# A Derivation of Ehrenfest's Theorem in a particular case

What are the missing lines in the integration?

$$\frac{\text d \langle {p} \rangle}{ \text{d} t}$$ $$= \frac{\text d}{\text d t} \int\limits_{-\infty}^{\infty} \Psi^* \left( \frac{\hbar}{i}\frac{\partial}{\partial x} \right) \Psi ~\text d x$$ $$= \frac{\hbar}{i}\int\frac{\partial }{\partial t} \left(\Psi^*\frac{\partial\Psi}{\partial x}\right)~\text d x$$ $$= \frac{\hbar}{i}\int \frac{\partial \Psi^*}{\partial t}\frac{\partial \Psi}{\partial x} +\Psi^*\frac{\partial }{\partial x}\frac{\partial \Psi }{ \partial t} ~\text {d} x$$ $$= \frac{\hbar}{i}\int\left( -\frac{i\hbar}{2m}\frac{\partial^2 \Psi^*}{\partial x^2}+\frac{i}{\hbar} V\Psi^*\right)\frac{\partial \Psi}{\partial x}+\Psi^* \frac{\partial}{\partial x} \left(\frac{i\hbar}{2m} \frac{\partial^2 \Psi }{\partial x^2} -\frac{i}{\hbar}V\Psi \right)~\text d x$$ $$=\int\left(V\Psi^*-\frac{\hbar^2}{2m}\frac{\partial^2 \Psi^*}{\partial x^2}\right) \frac{\partial \Psi}{\partial x}+\Psi^*\frac{\partial}{\partial x}\left(\frac{\hbar^2}{2m}\frac{\partial^2 \Psi}{\partial x^2}-V\Psi\right)\text d x$$

$$=\left. \left(V\Psi^*-\frac{\hbar^2}{2m}\frac{\partial^2 \Psi^*}{ \partial x^2} \right) \Psi \right|_{-\infty}^{\infty}-\int\left(\frac{\partial}{\partial x} (V\Psi^*)-\frac{\hbar^2}{2m}\frac{\partial^3 \Psi^*}{ \partial x^3}\right)\Psi \text d x+ \left.\Psi^*\left(\frac{\hbar^2}{2m}\frac{\partial^2 \Psi}{\partial x^2} - V\Psi\right)\right|_{-\infty}^\infty-\int\frac{\partial\Psi^*}{\partial x} \left( \frac{\hbar^2}{2m}\frac{\partial ^2 \Psi}{\partial x^2} - V \Psi \right)\text d x$$

$$=0 + \int\left(\frac{\hbar^2}{2m}\frac{\partial^3 \Psi^*}{ \partial x^3}-\frac{\partial}{\partial x} (V\Psi^*)\right) \Psi \text d x+0+\int\frac{\partial\Psi^*}{\partial x}\left(V\Psi - \frac{\hbar^2}{2m}\frac{\partial^2 \Psi}{\partial x^2}\right) \text d x$$ $$=\int\frac{\hbar^2}{2m} \left(\frac{\partial^3\Psi^*}{\partial x^3}\Psi -\frac{\partial\Psi^*}{\partial x} \frac{\partial^2\Psi}{\partial x^2}\right)+\frac{\partial\Psi^*}{\partial x}(V\Psi )-\frac{\partial}{\partial x}(V\Psi^*)\Psi\text d x$$

$$\vdots$$ $$=\int \frac{\hbar^2}{2m}\left( \Psi^* \frac{\partial^3 \Psi}{\partial x^3} -\frac {\partial^2 \Psi^*}{\partial x^2} \frac{\partial \Psi}{\partial x} \right)+\left( V \Psi^* \frac{\partial \Psi}{\partial x}-\Psi^* \frac{\partial}{\partial x} (V \Psi) \right)~\text d x$$ $$\vdots$$ $$= \int \left( V \Psi^* \frac{\partial \Psi}{\partial x}- \Psi^* \frac{\partial V}{\partial x} \Psi-\Psi^*V \frac{\partial \Psi}{\partial x} \right)~\text d x$$ $$= \int\limits_{-\infty}^{\infty} -\Psi^* \frac{\partial V}{\partial x} \Psi ~\text {d} x$$ $$=\left\langle - \frac{ \partial V }{\partial x} \right\rangle$$

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I can kinda understand that $\Psi$ must be zero at $\pm\infty$ because for the wave function to be normalized it must be localised, Is there similar reason that says that $\frac{\partial\Psi}{\partial x}$ must be zero at $\pm\infty$? –  Elements in Space Dec 30 '12 at 6:09

That's kind of horrible. But in the first three dots, the changed quantity is at least not wrong, as $$\Psi^*_{,x}(V\Psi) - (V\Psi^*)_{,x}\Psi = V\Psi\Psi^*_{.x} - (\Psi^*V_{,x} + V\Psi^*_{,x})\Psi = -\Psi^* V_{,x} \Psi$$ is definitely the same as $$V\Psi^*\Psi_{,x} - \Psi^*(V\Psi)_{,x} = \Psi^*V\Psi_{,x} - \Psi^*(V_{,x}\Psi + V\Psi_{,x}) = - \Psi^*V_{,x}\Psi.$$ The global conjugation flip over the entire integral is itself valid if that integral is real (since $\langle a|b \rangle = \langle b|a \rangle^*$), and we should expect it to be real on physical grounds, being $d\langle p\rangle/dt$. Though why would one want to do anything of the sort instead of directly proceeding to the goal, I really have no idea. In any case, this makes it clear that what's really missing is a proof of $$\int\left(\frac{\partial^3\Psi^*}{\partial x^3}\Psi -\frac{\partial\Psi^*}{\partial x} \frac{\partial^2\Psi}{\partial x^2}\right)~\mathrm{d}x = 0$$ or $$\int \left( \Psi^* \frac{\partial^3 \Psi}{\partial x^3} -\frac {\partial^2 \Psi^*}{\partial x^2} \frac{\partial \Psi}{\partial x} \right)~\mathrm{d}x = 0$$ These are also correct for nice enough boundary conditions, and can be shown using integration by parts twice on the first term.
Another method: by differentiation the definition of expectation value and substituting the Schrödinger equation for the time derivative of the state, $$\begin{eqnarray*} \frac{d\langle A\rangle}{dt} &=& \langle\dot{\psi}|A|\psi\rangle + \langle\psi|\dot{A}|\psi\rangle + \langle\psi|A|\dot{\psi}\rangle\\ &=& -\frac{1}{i\hbar}\langle\psi|H^\dagger A|\psi\rangle + \left\langle\frac{\partial A}{\partial t}\right\rangle + \frac{1}{i\hbar}\langle\psi|AH|\psi\rangle\\ &=& \left\langle\frac{\partial A}{\partial t}\right\rangle + \frac{1}{i\hbar}\left\langle[A,H]\right\rangle \end{eqnarray*}$$ Since $[p,H] = [p,V]$ is obvious, it is then easy to verify that $[p,V] = -i\hbar(\nabla V)$, and $p$ is not explicitly dependent on time, so that term doesn't even appear in that case. This scheme can be done without explicit bra-ket notation, but even the position-space integral representation is simpler if one leaves the Hamiltonian as an abstract operator for a while.