Picture Translation Table

This page is a convention-fixed translation sheet for closed-system quantum pictures. The general derivation is given in Picture Transformations. The broader Translation Table of Formulations compares pictures, kernels, path integrals, and phase-space methods conceptually; this page focuses narrowly on exact state-operator conversions and the signs that most often cause errors.

Unitary Picture Transformations

Let $R(t)$ be differentiable and unitary. The formulas hold directly for finite matrices; for unbounded observables require the transformed domains and derivatives to exist. Define picture $P$ from the Schrödinger picture by

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

To preserve matrix elements, transform observables and density operators as

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

The transformed state obeys a Schrödinger equation with

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

Every table below uses this dagger convention. Some sources instead define $\lvert\psi_P\rangle=R\lvert\psi_S\rangle$; their formulas are valid but have the daggers and connection-term sign reversed.

General Translation Table

Object	Schrödinger to picture $P$	Picture $P$ to Schrödinger
state vector	$\lvert\psi_P\rangle=R^\dagger\lvert\psi_S\rangle$	$\lvert\psi_S\rangle=R\lvert\psi_P\rangle$
bra	$\langle\psi_P\rvert=\langle\psi_S\rvert R$	$\langle\psi_S\rvert=\langle\psi_P\rvert R^\dagger$
observable	$A_P=R^\dagger A_SR$	$A_S=RA_PR^\dagger$
density operator	$\rho_P=R^\dagger\rho_SR$	$\rho_S=R\rho_PR^\dagger$
energy observable	$E_P=R^\dagger H_SR$	$H_S=RE_PR^\dagger$
state generator	$H_P=R^\dagger H_SR-i\hbar R^\dagger\dot R$	$H_S=RH_PR^\dagger+i\hbar\dot R R^\dagger$
evolution operator	$U_P(t,t_0)=R^\dagger(t)U_S(t,t_0)R(t_0)$	$U_S(t,t_0)=R(t)U_P(t,t_0)R^\dagger(t_0)$

The two endpoint factors in the evolution-operator row are essential. Setting $R(t_0)=I$ is convenient, but it is a convention rather than an identity that holds for every transformation.

Expectation-Value Invariance

For pure states,

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

For mixed states,

$$\begin{aligned} \langle A\rangle_t &= \operatorname{Tr} \left[ \rho_S(t)A_S(t) \right] \\ &= \operatorname{Tr} \left[ \rho_P(t)A_P(t) \right]. \end{aligned}$$

The second equality uses unitarity and cyclicity of the trace. A picture is a change of mathematical representatives, not a change in the predicted probability distribution.

Transition amplitudes are likewise invariant when endpoint states and the evolution operator are translated consistently:

$$\langle\phi_P(t)\rvert U_P(t,t_0) \lvert\psi_P(t_0)\rangle = \langle\phi_S(t)\rvert U_S(t,t_0) \lvert\psi_S(t_0)\rangle.$$

Schrödinger and Heisenberg Pictures

Let $U(t,t_0)$ be the full Schrödinger-picture evolution operator and choose

$$R(t)=U(t,t_0), \qquad R(t_0)=I.$$

Then

Object	Schrödinger picture	Heisenberg picture
state	$\lvert\psi_S(t)\rangle=U(t,t_0)\lvert\psi_S(t_0)\rangle$	$\lvert\psi_H\rangle=\lvert\psi_S(t_0)\rangle$
observable	$A_S(t)$	$A_H(t)=U^\dagger(t,t_0)A_S(t)U(t,t_0)$
density operator	$\rho_S(t)=U\rho_S(t_0)U^\dagger$	$\rho_H=\rho_S(t_0)$
state generator	$H_S(t)$	$H_H^{\rm state}=0$
state propagator	$U(t,t_0)$	$U_H(t,t_0)=I$

The notation $H_H^{\rm state}=0$ means that the transformed state equation has zero generator. The Hamiltonian can still be treated as an observable,

$$H_H^{\rm obs}(t) = U^\dagger(t,t_0)H_S(t)U(t,t_0),$$

and it need not vanish. Keeping these two roles distinct avoids the misleading statement that “the Hamiltonian is zero in the Heisenberg picture.”

