Pictures of Motion Overview

The Schrödinger, Heisenberg, and interaction pictures are unitarily equivalent descriptions of the same closed-system dynamics. They differ only in how they distribute time dependence among states, observables, and the generator of state evolution.

The organizing principle is simple: transform states and observables together, and every measurable prediction remains unchanged. A picture is therefore a choice of dynamical bookkeeping, not a different theory or interpretation.

This page provides the working dictionary needed throughout Core Formalism. The full derivation of arbitrary time-dependent changes of picture belongs to Picture Transformations, and the detailed uses of each picture live in the Quantum Dynamics volume.

Picture Versus Representation

A representation chooses coordinates or a basis for abstract states and operators. Position-space and momentum-space wavefunctions, for example, can both be used within the Schrödinger picture.

A picture is a time-dependent unitary change of descriptive variables. It moves dynamical time dependence between state representatives and operator representatives.

These choices are independent:

• one may use the Schrödinger picture in the position or energy representation;
• one may write Heisenberg operators as abstract operators or as matrices in a chosen basis;
• changing basis does not by itself turn one picture into another.

A time-independent unitary basis change has no moving-frame term. A genuinely time-dependent picture transformation does.

One Unitary Dictionary

Choose a unitary operator $R(t)$ and the convention

$$\begin{aligned} \lvert\psi_P(t)\rangle &= R(t)^\dagger\lvert\psi_S(t)\rangle,\\ A_P(t) &= R(t)^\dagger A_S(t)R(t). \end{aligned}$$

The subscript $S$ denotes Schrödinger-picture objects, while $P$ denotes the new picture. If $R(t_0)=I$, the representatives agree at the reference time.

Expectation values are invariant:

$$\begin{aligned} \langle\psi_P\rvert A_P\lvert\psi_P\rangle &= \langle\psi_S\rvert RR^\dagger A_S RR^\dagger \lvert\psi_S\rangle\\ &= \langle\psi_S\rvert A_S\lvert\psi_S\rangle. \end{aligned}$$

For a density operator,

$$\rho_P(t) = R(t)^\dagger\rho_S(t)R(t),$$

and the same statement is

$$\operatorname{Tr}\!\left(\rho_P A_P\right) = \operatorname{Tr}\!\left(\rho_S A_S\right).$$

Because $R(t)$ itself changes with time, the Hamiltonian driving the transformed state is not merely $R^\dagger H_SR$. With the convention above,

$$H_P^{\mathrm{state}}(t) = R^\dagger H_S R - i\hbar R^\dagger\dot R.$$

The second term is the generator of the moving picture. Its sign changes if the opposite convention for transforming states is adopted, which is why every calculation should state its convention.

The three standard pictures are special choices of $R(t)$:

Picture	Choice of $R(t)$	Where picture-induced time dependence lives	State generator
Schrödinger	$I$	states	$H_S(t)$
Heisenberg	full propagator $U(t,t_0)$	observables	zero
Interaction	reference propagator $U_0(t,t_0)$	both	transformed interaction $V_I(t)$

This table concerns picture-induced time dependence. An observable can still have explicit time dependence because the experimental quantity itself changes with time.

Schrödinger Picture

In the Schrödinger picture, $R=I$. States obey

$$i\hbar\frac{d}{dt}\lvert\psi_S(t)\rangle = H_S(t)\lvert\psi_S(t)\rangle,$$

and

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

An observable has no picture-induced motion:

$$A_S(t)=A_S$$

unless its definition has explicit time dependence. Its expectation value changes because the state changes:

$$\langle A\rangle_t = \langle\psi_S(t)\rvert A_S(t) \lvert\psi_S(t)\rangle.$$

This is the natural language when the evolving wavefunction, state populations, or numerical state propagation is the central object. It is also the default picture for most canonical wave-mechanics problems.

A fixed Schrödinger-picture operator is not automatically a conserved observable. Conservation concerns its measurement statistics under the evolving state, not whether the symbol $A_S$ is written without $t$. See Conservation Laws.

