Categories: Physics, Quantum mechanics.

# 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, 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.