Common Mistakes About Pictures

The pictures of quantum mechanics are equivalent descriptions of the same closed-system dynamics. Most mistakes come from transforming only part of a calculation, dropping a time-dependent term, or treating a convenient representation as a new physical theory.

The basic invariant is simple:

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

For density operators, the same statement is

$$\langle A\rangle_t = \operatorname{Tr}\bigl(\rho_S(t)A_S(t)\bigr) = \operatorname{Tr}\bigl(\rho_HA_H(t)\bigr).$$

When in doubt, reduce the calculation to one of these invariants and check that every object is in the same picture.

Picture equivalence and consistent observables

The Schrödinger, Heisenberg, and interaction pictures do not make different experimental predictions. They move time dependence between states, observables, and interactions.

Picture	State	Observable
Schrödinger	$\lvert\psi_S(t)\rangle$ evolves	$A_S$ usually fixed unless explicit
Heisenberg	$\lvert\psi_H\rangle$ fixed	$A_H(t)$ evolves
Interaction	$\lvert\psi_I(t)\rangle$ evolves under $V_I(t)$	$A_I(t)$ evolves under $H_0$

The pictures differ in convenience, not in content. If two picture calculations give different expectation values, something has been transformed inconsistently.

Mistake: Mixing States and Operators

A common error is to write a Heisenberg state with a Schrödinger operator and call the result time dependent:

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

This is generally the reference-time expectation value, not the time-dependent prediction. The Heisenberg expression should use the transformed operator:

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

Likewise, if a density operator has been transformed to a new picture, the observable must be transformed consistently before taking $\operatorname{Tr}(\rho A)$.

Mistake: Forgetting Explicit Time Dependence

The Heisenberg equation is

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

The final term is required whenever the Schrödinger-picture observable $A_S(t)$ explicitly depends on time. Examples include rotating measurement axes, time-dependent perturbing observables, and operators defined in a time-dependent frame.

For conservation laws, the correct criterion is

$$\frac{\partial A_S}{\partial t} + \frac{i}{\hbar}[H,A_S] =0,$$

not merely $[H,A_S]=0$.

Mistake: Losing the Hamiltonian Transformation Term

Under the convention

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

the transformed Hamiltonian is

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

The last term is easy to miss. It is the source of rotating-frame detunings and the cancellation that leaves only $V_I(t)$ in the interaction picture.

Some references use the opposite convention, $\lvert\psi_P\rangle=R\lvert\psi_S\rangle$. The formula then changes. Before importing a result from another source, identify where the dagger sits.

Mistake: Thinking the Interaction Picture Is Approximate

The interaction picture is an exact rewriting once a split is chosen:

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

The interaction-picture state obeys

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

Approximation enters only after one expands the interaction-picture evolution operator and truncates:

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

The picture is exact; the truncated Dyson series is approximate.

Mistake: Dropping Time Ordering

The compact expression

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

is not usually an ordinary exponential. A standard sufficient condition for removing time ordering is that the Hamiltonians or interaction Hamiltonians commute at different times:

$$[H(t_1),H(t_2)]=0$$

throughout the interval (with strong commutation and suitable domain control for unbounded operators). Without this condition one must check the ordered evolution explicitly; accidental equality at a particular endpoint is not a general justification.

Mistake: Thinking Fixed Heisenberg States Mean Nothing Evolves

In the Heisenberg picture, the state is fixed:

$$\lvert\psi_H\rangle=\lvert\psi_S(t_0)\rangle.$$

But observables evolve:

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

Time-dependent predictions are encoded in the operator. A fixed Heisenberg state does not mean fixed probabilities for all possible measurements.

Mistake: Confusing Picture Changes with Open-System Dynamics

A unitary picture change preserves expectation values and keeps closed-system equations in closed-system form. It can also be applied to an open-system master equation, where it transforms the dissipative operators rather than removing their physical effect.

Open-system dynamics is different. A reduced density operator may obey a master equation with non-Hamiltonian terms, such as a Lindblad equation. Those terms are not produced merely by changing from Schrödinger to Heisenberg or interaction picture.

Mistake: Treating Basis-Dependent Language as Picture-Invariant

Words such as “population” and “coherence” refer to matrix entries in a chosen basis. A change of basis, a change of picture, or an interaction-picture transformation can move phases between $\rho$ and $A$.

The invariant statement is not “this matrix element is physically the same in every representation.” The invariant statement is that probabilities and expectation values agree after all objects are transformed consistently.

Related calculations

Picture transformations derives the frame term and distinguishes the energy observable from the state generator. Explicit time-dependent operators works out the rotating-axis conservation test, and density operators in pictures gives the trace-rule version.

References

• 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. A student computes $\langle\psi_H\rvert A_S\lvert\psi_H\rangle$ and says it is the Heisenberg-picture expectation value at time $t$. What is missing?

Solution

The observable must be transformed:

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

The Heisenberg expectation value is

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

Using $A_S$ with $\lvert\psi_H\rangle$ generally gives the reference-time value, not the time-dependent prediction.

2. Under $\lvert\psi_P\rangle=R^\dagger\lvert\psi_S\rangle$, what term is missing from the incorrect formula $H_P=R^\dagger H_SR$?

Solution

The missing term is the generator of the moving picture:

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

It appears whenever $R(t)$ depends on time.

3. When may the time-ordered exponential reduce to an ordinary exponential?

Solution

It may reduce to an ordinary exponential when the Hamiltonians commute at different times:

$$[H(t_1),H(t_2)]=0$$

for all times in the interval. Then operator products can be rearranged, and time ordering no longer changes the series.

4. Is the interaction picture itself an approximation?

Solution

No. The interaction picture is an exact transformation after a split $H=H_0+V(t)$ has been chosen. Approximation enters when one expands the interaction-picture evolution operator and truncates the Dyson series or applies additional assumptions.