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

  1. Nov 27, 2013 #1
    1. The problem statement, all variables and given/known data
    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].


    2. Relevant 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?

    3. 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 doesnt seem right to me...

    (c) I have no idea!!!

    could you guys gimme a hand please?

    thanks a lot!
     
  2. jcsd
  3. Nov 27, 2013 #2
    Remember, the probabilities are from the modulus of the amplitudes squared.
    Would you be willing to show some work?
     
  4. Nov 27, 2013 #3

    vela

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    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.
     
  5. Nov 27, 2013 #4
    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 im a bit confused cos of the factors. I'll type it in word:
    http://imageshack.com/a/img21/6121/81jv.jpg [Broken]
     
    Last edited by a moderator: May 6, 2017
  6. Nov 27, 2013 #5
    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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