Normalizing Psi when it has no complex numbers

[DragonPetter]

Homework Statement

I need to square the magnitude of psi for each of my integrals

Homework Equations

for x between 0 and a, $$\psi(x,0)$$ = A(x/a), where A and a are constants

The Attempt at a Solution

So I take A(x/a) and square it since it is already positive. so $$A^2\ast(x^2/a^2)$$ . . but in the proof of my book, it makes use of the complex conjugate, which I am sure of how that fits for my particular psi function . . which makes me think that $$A^2\ast(x^2/a^2)$$ is not correct


so is $$|\psi|^2$$ = $$A^2\ast(x^2/a^2)$$ if $$\psi$$ = A(x/a)
in this example?

 

[kuruman]

Yes. The complex conjugate of a real function is the function itself.

 

[DragonPetter]

Hmm, so i am worrying over nothing. I am so new to this that I don't really understand why the wave function would have a complex number in it sometimes and other times not have it, since I don't even really know what reason a complex number is for in a wave function.