Calculating Expectation Value of Angular Momentum Squared for Hydrogen Atom

[sian130]

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=L_x²+L_y²+L_z²

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]

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?

 

[sian130]

Ok I know that:

〈H ̂ 〉= <S|H ̂|S>

which is:

=sum(a*_na_m<E_n|H|E_m>)
=sum(a*_na_m<E_n|E_m|E_m>)
=sum(a*_na_mE_m<E_n|E_m>)
=sum(a*_na_mE_mdelta_mn)
=sum(|a_m|² E_n)
=<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]

Sounds like you're on the right track.

 

[sian130]

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]

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.

 

[captainjack2000]

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?