# Homework Help: Normilzation of a wavefunction

1. Feb 13, 2010

### Nezva

Normalize this wavefunction
psi(x) = Acos(2x) from -pi/2 to pi/2

I'm unsure exactly hoe to 'normalize' the equation. I've heard a few conflicted explanations.

2. Feb 13, 2010

### Nezva

Ok, so the square of the wave function set equal to 1 (i.e. 100% probability) will give me the euqation I need to normalize the wavefunction, I believe.

So for this one,
int[A^2 * cos^2(2x)]=1, insert trig ident. for cos^2
int(A^2 * [1+cos(2x)/2])=1

set u = 2x and thus dx = 0.5du
rewrite:
1 = A^2/2 * int[1 + cos(u)*1/2du]

I get mathematically lost at this point. My calculus is not that strong. I'm continuing to work it and may be back soon if I figure it out.

3. Feb 13, 2010

### Phyisab****

Just integrate and you are done.

4. Feb 13, 2010

### Nezva

Ugh that was simple to figure out, I just separated the int[1+cos(2x) into two integrations prior to u substitution. This left me with this (Please check my work for errors):

1 = (A^2)/2 * (x) + ([A^2]/2 * int[cos(2x)])

note: x is the integral of 1 from the 1+cos(2x)

NOW I do the u-substitution to find the integral of cos(2x):

[A^2](x)/2 * [A^2]/2int[cos(2x)] = [A^2](x)/2 * -sin(2x)] = 1

Solving for A, A = sqrt[2/(x-0.5[sin(2x)])] from the -pi/2 to pi/2. So to find normalization A I subtract the large A from the small A?

5. Feb 13, 2010

### Phyisab****

I'm sorry but that is just really really super hard for me to read. Please try using latex. Click the capital sigma all the way to the right of the toolbar above the reply box.

I think you are right but I'm not sure what you mean by large A and small A. I did the problem, what is your final answer?

6. Feb 13, 2010

### Nezva

I've gotten to A=$$\sqrt{2/[x-0.5*sin(2x)]}$$. I've not yet run it for the interval -$$\Pi/2$$ to $$\Pi/2$$

edit: I was wondering if subtract $$x=-\Pi/2$$ from the value of $$x=\Pi/2$$ or vice versa?

7. Feb 13, 2010

### Phyisab****

The fundamental theorem of calculus states:

$$\int^{b}_{a}f(x)dx=F(b)-F(a)$$

Last edited: Feb 13, 2010
8. Feb 13, 2010

### Nezva

Trust me I'm working on it, I'm already in the class. Thank you for your time. I was simply unclear how that applied to a "normalization constant" since it's not an integral. Since it's a constant it's never actually integrated, yet it's subject to the A and B boundaries. Does that make any nonsense? Maybe I'm thinking on it too hard, need to take a step back.

9. Feb 13, 2010

### Phyisab****

Hmm I don't really understand, but I think you may be confusing yourself with the way you solved for A before evaluating the integral.

$$\int^{\pi/2}_{-\pi/2}\psi*\psi dx=1$$

$$A^{2}\int^{\pi/2}_{-\pi/2}cos^{2}(2x)dx=1$$

$$=A^{2}\int^{\pi/2}_{-\pi/2}(\frac{1}{2}+\frac{cos2x}{2})$$

$$=\frac{A^{2}}{2}(x+\frac{sin2x}{2})|^{\pi/2}_{-\pi/2}$$

$$=\frac{A^{2}}{2}(\frac{\pi}{2}-\frac{-\pi}{2})$$

$$=A^{2}\frac{\pi}{2}=1$$

10. Feb 13, 2010

### Nezva

Oh, I did make a math error on top of it all :S. Thank you! This was very enlightening.

11. Feb 13, 2010

### Phyisab****

I screwed up! When you reduce the power of the cosine, you should get cos(4x) right! Anyway I'm sure you can work that out.

12. Feb 14, 2010

### Nezva

Oh yah I noticed that when I was working it. I probably assisted your mistake in the way I wrote the problem. Thanks for teaching me this problem and how to use the forum boards more correctly.