# Finding the expectation value of the angular momentum squared for a wave function

## Homework Statement

Consider a hydrogen atom whose wave function at time t=0 is the following superposition of normalised energy eigenfunctions:

Ψ(r,t=0)=1/3 [2ϕ100(r) -2ϕ321(r) -ϕ430(r) ]

What is the expectation value of the angular momentum squared?

## Homework Equations

I know that L2 operator is:

-ℏ2 [1/sinθ d/dθ sinθ d/dθ+1/(sin2 θ) d2/dϕ2 ]

although I don't think I need to use it.

I know L2=Lx2+Ly2+Lz2

## The Attempt at a Solution

I am confused as to how to go about this. I don't think I need to be calculating an integral, as you would do to find the expectation value of, for example, x2 for a wavefunction. I think I need to calculate the number from squaring the coefficients of each part, and adding, but I'm not sure how to incorporate the L2 bit into this?

I would appreciate any help, I have been puzzling over this for ages now!

diazona
Homework Helper
Hopefully you remember that the expectation value of $L^2$ in a state $\vert\psi\rangle$ is
$$\langle\psi\vert L^2\vert\psi\rangle$$
When you plug in the given wavefunction, what do you get?

Now, what is the expectation value of $L^2$ in an eigenstate $\vert\psi_{nlm}\rangle$, in terms of the quantum numbers n,l,m?

Ok I know that:

〈H ̂ 〉= <S|H ̂|S>

which is:

=sum(a*nam<En|H|Em>)
=sum(a*nam<En|Em|Em>)
=sum(a*namEm<En|Em>)
=sum(a*namEmdeltamn)
=sum(|am|2 En)
=<E>

So am I right in thinking that I just have to do:

<L2> = sum(|coefficients|2 * L2)

If so, what do I use for L2?

Is it l(l+1)hbar2 ?

Thanks

diazona
Homework Helper
Sounds like you're on the right track.

Thank you for getting back to me so quickly.

I did as above, and got:

1/3(22l(l+1)hbar2 + 22l(l+1)hbar2 + l(l+1)hbar2)

Then used the values of l given in the subscript of each eigenfunction, and got an overall answer of 12hbar2. Does that sound about right?

Thanks again x

diazona
Homework Helper
1/3(22l(l+1)hbar2 + 22l(l+1)hbar2 + l(l+1)hbar2)
At the beginning, remember that you get a factor of 1/3 from each $\psi$ in
$$\langle\psi\vert L^2 \vert\psi\rangle$$
Other than that, it seems OK.

May I ask why you do not need to use the L^2 operator explicitly? How do you end up with your
sum(an* am <En|H|Em>) term?