# Help regarding normalization of wave functions

1. Jan 28, 2010

### legend

Hi, i need some help regarding normalization of a wave function, i feel it is a very simple problem, but i am having a hard time figuring it out. I would really appreciate it if anybody could help me out a bit regarding this.

I need to normalize the following wavefunctions by figuring out the constant B.

(1) ψ(x) = Bexp(ikx) where ψ(x) is non‐zero between x=0 and x=L ;
(2) ψ(x) = Bexp(−kx) where ψ(x) is non‐zero between x=0 and x=∞

2. Jan 28, 2010

### Staff: Mentor

Do you know what it means for a function to be normalized? What equation or property does a normalized function have to satisfy?

3. Jan 28, 2010

### dacruick

|ψ(x)|² = 1

So legend square both sides, set |ψ(x)|² equal to one. then solve for B

4. Jan 28, 2010

### legend

Thanks a lot dacruick.

@jtbell... actually i have some vague idea, i am still in the process of coming to terms with quantum mechanics (i don't have a physics background :-( )

5. Jan 28, 2010

### Avodyne

Wrong.

6. Jan 28, 2010

### Staff: Mentor

No, you also have to integrate. In general:

$$\int^{+\infty}_{-\infty}{|\psi(x)|^2 dx} = \int^{+\infty}_{-\infty}{\psi^*(x)\psi(x) dx} = 1$$

In this case, one actually has to integrate only over the region where $\psi(x)$ is non-zero.

7. Jan 29, 2010

### Halcyon-on

If $$\psi(x)$$ is a period wave with period L or it is in a box

$$\int^{+L}_{0}{|\psi(x)|^2 dx} = \int^{+L}_{0}{\psi^*(x)\psi(x) dx} = 1$$

8. Jan 29, 2010

### legend

How do i go about the second function i.e Bexp(−kx)? Do i replace "-k" by "i²k" and proceed?

9. Jan 29, 2010

### Truecrimson

The wave function is real so $$\psi=\psi^*=Be^{-kx}$$. Just integrate as it is and you'll be fine. (Replacing $$-k\to i^2k$$ will also give the correct answer because it actually does nothing. You don't have to do that.)

10. Jan 31, 2010

### legend

Thanks a lot to all of you guys for your help. So i solved them and got the following results. Could anyone please let me know if they are correct?
For number 1 : B = 1/(√L)
For number 2 : B = √(2k)

11. Jan 31, 2010

### Truecrimson

Yes, they are correct.

12. Jan 31, 2010

Thanks a lot