# Show that the Hydrogen wave functions are normalized

• kneesarethebees
You don't need an equation for ## \Phi (\phi) ##. You just need to integrate ## d \phi ##. It's a little bit like integrating ## d \phi /d \phi ##. That gives ## \phi +C ##, but ## C ## is a constant, which we can ignore. So ## \int \Phi (\phi) \, d \phi=\int 1 \, d \phi= \phi |_{0}^{2 \pi}= \left( 2 \pi \right) ##.In summary, the wave functions (1,0,0) and (2,0,0) can be properly normalized by using the equation (2
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?

## r ## goes from 0 to ## +\infty ##. ## \theta ## goes from ## 0 ## to ## \pi ##, and ## \phi ## goes from 0 to ## 2 \pi ##.

kneesarethebees
You need to integrate over all space, so it's going to be a three-dimensional integral. You need to use (2).

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 :)

Because you’re doing it wrong. You need to show your work. We can’t tell what you did otherwise.

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

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.

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.

kneesarethebees said:
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:
$$\int_0^{2\pi} \int_0^\pi \int_0^\infty \left| \psi \right| ^2 ~ r^2 ~ \sinθ ~ dr ~ dθ ~ d\phi = 1$$

kneesarethebees said:
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.)

## 1. What does it mean for a wave function to be normalized?

Normalization of a wave function means that the total probability of finding the particle in any location within the given space is equal to 1. This ensures that the wave function is a valid representation of the particle's position.

## 2. How is the normalization condition expressed mathematically for the Hydrogen wave function?

The normalization condition for the Hydrogen wave function is expressed as 0 |Rn,l(r)|2 r2 dr = 1, where Rn,l(r) is the radial wave function, n is the principal quantum number, and l is the orbital angular momentum quantum number.

## 3. What is the significance of the normalization condition for the Hydrogen wave function?

The normalization condition ensures that the wave function is a valid probability distribution function. It also allows for the calculation of physical quantities, such as the expectation value of the position, momentum, and energy of the particle.

## 4. How is the normalization condition for the Hydrogen wave function verified?

The normalization condition can be verified by evaluating the integral of the squared radial wave function over the specified range and confirming that it equals 1. This can be done analytically or numerically using computational methods.

## 5. Can the normalization condition be applied to other wave functions besides the Hydrogen wave function?

Yes, the normalization condition is a fundamental concept in quantum mechanics and applies to all wave functions. However, the specific form of the integral may differ depending on the system and the type of wave function being normalized.

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