# Having trouble with this normalization

1.Question.

the unnormalized excited state wavefuction of the H atom is:
$$\psi = ( 2 - r/a_0 ) e^a$$

where a = $$-r/a_0$$

Normalize the function to one.

2. My attempts.

I tried 'integrating' the psi*psi, i.e. I squared the above wavefuction.

$$N^2\int_{0}^{\infty} R^2e^{2a}\int_{0}^{\pi}sin \theta d\theta\int_{0}^{2\pi}d\phi =1$$

From here, I got confused with the examples. The second and third integrations were obviously 2 and $$2\pi$$, but the first integration I'm getting confused as to what to put as $$R^2$$ and I'm getting the same answer as the examples because when I done the calculation, the power to which R is the same with every question, which was $$a^3_0/4$$

Hence: $$a^3_0/4*2*2\pi$$ =$$1/N^2$$

Hence I worked out N from this equation, but the answer I got was:

$$\psi =(1/\pi a^3_0)^{1/2}e^{-r/a_0}$$

Which is the same as every other damn normalized wavefuction in the book.

I think I'm having trouble actually understanding how $$R^2$$ is translated from the original wavefuction to the actual integration, hence my immense trouble to working out what R is. Can anyone help with this?

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quantumdude
Staff Emeritus
Gold Member
I'm afraid I can't parse what you're saying. Could you try to clean it up?

Sorry, Had a lot of trouble getting used to the Latex type setting. Updated it so you can read it a bit.. hopefully.

Dr Transport
Gold Member
$$\int_{0} ^{\infty} (4 -4a+a^{2}) e^{-2a} a^{2} da /a_{0} ^{2}$$
where $a = r/a_{0}$. These are straightforward integrals that are found in any integral table.

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thank you!

Wait, when I calculated the first integral, I got:

$$\psi = (2 - a)e^{-a}$$

hence:

First integral: $$\int_{0}^{\infty}(4-4a+a^2)(e^{-2a})a_0da$$
=$$a_0\int_{0}^{\infty}(4-4a+a^2)(e^{-2a})da$$

since: $$a = -r/a_0$$

$$da/dr = 1/a_0$$
$$da = (1/a_0)dr$$
$$(a_0)da = dr$$
I'm not sure I understand how you got the $$a^2/a_0^2$$ Could you, if you have the time, care to elaborate?

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Dr Transport
Gold Member
Change variables from $r$ to $a = r/a_{0}$, you have to remember that in sphereical coordinates you have $r^{2} dr$ for the integration over all r space. Check my original work because I forget to put in factors normalization until I check my units at the end. and I usually screw up early in the am.....

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Still having trouble with the first integration part.. I'll rewrite the original equation:

$$\psi = (2-r/a_0)e^{-r/a_0}$$

Dr Transport
Gold Member
Let $\alpha = r/a_{0}$, then

$$|\psi |^{2} = (2-\alpha)^{2} e^{-2\alpha}$$ so the integral becomes

$$\int_{0} ^{\infty} (4-4\alpha+\alpha^{2})e^{-2\alpha} a_{0} ^{3} \alpha^{2} d\alpha$$

From here you have yo do the integrals, the can be found in an integral table.... In general you have

$$\int _{0} ^{\infty} x^{n} e^{-ax} dx = \frac{n!}{a^{n-1}}$$ for $a > 0, n$ an integer. From here you can find everything you need to calculate the normalization factor.

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I have nothing helpful to add. Just two (maybe stupid) questions as I try to follow along:
In the original equation should the exp. of e be a/2?
Should the integrals of the theta and Phi functions be included in the normalization?

Dr Transport
Now as for the angular integrals, the normnalization of them are included in the integral in the form of factors of $\pi$ which up to this point have been ignored. The part of the original posters problem is that they are forgeting that the form of the volume element in spherical coordinates is $r^{2} dr \cos(\theta) d \theta d \phi$.