-2
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

Your equation right after "Then, using time-dependent Schrodinger equation:" is not correct.

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.

$\endgroup$
  • $\begingroup$ Thanks! Though I think that was just a typo, the third line under that is still correct, which leaves me with zero still. $\endgroup$ – Matt Chu Sep 21 '18 at 2:27
  • $\begingroup$ @MattChu You're still having sign problems though. Why are there TWO minus signs in the first term of your second-to-last equation? In general, the only way you're going to get zero here is with sign confusion, so if you still are getting zero, check for more sign errors! $\endgroup$ – Jahan Claes Sep 21 '18 at 3:28
  • $\begingroup$ @JahanClaes I have fixed it, yet I'm still not sure how it wouldn't equal zero. If every term is real, its complex conjugate is itself and it gets canceled out. $\endgroup$ – Matt Chu Sep 21 '18 at 3:37
  • 2
    $\begingroup$ @MattChu I'm not sure how you've come to the conclusion that everything is real. Clearly, in your last equation, the first terms in each of the two parenthesis are zero. The last terms in the two parenthesis exactly cancel. And the middle two terms are NOT equal to each other, are NOT real, and do NOT cancel! They're the important terms you need to think about harder. $\endgroup$ – Jahan Claes Sep 21 '18 at 4:01
  • $\begingroup$ To add to what @JahanClaes just said. Remember that the middle terms still have those factors out front. They actually are real, but they don't cancel. They are actually the same thing, and they add to give you exactly what you want. $\endgroup$ – Aaron Stevens Sep 21 '18 at 4:07

Not the answer you're looking for? Browse other questions tagged or ask your own question.