Ehrenfest theorem

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The Ehrenfest theorem, named after Paul Ehrenfest, the Austrian physicist and mathematician, relates the time derivative of the expectation value for a quantum mechanical operator to the expectation of the commutator of that operator with the Hamiltonian of the system, expectations which are connected to classical mechanics.

It reads

$\frac{d}{dt}\langle A\rangle = \frac{1}{i\hbar}\langle [A,H] \rangle+ \left\langle \frac{\partial A}{\partial t}\right\rangle ~,$

where A is some QM operator and $\langle A\rangle$ is its expectation value.

Ehrenfest's theorem is most apparent in the Heisenberg picture of quantum mechanics, where it is just the expectation value of the Heisenberg equation of motion. It provides mathematical support to the correspondence principle. The reason is that Ehrenfest's theorem is closely related to Liouville's theorem of Hamiltonian mechanics, which involves the Poisson bracket instead of a commutator. Dirac's rule of thumb indicates that statements in quantum mechanics which contain a commutator correspond to statements in classical mechanics where the commutator is supplanted by a Poisson bracket multiplied by iħ. This makes the operator expectation values obey corresponding classical equations of motion, provided the Hamiltonian is at most quadratic in the coordinates and momenta. Otherwise, the evolution equations still may hold approximately, provided fluctuations are small.

Derivation in the Schrödinger Picture

Suppose some system is presently in a quantum state Φ. If we want to know the instantaneous time derivative of the expectation value of A, that is, by definition

$\frac{d}{dt}\langle A\rangle = \frac{d}{dt}\int \Phi^* A \Phi~dx^3 = \int \left( \frac{\partial \Phi^*}{\partial t} \right) A\Phi~dx^3 + \int \Phi^* \left( \frac{\partial A}{\partial t}\right) \Phi~dx^3 +\int \Phi^* A \left( \frac{\partial \Phi}{\partial t} \right) ~dx^3$

$= \int \left( \frac{\partial \Phi^*}{\partial t} \right) A\Phi~dx^3 + \left\langle \frac{\partial A}{\partial t}\right\rangle + \int \Phi^* A \left( \frac{\partial \Phi}{\partial t} \right) ~dx^3,$

where we are integrating over all space. If we apply the Schrödinger equation, we find that

$\frac{\partial \Phi}{\partial t} = \frac{1}{i\hbar}H\Phi$

and

$\frac{\partial \Phi^*}{\partial t} = \frac{-1}{i\hbar}\Phi^*H^* = \frac{-1}{i\hbar}\Phi^*H.$

Note $H=H^*$, because the Hamiltonian is Hermitian. Placing this into the above equation we have

$\frac{d}{dt}\langle A\rangle = \frac{1}{i\hbar}\int \Phi^* (AH-HA) \Phi~dx^3 + \left\langle \frac{\partial A}{\partial t}\right\rangle = \frac{1}{i\hbar}\langle [A,H]\rangle + \left\langle \frac{\partial A}{\partial t}\right\rangle.$

Often (but not always) the operator A is time independent, so that its derivative is zero and we can ignore the last term.

Derivation in the Heisenberg Picture

In the Heisenberg Picture, the derivation is trivial. The Heisenberg picture moves the time dependence of the system to operators instead of state vector. Starting with the Heisenberg equation of motion

$\frac{d}{dt}A(t) = \frac{\partial A(t)}{\partial t} + \frac{1}{i \hbar}[A(t),H],$

we can derive Ehrenfest's Theorem simply by projecting the Heisenberg Equation onto $|\Psi\rangle$ from the right and $\langle\Psi|$ from the left, or taking the expectation value, so

$\langle\Psi|\frac{d}{dt}A(t)|\Psi\rangle = \langle\Psi|\frac{\partial A(t)}{\partial t}|\Psi\rangle + \langle\Psi|\frac{1}{i \hbar}[A(t),H)]|\Psi\rangle,$

We can pull the $\frac{d}{dt}$ out of the first term since the state vectors are no longer time dependent in the Heisenberg Picture. Therefore,

$\frac{d}{dt}\langle A(t)\rangle = \left\langle\frac{\partial A(t)}{\partial t}\right\rangle + \frac{1}{i \hbar}\langle[A(t),H)]\rangle$

General example

For the very general example of a massive particle moving in a potential, the Hamiltonian is simply

$H(x,p,t) = \frac{p^2}{2m} + V(x,t)$

where $x$ is just the location of the particle. Suppose we wanted to know the instantaneous change in momentum $p$. Using Ehrenfest's theorem, we have

