)=H(t)U(t,t_0), \qquad U(t_0,t_0)=I.$$

For an unbounded Hamiltonian, self-adjointness at each instant is not by itself a complete existence theorem: one also needs compatible domains and enough regularity in $t$. The formulas below are exact in finite dimensions and remain valid for infinite-dimensional systems whenever the stated products and derivatives exist on an invariant domain. This is a qualitative warning rather than a sufficient theorem for nonautonomous unbounded evolution.

Differentiate $U^\dagger U$ and use $\partial_tU=-iHU/\hbar$ and $\partial_tU^\dagger=iU^\dagger H/\hbar$:

$$\partial_t(U^\dagger U) =\frac{i}{\hbar}U^\dagger HU -\frac{i}{\hbar}U^\dagger HU=0.$$

Since $U(t_0,t_0)=I$, the propagator is unitary. It therefore preserves inner products and composes as

$$U(t_2,t_0)=U(t_2,t_1)U(t_1,t_0), \qquad U(t_0,t)=U(t,t_0)^\dagger.$$

Pure states evolve as $|\psi(t)\rangle=U|\psi(t_0)\rangle$; density operators evolve as

$$\rho(t)=U\rho(t_0)U^\dagger, \qquad i\hbar\,\dot\rho=[H,\rho].$$

Both trace and positivity are preserved. Preskill derives closed-system unitarity from infinitesimal Schrödinger evolution in Preskill 2015, §2.1, Eqs. (2.10)–(2.12), printed pp. 5–6.

Time ordering is part of the definition

For time-independent $H$, functional calculus gives

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

For $t\geq t_0$, if $[H(t_1),H(t_2)]=0$ for every pair of times, the same formula holds with $H(t-t_0)$ replaced by $\int_{t_0}^{t}H(t')dt'$. For unbounded self-adjoint operators this statement requires strong commutativity and a well-defined self-adjoint time integral, not merely a formal commutator on a common core. Otherwise, later-time operators must stand to the left. The notation

$$U(t,t_0)=\mathcal T\exp\!\left[ -\frac{i}{\hbar}\int_{t_0}^{t}H(t')\,dt' \right]$$

means the ordered Dyson series

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

Thus $\mathcal T$ is not an extra dynamical operation; it records the order already imposed by composition. Turner constructs the ordered product and Dyson series in Turner 2017, Lecture 8, §8.1, Eqs. (8.9)–(8.22), printed pp. 41–42. For $t\lt t_0$, use $U(t,t_0)=U(t_0,t)^\dagger$; the corresponding explicit series is anti-time-ordered.

For a state obeying unitary dynamics, explicit time dependence is the only source of change in the mean energy:

$$\frac{d}{dt}\operatorname{Tr}(\rho H) =\operatorname{Tr}(\rho\,\partial_tH),$$

provided the traces and derivatives exist. A time-dependent Hamiltonian can therefore generate unitary motion without conserving energy.

One unitary transformation contains all pictures

Let $W(t)$ be any differentiable unitary with $W(t_0)=I$, and define

$$|\psi_W\rangle=W^\dagger|\psi_S\rangle, \qquad \rho_W=W^\dagger\rho_SW, \qquad A_W=W^\dagger A_SW.$$

Direct differentiation gives the generator of transformed-state evolution,

$$\boxed{ K_W=W^\dagger H_SW-i\hbar W^\dagger\dot W. }$$

The symbol $K_W$ is deliberate: it need not equal the transformed energy observable $H_W^{\mathrm{obs}}=W^\dagger H_SW$. The second term is the generator contributed by the moving representation. Every expectation value is unchanged,

$$\operatorname{Tr}(\rho_SA_S) =\operatorname{Tr}(W^\dagger\rho_SWW^\dagger A_SW) =\operatorname{Tr}(\rho_WA_W),$$

and the same cancellation preserves Born probabilities and transition amplitudes. Turner derives the equal-matrix-element statement and the Heisenberg equation in Turner 2017, Lecture 6, §6.2.2, Eqs. (6.27)–(6.44), printed pp. 36–38.

