# Normalization of wave function in x, y and z

1. Oct 27, 2007

### Lorna

Hello,

How do I find the normalization constant for psi(x,y,z) = N exp -(x/2+y/2+z/2) ??

I did the following:

$$\int(psi^* psi)dx dy dz = 1$$

the integral bounds are from -infinity to infinity and the * means the complex conjugate.The integral is so weird that I couldn't find N. I used maple to evalulate the integral and it gave me something like infinity multiplied by some other things.

thanks

2. Oct 27, 2007

### nrqed

Welcome to the forums!

The equation for the normalization you wrote is entirely correct.
However, I am sure that you have the wrong expression for the wavefunction because the wavefunction you wrote down is non normalizable. Please double check the problem because I am sure this is not the correct wavefunction. Are you sure it was not $$N exp -(\frac{x^2}{2}+\frac{y^2}{2}+\frac{z^2}{2})$$

3. Oct 27, 2007

### Lorna

That's what I thought too, but I double checked and I have it written in two sources. The only difference between what I typed here and the original is that the x, y and z are in absolute, that is: |x|, |y|, and |z| and that each is divided by a constant, a, b and c respectiviely.

4. Oct 28, 2007

### dextercioby

I see. What does $\int_{-\infty}^{\infty} e^{-|x|} \ {} dx$ equal to ?

5. Oct 28, 2007

### Lorna

$$-e^(-|x|)$$ evaluated from -infinity to infinity which is zero?

6. Oct 28, 2007

### Lorna

Oops, I guess the derivative of |x| is : |x|/x
But that gives an infinity/infinity which is not defined!

Last edited: Oct 28, 2007
7. Oct 28, 2007

### dextercioby

Do you know how to explicitate the || ?

8. Oct 28, 2007

### Lorna

I am not sure what you mean by explicitate.

9. Oct 28, 2007

### dextercioby

Simply means to write what |x| is equal to on $\mathbb{R}$.

10. Oct 28, 2007

### Lorna

yes. the value inside the absolute can be +ve or -ve, but the absolute value makes it always +ve regardless of the sign of x, so :

|x| = x when x >=0 and -x when x <0

is that what you meant>?

11. Oct 28, 2007

### dextercioby

Yes, now can you compute the integral i wrote above ?

12. Oct 28, 2007

### Lorna

I get -infinity or :

- (|x|/x * e^ -|x|) eval. bw infinity and -infinity so =
- (1*e^-infinity - {- e^infinity}) = infinity

13. Oct 28, 2007

### Manchot

Lorna, try splitting the integral that dextercioby gave you into two pieces.

14. Oct 28, 2007

### Lorna

I am not sure I know how to do it, that's why I'm asking. I get:

answer (if x >=0) = -e^-x (eval from -inf to +inf)
answer (if x < 0) = e^x (eval from -inf to +inf)

I think I'm missing something here

15. Oct 29, 2007

### Manchot

Why are you integrating both halves of the function from -infinity to +infinity? The negative side should be integrated from -infinity to 0, and the positive side should be integrated from 0 to +infinity.

16. Oct 29, 2007

### Lorna

Wow Manchot, now I see how it works! Thanks a lot for all who helped.

17. Oct 29, 2007

### nrqed

Ahhh! You did not mention absolute values in your initial post!!!!!

It's simply this:

for x>0, we have $$e^{- |x|} = e^{-x}$$

and for x<0, we have $$e^{- |x|} = e^{x}$$

Now you simply have to break up the x integral into
$$\int_{\infty}^{\infty} = \int_{-\infty}^0 + \int_0^{\infty}$$ and in each integral replace the exponential of the absolute value by its expression given above.