The inverse observable translation is

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

The Heisenberg operator equation is

$$\frac{dA_H}{dt} = \frac{i}{\hbar} [H_H^{\rm obs}(t),A_H(t)] + U^\dagger(t,t_0) \frac{\partial A_S}{\partial t} U(t,t_0).$$

The last term is present when the Schrödinger-picture observable has explicit time dependence.

Schrödinger and Interaction Pictures

Split the Hamiltonian as

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

Let $U_0(t,t_0)$ solve

$$i\hbar\frac{\partial U_0}{\partial t} = H_0(t)U_0(t,t_0), \qquad U_0(t_0,t_0)=I.$$

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

Object	Schrödinger picture	Interaction picture
state	$\lvert\psi_S(t)\rangle$	$\lvert\psi_I(t)\rangle=U_0^\dagger(t,t_0)\lvert\psi_S(t)\rangle$
observable	$A_S(t)$	$A_I(t)=U_0^\dagger(t,t_0)A_S(t)U_0(t,t_0)$
density operator	$\rho_S(t)$	$\rho_I(t)=U_0^\dagger(t,t_0)\rho_S(t)U_0(t,t_0)$
state generator	$H_0(t)+V(t)$	$V_I(t)=U_0^\dagger(t,t_0)V(t)U_0(t,t_0)$
evolution operator	$U(t,t_0)$	$U_I(t,t_0)=U_0^\dagger(t,t_0)U(t,t_0)$

The inverse translations are

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

and similarly for $\rho_S$. The interaction-picture evolution satisfies

$$i\hbar \frac{\partial U_I}{\partial t} = V_I(t)U_I(t,t_0).$$

No approximation has been made. Approximation enters only when $U_I$, $V_I$, or the Dyson series is truncated or simplified.

Interaction and Heisenberg Pictures

At $t_0$, choose all three pictures to coincide. Since

$$U(t,t_0) = U_0(t,t_0)U_I(t,t_0),$$

the interaction and Heisenberg representatives are related by $U_I$:

Object	Interaction to Heisenberg	Heisenberg to interaction
state	$\lvert\psi_H\rangle=U_I^\dagger(t,t_0)\lvert\psi_I(t)\rangle$	$\lvert\psi_I(t)\rangle=U_I(t,t_0)\lvert\psi_H\rangle$
observable	$A_H(t)=U_I^\dagger(t,t_0)A_I(t)U_I(t,t_0)$	$A_I(t)=U_I(t,t_0)A_H(t)U_I^\dagger(t,t_0)$
density operator	$\rho_H=U_I^\dagger(t,t_0)\rho_I(t)U_I(t,t_0)$	$\rho_I(t)=U_I(t,t_0)\rho_HU_I^\dagger(t,t_0)$

This direct conversion is often cleaner than translating back through the Schrödinger picture, especially in perturbative calculations.

Density-Operator Equations

For closed Schrödinger-picture evolution,

$$i\hbar\frac{d\rho_S}{dt} = [H_S,\rho_S].$$

Under the general picture transformation,

$$i\hbar\frac{d\rho_P}{dt} = [H_P,\rho_P].$$

The special cases are

$$\frac{d\rho_H}{dt}=0$$

and

$$i\hbar\frac{d\rho_I}{dt} = [V_I(t),\rho_I(t)].$$

These formulas apply to closed-system unitary evolution. A reduced open-system state generally has additional dissipative or memory terms; changing picture then requires transforming the full generator, not only its Hamiltonian part.

Explicit Time Dependence

For

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

the full derivative is

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

Define $K_P=i\hbar R^\dagger\dot R$. Then $H_P=E_P-K_P$, whereas

$$\dot A_P=\frac{i}{\hbar}[K_P,A_P]+R^\dagger(\partial_t A_S)R.$$

The operator’s picture motion is generated by $K_P$, not by the transformed state generator $H_P$. This distinction recovers fixed Schrödinger observables when $R=I$ and moving Heisenberg observables when $R=U$.

Convention Translation

If another source uses

$$\lvert\widetilde\psi_P\rangle = R\lvert\psi_S\rangle,$$

then consistency requires

$$\widetilde A_P = RA_SR^\dagger, \qquad \widetilde\rho_P = R\rho_SR^\dagger,$$

and

$$\widetilde H_P = RH_SR^\dagger + i\hbar\dot R R^\dagger.$$

Do not borrow the state definition from one convention and the Hamiltonian formula from the other. A quick diagnostic is expectation-value invariance: inconsistent dagger placement fails immediately.

