Wave Function: Normalization Constant

[George Jones]

I'm getting confused over what $(Ae^{-i{p_0}x/\hbar})^2$ is. And then the integrating of that.

You are not supposed to integrate $(Ae^{-i{p_0}x/\hbar})^2$, you are supposed to integrate $|Ae^{-i{p_0}x/\hbar}|^2$.

 

[teme92]

So would that be $A^2(1)$ as $|e^{-i{p_0}x/\hbar}|^2=1$?

 

[George Jones]

Yes.

 

[teme92]

So $A=1$? Thanks for all the help George. How do I proceed to sketch the items asked? I assume $A$ is the real part and $e^{-i{p_0}x/\hbar}$ the imaginary part and $|\psi(x)|^2=1$. How do these look when sketched?

 

[George Jones]

So $A=1$?

No. Remember, you have to calculate

$$\int^{a/2}_{-a/2} \left| \psi \left(x\right) \right|^2 dx.$$

 

[teme92]

Ah there's still the $dx$ left:

$\frac{a}{2}+\frac{a}{2}=a$, therefore $A=\frac{1}{a}$

 

[George Jones]

Not quite. There is an A from psi, and another A from \bar{psi}.

 

[teme92]

$A=\frac{1}{\sqrt{a}}$?

 

[George Jones]

Looks okay.

 

[teme92]

Brilliant, and regarding the sketch part?

 

[George Jones]

What are $Re(e^{-i\theta})$ and $Im(e^{-i\theta})$?

 

[teme92]

$Re=\theta$ and $Im=-i$?

 

[George Jones]

No. Recall how $e^{i\theta}$ was expressed earlier in the thread.

 

[teme92]

Oh I get you. $Re=cos\theta$ and $Im=isin\theta$?