Picture Transformations

A picture transformation is a time-dependent unitary change of variables that moves time dependence between states, observables, and Hamiltonians without changing physical predictions.

This page uses the convention

$$\lvert\psi_P(t)\rangle = R^\dagger(t)\lvert\psi_S(t)\rangle, \qquad A_P(t) = R^\dagger(t)A_S(t)R(t),$$

where $R(t)$ is unitary. With this convention, the transformed Hamiltonian is

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

The second term is the common source of sign mistakes. It appears whenever the transformation itself depends on time.

General Change of Picture

Let the Schrödinger-picture state obey

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

Choose a unitary $R(t)$ and define the transformed state by

$$\lvert\psi_P(t)\rangle = R^\dagger(t)\lvert\psi_S(t)\rangle.$$

The transformed observable is chosen so that expectation values are unchanged:

$$A_P(t)=R^\dagger(t)A_S(t)R(t).$$

Then

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

The picture has changed; the prediction has not.

Hamiltonian Transformation

Differentiate the transformed state:

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

Substitute $\lvert\psi_S\rangle=R\lvert\psi_P\rangle$ and the Schrödinger equation:

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

Unitarity gives

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

so

$$H_P = R^\dagger H_SR - i\hbar R^\dagger\dot R.$$

The first term is the ordinary transformed Hamiltonian. The second term is the generator of the moving picture itself.

Heisenberg Picture as a Special Case

Set

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

where $U(t,t_0)$ is the full time-evolution operator. Then

$$\lvert\psi_H\rangle = U^\dagger(t,t_0)\lvert\psi_S(t)\rangle = \lvert\psi_S(t_0)\rangle,$$

and

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

The transformed Hamiltonian vanishes for the state equation:

$$H_H^{\mathrm{state}} = U^\dagger HU - i\hbar U^\dagger\dot U =0,$$

because $i\hbar\dot U=HU$. This is why the Heisenberg state is fixed. The dynamics reappears in the operator equation of motion.

Interaction Picture as a Special Case

For a split

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

choose

$$R(t)=U_0(t,t_0)=e^{-iH_0(t-t_0)/\hbar}.$$

Then

$$\lvert\psi_I(t)\rangle = U_0^\dagger(t,t_0)\lvert\psi_S(t)\rangle, \qquad A_I(t) = U_0^\dagger(t,t_0)A_SU_0(t,t_0).$$

The transformed Hamiltonian is

$$\begin{aligned} H_I(t) &= U_0^\dagger(t,t_0)\bigl(H_0+V(t)\bigr)U_0(t,t_0) - i\hbar U_0^\dagger(t,t_0)\dot U_0(t,t_0) \\ &= U_0^\dagger(t,t_0)V(t)U_0(t,t_0) = V_I(t). \end{aligned}$$

The $H_0$ part cancels against the extra generator term. This cancellation is the algebraic heart of the interaction picture.

Rotating Frames

A rotating frame is another time-dependent picture. Let

$$R(t)=e^{-i\omega tG/\hbar},$$

where $G$ generates the rotation. Since

$$\dot R(t) = -\frac{i\omega}{\hbar}GR(t),$$

the extra term is

$$-i\hbar R^\dagger\dot R = -\omega G.$$

If $H_S$ commutes with $G$, then

$$H_P=H_S-\omega G.$$

For a spin with $H_S=\omega_0S_z$ and $G=S_z$, a frame rotating about $z$ has

$$H_P=(\omega_0-\omega)S_z.$$

This subtraction is why rotating frames expose detunings in driven two-level systems. More detailed driven-system approximations, such as the rotating-wave approximation, belong outside this page.

Operator Equations Under a Picture Change

The transformed operator is

$$A_P(t)=R^\dagger(t)A_S(t)R(t).$$

Differentiating gives both explicit time dependence and picture-induced time dependence:

$$\frac{dA_P}{dt} = R^\dagger\frac{\partial A_S}{\partial t}R + \dot R^\dagger A_SR + R^\dagger A_S\dot R.$$

When $R$ is chosen to be a time-evolution operator, this becomes the familiar Heisenberg equation. For a general $R(t)$, it is better to derive the operator equation from the transformed Hamiltonian and the transformed explicit derivative rather than guess signs.

