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.

Helpful background. The Time-Evolution Operator supplies $U(t,t_0)$, and Constants of Motion explains what it means for an observable to be conserved.

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, choose the state at the reference time:

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

and move the time dependence into observables:

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

Expectation values agree:

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

This equality gives a practical check on every picture change: transform the state and the measured operator together. The explicit dependence in $A_S(t)$, such as a changing laboratory measurement axis, is distinct from the dependence introduced by conjugating with $U$.

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.

For a harmonic oscillator, however, the equations for $x_H$ and $p_H$ close on just those two operators. If only their means are required, solving these equations can be shorter than evolving an entire wavefunction.

Heisenberg Picture

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

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

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 $H_0$:

$$A_I(t) = U_0^\dagger(t,t_0)A_S(t)U_0(t,t_0),$$

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 exact. Expanding and truncating its evolution is a separate perturbative step. For mixed states, transform $\rho_I=U_0^\dagger\rho_SU_0$ as well as $A_I$; then $\operatorname{Tr}(\rho_IA_I)=\operatorname{Tr}(\rho_SA_S)$.

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.

References

• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977 — states and observables in different pictures.
• P. A. M. Dirac, The Principles of Quantum Mechanics, 4th ed., Oxford University Press, 1958 — equivalent representations of dynamics.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020 — picture transformations and examples.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994 — operator equations of motion.

Exercises

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

Solution

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

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

The two expressions are different representations of the same number.

2. 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.

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

Solution

The Heisenberg state is fixed, but observables evolve:

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

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.