Normalization of wave function in x, y and z

[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.

Please help.
thanks

 

[nrqed]

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.

Please help.
thanks

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 $$N exp -(\frac{x^2}{2}+\frac{y^2}{2}+\frac{z^2}{2})$$
instead?

 

[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.

 

[dextercioby]

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

 

[Lorna]

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

 

[Lorna]

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

 

[dextercioby]

Do you know how to explicitate the || ?

 

[Lorna]

I am not sure what you mean by explicitate.

 

[dextercioby]

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

 

[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>?

 

[dextercioby]

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

 

[Lorna]

I get -infinity or :

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

 

[Manchot]

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

 

[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

 

[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.

 

[Lorna]

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

 

[nrqed]

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

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

You should not have an overall minus sign in front.

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.