The detailed treatment is Schrödinger Picture.

Heisenberg Picture

Choose the full propagator,

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

The transformed state is fixed:

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

Observables carry the dynamical time dependence:

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

For an observable with possible explicit time dependence,

$$\frac{dA_H}{dt} = \frac{i}{\hbar}[H_H,A_H] + \left( \frac{\partial A_S}{\partial t} \right)_H,$$

where

$$\begin{aligned} H_H(t) &= U^\dagger H_S(t)U,\\ \left( \frac{\partial A_S}{\partial t} \right)_H &= U^\dagger \frac{\partial A_S}{\partial t} U. \end{aligned}$$

The state generator vanishes:

$$\begin{aligned} H_H^{\mathrm{state}} &= U^\dagger H_SU - i\hbar U^\dagger\dot U\\ &=0. \end{aligned}$$

This zero does not mean that energy vanishes. The transformed energy observable $U^\dagger H_SU$ remains a physical observable. The zero refers only to the Hamiltonian governing the already-fixed Heisenberg state.

Expectation values agree with the Schrödinger picture:

$$\langle A\rangle_t = \langle\psi_H\rvert A_H(t) \lvert\psi_H\rangle.$$

The Heisenberg picture is especially useful for operator equations, conserved quantities, commutator algebras, response functions, and multi-time correlation functions. It is also the standard bridge to time-dependent quantum fields.

The detailed pages are Heisenberg Picture and Heisenberg Equations of Motion.

Interaction Picture

Split the Hamiltonian into a reference part and a remainder:

$$H_S(t)=H_0(t)+V(t).$$

Let $U_0(t,t_0)$ be the exact propagator generated by $H_0(t)$:

$$\begin{aligned} i\hbar\frac{\partial U_0(t,t_0)}{\partial t} &= H_0(t)U_0(t,t_0),\\ U_0(t_0,t_0)&=I. \end{aligned}$$

Choose

$$R(t)=U_0(t,t_0).$$

The interaction-picture state and observable are

$$\begin{aligned} \lvert\psi_I(t)\rangle &= U_0(t,t_0)^\dagger \lvert\psi_S(t)\rangle,\\ A_I(t) &= U_0(t,t_0)^\dagger A_S(t) U_0(t,t_0). \end{aligned}$$

The reference contribution cancels from the state generator, leaving

$$V_I(t) = U_0(t,t_0)^\dagger V(t) U_0(t,t_0),$$

and

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

If $U_I(t,t_0)$ propagates the interaction-picture state, then

$$\begin{aligned} U_I(t,t_0) &= U_0(t,t_0)^\dagger U(t,t_0),\\ U(t,t_0) &= U_0(t,t_0)U_I(t,t_0). \end{aligned}$$

The interaction picture is exact. Approximation enters only when $U_I$ is truncated, expanded, or otherwise estimated. The split is not unique:

• if $H_0=0$, the interaction picture reduces to the Schrödinger picture;
• if $H_0=H$, it reduces to the Heisenberg allocation of time dependence;
• useful choices place the exactly solvable or dominant motion in $H_0$ and leave a simpler $V$.

This picture is the standard starting point for time-dependent perturbation theory, transition amplitudes, Dyson series, scattering expansions, and perturbative QFT.

The exact definition is developed in Interaction Picture. Its perturbative use belongs to Interaction Picture for Perturbation Theory.

Same Physics, Different Bookkeeping

For every picture related by $R(t)$,

$$\begin{aligned} \langle A\rangle_t &= \langle\psi_S(t)\rvert A_S(t) \lvert\psi_S(t)\rangle\\ &= \langle\psi_P(t)\rvert A_P(t) \lvert\psi_P(t)\rangle. \end{aligned}$$

The equality extends beyond means. If $P_{a,S}(t)$ is a spectral projector or measurement effect, define

$$P_{a,P}(t) = R(t)^\dagger P_{a,S}(t)R(t).$$

Then

$$\begin{aligned} p(a,t) &= \operatorname{Tr}\!\left[ \rho_S(t)P_{a,S}(t) \right]\\ &= \operatorname{Tr}\!\left[ \rho_P(t)P_{a,P}(t) \right]. \end{aligned}$$

All single-time probabilities are therefore picture independent. Multi-time amplitudes and correlation functions also agree when every state, operator, propagator, and ordering prescription is translated consistently.

