Show that the Hydrogen wave functions are normalized

  • #1
kneesarethebees

Homework Statement


Show that the (1,0,0) and (2,0,0) wave functions are properly normalized.
We know that:

Ψ(1,0,0) = (2/(a0^(3/2))*e^(-r/a0)*(1/sqrt(2))*(1/sqrt(2*pi))

where:
R(r) = (2/(a0^(3/2))*e^(-r/a0)
Θ(θ) = (1/sqrt(2))
Φ(φ) = (1/sqrt(2*pi))

Homework Equations


(1) ∫|Ψ|^2 dx = 1
(2) |R(r)|^2*|Θ(θ)|^2*|Φ(φ)|^2*r^2*sinθ dr dθ dφ ........ (? might be useful)


The Attempt at a Solution


So I know that in order for a function to be properly normalized, it has to have the absolute value of the sine wave squared equal to 1. I originally integrated from negative inf to positive inf, but that did not give me 1 = 1. I tried looking to see where I went wrong but I found equation (2) in my book, but wasn't sure how to integrate that. What am I doing wrong? I think my limits of integration might be wrong because I was doing:

(3) ∫ ((2/(a0^(3/2))*e^(-r/a0)*(1/sqrt(2))*(1/sqrt(2*pi)))^2 dr = 1 from - inf to + inf

and not worrying about that sin θ thing I mentioned earlier.

Do I need to be using equation (2)? How do I find limits of integration? I know that - infinity doesn't make sense for r, so maybe it is just from 0 to infinity with equation 3 maybe?
 

Answers and Replies

  • #2
Charles Link
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## r ## goes from 0 to ## +\infty ##. ## \theta ## goes from ## 0 ## to ## \pi ##, and ## \phi ## goes from 0 to ## 2 \pi ##.
 
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  • #3
vela
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You need to integrate over all space, so it's going to be a three-dimensional integral. You need to use (2).
 
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  • #4
kneesarethebees
Okay, so since I was given what R(r), Θ(θ), and Φ(φ) is, I came up with this equation from what vela has suggested and integrating over what Charles said:

∫∫|R(r)|^2*r^2*sinθ dr dθ for r from 0 to inf and theta from 0 to pi

(1/pi*a0^3) ∫∫ e^(-2r/a0)*r^2*sinθ dr dθ

When evaluating the dr integral using a table of integrals, I get the value of 0. Now when evaluating the dθ integral I get -cosθ from 0 to pi, which results in 2. Why is this still not giving me 1?

Thanks guys :)
 
  • #5
vela
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Because you’re doing it wrong. You need to show your work. We can’t tell what you did otherwise.
 
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  • #6
kneesarethebees
Alright, so the table in my book gives me the ##R(r)##, ##Θ(θ)##, and ##Φ(φ)## for (1,0,0). We know that
$$\int \left | \psi \right| ^2 = 1 $$ and
$$\left| R(r) \right| ^2 ~ \left| Θ(θ) \right| ^2 ~ \left| Φ(φ) \right| ^2 ~ r^2 ~ sinθ ~ dr ~ dθ ~ dφ $$ and
$$\psi = \left| R(r) \right| ^2 ~ \left| Θ(θ) \right| ^2 ~ \left| Φ(φ) \right| ^2$$
Therefore, we can say that
$$\int_0^\infty \int_0^\pi \left| \psi \right| ^2 ~ r^2 ~ sinθ ~ dr ~ dθ = 1$$

Okay, so now I just plug in my values for ##\psi## which gave me the integral:
$$\int_0^\infty \int_0^\pi \frac {4} {4 \pi a_0^3} e^{-2r/a_0} r^2 sin \theta ~ dr ~ d\theta$$
Pull the constants out:
$$\frac {1} {\pi a_0^3} \int_0^\infty \int_0^\pi r^2 e^{-2r/a_0} sin \theta ~ dr ~ d\theta$$

Evaluating the ##dr## integral by using a table of integrals:
$$\int_0^\infty r^2 e^{-2r/a_0} ~ dr = \left[ \frac {-e^{-2r/a_0}} {2/a_0} \left( r^2 + \frac {2r} {2/a_0} + \frac {2} {4/a_0^2} \right) \right]_0^\infty$$
##0 - 3.7*10^{-5} = 3.7*10^{-5}## when evaluating

Evaluating ##d\theta##
$$\int_0^\pi sin\theta ~ d\theta = \left[ -cos\theta \right]_0^\pi$$
##1 - \left( -1 \right) = 2##

Using values now:
$$\frac {1} {\pi a_0^3} \left[ 3.7*10^{-4} * 2 \right] = 4.5*10^{-4} \neq 1$$

I took a couple of hours to learn LaTeX hopefully that outlines my steps more clearly? I think I might be integrating the double integral wrong with my two steps which is why I can't get it to equal 1
 
  • #7
Charles Link
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If you wrote it correctly, I get ## (1/4 )a_o^3 ## for the ## dr ## integral. Also the ## \phi ## integral gives ## 2 \pi ##. I think the results work.
 
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  • #8
kneesarethebees
Charles, how are you integrating for the ##\phi## integral? I don't have an equation for that, just a value for Φ(##\phi##) as seen from my response right above yours.
 
  • #9
DrClaude
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Therefore, we can say that
$$\int_0^\infty \int_0^\pi \left| \psi \right| ^2 ~ r^2 ~ sinθ ~ dr ~ dθ = 1$$
That is not correct. First, the order of the integral signs and the integration elements should be consistent. Second, this has to be a three-dimensional integral:
$$
\int_0^{2\pi} \int_0^\pi \int_0^\infty \left| \psi \right| ^2 ~ r^2 ~ \sinθ ~ dr ~ dθ ~ d\phi = 1
$$
 
  • #10
Charles Link
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Charles, how are you integrating for the ##\phi## integral? I don't have an equation for that, just a value for Φ(##\phi##) as seen from my response right above yours.
For ## A ## is a constant, ## \int\limits_{0}^{2 \pi} A \, d \phi=A \, \phi |_{0}^{2 \pi}=A [2 \pi-0]=A (2 \pi) ##. (It's as simple as they come, but you might be somewhat new to calculus, so I spelled it out in detail.)
 

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