# Complex Conjugate of Fourier Transform

## Main Question or Discussion Point

Hello All,

As I understand it, the wavefunction Psi(x) can be written as a sum of all the particle's momentum basis states (which is the Fourier transform of Psi(x)). I was woundering if the wavefunction's complex conjugate Psi*(x) can be written out in terms of momentum basis states, similar to the transform of Psi(x), and if so, what form would it take in order to make make the probability density of Psi(x)Psi*(x) the correct real-valued number.

In other words, if both Psi(x) and Psi*(x) were written out in their Fourier transform forms, and multiplied together in these forms, what would the transform of Psi*(x) look like?

Any insight into this would be greatly appriciated.

-Andrew

## Answers and Replies

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No magic here. If
$$\psi(x)= \frac{1}{\sqrt{2\pi}} \int\!dk\, \psi(k) e^{\frac{i k x}{\hbar}$$,
then
$$\psi^*(x)= \frac{1}{\sqrt{2\pi}} \int\!dk\, \psi^*(k) e^{\frac{-i k x}{\hbar}$$,

Last edited:
Thanks for the help, lbrits!

I figured the complex conjugate would involve making negative all the imaginary parts of the transform, but it seemed like doing the ol' FOIL method when multiplying Psi(x) and Psi*(x) would result in a number that would generally be complex, not the real number that the probability density should be. I guess I'm a bit rusty on my math :-)

Thanks again for the help,
-droo

Yeah I was a bit rusty on the whole distributivity property of products over sums. Anyway, things are quite forgiving:
$$\psi^*(x) \psi(x) = \frac{1}{2\pi \hbar} \int\!dk\,dq\, \psi^*(k) \psi(q) e^{i (k-q)x/hbar}$$
I believe you can show this to be the Fourier transform of $$f(k) = \int\!dq\, \psi^*(k) \psi(k-q)$$, and to show that the thing is real, is to remember that in the fourier transform things always come in pairs, one multiplied by $$e^{ikx}$$ and one by $$e^{-ikx}$$. The previous expression exhibits a symmetry under k -> -k.