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 [itex]\psi_{n,l,m}[/itex] 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 [itex]t=0[/itex] given by a linear combination of states:
[itex]\Psi=\frac{1}{3}(2\psi_{0,0,0}+2\psi_{2,1,0}+\psi_{3,2,2})[/itex]

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

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

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

Homework Equations

So here's what we need I think:

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

The Attempt at a Solution

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

(b) I've got something pretty weird...like [itex]<\vec{\hat{L}}^{2}>=\frac{8}{3}\hbar^{2}[/itex] and [itex]<\hat{L}_{z}>=\frac{2}{3}\hbar[/itex] 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 [itex]E_{1}=\frac{2}{3}, E_{2}=\frac{2}{3}, E_{3}=\frac{1}{3}[/itex] or am I missing something?

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

(b) I've got something pretty weird...like [itex]<\vec{\hat{L}}^{2}>=\frac{8}{3}\hbar^{2}[/itex] and [itex]<\hat{L}_{z}>=\frac{2}{3}\hbar[/itex] 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:
[itex]E_{1}=\frac{4}{9}, E_{2}=\frac{4}{9}, E_{3}=\frac{1}{9}[/itex]?

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 [itex]\psi[/itex]'s...