---
title: "Heisenberg picture"
sort_title: "Heisenberg picture"
date: 2021-02-24
categories:
- Quantum mechanics
- Physics
layout: "concept"
---

The **Heisenberg picture** is an alternative formulation of quantum mechanics,
and is equivalent to the traditional Schrödinger equation.

In the Schrödinger picture,
time-independent operators are constant by definition,
and the state $$\Ket{\psi_S(t)}$$ varies as follows, where $$\hat{U}(t)$$
is the [time evolution operator](/know/concept/time-evolution-operator/):

$$\begin{aligned}
    \Ket{\psi_S(t)}
    = \hat{U}(t) \Ket{\psi_S(0)}
\end{aligned}$$

In the Heisenberg picture, the roles are reversed:
the states $$\Ket{\psi_H}$$ are constants,
and instead the operators vary in time.
In some situations this approach can be more convenient,
and since we usually care about the evolution of observable quantities,
studying the corresponding operators directly
may make more sense than finding abstract quantum states.
Another advantage is that basis states remain fixed,
which can simplify calculations.

Given a Schrödinger-picture state $$\Ket{\psi_S(t)}$$
and an operator $$\hat{L}_S(t)$$ that may or may not depend on time,
they can be converted to the Heisenberg picture by the following transformation:

$$\begin{aligned}
    \boxed{
        \Ket{\psi_H}
        \equiv \Ket{\psi_S(0)}
    }
    \qquad\qquad
    \boxed{
        \hat{L}_H(t)
        \equiv \hat{U}^\dagger(t) \: \hat{L}_S(t) \: \hat{U}(t)
    }
\end{aligned}$$

Note that if $$\hat{H}_S$$ is time-independent,
then it commutes with $$\hat{U}(t)$$,
meaning $$\hat{H}_H = \hat{H}_S$$,
so it can simply be labelled $$\hat{H}$$.
This is not true for time-dependent Hamiltonians.

Since $$\hat{U}(t)$$ is unitary,
the expectation value of a given operator is unchanged:

$$\begin{aligned}
    \expval{\hat{L}_H}
    &= \matrixel{\psi_H}{\hat{L}_H(t)}{\psi_H}
    \\
    &= \matrixel{\psi_S(0)}{\hat{U}^\dagger(t) \: \hat{L}_S(t) \: \hat{U}(t)}{\psi_S(0)}
    \\
    &= \matrixel{\hat{U}(t) \psi_S(0)}{\hat{L}_S(t)}{\hat{U}(t) \psi_S(0)}
    \\
    &= \matrixel{\psi_S(t)}{\hat{L}_S}{\psi_S(t)}
    \\
    &= \expval{\hat{L}_S}
\end{aligned}$$

The Schrödinger and Heisenberg pictures therefore respectively
correspond to active and passive transformations by $$\hat{U}(t)$$
in [Hilbert space](/know/concept/hilbert-space/).
The two formulations are entirely equivalent,
and can be derived from one another,
as we will show shortly.

In the Heisenberg picture, the states are constant,
so the time-dependent Schrödinger equation is not directly useful.
Instead, we use it derive a new equation for $$\hat{L}_H(t)$$,
with the key being that $$\hat{U}(t)$$ itself
satisfies the Schrödinger equation by definition:

$$\begin{aligned}
    \dv{}{t} \hat{U}(t)
    = - \frac{i}{\hbar} \hat{H}_S(t) \: \hat{U}(t)
\end{aligned}$$

Where $$\hat{H}_S(t)$$ may depend on time. We differentiate the definition of
$$\hat{L}_H(t)$$ and insert the other side of the Schrödinger equation when necessary:

$$\begin{aligned}
    \dv{\hat{L}_H}{t}
    &= \dv{\hat{U}^\dagger}{t} \hat{L}_S \hat{U}
    + \hat{U}^\dagger \hat{L}_S \dv{\hat{U}}{t}
    + \hat{U}^\dagger \dv{\hat{L}_S}{t} \hat{U}
    \\
    &= \frac{i}{\hbar} \hat{U}^\dagger \hat{H}_S (\hat{U} \hat{U}^\dagger) \hat{L}_S \hat{U}
    - \frac{i}{\hbar} \hat{U}^\dagger \hat{L}_S (\hat{U} \hat{U}^\dagger) \hat{H}_S \hat{U}
    + \bigg( \dv{\hat{L}_S}{t} \bigg)_H
    \\
    &= \frac{i}{\hbar} \hat{H}_H \hat{L}_H
    - \frac{i}{\hbar} \hat{L}_H \hat{H}_H
    + \bigg( \dv{\hat{L}_S}{t} \bigg)_H
\end{aligned}$$

We thus get the following equation of motion for operators in the Heisenberg picture:

$$\begin{aligned}
    \boxed{
        \dv{}{t}\hat{L}_H(t)
        = \frac{i}{\hbar} \comm{\hat{H}_H(t)}{\hat{L}_H(t)} + \bigg( \dv{}{t}\hat{L}_S(t) \bigg)_H
    }
\end{aligned}$$

This result is arguably more intuitive than the Schrödinger picture,
because it allows us to think about observables (i.e. operators) in a more classical way.
For example, inserting the position $$\hat{X}$$
and momentum $$\hat{P} = - i \hbar \: \idv{}{\hat{X}}$$
gives the following Newton-style relations:

$$\begin{aligned}
    \dv{\hat{X}}{t}
    &= \frac{i}{\hbar} \comm{\hat{H}}{\hat{X}}
    = \frac{\hat{P}}{m}
    \\
    \dv{\hat{P}}{t}
    &= \frac{i}{\hbar} \comm{\hat{H}}{\hat{P}}
    = - \dv{V(\hat{X})}{\hat{X}}
\end{aligned}$$

Where the commutators have been treated as known.
These equations would not be valid in the Schrödinger picture,
unless we took their expectation value
to get [Ehrenfest's theorem](/know/concept/ehrenfests-theorem/).