---
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 traditionally-taught Schrödinger equation.

In the Schrödinger picture, the operators (observables) are fixed
(as long as they do not depend on time), while the state
$\Ket{\psi_S(t)}$ changes according to the Schrödinger equation,
which can be written using the generator of translations $\hat{U}(t)$ like so,
for a time-independent $\hat{H}_S$:

$$\begin{aligned}
    \Ket{\psi_S(t)} = \hat{U}(t) \Ket{\psi_S(0)}
    \qquad \quad
    \boxed{
        \hat{U}(t) \equiv \exp\!\bigg(\!-\! i \frac{\hat{H}_S t}{\hbar} \bigg)
    }
\end{aligned}$$

In contrast, the Heisenberg picture reverses the roles:
the states $\Ket{\psi_H}$ are invariant,
and instead the operators vary with time.
An advantage of this is that the basis states remain the same.

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

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

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 thus entirely equivalent,
and can be derived from one another,
as will be shown shortly.

In the Heisenberg picture, the states are constant,
so the time-dependent Schrödinger equation is not directly useful.
Instead, we will use it derive a new equation for $\hat{L}_H(t)$.
The key is that the generator $\hat{U}(t)$ is defined from the Schrödinger equation:

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

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:

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

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

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

This equation is closer to classical mechanics than the Schrödinger picture:
inserting the position $\hat{X}$ and momentum $\hat{P} = - i \hbar \: \idv{}{\hat{X}}$
gives the following Newton-style equations:

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

For a proof, see [Ehrenfest's theorem](/know/concept/ehrenfests-theorem/),
which is closely related to the Heisenberg picture.