Pictures of Quantum Mechanics

A picture of quantum mechanics is a convention for where the time dependence is placed. The physical predictions do not change when states, observables, and Hamiltonians are transformed consistently.

The three basic pictures are:

Picture	What carries the main time dependence	Most useful for
Schrödinger	states and wavefunctions	state propagation, wave mechanics, numerical evolution
Heisenberg	observables	operator equations, symmetries, constants of motion
Interaction	states under the transformed interaction, operators under $H_0$	perturbation setup, driven systems, QFT bridge

The word “picture” does not mean a different theory. It means a different representation of the same time evolution.

Core Equivalence

Let $U(t,t_0)$ be the closed-system time-evolution operator. In the Schrödinger picture,

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

In the Heisenberg picture, one usually fixes

\lvert\psi_H\rangle = \lvert\psi_S(t_0)\rangle,

and moves the time dependence into observables:

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

Expectation values agree:

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

This equality is the anchor of the chapter. Any calculation that mixes pictures must preserve it.

Reading Path

Question	Start here	What to watch
How do states carry time dependence?	Schrödinger Picture	Observables may still have explicit time dependence.
How do observables carry time dependence?	Heisenberg Picture	Fixed states do not mean fixed physics.
How do we split solvable and perturbing motion?	Interaction Picture	The picture is exact before any approximation is made.
How do general time-dependent frame changes work?	Picture Transformations	The transformed Hamiltonian includes an extra generator term.
What equation moves Heisenberg operators?	Heisenberg Equations of Motion	Keep the explicit time-derivative term.
When do expectation values look classical?	Ehrenfest Theorem	$\langle V'(x)\rangle$ is not generally $V'(\langle x\rangle)$ .
What does explicit operator time dependence mean?	Operators with Explicit Time Dependence	Separate a changing observable from picture-induced evolution.
How do mixed states transform between pictures?	Density Operators in Different Pictures	Trace expectation values are the invariant object.
What usually goes wrong in picture calculations?	Common Mistakes About Pictures	Check consistency before trusting a result.

For a compact cross-volume table, see Translation Table of Formulations.

Schrödinger Picture

The Schrödinger picture is closest to the wavefunction-first presentation of quantum mechanics. The state obeys

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

This picture is natural when the state itself is the object to propagate or visualize: wave packets, finite-dimensional state vectors, numerical time evolution, and boundary-value wave mechanics.

Its main risk is overuse. Some problems become simpler when one asks how observables move instead of how the entire state vector moves.

Heisenberg Picture

The Heisenberg picture fixes the state at a reference time and evolves observables:

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

For an operator with possible explicit time dependence, the equation of motion is

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

This is often the best language for constants of motion. If the right-hand side vanishes, the operator is conserved. It is also the natural bridge to field theory, where local Heisenberg operators become central objects.

Interaction Picture

The interaction picture begins with a split

H(t)=H_0+V(t),

where $H_0$ is chosen to be exactly solvable or structurally simple. Operators evolve with $H_0$ :

A_I(t) = U_0^\dagger(t,t_0)A_SU_0(t,t_0),

while states evolve under the transformed interaction:

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

The interaction picture is not an approximation by itself. It becomes a perturbative tool only after the interaction-picture evolution operator is expanded and truncated.

What the Pictures Make Easier

Task	Often easiest picture	Reason
Propagate a wavefunction	Schrödinger	the state is the computed object
Derive operator equations	Heisenberg	commutators directly generate motion
Identify conserved observables	Heisenberg	constants are time-independent operators
Set up time-dependent perturbation theory	Interaction	solvable motion is factored out
Prepare for QFT perturbation theory	Interaction and Heisenberg	time ordering and operator products are explicit
Compare with classical equations	Heisenberg or Schrödinger expectation values	Ehrenfest theorem follows from either language

These are pragmatic preferences, not rules of physics.

Boundaries

This chapter is about exact picture transformations for closed-system quantum mechanics. It does not own detailed perturbation theory, scattering amplitudes, decoherence, or Lindblad evolution. Those topics may use a picture, especially the interaction picture, but their canonical homes are later volumes or other chapters.

Density operators can be transformed between pictures too:

\rho_H = \rho_S(t_0), \qquad A_H(t)=U^\dagger A_SU.

The full density-operator treatment belongs with closed-system density-matrix dynamics and open-system bridges; this chapter introduces only the picture logic needed to read those pages.

QFT Bridge

The Heisenberg and interaction pictures become especially important in field theory. Heisenberg fields encode operator time dependence, while interaction-picture perturbation theory leads to Dyson expansions, time-ordered products, correlation functions, and the S-matrix.

The bridge begins inside this volume with Time Ordering, Dyson Expansion as Formal Evolution, and Why Dynamics Matters for QFT.

Common Mistakes

Treating one picture as physically more real than the others.
Mixing Schrödinger states with Heisenberg operators without applying the transformation.
Forgetting explicit time dependence in an operator.
Thinking the interaction picture is automatically perturbative.
Choosing an $H_0$ split without checking whether it simplifies the problem.
Interpreting fixed Heisenberg states as absence of time-dependent predictions.
Dropping picture labels in a calculation where ambiguity matters.

Cross-Links

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.
M. E. Peskin and D. V. Schroeder, An Introduction to Quantum Field Theory, Addison-Wesley, 1995.

Exercises

Show that the Schrödinger and Heisenberg pictures give the same expectation value when $A_H(t)=U^\dagger A_SU$ .

Solution

Use $\lvert\psi_S(t)\rangle=U\lvert\psi_H\rangle$ . Then

\langle\psi_S(t)\rvert A_S\lvert\psi_S(t)\rangle = \langle\psi_H\rvert U^\dagger A_SU\lvert\psi_H\rangle = \langle\psi_H\rvert A_H(t)\lvert\psi_H\rangle.

The two expressions are different representations of the same number.

If $H=H_0+V(t)$ and $V(t)=0$ , what happens to the interaction-picture state?

Solution

The interaction-picture equation is

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

If $V(t)=0$ , then $V_I(t)=0$ , so

\frac{d}{dt}\lvert\psi_I(t)\rangle=0.

The state is fixed in the interaction picture because all $H_0$ motion has been moved into the operators.

Why is it misleading to say the Heisenberg picture has “no dynamics”?

Solution

The Heisenberg state is fixed, but observables evolve:

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

Time-dependent predictions come from expectation values such as

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

The dynamics has moved from the state to the operator; it has not disappeared.