For the interaction-picture row below, choose a split $H=H_0+V$, let $U_0$ be the propagator generated by $H_0$, and define $|\psi_I\rangle=U_0^\dagger|\psi_S\rangle$, $A_I=U_0^\dagger A_SU_0$, and $V_I=U_0^\dagger VU_0$. The next section derives their evolution.

Picture	Choice of $W$	State	Operator	State generator
Schrödinger	$I$	Evolves with $H$	Fixed unless explicitly time dependent	$H$
Heisenberg	Full propagator $U$	Fixed	$A_H=U^\dagger A_SU$	$0$
Interaction	Reference propagator $U_0$	Evolves with $V_I$	Transformed by $U_0$	$V_I$

In the Heisenberg picture, $W=U$, so the state generator $K_H$ vanishes. The energy observable does not vanish: it is $H_H^{\mathrm{obs}}=U^\dagger H_SU$. The operator equation is

$$\frac{dA_H}{dt} =\frac{i}{\hbar}[H_H^{\mathrm{obs}},A_H] +U^\dagger\frac{\partial A_S}{\partial t}U,$$

An operator with no explicit time dependence is conserved if it commutes with the Hamiltonian throughout the interval. This is the quantum form of the symmetry–generator connection, but it is a statement about the full Hamiltonian and its domain, not merely a formal commutator of differential expressions.

For a continuous symmetry generated by a self-adjoint $Q$, suppose the Hamiltonian is invariant for every $\epsilon$ near zero and every relevant time:

$$R(\epsilon)=e^{-i\epsilon Q/\hbar}, \qquad R(\epsilon)^\dagger H(t)R(\epsilon)=H(t) \quad\Longrightarrow\quad [H(t),Q]=0 \quad\Longrightarrow\quad \frac{d}{dt}\langle Q\rangle=0,$$

assuming $Q$ has no explicit time dependence and all products share an appropriate domain.

The interaction picture is exact before any expansion

Split $H(t)=H_0(t)+V(t)$ and let $U_0$ be the exact propagator generated by $H_0$. With $W=U_0$,

$$|\psi_I\rangle=U_0^\dagger|\psi_S\rangle, \qquad A_I=U_0^\dagger A_SU_0,$$

and

$$U_I(t,t_0)=U_0(t,t_0)^\dagger U(t,t_0), \qquad V_I(t)=U_0^\dagger V(t)U_0.$$

Using the equations for $U$ and $U_0$ gives

$$i\hbar\,\partial_tU_I(t,t_0)=V_I(t)U_I(t,t_0), \qquad U_I(t_0,t_0)=I.$$

This factorization and its infinite Dyson series are exact. Approximation begins only when that series is truncated or otherwise estimated, and the size of the omitted terms must then be controlled. Turner gives the factorization and transformed interaction in Turner 2017, Lecture 8, §8.2, Eqs. (8.23)–(8.36), printed pp. 42–43.

Worked example: one precessing spin

Let

$$H=\frac{\hbar\Omega}{2}\sigma_z, \qquad |\psi(0)\rangle=|+x\rangle,$$

where $|+x\rangle=(|+z\rangle+|-z\rangle)/\sqrt2$. In the Schrödinger picture, use the Pauli identity $[\sigma_z,\sigma_x]=2i\sigma_y$. The evolving state is

$$|\psi_S(t)\rangle =\frac{e^{-i\Omega t/2}|+z\rangle +e^{i\Omega t/2}|-z\rangle}{\sqrt2},$$

so $\langle\sigma_x\rangle_S=\cos(\Omega t)$, $\langle\sigma_y\rangle_S=\sin(\Omega t)$, and $\langle\sigma_z\rangle_S=0$. The three component squares sum to one, as required for the rotating unit Bloch vector. In the Heisenberg picture the state is fixed and

$$\sigma_{x,H}(t) =U^\dagger\sigma_xU =\sigma_x\cos(\Omega t)-\sigma_y\sin(\Omega t).$$

Its expectation in $|+x\rangle$ is again $\cos(\Omega t)$. The sign is checked without memorizing a rotation matrix:

$$\dot\sigma_{x,H}(0) =\frac{i}{\hbar}[H,\sigma_x] =-\Omega\sigma_y.$$

The two calculations distribute time dependence differently but compute the same detector statistic.