$\frac{d}{dt}\langle p\rangle = \frac{1}{i\hbar}\langle [p,H]\rangle + \left\langle \frac{\partial p}{\partial t}\right\rangle = \frac{1}{i\hbar}\langle [p,V(x,t)]\rangle$

since the operator $p$ commutes with itself and has no time dependence. By expanding the right-hand-side, replacing p by $-i\hbar \nabla$, we get

$\frac{d}{dt}\langle p\rangle = \int \Phi^* V(x,t)\nabla\Phi~dx^3 - \int \Phi^* \nabla (V(x,t)\Phi)~dx^3.$

After applying the product rule on the second term, we have

$\begin{align} \frac{d}{dt}\langle p\rangle =& \int \Phi^* V(x,t)\nabla\Phi~dx^3 - \int \Phi^* (\nabla V(x,t))\Phi ~dx^3 - \int \Phi^* V(x,t)\nabla\Phi~dx^3 \\ =& - \int \Phi^* (\nabla V(x,t))\Phi ~dx^3 \\ =& \langle -\nabla V(x,t)\rangle = \langle F \rangle, \end{align}$

but we recognize this as Newton's second law. This is an example of the correspondence principle, the result manifests as Newton's second law in the case of having so many particles that the net motion is given exactly by the expectation value of a single particle.

Similarly we can obtain the instantaneous change in the position expectation value.

$\begin{align} \frac{d}{dt}\langle x\rangle =& \frac{1}{i\hbar}\langle [x,H]\rangle + \left\langle \frac{\partial x}{\partial t}\right\rangle \\ =& \frac{1}{i\hbar}\langle [x,\frac{p^2}{2m} + V(x,t)]\rangle + 0 = \frac{1}{i\hbar}\langle [x,\frac{p^2}{2m}]\rangle \\ =& \frac{1}{i\hbar}\langle [x,\frac{p^2}{2m}]\rangle = \frac{1}{i\hbar 2 m}\langle [x,p] \frac{d}{dp} p^2\rangle \\ =& \frac{1}{i\hbar 2 m}\langle i \hbar 2 p\rangle = \frac{1}{m}\langle p\rangle \end{align}$

This result is again in accord with the classical equation.

Derivation of the Schrödinger equation from the Ehrenfest theorems

It was established above that the Ehrenfest theorems are consequences of the Schrödinger equation. However, the converse is also true: the Schrödinger equation can be inferred from the Ehrenfest theorems. We begin from

$\begin{align} m\frac{d}{dt} \langle \Psi(t) | \hat{x} | \Psi(t) \rangle &= \langle \Psi(t) | \hat{p} | \Psi(t) \rangle, \\ \frac{d}{dt} \langle \Psi(t) | \hat{p} | \Psi(t) \rangle &= \langle \Psi(t) | -V'(\hat{x}) | \Psi(t) \rangle. \end{align}$

Applications of the product rule leads to

$\begin{align} \langle d\Psi/dt | \hat{x} | \Psi \rangle + \langle \Psi | \hat{x} | d\Psi/dt \rangle &= \langle \Psi | \hat{p}/m | \Psi \rangle, \\ \langle d\Psi/dt | \hat{p} | \Psi \rangle + \langle \Psi | \hat{p} | d\Psi/dt \rangle & = \langle \Psi | -V'(\hat{x}) | \Psi \rangle, \end{align}$

into which we substitute a consequence of Stone's theorem

$i\hbar | d \Psi(t)/dt \rangle = \hat{H} | \Psi(t) \rangle,$

where $\hbar$ was introduced as a normalization constant to the balance dimensionality. Since these identities must be valid for any initial state, the averaging can be dropped and the system of commutator equations for the unknown quantum generator of motion $\hat{H}$ are derived

$im [\hat{H}, \hat{x}] = \hbar \hat{p}, \qquad i [\hat{H}, \hat{p}] = -\hbar V'(\hat{x}).$

Assuming that observables of the coordinate and momentum obey the canonical commutation relation $[ \hat{x}, \hat{p} ] = i\hbar$. Setting $\hat{H} = H(\hat{x}, \hat{p})$, the commutator equations can be converted into the differential equations

$m \frac{\partial H (x,p)}{\partial p} = p, \qquad \frac{\partial H(x,p)}{\partial x} = V'(x),$

whose solution is the familiar quantum Hamiltonian

$\hat{H} = \frac{\hat{p}^2}{2m} + V(\hat{x}).$

Whence, the Schrödinger equation was derived from the Ehrenfest theorems by assuming the canonical commutation relation between the coordinate and momentum. If one assumes that the coordinate and momentum commute, the same computational method leads to the Koopman–von Neumann classical mechanics, which is the Hilbert space formulation of classical mechanics. Therefore, this derivation as well as the derivation of the Koopman–von Neumann mechanics shows that the essential difference between quantum and classical mechanics reduces to the value of the commutator $[ \hat{x}, \hat{p} ]$.