Probability of measuring E in a Hydrogen atom, and expectation values

[Dixanadu]

Homework Statement

Hey guys, so here's the question:
The energy eigenstates of the hydrogen atom \psi_{n,l,m} are orthonormal and labeled by three quantum numbers: the principle quantum number n and the orbital angular momentum eigenvalues l and m. Consider the state of a hydrogen atom at t=0 given by a linear combination of states:
\Psi=\frac{1}{3}(2\psi_{0,0,0}+2\psi_{2,1,0}+\psi_{3,2,2})

(a) What is the probability to find in a measurement of energy E_{1}, E_{2}, E_{3}?

(b) Find the expectation values of the energy \vec{\hat{L}}^{2} and L_{z}.

(c) Does this state have definite parity? (HINT: use orthonormality of the \psi_{n,l,m} and the known eigenvalues of \psi_{n,l,m} with respect to \hat{H}, \vec{\hat{L}}^{2}, \hat{L}_{z}.

Homework Equations

So here's what we need I think:

Eigenvalues of \vec{\hat{L}}^{2} = \hbar^{2}l(l+1)
Eigenvalues of \hat{L}_{z} = \hbar m
Eigenvalues of \hat{H} = E_{n}..right?

The Attempt at a Solution

so for part (a)...is this just really trivial, that the E_{1}=\frac{2}{3}, E_{2}=\frac{2}{3}, E_{3}=\frac{1}{3} or am I missing something?

(b) I've got something pretty weird...like <\vec{\hat{L}}^{2}>=\frac{8}{3}\hbar^{2} and <\hat{L}_{z}>=\frac{2}{3}\hbar which doesn't seem right to me...

(c) I have no idea!

could you guys gimme a hand please?

thanks a lot!

 

[MisterX]

so for part (a)...is this just really trivial, that the E_{1}=\frac{2}{3}, E_{2}=\frac{2}{3}, E_{3}=\frac{1}{3} or am I missing something?

Remember, the probabilities are from the modulus of the amplitudes squared.

(b) I've got something pretty weird...like <\vec{\hat{L}}^{2}>=\frac{8}{3}\hbar^{2} and <\hat{L}_{z}>=\frac{2}{3}\hbar which doesn't seem right to me...

Would you be willing to show some work?

 

[vela]

You might also have noticed that if you added up the three probabilities you got, they sum to more than 1, which should strike you as wrong.

 

[Dixanadu]

as far as the probabilities go, is the correct answer then:
E_{1}=\frac{4}{9}, E_{2}=\frac{4}{9}, E_{3}=\frac{1}{9}?

And yea, i'll show u my working for part (b), but I am a bit confused cos of the factors. I'll type it in word:
http://imageshack.com/a/img21/6121/81jv.jpg

 

[Dixanadu]

I have a hunch that I'm not meant to multiply by the factors already in front of the \psi's...