Ehrenfest theorem: Definition from Answers.com

Quantum mechanics

$\Delta x\, \Delta p \ge \frac{\hbar}{2}$

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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 commutator of that operator with the Hamiltonian of the system. It is

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

where A is some QM operator and $\langle A\rangle$ is its expectation value. Ehrenfest's theorem is obvious in the Heisenberg picture of quantum mechanics, where it is just the expectation value of the Heisenberg equation of motion.

Ehrenfest's theorem is closely related to Liouville's theorem from Hamiltonian mechanics, which involves the Poisson bracket instead of a commutator. In fact, it is a rule of thumb that a theorem in quantum mechanics which contains a commutator can be turned into a theorem in classical mechanics by changing the commutator into a Poisson bracket and multiplying by $i\hbar$ .

Derivation

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

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

Notice $H = H *$ 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.

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.^[2] 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

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

$= - \int \Phi^* (\nabla V(x,t))\Phi ~dx^3$

$= \langle -\nabla V(x,t)\rangle = \langle F \rangle,$

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.

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

$= \frac{1}{i\hbar}\langle [x,\frac{p^2}{2m} + V(x,t)]\rangle + 0 = \frac{1}{i\hbar}\langle [x,\frac{p^2}{2m}]\rangle =$

$= \frac{1}{i\hbar}\langle [x,\frac{p^2}{2m} + V(x,t)]\rangle = \frac{1}{i\hbar 2 m}\langle [x,p] \frac{d}{dp} p^2\rangle =$

$= \frac{1}{i\hbar 2 m}\langle i \hbar 2 p\rangle = \frac{1}{m}\langle p\rangle$

This result is again in accord with the classical equation.

Notes

^ In bra-ket notation

$\frac{\partial}{\partial t}\langle \phi |x\rangle =\frac{-1}{i\hbar}\langle \phi |\hat{H}|x\rangle =\frac{-1}{i\hbar}\langle \phi |x \rangle H=\frac{-1}{i\hbar}\Phi^*H$

where $\hat{H}$ is the Hamiltonian operator, and H is the Hamiltonian represented in coordinate space (as is the case in the derivation above). In other words, we applied the adjoint operation to the entire Schrödinger equation, which flipped the order of operations for H and $Φ$ .
^ Although the expectation value of the momentum $\scriptstyle \langle p \rangle$ , which is a real-number-valued function of time, will have time dependence, the momentum operator $\scriptstyle p$ does not. Rather, the momentum operator is a constant linear operator on the Hilbert space of the system. The time dependence of the expectation value is due to the time evolution of the wavefunction for which the expectation value is calculated. An Ad hoc example of an operator which does have time dependence is $\scriptstyle x t^2$ , where $\scriptstyle x$ is the ordinary position operator and $\scriptstyle t$ is just the (non-operator) time.

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