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 ⟨L^⟩\expval{\hat{L}}.

The time-dependent Schrödinger equation is as follows, where prime denotes differentiation with respect to time tt:

∣ψ′⟩=1iℏH^∣ψ⟩⟨ψ′∣=−1iℏ⟨ψ∣H^\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 L^\hat{L} and a state ∣ψ⟩\Ket{\psi}, the time-derivative of the expectation value ⟨L^⟩\expval{\hat{L}} is as follows (due to the product rule of differentiation):

d⟨L^⟩dt=⟨ψ∣L^∣ψ′⟩+⟨ψ′∣L^∣ψ⟩+⟨ψ∣L^′∣ψ⟩=1iℏ⟨ψ∣L^H^∣ψ⟩−1iℏ⟨ψ∣H^L^∣ψ⟩+⟨dL^dt⟩\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:

d⟨L^⟩dt=1iℏ⟨[L^,H^]⟩+⟨dL^dt⟩\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 ⟨ψ∣\Bra{\psi} and ∣ψ⟩\Ket{\psi}.

Two observables of particular interest are the position X^\hat{X} and momentum P^\hat{P}. Applying the above theorem to X^\hat{X} yields the following, which we reduce using the fact that X^\hat{X} commutes with the potential V(X^)V(\hat{X}), because one is a function of the other:

d⟨X^⟩dt=1iℏ⟨[X^,H^]⟩=12iℏm⟨[X^,P^2]+2m[X^,V(X^)]⟩=12iℏm⟨[X^,P^2]⟩=12iℏm⟨P^[X^,P^]+[X^,P^]P^⟩=2iℏ2iℏm⟨P^⟩=⟨P^⟩m\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:

d⟨X^⟩dt=⟨P^⟩m\begin{gathered} \boxed{ \dv{\expval{\hat{X}}}{t} = \frac{\expval{\hat{P}}}{m} } \end{gathered}

Next, applying the general formula to the expected momentum ⟨P^⟩\expval{\hat{P}} gives us:

d⟨P^⟩dt=1iℏ⟨[P^,H^]⟩=12iℏm⟨[P^,P^2]+2m[P^,V(X^)]⟩=1iℏ⟨[P^,V(X^)]⟩\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 X^\hat{X}-basis and use a test function f(x)f(x):

[−iℏddx,V(x)] f(x)=−iℏdVdxf(x)−iℏV(x)dfdx+iℏV(x)dfdx=−iℏdVdxf(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:

d⟨P^⟩dt=−iℏiℏ⟨dVdX^⟩=−⟨dVdX^⟩\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:

d⟨P^⟩dt=−⟨∂V∂X^⟩\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 H^\hat{H} with respect to X^\hat{X} and P^\hat{P}:

⟨∂H^∂P^⟩=d⟨X^⟩dt−⟨∂H^∂X^⟩=d⟨P^⟩dt\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.