Normalizing Wave Function of A Ring

[mdmman]

Homework Statement

<br /> \psi_{n}(\theta)=A_{n} \exp(\imath n \theta)<br />
where n is an integer

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

Homework Equations

NA

The Attempt at a Solution

<br /> 1=\int_0^{2\pi} |\psi_{n}(\theta)|^2 d\theta<br />

<br /> 1=|A_{n}|^2\int_0^{2\pi} \exp(2\imath n \theta) d\theta<br />

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

<br /> 1=|A_{n}|^2 [0]<br />

<br /> 1=0<br />

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

 

[George Jones]

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

 

[mdmman]

<br /> \| \exp(\imath n \theta) \|^2 = \| \exp(2\imath n \theta) \| = \|cos(2n\theta)+sin(2n\theta)\imath \|<br />

correct?

 

[borgwal]

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

 

[George Jones]

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

 

[mdmman]

<br /> \| \exp(\imath n \theta) \|^2 = 1<br />

Man, I can't believe I missed that!

<br /> 1=|A_{n}|^2\int_0^{2\pi} \|\exp(\imath n \theta)\|^2 d\theta<br />

<br /> 1=|A_{n}|^2\int_0^{2\pi} 1 d\theta<br />

<br /> 1=|A_{n}|^2[\theta]_0^{2\pi}<br />

<br /> 1=|A_{n}|^2*2\pi<br />

<br /> A_{n}=\frac{1}{\sqrt{2\pi}}<br />

This is the correct solution, right?

 

[George Jones]

This is the correct solution, right?

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.