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.

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