A common failure is to transform only half of an expression. For example,

$$\langle\psi_H\rvert A_S\lvert\psi_H\rangle$$

is generally not the expectation value at time $t$. The fixed Heisenberg state must be paired with $A_H(t)$, not with an untransformed Schrödinger operator.

Worked Dictionary: Harmonic Oscillator

Consider

$$H= \hbar\omega \left( a^\dagger a+\frac{1}{2} \right),$$

and write $\tau=t-t_0$. If

$$\lvert\psi(t_0)\rangle = \sum_{n=0}^{\infty}c_n\lvert n\rangle,$$

then the Schrödinger-picture state is

$$\lvert\psi_S(t)\rangle = \sum_{n=0}^{\infty} c_n e^{-i\omega(n+1/2)\tau} \lvert n\rangle.$$

The operator $a_S=a$ is fixed. In the Heisenberg picture the state is fixed, while

$$\begin{aligned} a_H(t) &= e^{-i\omega\tau}a,\\ a_H^\dagger(t) &= e^{i\omega\tau}a^\dagger. \end{aligned}$$

Consequently,

$$\begin{aligned} x_H(t) &= \sqrt{\frac{\hbar}{2m\omega}} \left[ e^{-i\omega\tau}a + e^{i\omega\tau}a^\dagger \right]. \end{aligned}$$

The two calculations give the same number:

$$\langle\psi_S(t)\rvert x\lvert\psi_S(t)\rangle = \langle\psi_H\rvert x_H(t)\lvert\psi_H\rangle.$$

If the interaction-picture choice is $H_0=H$ and $V=0$, then $\lvert\psi_I\rangle$ is fixed and $x_I(t)=x_H(t)$. If instead $H_0=0$, then $\lvert\psi_I(t)\rangle=\lvert\psi_S(t)\rangle$ and $x_I=x_S$. The interaction picture interpolates between the two familiar allocations according to the chosen split.

Choosing a Picture

Use the Schrödinger picture when:

• the state vector, wavefunction, or population transfer is the primary object;
• diagonalizing or numerically propagating the Hamiltonian is direct;
• boundary conditions and spatial wave mechanics dominate the calculation;
• only a small number of state amplitudes are needed.

Use the Heisenberg picture when:

• operator equations close on a small algebra;
• conserved quantities and commutators are central;
• multi-time observables, response functions, or field operators are primary;
• the state is complicated but operator motion is simple.

Use the interaction picture when:

• $H=H_0+V$ separates solvable background motion from a residual interaction;
• transitions driven by $V$ are the target;
• perturbation theory, scattering, or time-ordered expansions are required;
• a rotating or moving frame removes a large, known piece of the motion.

No choice is universally superior. The best picture makes the quantities needed for the calculation evolve as simply as possible.

Explicit Time Dependence and Reference Time

An observable can have explicit time dependence in every picture. For example, a detector orientation, control parameter, or externally defined measurement may change with laboratory time. In the Heisenberg picture,

$$A_H(t) = U^\dagger(t,t_0) A_S(t) U(t,t_0)$$

contains both explicit dependence from $A_S(t)$ and picture-induced dependence from $U(t,t_0)$. Dropping the explicit derivative produces an incomplete equation of motion. See Explicitly Time-Dependent Operators.

The reference time $t_0$ is also part of the dictionary. With

$$R(t_0)=I,$$

all picture representatives coincide at $t_0$. Choosing another reference time changes the representatives and intermediate formulas, not the predictions.

