# How to prove that the time derivative of the expectation of position equals momentum over mass? [closed]

So given the expectation value of position,

$$\langle x \rangle = \int_{-\infty}^\infty \Psi^* x \Psi \ dx$$

I'm trying to show that the time derivative of this is equal to $$\frac{\langle p_x \rangle}{m}$$.

I started by using the product rule, which gave:

$$\frac{\partial \langle x \rangle}{\partial t} = \int_{-\infty}^\infty \left[ \Psi^* x \frac{\partial \Psi}{\partial t} + \frac{\partial \Psi^*}{\partial t} x \Psi \right] dx$$

Then, using the time-dependent Schrodinger equation:

$$\frac{\partial \langle x \rangle}{\partial t} = \frac{1}{i\hbar} \int_{-\infty}^\infty \Psi^* x (H \Psi) dx - \frac{1}{i\hbar} \int_{-\infty}^\infty (H \Psi)^* x \Psi dx$$ $$= \frac{1}{i\hbar} \int_{-\infty}^\infty \Psi^* x \left( - \frac{\hbar^2}{2m} \frac{\partial^2 \Psi}{\partial x^2} + V \Psi \right) dx - \frac{1}{i\hbar} \int_{-\infty}^\infty \left( - \frac{\hbar^2}{2m} \frac{\partial^2 \Psi^*}{\partial x^2} + V \Psi^* \right) x \Psi dx$$

(The $$V$$ components cancel out.)

$$= -\frac{i \hbar}{2m} \int_{\infty}^\infty \left[ \Psi^* x \frac{\partial^2 \Psi}{\partial x^2} \right] dx + \frac{i\hbar}{2m} \int_{-\infty}^\infty \left[\frac{\partial^2 \Psi^*}{\partial x^2} x \Psi \right] dx$$

I then tried integrating by parts, which gives:

$$= \frac{i\hbar}{2m} \left( \left. -x\Psi^* \frac{\partial \Psi}{\partial x} \right|_{-\infty}^\infty + \int_{-\infty}^\infty \Psi^* \frac{\partial \Psi}{\partial x} dx + \int_{\infty}^\infty x \left| \frac{\partial \Psi}{\partial x} \right|^2 dx \right) + \frac{i\hbar}{2m} \left( \left. x\Psi \frac{\partial \Psi^*}{\partial x} \right|_{-\infty}^\infty - \int_{-\infty}^\infty \Psi \frac{\partial \Psi^*}{\partial x} dx - \int_{-\infty}^\infty x \left| \frac{\partial \Psi}{\partial x} \right|^2 dx \right)$$

However, at this point, every part in the above equation is real, so the equation ends up equalling zero. Where did I go wrong?

Hint: \begin{align} H \Psi \; &= \; i \hbar \frac{\partial \Psi}{\partial t} \\ \implies H \Psi^{*} \; &= \; \color{red}-i \hbar \frac{\partial \Psi^{*}}{\partial t} \end{align} You miss the minus sign.