Quantum Mechanics: total angular momentum of an electron in a hydrogen atom

[bdizzle329]

An electron in a hydrogen atom occupies the combined position and spin state.

\Psi\left(\vec{r},\xi\right)=\left(\sqrt{1/3}Y^{1}_{0}\xi_{+}+\sqrt{2/3}Y^{1}_{1}\xi_{-}\right)

What are the possible measured values of J^{2} (where J is the total angular momentum of the electron L + S) and with what probability will each be found?


J^{2} = \hbar^{2}j\left(j+1\right)

\left|l-s\right|\geq j \geq l+s , where l and s are the orbital angular momentum and spin angular momentum quantum numbers, respectively.

I know that, according to the given position state of the electron and the fact that it is an electron, l = 1 and s = 1/2.

I know that j will be 1/2 or 3/2. Therefore, J^{2}, when measured, will be either 3/4\hbar^{2} or 15/4\hbar^{2}.

I am having trouble determining the associated probabilities of the possible measurement values for J^{2}. From what I have read about addition of angular momentum, it would seem that I would need to calculate the Clebsch-Gordon coefficients for the total angular momentum. I am not really sure where to start. I have read Griffiths' explanation for angular momentum and the Clebsch-Gordon coefficients but he doesn't explain how to use them for total angular momentum.

I feel dumb asking this kind of question but I am having trouble understanding Quantum Mechanics.

 

[bdizzle329]

The wavefunction should be:

\Psi\left(\vec{r},\xi\right)=\left(\sqrt{1/3}Y^{0}_{1}\xi_{+}+\sqrt{2/3}Y^{1}_{1}\xi_{-}\right)

 

[Tvdmeer]

In order to get the probabilities you need to use the Clebsch-Gordan table.

From the original equation we know that Y⁰₁ is the same as |10> and Y¹₁ is the same as |11>. We also know that electrons have an inherent spin of 1/2, so the chi (+) is |1/2 1/2> and the chi (-) is |1/2 -1/2>.

This is evident because chi + is spin up and chi - is spin down and we are working with an electron.

So at this point we have:
Y⁰₁ = |1 0>

Y¹₁ = |1 1>

Chi (+) = |1/2 1/2>

Chi (-) = |1/2 -1/2>

So now we plug this back into the first equation and for now we will ignore the radial part (R₂₁). When we do this we get:

sqrt(1/3) |10> |1/2 1/2> +sqrt(2/3) |11> |1/2 -1/2>

This is where the Clebsch-Godradn table comes into play.

We know that the spin of electron is 1/2, and we also know that l is one (we got that because the lower part of the angular equation is on for both Y⁰₁, and Y¹₁, and also in the radial equation R_nl)

So we go to the 1 X 1/2 clebsch-Gordan table.

so |10> |1/2 1/2> = sqrt(2/3) |3/2 1/2> - sqrt(1/3) |1/2 1/2>
This one we got from the line highlighted in yellow
and |11> |1/2 -1/2> = sqrt(1/3) |3/2 1/2> + sqrt(2/3) |1/2 1/2>
This one we got from the line highlighted in green

Now we plug this back in and get
sqrt(1/3)*(sqrt(2/3) |3/2 1/2> - sqrt(1/3) |1/2 1/2>)+ sqrt(2/3)(sqrt(1/3) |3/2 1/2> + sqrt(2/3) |1/2 1/2>)

This equals sqrt(2)/3|3/2 1/2>+sqrt(2)/3|3/2 1/2>+(2/3)|1/2 1/2> - (1/3)|1/2 1/2>
Which simplifies to 2*sqrt(2)/3|3/2 1/2>+(1/3)|1/2 1/2>

So J² = (3/2)(3/2+1)hbar² with a probability of 2*2/9
= (1/2)(1/2+1)hbar² with a probability of 1/9

And J_z = (1/2)hbar with a probability of 1