Open-System Boundary

For reduced open-system dynamics, the Schrödinger-picture state may evolve through a nonunitary channel. A corresponding Heisenberg description still exists through the channel’s adjoint acting on observables, but it is not generally obtained by conjugation with one unitary operator on the reduced Hilbert space.

The unitary formulas on this page apply directly to closed systems. Their open-system extension belongs with quantum channels and master equations; see Density Operators in Different Pictures and the Lindblad–GKSL Equation.

Where Detailed Treatment Lives

Use this overview as a translation key. For derivations and advanced applications, continue to:

• Schrödinger Picture
• Heisenberg Picture
• Interaction Picture
• Picture Transformations
• Density Operators in Different Pictures
• Translation Table of Formulations
• Which Formulation Should I Use?
• Time Ordering

Common Mistakes

• Treating the pictures as different physical theories or interpretations.
• Confusing a picture with a position-, momentum-, or energy-basis representation.
• Transforming the state but not the observable, or vice versa.
• Forgetting the moving-frame term $-i\hbar R^\dagger\dot R$ in the transformed state generator.
• Using the wrong sign because two sources adopt opposite conventions for $R$ and $R^\dagger$.
• Assuming a fixed Schrödinger-picture operator is automatically conserved.
• Interpreting the zero Heisenberg state generator as a zero energy observable.
• Forgetting explicit time dependence already present in $A_S(t)$.
• Assuming the interaction picture is approximate by definition.
• Choosing $H_0$ without checking whether it actually simplifies $V_I(t)$.
• Dropping time ordering when $V_I(t)$ fails to commute with itself at different times.
• Changing the reference time in only part of a calculation.

Cross-Links

• Time-Evolution Operator
• Unitary Time Evolution
• Time-Dependent Hamiltonians
• Conservation Laws
• Heisenberg Equations of Motion
• From Evolution Operators to Time-Ordered Products

References

• P. A. M. Dirac, The Principles of Quantum Mechanics, 4th ed., Oxford University Press, 1958, secs. 26–28.
• A. Messiah, Quantum Mechanics, Dover, 1999, vol. 1, chs. 8–9.
• 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 18.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020, chs. 2 and 5.
• L. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd ed., World Scientific, 2014, chs. 3–4.

Exercises

1. Prove directly that the expectation value is invariant under the picture convention

$$\lvert\psi_P\rangle=R^\dagger\lvert\psi_S\rangle, \qquad A_P=R^\dagger A_SR.$$

Solution

The transformed bra is

$$\langle\psi_P\rvert = \langle\psi_S\rvert R.$$

Therefore

$$\begin{aligned} \langle\psi_P\rvert A_P\lvert\psi_P\rangle &= \langle\psi_S\rvert R(R^\dagger A_SR)R^\dagger \lvert\psi_S\rangle\\ &= \langle\psi_S\rvert A_S \lvert\psi_S\rangle, \end{aligned}$$

because $RR^\dagger=I$. The same cancellation proves equality of all Born probabilities when the measurement effects are transformed in the same way.

2. Starting from $\lvert\psi_P\rangle=R^\dagger\lvert\psi_S\rangle$, derive the Hamiltonian that generates the transformed state.

Solution

Differentiate the state:

$$\frac{d}{dt}\lvert\psi_P\rangle = \dot R^\dagger\lvert\psi_S\rangle + R^\dagger\frac{d}{dt}\lvert\psi_S\rangle.$$

Use $\lvert\psi_S\rangle=R\lvert\psi_P\rangle$ and

$$i\hbar\frac{d}{dt}\lvert\psi_S\rangle = H_S\lvert\psi_S\rangle.$$

Then

$$\begin{aligned} i\hbar\frac{d}{dt}\lvert\psi_P\rangle &= \left( i\hbar\dot R^\dagger R + R^\dagger H_SR \right) \lvert\psi_P\rangle. \end{aligned}$$

