# Wave Function: Normalization Constant

1. Aug 5, 2015

### teme92

1. The problem statement, all variables and given/known data
Consider a free particle, initially with a well defined momentum $p_0$, whose wave function is well approximated by a plane wave. At $t=0$, the particle is localized in a region $-\frac{a}{2}\leq x \leq\frac{a}{2}$, so that its wave function is

$\psi(x)=\begin{cases} Ae^{-ip_0x/\hbar} & if -\frac{a}{2}\leq x \leq\frac{a}{2} \\0 & \text{otherwise} \end{cases}$

Find the normalization constant $A$ and sketch $Re(\psi(x))$, $Im(\psi(x))$ and $|\psi(x)|^2$.

2. Relevant equations

3. The attempt at a solution
So here's what I done:

$A^2\int_{-\frac{a}{2}}^\frac{a}{2} e^{-ip_0x/\hbar}dx=1$

$A^2.-\frac{\hbar}{ip_0}.e^{-ip_0x/\hbar}=1$

$A^2=-\frac{ip_0}{\hbar}.\frac{1}{e^{-ip_0a/2\hbar}-e^{-ip_0a/2\hbar}}$

Is this the correct method? Also I have no idea how to sketch the function asked. Any help would be greatly appreciated.

2. Aug 5, 2015

### George Jones

Staff Emeritus
How did you get this?

3. Aug 5, 2015

### teme92

I took it from $P(t)=\int_{-\frac{a}{2}}^\frac{a}{2}|\psi(x,t)|^2=1$ subbed my wave function into that then. Is this method wrong?

4. Aug 5, 2015

### George Jones

Staff Emeritus
If $\psi \left( x \right) =Ae^{-ip_0x/\hbar}$, what is $\left| \psi \left( x \right) \right|^2$?

5. Aug 5, 2015

### teme92

$|\psi(x)|^2=A^2e^{-2ip_0z/\hbar}$?

6. Aug 5, 2015

### George Jones

Staff Emeritus
Not quite. If $Z$ is a complex number, are $Z^2$ and $\left|Z\right|^2$ the same?

7. Aug 5, 2015

### teme92

No, $|Z|^2$ is the modulus squared ie. $(\sqrt{a^2+b^2})^2$

8. Aug 5, 2015

### George Jones

Staff Emeritus
Yes. What if the complex value $Z$ is expressed in polar notation?

9. Aug 5, 2015

### teme92

Is it $cos\theta +isin\theta$?

10. Aug 5, 2015

### George Jones

Staff Emeritus
Well, $e^{i\theta} = \cos \theta + i \sin \theta$. What is $\left| e^{i\theta} \right|^2$ for real $\theta$?

11. Aug 5, 2015

### teme92

$sin2\theta + 1$?

12. Aug 5, 2015

### George Jones

Staff Emeritus
No. You're making this more difficult than it actually is. Forget about a and b and theta. Can you express $\left| Z \right|^2$ in terms of $Z$ and the complex conjugate of $Z$?

13. Aug 5, 2015

### kq6up

Remember, also. The complex conjugate of $\psi$, and $|\psi|^2=\psi^{*}\psi$.

Chris

14. Aug 5, 2015

### teme92

Oh $|\psi|^2=z\bar{z}$

15. Aug 5, 2015

### George Jones

Staff Emeritus
Yes! And if $z = e^{i\theta}$, then $\bar{z} =$?

16. Aug 5, 2015

### teme92

Is it $\bar{z}=e^{-i\theta}$?

Last edited: Aug 5, 2015
17. Aug 5, 2015

### George Jones

Staff Emeritus
When complex conjugating, what is done to every $i$?

18. Aug 5, 2015

### teme92

Sorry I was supposed to put a minus in that I'll edit

19. Aug 5, 2015

### George Jones

Staff Emeritus
And $z \bar{z} =$?

20. Aug 5, 2015

### teme92

They $i\theta$ and $-i\theta$ cancel each other out so $e^0=1$