# W is real valued?

## Homework Statement

The question is in the post below.

Show the equation W posted is real valued.

## The Attempt at a Solution

The idea is to show that W conjugate = W but there is a complex exponential in W which makes things tricky?

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## Answers and Replies

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HallsofIvy
Homework Helper
You have posted a link to a restricted web-site.

Okay. I have put the equation up in the document. Show W is real valued.

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Is it because psi conjugated produces a negative in the y value so y has only an imaginary component. mod spi sqaured is real. since y has a factor or i, it will cancel with the i already apparent in the exponential. So everything in the integral is positive hence W is positive.

HallsofIvy
Homework Helper
Yes. The imaginary part of the integrand is an odd function ($e^{iy}= cos(y)+ i sin(y)$ and sin is odd) so its integral over a region symmetric about 0 ($-\infty$ to $\infty$ is 0.

That is very neat. But is my long explanation also correct? Although in my explantion, I said that y is purely imaginary. Is that correct would sin(y) make sense then?

In that integral y is claimed to be the integration variable but how does that make sense? Why do you need an integration variable? why not integrate wrt p or x?

This means y can't be complex valued which raises the question why are the arguments in psi have plus and minus y/2 for non conjugate and conjugate psi.

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That is very neat. But is my long explanation also correct? Although in my explantion, I said that y is purely imaginary. Is that correct would sin(y) make sense then?

In that integral y is claimed to be the integration variable but how does that make sense? Why do you need an integration variable? why not integrate wrt p or x?

This means y can't be complex valued which raises the question why are the arguments in psi have plus and minus y/2 for non conjugate and conjugate psi.
$y$ is a dummy variable. Imagine if you had a function:

$$f(x) = \sum_{n=1}^x 1$$

The integral is no different, $y$ is used to "increment" (so to speak), just as $n$ is.

$y$ is a dummy variable. Imagine if you had a function:

$$f(x) = \sum_{n=1}^x 1$$

The integral is no different, $y$ is used to "increment" (so to speak), just as $n$ is.
Right, W is a function of x and p and the integral has nothing to do with x or p, it is there to evaluate a number with constants x and p.