# Normalizing the Wave Function

## Homework Statement

Normalize the wave function $Are^{-r/\alpha}$ from $r=0$ to $r=\infty$ where $A$ and $\alpha$ are constants.

## The Attempt at a Solution

Beware, this is my first actual normalization problem! This thread could turn out to be pointless!

I started by integrating the function given from r=0 to r=infinity with respect to r. I then set this value equal to 1. This is what I got:

$A\alpha^2=1$

I'm just not sure if this is right because in all the example problems in class he actually solved for the variables.. Whereas in this one, I have two variables and only one equation so I'm unable to solve for the variables.. Is this solution right? Do I just say that A times alpha squared must be equal to 1 and leave it at that?

Should the integral of the wave function be equal to 1 or should the integral of probability density be equal to 1?

Also, remember you are trying to find the value of A that will make the initial WF normalized, so it's alright if there are some unknown constants in the final expression of A.

Should the integral of the wave function be equal to 1 or should the integral of probability density be equal to 1?

That was a fast reply! Usually takes a while on here.

Thank you so much! I was actually in the middle of eating and I realized that. Rookie mistake! I'll return to this thread after I eat and retry the problem

That was a fast reply! Usually takes a while on here.

Thank you so much! I was actually in the middle of eating and I realized that. Rookie mistake! I'll return to this thread after I eat and retry the problem

Ok, so here is what I got for my probability distribution:

$P(r)=A^{\star}Ar^2e^{-2r/\alpha}$

Do I also need to consider the possibility that $\alpha$ could be complex? Please tell me no! Anyways, integrating this from zero to infinity yeilds:
$A^{\star}A(\frac{\alpha^3}{4})$

Setting it equal to 1 clearly still gives an equation of multiple variables.. (3 variables since A has two components? Or is it still just 2 variables?)

How do I know if one of the unknown constants is complex or not? Also, assuming that I'm right, do I just leave my answer as stated above (set equal to 1) or do I have to do something else? Ok, so here is what I got for my probability distribution:

$P(r)=A^{\star}Ar^2e^{-2r/\alpha}$

Do I also need to consider the possibility that $\alpha$ could be complex? Please tell me no! Anyways, integrating this from zero to infinity yeilds:
$A^{\star}A(\frac{\alpha^3}{4})$

Setting it equal to 1 clearly still gives an equation of multiple variables.. (3 variables since A has two components? Or is it still just 2 variables?)

How do I know if one of the unknown constants is complex or not? Also, assuming that I'm right, do I just leave my answer as stated above (set equal to 1) or do I have to do something else? What is another way to write $A^{\star}A$? Think back to complex numbers...

Your probability density looks correct, EDIT and the integral is as well.

I'm in QM 1 at the moment so I don't feel the most qualified to answer your question whether alpha can be complex, but it shouldn't matter in this problem of normalizing the wave function.

Edit:

In normalization problems once you have solved the integral it's just a matter of solving for A, ie:

$\left | A \right |^2(stuff)=1$

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What is another way to write $A^{\star}A$? Think back to complex numbers...

Your probability density looks correct, however the integral is not. The integral of the prob density is a gaussian integral and cannot be evaluated with normal analytical techniques. Your qm textbook should have tables with formulas to compute gaussian integrals.

I'm in QM 1 at the moment so I don't feel the most qualified to answer your question whether alpha can be complex, but it shouldn't matter in this problem of normalizing the wave function.

Well, the only thing I can think of is $A^2$, but I'm in complex analysis right now and thus far I've treated $z^2$ as $(re^{i\theta})^2=re^{i\theta}re^{i\theta}=z*z$ and I'm pretty sure that's right, so I'm not sure what you're implying.

The integral is a double integration by parts edit-
I saw your edit and whipped out some paper to try to see what you mean by $|A|^2=A^{\star}A$ and I got it, thank you :) So you can only solve for the magnitude of A, you can't actually solve for A? What do you plug back into the wave function? :P

king vitamin
Gold Member
If I had to guess, I would assume that $\alpha$ is not supposed to be complex here, but in principle I suppose it could be. Is there any context to clarify?

Regarding determining coefficients, it's really up to you. First of all, note that every wave function is ambiguous in that, given a normalized wave function $\psi$, the wave function $e^{i \phi}\psi$ is also normalized, so you really do have a choice here. Most people would choose the phase here so that A is a real number.

Now you have a choice between solving A in terms of $\alpha$ and vice-versa. Once again, in principle it doesn't matter, but usually context dictates which is best. Almost always in quantum problems, you find an eigenfunction which takes some functional form, $f(x)$, but of course we are interested in normalizing the eigenfunction. So usually one multiplies the function by a constant, so you have $A*f(x)$, and then normalizes the wave function. Since A was just an added proportionality constant, it's sensible to get rid of it instead of any other given variable in the problem (which will in general be physical).

So if I had to guess, you probably want to solve A in terms of $\alpha$, but I don't have context!

So you can only solve for the magnitude of A, you can't actually solve for A? What do you plug back into the wave function? :P

No, you can solve for A from this expression

$\left | A \right |^2(stuff)=1$

king vitamin
Gold Member
No, you can solve for A from this expression

$\left | A \right |^2(stuff)=1$

But knowing $\left | A \right |^2$ does not give you A, only its magnitude. Any number $e^{i \phi}A$ where $\phi$ is real will give the same value for $\left | A \right |^2$.

But knowing $\left | A \right |^2$ does not give you A, only its magnitude. Any number $e^{i \phi}A$ where $\phi$ is real will give the same value for $\left | A \right |^2$.

Thanks :) The way I worded the question above is the exact same as it was worded in the problem. Sorry that I'm unable to give you a more specific context.

I just left my answer as $|A|=$. I'm fairly certain he does not want us to consider complex values for A just yet, so I'm fairly sure this is the answer he's looking for.

But knowing $\left | A \right |^2$ does not give you A, only its magnitude. Any number $e^{i \phi}A$ where $\phi$ is real will give the same value for $\left | A \right |^2$.

While that is true, in the context of this problem (which I assume is from an intro qm course) isn't it usually implied that the phase is 1? EDIT: Sorry, I'm not doubting anything you've said king vitamin. So far from my experience in QM 1 we've normalized WF's where A is taken to be real, so |A|=A.

king vitamin