Common pitfalls

Treating a picture as an approximation. A picture change is an exact unitary transformation. Error enters only after a series, basis, or effective Hamiltonian is truncated.

Dropping time ordering. The ordinary exponential of $\int H(t)dt$ is generally wrong when Hamiltonians at different times do not commute. The ordered product fixes the operator sequence.

Forgetting the moving-frame term. Under a time-dependent $W$, the state generator $K_W$ is not just the transformed energy observable $W^\dagger HW$; the term $-i\hbar W^\dagger\dot W$ is required for equivalent dynamics.

Confusing unitarity with energy conservation. Unitarity preserves the norm and total probability. Energy is conserved only under the additional conditions stated above.

The bounded bridge to gravitation

On a stationary spacetime, a chosen time-translation symmetry can provide a conserved frequency label. Positivity of its generator and a preferred particle interpretation require additional global conditions. On a genuinely time-dependent background, there need not be a globally conserved energy. Neither statement is changed by moving between quantum pictures.

A picture transformation is also not a spacetime coordinate change or a gravitational gauge transformation: $W$ acts on the quantum description, whereas coordinates relabel spacetime events. Quantum field theory introduces further questions—field domains, vacuum choice, and whether evolution is unitarily implementable in a selected representation—that are outside this one-system bridge. The next page develops the rotation generators used to label gravitational modes in Angular Momentum, Spin, and Rotations.

Exercises

A rotating representation

For the precessing spin above, choose $W(t)=e^{-i\omega t\sigma_z/2}$. Find $K_W$, the transformed state, and $\sigma_{x,W}$. Show directly that $\langle\sigma_{x,W}\rangle_W=\cos(\Omega t)$.

Solution

Because $W$ commutes with $H$ and $W^\dagger\dot W=-i\omega\sigma_z/2$,

$$K_W=W^\dagger HW-i\hbar W^\dagger\dot W =\frac{\hbar(\Omega-\omega)}{2}\sigma_z.$$

The transformed state is

$$|\psi_W(t)\rangle =W^\dagger U|+x\rangle =e^{-i(\Omega-\omega)t\sigma_z/2}|+x\rangle.$$

The transformed observable is

$$\sigma_{x,W} =W^\dagger\sigma_xW =\sigma_x\cos(\omega t)-\sigma_y\sin(\omega t).$$

In the transformed state, $\langle\sigma_x\rangle_W=\cos[(\Omega-\omega)t]$ and $\langle\sigma_y\rangle_W=\sin[(\Omega-\omega)t]$. Therefore

$$\langle\sigma_{x,W}\rangle_W =\cos(\omega t)\cos[(\Omega-\omega)t] -\sin(\omega t)\sin[(\Omega-\omega)t] =\cos(\Omega t).$$

At $\omega=\Omega$ the transformed state is fixed, but the observable carries all the time dependence. This is representation bookkeeping, not frozen physics.

References

• Preskill, John. Quantum Information, Chapter 2, revised July 2015, §2.1, printed pp. 3–7. California Institute of Technology lecture notes. Official PDF.
• Turner, Andrew. Quantum Theory I, Lecture 6, prepared for instructor Senthil Todadri, MIT 8.321, Fall 2017, printed pp. 33–38. MIT OpenCourseWare PDF.
• Turner, Andrew. Quantum Theory I, Lecture 8, prepared for instructor Senthil Todadri, MIT 8.321, Fall 2017, printed pp. 40–43. MIT OpenCourseWare PDF.