# Normalizing Wave Function of A Ring

1. Nov 22, 2008

### mdmman

1. The problem statement, all variables and given/known data

$$\psi_{n}(\theta)=A_{n} \exp(\imath n \theta)$$
where n is an integer

Calculate the factor $$A_{n}$$ if the wave function is normalized between
$$\theta = 0$$ and $$\theta = 2\pi$$.

2. Relevant equations
NA

3. The attempt at a solution

$$1=\int_0^{2\pi} |\psi_{n}(\theta)|^2 d\theta$$

$$1=|A_{n}|^2\int_0^{2\pi} \exp(2\imath n \theta) d\theta$$

$$1=|A_{n}|^2 [\frac{.5sin(2n\theta)}{n} - \frac{.5cos(2n\theta)}{n} \imath]_0^{2\pi}$$

$$1=|A_{n}|^2 [0]$$

$$1=0$$

Obviously 1 does not equal 0 :) . Am I missing something?

2. Nov 22, 2008

### George Jones

Staff Emeritus
What does $\| \exp(\imath n \theta) \|^2$ equal?

3. Nov 22, 2008

### mdmman

$\| \exp(\imath n \theta) \|^2 = \| \exp(2\imath n \theta) \| = \|cos(2n\theta)+sin(2n\theta)\imath \|$

correct?

4. Nov 22, 2008

### borgwal

You just took the square of the wavefunction, instead of the absolute value squared.

5. Nov 22, 2008

### George Jones

Staff Emeritus
If $z = a + \imath b$, where $a$ and $b$ are both real, what does $|z|^2$ equal?

6. Nov 22, 2008

### mdmman

$\| \exp(\imath n \theta) \|^2 = 1$

Man, I can't believe I missed that!

$$1=|A_{n}|^2\int_0^{2\pi} \|\exp(\imath n \theta)\|^2 d\theta$$

$$1=|A_{n}|^2\int_0^{2\pi} 1 d\theta$$

$$1=|A_{n}|^2[\theta]_0^{2\pi}$$

$$1=|A_{n}|^2*2\pi$$

$$A_{n}=\frac{1}{\sqrt{2\pi}}$$

This is the correct solution, right?

7. Nov 22, 2008

### George Jones

Staff Emeritus
This looks good and is probably the expected answer, but note that multiplying your $A_n$ by any phase factor would give something that also works.