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I was doing a problem today in the book that deals with Schrodinger's equation for a potential well. Basically a potential well is when the potential energy is $0$ for some interval and then $V_1$ everywhere else, which is used for trapping electrons.

The problem I am having is in calculating the average value for

$$(z-\langle z \rangle )^2$$

I know that to calculate the average value, the expectation formula must be used which is,

$$\langle A \rangle = \frac{\int_x \psi^* A \psi dx}{\int_x |\psi|^2dx}$$

I am having issues with this because the book does not say what $\psi ^*$ is. However they do differentiate between $\Psi$ and $\psi$. In the book, $\Psi$ is a solution to Schrodinger's equation while $\psi$ is the position component to Schrodinger's equation. Basically $\Psi = r(t)\psi$. Because this is a potential well, there is no reflected wave which leads to the

$$\psi = A\exp(ikz)$$

So given this, does that really mean then that the complex conjugate is

$$\psi ^* = A\exp(-ikz)$$

The problem here is that this leads to a computation simply of calculating

$$\langle A \rangle = \int_x Axdx$$

Note this comes from the fact that $$\langle x \rangle = 0$$ Given that the lower limit is $-L$ and upper limit is $L$. It is almost no surprise to me that this is the case because the midpoint would be the average in many symmetric distributions. But $z^2$ doesn't seem as obvious

I do know the full solution $\Psi$ as well although the time component is also complex.

To me this seems to not be possible. I know that in probability theory the each individual value depends on the probability density function. So how is it possible that the PDF is eliminated?

Any advice for this problem? I would like to learn more about dealing with expected values of Schrodinger's wave function.

Jonathan Aguilera
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    $\psi$ is a function such that for all $(x,t)$, $\psi^(x,t) = (\psi(x,t))^$. The '' on RHS is complex conjugation as a complex number. – achille hui Mar 11 '19 at 04:55
  • Okay so if I have $\psi = exp(-ix)$ is this the complex conjugate, $\psi ^* = exp(ix)$. I understand the complex conjugate as that number which gets rid of the imaginary component of a number and transforms two complex numbers into a real number. I have looked at two sources which describe the expectation of some operator just like I defined it in my question, but none mention what $\psi ^*$ even is. – Jonathan Aguilera Mar 11 '19 at 06:34
  • Yes. if $\psi = \exp(-ix)$, then $\psi^* = \exp(ix)$. – achille hui Mar 11 '19 at 09:17
  • Thank you! As an additional question, do you have any clue as to why the imaginary component of a complex wave is picked instead of the whole wave? I am asking because in the example I am working, it seems like rather than choosing the whole complex exponential. – Jonathan Aguilera Mar 11 '19 at 09:23

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