Differentiating $R^\dagger R=I$ gives

$$\dot R^\dagger R = -R^\dagger\dot R.$$

Hence

$$H_P^{\mathrm{state}} = R^\dagger H_SR - i\hbar R^\dagger\dot R.$$

3. Derive the Heisenberg equation for an observable $A_S(t)$ with explicit time dependence.

Solution

Start with

$$A_H(t)=U^\dagger A_S(t)U.$$

The propagator equations imply

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

Differentiate all three factors:

$$\begin{aligned} \frac{dA_H}{dt} &= \frac{i}{\hbar}U^\dagger H_SA_SU - \frac{i}{\hbar}U^\dagger A_SH_SU\\ &\quad+ U^\dagger \frac{\partial A_S}{\partial t} U. \end{aligned}$$

Recognizing the transformed operators gives

$$\frac{dA_H}{dt} = \frac{i}{\hbar}[H_H,A_H] + \left( \frac{\partial A_S}{\partial t} \right)_H.$$

4. For $H=H_0(t)+V(t)$, show that $U_I=U_0^\dagger U$ is generated by $V_I=U_0^\dagger VU_0$ and that $U=U_0U_I$.

Solution

Differentiate $U_I=U_0^\dagger U$:

$$\dot U_I = \dot U_0^\dagger U + U_0^\dagger\dot U.$$

Using

$$\begin{aligned} i\hbar\dot U &= (H_0+V)U,\\ i\hbar\dot U_0 &= H_0U_0, \end{aligned}$$

gives

$$\begin{aligned} i\hbar\dot U_I &= -U_0^\dagger H_0U + U_0^\dagger(H_0+V)U\\ &= U_0^\dagger VU. \end{aligned}$$

Since $U=U_0U_I$,

$$i\hbar\dot U_I = \left(U_0^\dagger VU_0\right)U_I = V_IU_I.$$

Multiplying $U_I=U_0^\dagger U$ on the left by $U_0$ gives the exact factorization

$$U=U_0U_I.$$

5. For the harmonic oscillator, use $[H,a]=-\hbar\omega a$ to find $a_H(t)$ and show that the number operator $N=a^\dagger a$ is conserved.

Solution

The Heisenberg equation gives

$$\frac{da_H}{dt} = \frac{i}{\hbar}[H,a_H] = -i\omega a_H.$$

With $a_H(t_0)=a$,

$$a_H(t) = e^{-i\omega(t-t_0)}a.$$

Similarly,

$$a_H^\dagger(t) = e^{i\omega(t-t_0)}a^\dagger.$$

Therefore

$$\begin{aligned} N_H(t) &= a_H^\dagger(t)a_H(t)\\ &= a^\dagger a =N. \end{aligned}$$

Equivalently, $[H,N]=0$. The ladder operators rotate in phase while the occupation-number distribution remains fixed.

6. Classify each statement as correct or incorrect, and repair the incorrect ones.

   1. A position-space wavefunction is necessarily in the Schrödinger picture.
   2. The interaction picture is an approximation.
   3. The Heisenberg-picture energy observable vanishes because the Heisenberg state is fixed.
   4. One may compute a time-dependent mean using $\langle\psi_H\rvert A_S\lvert\psi_H\rangle$.

Solution

All four statements are incorrect.

1. Position space is a representation. Schrödinger, Heisenberg, or interaction-picture objects can all be represented in a position basis.
2. The interaction-picture transformation is exact. Approximation enters only when its evolution operator or transformed interaction is approximated.
3. The generator of the fixed Heisenberg state is zero, but the energy observable is $H_H=U^\dagger H_SU$ and need not vanish.
4. The Heisenberg state must be paired with the Heisenberg operator:

$$\langle A\rangle_t = \langle\psi_H\rvert A_H(t) \lvert\psi_H\rangle.$$

Equivalently, use the evolving Schrödinger state with $A_S(t)$. Mixing representatives from different pictures generally gives the wrong time dependence.