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

Dixanadu
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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!
 
on Phys.org
Dixanadu said:
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.
Dixanadu said:
(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?
 
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.
 
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
 
Last edited by a moderator:
I have a hunch that I'm not meant to multiply by the factors already in front of the [itex]\psi[/itex]'s...
 

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