I'm trying to prove Ehrenfest Theorem:

$$m\frac{d}{dt}\langle \vec{\hat x}\rangle\;=\; \langle\vec {\hat{p}}\rangle$$

We can just consider one component of $\vec x$, say $x$.

$$m\frac{d}{dt}\langle\hat{x}\rangle\; =m\int x\left (\frac{d\rho}{dt}\right )\,d^3\vec x$$

Now I can reduce the result in question to

$$m\frac{d}{dt}\langle\hat{x}\rangle\; =\frac{i\hbar}{2}\int \left( \frac{d\Psi^*}{dx}\Psi-\Psi^*\frac{d\Psi}{dx}\right)\,d^3\vec x$$

Which I understand the right hand side is:


However I don't understand how this equates to $\langle\hat{p_x}\rangle$?

I can prove for normalizable states the expectation of the momentum operator is real, which would imply that this gives zero?

  • 3
    $\begingroup$ The RHS is $\frac{\langle\hat{p_x}\rangle^*\color{red}{+}\langle\hat{p_x}\rangle}{2}$. In coordinate representation, $\vec{p} = -i\hbar\vec{\nabla}$ $\endgroup$ Jan 12, 2016 at 17:02
  • $\begingroup$ To add on to achille's comment: And then use the fact that momentum is a hermitian operator. $\endgroup$ Jan 12, 2016 at 17:03

1 Answer 1


The time-dependent Schrodinger Equation is given by

$$i \bar h\frac{\partial \psi(\vec r, t)}{\partial t}=H\{\psi(\vec r, t)\}$$

where the Hamiltonial operator is

$$H\{\cdot \}=-\frac{\bar h^2}{2m}\nabla^2\{\cdot\}+V(\vec r,t)\{\cdot\}$$

Therefore, we can write

$$\begin{align} m\frac d{dt}\int_V \psi^*(\vec r, t)\vec r \psi(\vec r, t)\,dV&=\frac{im}{\bar h}\int_V \vec r\left(H\{\psi^*(\vec r, t)\} \psi(\vec r, t)-H\{\psi(\vec r, t)\} \psi^*(\vec r, t)\right)\,dV\\\\ &=-\frac{i\bar h}{2}\int_V \vec r\left(\psi(\vec r, t)\nabla^2\psi^*(\vec r, t)-\psi^*(\vec r, t)\nabla^2\psi(\vec r, t)\right)\,dV\\\\ &=-\frac{i\bar h}{2}\int_V \vec r\left(\nabla \cdot \left(\psi(\vec r, t)\nabla\psi^*(\vec r, t)-\psi^*(\vec r, t)\nabla\psi(\vec r, t)\right)\right)\,dV\\\\ &=-\frac{i\bar h}{2}\oint_S \vec r\left(\psi(\vec r, t)\frac{\partial \psi^*(\vec r, t)}{\partial n}-\psi^*(\vec r, t)\frac{\partial \psi(\vec r, t)}{\partial n}\right)\,dS\\\\ &+\bbox[5px,border:2px solid #C0A000]{\frac{i\bar h}{2}\int_V\left(\psi(\vec r, t)\nabla\psi^*(\vec r, t)-\psi^*(\vec r, t)\nabla\psi(\vec r, t)\right)\,dV}\,\,\dots \text{OP Starting Point}\\\\ &=-\frac{i\bar h}{2}\oint_S \left(\vec r \psi(\vec r, t)\frac{\partial \psi^*(\vec r, t)}{\partial n}-\vec r \psi^*(\vec r, t)\frac{\partial \psi(\vec r, t)}{\partial n} -\psi^*(\vec r, t)\psi(\vec r, t)\right)\,dS\\\\ &-\frac{i\bar h}{2}\int_V 2 \psi^*(\vec r, t)\nabla \psi(\vec r, t)\,dV \end{align}$$

If we take $V$ to be all of space, then the surface integral vanishes and we are left with

$$\begin{align} m\frac d{dt}\int_V \psi^*(\vec r, t)\vec r \psi(\vec r, t)\,dV&=\int_V \psi^*(\vec r, t)\left(-i\bar h \nabla \psi(\vec r, t)\right)\,dV\\\\ &=\langle \vec p \rangle \end{align}$$


$$\langle \vec p \rangle=\int_V \psi^*(\vec r, t)\left(-i\bar h \nabla \psi(\vec r, t)\right)\,dV$$

since the momentum operator in the spatial representation is given by

$$\vec P_{op}\{ \psi(\vec r, t)\}=-i\bar h\nabla \psi(\vec r, t)$$


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