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I've got the wave function

$$\Psi(x,t) = \frac{e^{-x^2/4(a^2+i\hbar t/2m)}}{\sqrt[4]{2\pi}\sqrt{a+i\hbar t/2ma}}$$

and am trying to find the expectation value of $p^2$. So I simplify things a bit by setting

$$\Psi(x,t) = \frac{e^{-x^2/4(a^2+i\hbar t/2m)}}{\sqrt[4]{2\pi}\sqrt{a+i\hbar t/2ma}} = \frac{e^{-x^2/A}}{B},$$

and then evaluating the integral $\int_{-\infty}^{\infty} dx\ \Psi^*(-\hbar^2\frac{\partial^2}{\partial x^2})\Psi$, and simplifying I'm getting

$$\frac{2\hbar^2}{|A|^2|B|^2}\frac{\sqrt{\pi}|A|\operatorname{Im}(A)\color{red}{iA^*}}{[\operatorname{Re}(A)]^{3/2}}$$

Before even plugging in my expressions for $A$ and $B$ I see something wrong: this is the expectation value of an observable and thus should be real. But all of the numbers in this last expression are real except the part in red which will be complex. Thus I've made some mistake.

I can't see a mistake in my calculations. Can anyone see where I am going wrong here?


Edit: Here is what I did to get that answer:

$$\begin{gather}\frac{\partial}{\partial x}\frac{e^{-x^2/A}}{B} = \frac{-2x}{A}\frac{e^{-x^2/A}}{B} \\ \implies \frac{\partial^2}{\partial x^2}\frac{e^{-x^2/A}}{B} = \left[\frac{-2}{A}+\frac{4x^2}{A^2}\right]\frac{e^{-x^2/A}}{B} = \frac{-2}{A}\left[1-\frac{2x^2}{A}\right]\frac{e^{-x^2/A}}{B} \\ \implies \hat p\frac{e^{-x^2/A}}{B} = \frac{2\hbar^2}{A}\left[1-\frac{2x^2}{A}\right]\frac{e^{-x^2/A}}{B} \\ \implies \frac{e^{-x^2/A^*}}{B^*}\hat p\frac{e^{-x^2/A}}{B} = \frac{e^{-x^2/A^*}}{B^*}\frac{2\hbar^2}{A}\left[1-\frac{2x^2}{A}\right]\frac{e^{-x^2/A}}{B} = \frac{2\hbar^2}{A|B|^2}\left[1-\frac{2x^2}{A}\right]e^{-x^2(\frac 1A+\frac 1{A^*})}\end{gather}$$

Then I use Mathematica to compute $$\int_{-\infty}^{\infty}dx\ \left[1-\frac{2x^2}{A}\right]e^{-x^2\operatorname{Re}(A)/|A|^2} = \frac{\sqrt{\pi}|A|(A-\operatorname{Re}(A))}{\operatorname{Re}(A)^{3/2}}$$

From there I multiply the "constant" back in and rearrange:

$$\frac{2\hbar^2}{A|B|^2}\frac{\sqrt{\pi}|A|(A-\operatorname{Re}(A))}{\operatorname{Re}(A)^{3/2}} = \frac{2\hbar^2}{A|B|^2}\frac{\sqrt{\pi}|A|(i\operatorname{Im}(A))}{\operatorname{Re}(A)^{3/2}}\frac{A^*}{A^*} = \frac{2\hbar^2}{|A|^2|B|^2}\frac{\sqrt{\pi}|A|\operatorname{Im}(A)iA^*}{\operatorname{Re}(A)^{3/2}}$$

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    $\begingroup$ You've calculated the integral incorrectly. Write your computation in more detail $\endgroup$ – OON Sep 2 '16 at 21:34
  • $\begingroup$ Just in case, I've computed it myself and know that the result is different and real. $\endgroup$ – OON Sep 2 '16 at 21:36
  • $\begingroup$ @OON Done.$\ \ $ $\endgroup$ – Bobbie D Sep 2 '16 at 22:21
  • $\begingroup$ BTW, I also worked out the integral that I did with Mathematica by integral table just to make sure. And I got the same result. $\endgroup$ – Bobbie D Sep 2 '16 at 22:34
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$$z=\operatorname{Re}(z)+i\operatorname{Im}(z),\quad z^\ast=\operatorname{Re}(z)-i\operatorname{Im}(z)$$ therefore $$z+z^\ast=2\operatorname{Re}(z)$$ You lost factor of $2$ in $$\frac{1}{A}+\frac{1}{A^\ast}=2\frac{\operatorname{Re}(A)}{|A|^2}$$

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  • $\begingroup$ Wow. I can't believe that little mistake had such a drastic effect on the integral. Thanks for spotting my error! $\endgroup$ – Bobbie D Sep 2 '16 at 22:50

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