Expected value of p, example of Ehrenfest theorem

\displaystyle  \hat{p}=\frac{\hbar}{i}\frac{\partial}{\partial x}

\displaystyle  \left< p\right> =\frac{\hbar}{i} \int \Psi^* (\frac{\partial}{\partial x}\Psi) \, dx

\displaystyle  \frac{d\left< p\right>}{dt}  = \frac{\hbar}{i}\int \frac{d}{dt} \left(  \Psi^* (\frac{\partial}{\partial x}\Psi)\right ) \, dx

\displaystyle  H= \frac{p^2}{2m}+V

\displaystyle  \hat{H}=\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2}+\hat{V}

\displaystyle  i\hbar \frac{\partial \Psi}{\partial t} = \hat{H}\Psi

\displaystyle  \frac{\partial \Psi}{\partial t}=\frac{\hbar i}{2m}\frac{\partial^2 \Psi}{\partial x^2}-\frac{i }{\hbar}\hat{V}\Psi

\displaystyle  \frac{\partial \Psi^*}{\partial t}=-\frac{\hbar i}{2m}\frac{\partial^2 \Psi^*}{\partial x^2}+\frac{i }{\hbar}\hat{V}\Psi^*

\displaystyle    \frac{d}{dt} \left( \Psi^* (\frac{\partial}{\partial x}\Psi)\right ) =  (-\frac{\hbar i}{2m}\frac{\partial^2 \Psi^*}{\partial x^2}+\frac{i }{\hbar}\hat{V}\Psi^*)\frac{\partial \Psi}{\partial x}+\Psi^*\frac{\partial}{\partial x}(\frac{\hbar i}{2m}\frac{\partial^2 \Psi}{\partial x^2}-\frac{i }{\hbar}\hat{V}\Psi)

\displaystyle    \frac{d}{dt} \left( \Psi^* (\frac{\partial}{\partial x}\Psi)\right ) =(-\frac{\hbar i}{2m}\frac{\partial^2 \Psi^*}{\partial x^2}+\frac{i }{\hbar}\hat{V}\Psi^*)\frac{\partial \Psi}{\partial x}+\Psi^*\frac{\partial}{\partial x}(\frac{\hbar i}{2m}\frac{\partial^2 \Psi}{\partial x^2}-\frac{i }{\hbar}\hat{V}\Psi)+\frac{\partial \Psi^*}{\partial x}(\frac{\hbar i}{2m}\frac{\partial^2 \Psi}{\partial x^2})-\frac{\partial \Psi^*}{\partial x}(\frac{\hbar i}{2m}\frac{\partial^2 \Psi}{\partial x^2})

\displaystyle    \frac{d}{dt} \left( \Psi^* (\frac{\partial}{\partial x}\Psi)\right ) =(-\frac{\hbar i}{2m}\frac{\partial^2 \Psi^*}{\partial x^2}+\frac{i }{\hbar}\hat{V}\Psi^*)\frac{\partial \Psi}{\partial x}+\frac{\partial}{\partial x}(\Psi^* \frac{\hbar i}{2m}\frac{\partial^2 \Psi}{\partial x^2})-\Psi^*\frac{\partial}{\partial x}(\frac{i }{\hbar}\hat{V}\Psi)-\frac{\partial \Psi^*}{\partial x}(\frac{\hbar i}{2m}\frac{\partial^2 \Psi}{\partial x^2})

\displaystyle    \int \frac{d}{dt} \left( \Psi^* (\frac{\partial}{\partial x}\Psi)\right )\,dx =\int \left[ (-\frac{\hbar i}{2m}\frac{\partial^2 \Psi^*}{\partial x^2}+\frac{i }{\hbar}\hat{V}\Psi^*)\frac{\partial \Psi}{\partial x}-\Psi^*\frac{\partial}{\partial x}(\frac{i }{\hbar}\hat{V}\Psi)-\frac{\partial \Psi^*}{\partial x}(\frac{\hbar i}{2m}\frac{\partial^2 \Psi}{\partial x^2})\right]\,dx

\displaystyle  \int \frac{d}{dt} \left(\Psi^* (\frac{\partial}{\partial x}\Psi)\right)  \,dx =  \int    \left[(-\frac{\partial}{\partial x}(\frac{\hbar i}{2m}\frac{\partial \Psi^*}{\partial x}\frac{\partial \Psi}{\partial x})+\frac{i }{\hbar}\hat{V}\Psi^*\frac{\partial \Psi}{\partial x}-\Psi^*\frac{\partial}{\partial x}(\frac{i }{\hbar}\hat{V}\Psi)\right]  \,dx

\displaystyle  \int \frac{d}{dt} \left(\Psi^* (\frac{\partial}{\partial x}\Psi)\right)  \,dx =  \int    \left[\frac{i }{\hbar}\hat{V}\Psi^*\frac{\partial \Psi}{\partial x}-\Psi^*\frac{\partial}{\partial x}(\frac{i }{\hbar}\hat{V}\Psi)\right]  \,dx

\displaystyle  \int \frac{d}{dt} \left(\Psi^* (\frac{\partial}{\partial x}\Psi)\right)  \,dx =  -\frac{i }{\hbar}  \int  \Psi^*\frac{\partial V}{\partial x}\Psi  \,dx

\displaystyle  \frac{d\left< p\right>}{dt}  =  -\left< \frac{\partial V}{\partial x}\right>

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.

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.

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.

http://en.wikipedia.org/wiki/Ehrenfest_theorem

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