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---
title: "Ehrenfest's theorem"
firstLetter: "E"
publishDate: 2021-02-24
categories:
- Quantum mechanics
- Physics
date: 2021-02-24T14:53:13+01:00
draft: false
markup: pandoc
---
# Ehrenfest's theorem
In quantum mechanics, **Ehrenfest's theorem** gives a general expression for the
time evolution of an observable's expectation value $\expval*{\hat{L}}$.
The time-dependent Schrödinger equation is as follows,
where prime denotes differentiation with respect to time $t$:
$$\begin{aligned}
\ket{\psi'} = \frac{1}{i \hbar} \hat{H} \ket{\psi}
\qquad
\bra{\psi'} = - \frac{1}{i \hbar} \bra{\psi} \hat{H}
\end{aligned}$$
Given an observable operator $\hat{L}$ and a state $\ket{\psi}$,
the time-derivative of the expectation value $\expval*{\hat{L}}$ is as follows
(due to the product rule of differentiation):
$$\begin{aligned}
\dv{\expval*{\hat{L}}}{t}
&= \matrixel{\psi}{\hat{L}}{\psi'} + \matrixel{\psi'}{\hat{L}}{\psi} + \matrixel{\psi}{\hat{L}'}{\psi}
\\
&= \frac{1}{i \hbar} \matrixel{\psi}{\hat{L}\hat{H}}{\psi}
- \frac{1}{i \hbar} \matrixel{\psi}{\hat{H}\hat{L}}{\psi}
+ \expval{\dv{\hat{L}}{t}}
\end{aligned}$$
The first two terms on the right can be rewritten using a commutator,
yielding the general form of Ehrenfest's theorem:
$$\begin{aligned}
\boxed{
\dv{\expval*{\hat{L}}}{t}
= \frac{1}{i \hbar} \expval{[\hat{L}, \hat{H}]} + \expval{\dv{\hat{L}}{t}}
}
\end{aligned}$$
In practice, since most operators are time-independent,
the last term often vanishes.
As a interesting side note, in the [Heisenberg picture](/know/concept/heisenberg-picture/),
this relation proves itself,
when one simply wraps all terms in $\bra{\psi}$ and $\ket{\psi}$.
Two observables of particular interest are the position $\hat{X}$ and momentum $\hat{P}$.
Applying the above theorem to $\hat{X}$ yields the following,
which we reduce using the fact that $\hat{X}$ commutes
with the potential $V(\hat{X})$,
because one is a function of the other:
$$\begin{aligned}
\dv{\expval*{\hat{X}}}{t}
&= \frac{1}{i \hbar} \expval{[\hat{X}, \hat{H}]}
= \frac{1}{2 i \hbar m} \expval{[\hat{X}, \hat{P}^2] + 2 m [\hat{X}, V(\hat{X})]}
= \frac{1}{2 i \hbar m} \expval{[\hat{X}, \hat{P}^2]}
\\
&= \frac{1}{2 i \hbar m} \expval{\hat{P} [\hat{X}, \hat{P}] + [\hat{X}, \hat{P}] \hat{P}}
= \frac{2 i \hbar}{2 i \hbar m} \expval*{\hat{P}}
= \frac{\expval*{\hat{P}}}{m}
\end{aligned}$$
This is the first part of the "original" form of Ehrenfest's theorem,
which is reminiscent of classical Newtonian mechanics:
$$\begin{gathered}
\boxed{
\dv{\expval*{\hat{X}}}{t} = \frac{\expval*{\hat{P}}}{m}
}
\end{gathered}$$
Next, applying the general formula to the expected momentum $\expval*{\hat{P}}$
gives us:
$$\begin{aligned}
\dv{\expval*{\hat{P}}}{t}
&= \frac{1}{i \hbar} \expval{[\hat{P}, \hat{H}]}
= \frac{1}{2 i \hbar m} \expval{[\hat{P}, \hat{P}^2] + 2 m [\hat{P}, V(\hat{X})]}
= \frac{1}{i \hbar} \expval{[\hat{P}, V(\hat{X})]}
\end{aligned}$$
To find the commutator, we go to the $\hat{X}$-basis and use a test
function $f(x)$:
$$\begin{aligned}
\comm{- i \hbar \dv{x}}{V(x)} \: f(x)
&= - i \hbar \frac{dV}{dx} f(x) - i \hbar V(x) \frac{df}{dx} + i \hbar V(x) \frac{df}{dx}
= - i \hbar \frac{dV}{dx} f(x)
\end{aligned}$$
By inserting this result back into the previous equation, we find the following:
$$\begin{aligned}
\dv{\expval*{\hat{P}}}{t}
&= - \frac{i \hbar}{i \hbar} \expval{\frac{d V}{d \hat{X}}}
= - \expval{\frac{d V}{d \hat{X}}}
\end{aligned}$$
This is the second part of Ehrenfest's theorem,
which is also similar to Newtonian mechanics:
$$\begin{gathered}
\boxed{
\dv{\expval*{\hat{P}}}{t} = - \expval{\pdv{V}{\hat{X}}}
}
\end{gathered}$$
There is an important consequence of Ehrenfest's original theorems
for the symbolic derivatives of the Hamiltonian $\hat{H}$
with respect to $\hat{X}$ and $\hat{P}$:
$$\begin{gathered}
\boxed{
\expval{\pdv{\hat{H}}{\hat{P}}}
= \dv{\expval*{\hat{X}}}{t}
}
\qquad \quad
\boxed{
- \expval{\pdv{\hat{H}}{\hat{X}}}
= \dv{\expval*{\hat{P}}}{t}
}
\end{gathered}$$
These are easy to prove yourself,
and are analogous to Hamilton's canonical equations.
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