Common Mistakes

• Transforming only the state. Observables or density operators must be translated consistently for predictions to agree.
• Forgetting $-i\hbar R^\dagger\dot R$. A time-dependent frame has its own generator.
• Using $R(t)$ at both propagator endpoints. The correct factors are $R^\dagger(t)$ and $R(t_0)$.
• Assuming $R(t_0)=I$ silently. Include the initial factor unless the convention is declared.
• Saying the Heisenberg Hamiltonian is zero without qualification. The state generator vanishes; the Hamiltonian observable need not.
• Dropping explicit operator time dependence. It survives as a separately transformed derivative term.
• Treating $H_0$ as time independent by definition. A time-dependent solvable $H_0(t)$ is allowed if its propagator $U_0$ is defined correctly.
• Calling the interaction picture approximate. The picture is exact before any perturbative truncation.
• Mixing the two common $R$ conventions. This reverses daggers and the connection-term sign.
• Applying closed-system density formulas to reduced open dynamics. Nonunitary terms must also be transformed.

References

• L. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd ed., World Scientific, 2014.
• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
• P. A. M. Dirac, The Principles of Quantum Mechanics, 4th ed., Oxford University Press, 1958.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.

Exercises

1. Derive the transformed Hamiltonian

Starting from $\lvert\psi_P\rangle=R^\dagger\lvert\psi_S\rangle$, derive $H_P$.

Solution

Differentiate the transformed state and use the Schrödinger equation:

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

Differentiating $R^\dagger R=I$ gives $\dot R^\dagger R=-R^\dagger\dot R$. Hence

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

2. Verify the transformed propagator

Show that

$$U_P(t,t_0) = R^\dagger(t)U_S(t,t_0)R(t_0)$$

maps the transformed initial state to the transformed final state.

Solution

The transformed initial state is

$$\lvert\psi_P(t_0)\rangle = R^\dagger(t_0) \lvert\psi_S(t_0)\rangle.$$

Then

$$\begin{aligned} U_P(t,t_0) \lvert\psi_P(t_0)\rangle &= R^\dagger(t)U_S(t,t_0) R(t_0)R^\dagger(t_0) \lvert\psi_S(t_0)\rangle \\ &= R^\dagger(t) \lvert\psi_S(t)\rangle \\ &= \lvert\psi_P(t)\rangle. \end{aligned}$$

3. Prove trace-rule invariance

Show that $\operatorname{Tr}(\rho_PA_P)=\operatorname{Tr}(\rho_SA_S)$.

Solution

Substitute the transformed operators:

$$\begin{aligned} \operatorname{Tr}(\rho_PA_P) &= \operatorname{Tr} \left( R^\dagger\rho_SR R^\dagger A_SR \right) \\ &= \operatorname{Tr} \left( R^\dagger\rho_SA_SR \right) \\ &= \operatorname{Tr} \left( \rho_SA_SRR^\dagger \right) \\ &= \operatorname{Tr}(\rho_SA_S). \end{aligned}$$

The third line uses cyclicity of the trace.

4. Translate directly from interaction to Heisenberg

Derive $A_H=U_I^\dagger A_IU_I$ from $U=U_0U_I$.

Solution

By definition,

$$A_H = U^\dagger A_SU$$

and

$$A_I = U_0^\dagger A_SU_0.$$

Using $U=U_0U_I$ gives

$$\begin{aligned} A_H &= U_I^\dagger U_0^\dagger A_S U_0U_I \\ &= U_I^\dagger A_IU_I. \end{aligned}$$

5. Diagnose a mixed convention

A source defines $\lvert\psi_P\rangle=R\lvert\psi_S\rangle$ but writes $A_P=R^\dagger A_SR$. Explain the failure.

Solution

With the stated state convention, the transformed bra is $\langle\psi_P\rvert=\langle\psi_S\rvert R^\dagger$. Expectation invariance requires

$$A_P = RA_SR^\dagger,$$

because then

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

Using $R^\dagger A_SR$ with the same state definition generally inserts extra powers of $R$ and fails this test.