Gauge-Like Aspects

A picture transformation resembles a gauge choice in a limited sense: the mathematical representatives change while physical expectation values remain invariant. The extra Hamiltonian term

$$-i\hbar R^\dagger\dot R$$

acts like a connection term associated with the moving frame.

This analogy is useful but should not be overread. A change of quantum picture is not automatically an electromagnetic gauge transformation, and a gauge transformation may also change potentials and wavefunction phases in a problem-specific way.

Convention Warning

Some books define the transformed state as $R(t)\lvert\psi_S(t)\rangle$ instead of $R^\dagger(t)\lvert\psi_S(t)\rangle$. That convention is equally valid, but the transformed operator and Hamiltonian formulas change:

$$\lvert\psi_P\rangle=R\lvert\psi_S\rangle \quad \Longrightarrow \quad H_P=RH_SR^\dagger+i\hbar\dot R R^\dagger.$$

Before comparing formulas across sources, identify which side carries the dagger.

Common Mistakes

• Transforming states but leaving observables in the old picture.
• Forgetting the extra term $-i\hbar R^\dagger\dot R$ for time-dependent $R(t)$.
• Using the right formula with the opposite convention for $R(t)$.
• Treating the interaction picture as approximate before any expansion is truncated.
• Confusing explicit time dependence in $A_S(t)$ with time dependence induced by $R(t)$.
• Calling a nonunitary similarity transformation a change of picture without checking the inner product and expectation values.
• Dropping picture labels in a calculation where several transformed objects appear.

Cross-Links

• Pictures of Quantum Mechanics
• Schrödinger Picture
• Heisenberg Picture
• Interaction Picture
• Heisenberg Equations of Motion
• Operators with Explicit Time Dependence
• Density Operators in Different Pictures
• Time-Evolution Operator
• Dyson Expansion as Formal Evolution
• Translation Table of Formulations
• Pictures of Motion Overview

References

• P. A. M. Dirac, The Principles of Quantum Mechanics, 4th ed., Oxford University Press, 1958.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
• H. Goldstein, C. Poole, and J. Safko, Classical Mechanics, 3rd ed., Addison-Wesley, 2002.

Exercises

1. Derive the transformed Hamiltonian for the convention $\lvert\psi_P\rangle=R^\dagger\lvert\psi_S\rangle$.

Solution

Differentiate:

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

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

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

Since $\dot R^\dagger R=-R^\dagger\dot R$,

$$H_P=R^\dagger H_SR-i\hbar R^\dagger\dot R.$$

2. Show that choosing $R(t)=U(t,t_0)$ makes the Heisenberg state time independent.

Solution

With $R=U$,

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

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

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

This fixed state is the Heisenberg state.

3. Let $R(t)=e^{-i\omega tG/\hbar}$. Compute the extra term in the transformed Hamiltonian.

Solution

Differentiate:

$$\dot R(t) = -\frac{i\omega}{\hbar}GR(t).$$

Then

$$R^\dagger\dot R = -\frac{i\omega}{\hbar}G,$$

assuming $G$ commutes with its own exponential. Therefore

$$-i\hbar R^\dagger\dot R = -\omega G.$$

4. Under a unitary picture transformation, why must observables transform as $A_P=R^\dagger A_SR$ if states transform as $\lvert\psi_P\rangle=R^\dagger\lvert\psi_S\rangle$?

Solution

The expectation value should be invariant:

$$\langle\psi_P\rvert A_P\lvert\psi_P\rangle = \langle\psi_S\rvert A_S\lvert\psi_S\rangle.$$

Since $\lvert\psi_P\rangle=R^\dagger\lvert\psi_S\rangle$, one has $\langle\psi_P\rvert=\langle\psi_S\rvert R$. Choosing

$$A_P=R^\dagger A_SR$$

gives

$$\langle\psi_P\rvert A_P\lvert\psi_P\rangle = \langle\psi_S\rvert RR^\dagger A_SRR^\dagger\lvert\psi_S\rangle = \langle\psi_S\rvert A_S\lvert\psi_S\rangle.$$