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Homework Help: Quantum Mechanics , bra-ket , angular momentum eigenkets, eigenvalues

  1. Mar 23, 2014 #1
    I have a question on the algebra involved in bra-ket notation, eigenvalues of [itex]\hat{J}[/itex][itex]_{z}[/itex], [itex]\hat{J}[/itex][itex]^{2}[/itex] and the ladder operators [itex]\hat{J}[/itex][itex]_{\pm}[/itex]

    The question has asked me to neglect terms from O(ε[itex]^{4}[/itex])

    I am using the following eigenvalue, eigenfunction results, where l[itex]jm\rangle[/itex] is a simultaneous eignenket of [itex]\hat{J}[/itex][itex]^{2}[/itex] and [itex]\hat{J}[/itex][itex]_{z}[/itex]:

    1)[itex]\hat{J}[/itex][itex]^{2}[/itex] [itex]|[/itex][itex]jm\rangle[/itex]=j(j+1)ℏ[itex]^{2}[/itex][itex]|[/itex][itex]jm\rangle[/itex]

    So far the working is:(we are told j is fixed at j=1)

    [itex]\langle1m'[/itex][itex]|[/itex] ([itex]\hat{1}[/itex]-[itex]\frac{ε}{2ℏ}[/itex] ([itex]\hat{J}[/itex][itex]_{+}[/itex] [itex] -[/itex] [itex]\hat{J}[/itex][itex]_{-}[/itex])[itex] + [/itex] [itex]\frac{ε^{2}}{8ℏ}[/itex]( [itex]\hat{J}[/itex][itex]_{+}[/itex][itex]^{2}[/itex][itex]+[/itex][itex]\hat{J}[/itex][itex]_{-}[/itex][itex]^{2}[/itex][itex] - [/itex][itex]2[/itex][itex]\hat{J}[/itex][itex]^{2}[/itex][itex] + [/itex][itex]2[/itex][itex]\hat{J}[/itex][itex]_{z}[/itex][itex]^{2}[/itex])) [itex]|[/itex] [itex]1m\rangle[/itex] = [itex]\langle1m'[/itex] [itex]|[/itex] [itex]1m\rangle[/itex][itex] -[/itex][itex]\frac{ε}{2}[/itex]([itex]\sqrt{(1-m)(2+m)}[/itex][itex]\langle1m'[/itex][itex]|[/itex] [itex]1(m+1)\rangle[/itex] [itex]+[/itex] [itex]\sqrt{(1+m)(2-m)}[/itex][itex]\langle1m'[/itex][itex]|[/itex] [itex]1(m-1)\rangle[/itex] [itex]+[/itex][itex]\frac{ε^{2}}{4}[/itex]((m[itex]^{2}[/itex]-4)[itex]\langle1m'[/itex] [itex]|[/itex] [itex]1m\rangle[/itex] +[itex]\frac{1}{2ℏ^{2}}[/itex][itex]\langle1m'[/itex][itex]|[/itex] [itex]\hat{J}[/itex][itex]_{+}[/itex][itex]^{2}[/itex] [itex]+[/itex] [itex]\hat{J}[/itex][itex]_{-}[/itex][itex]^{2}[/itex][itex]|[/itex][itex]1m\rangle[/itex])

    My Questions:

    - looking at the [itex]\hat{J}[/itex][itex]_{z}[/itex] operator, when it is squared, this has kept the same eigenkets, but squared the eigenvalues. Is this a general result, for eigenvalues and eigenkets? (I have seen this many times and have not gave it a second thought but see next question).
    - Using result 3, i would do the same with [itex]\hat{J}[/itex][itex]_{\pm}[/itex] . However my solution says that terms proportional to ( [itex]\hat{J}[/itex][itex]_{+}[/itex][itex]^{2}[/itex] [itex]+[/itex] [itex]\hat{J}[/itex][itex]_{-}[/itex][itex]^{2}[/itex]) should be neglected as they will yield only contributions of O(ε[itex]^{4}[/itex]).

    So for this term I would get (including the constants it is multiplied by) :
    [itex]\frac{ε^{2}}{8ℏ^{2}}[/itex][itex]\langle1m'[/itex][itex]|[/itex] [itex]\hat{J}[/itex][itex]_{+}[/itex][itex]^{2}[/itex] [itex]+[/itex] [itex]\hat{J}[/itex][itex]_{-}[/itex][itex]^{2}[/itex][itex]|[/itex][itex]1m\rangle[/itex] = [itex]\frac{ε^{2}}{8ℏ^{2}}[/itex]((1-m)(2+m)ℏ[itex]|[/itex][itex]1(m+1)\rangle[/itex]+(1+m)(-m)ℏ[itex]|[/itex][itex]1(m-1)\rangle[/itex]

    And so I can not see where the extra ε[itex]^{2}[/itex] is coming from such that a ε[itex]^{4}[/itex] is yielded that should be neglected.

    Many Thanks to anyone who can help shed some light on this, greatly appeciated !
  2. jcsd
  3. Mar 29, 2014 #2
    Yes if A has eigenvalue a and eigenket |a> we get A2|a> = (AA)|a> = A(A|a>) = A(a|a>) = (Aa)|a> = (aA)|a> = a(A|a>) = a(a|a>) = (aa)|a> = a2|a>. But note that |jm> are NOT eigenkets of J+ or J-. I also found a mistake in your calculation of J- |1m>
  4. Mar 30, 2014 #3
    Thanks for your reply. I thought that I have not used |jm> as eigenkets of J+ or J- , as by 3) i have instead used the eigenkets |jm+1> and |jm-1> ?

    Thanks, I see the mistake (can not seem to edit original post) it should be J-|jm>=√(1+m)(-m)|jm-1> instead of J-|jm>=√(1+m)(2-m)|jm-1>.
  5. Mar 30, 2014 #4
    They are kets but they are not eigenkets. To be an eigenket the equation looks like A|a> = a|a>. The same ket on both sides of the equation, then the ket is called an eigenket.
  6. Apr 2, 2014 #5
    Okay thanks , I see.

    I see my mistake, these are only the probability amplitudes ! Not the probabilities. So the ε[itex]^{4}[/itex] will be yielded by squaring a term multiplied by ε[itex]^{2}[/itex].

    But, I would then neglect [itex]\frac{ε^{2}}{4}[/itex]((m[itex]^{2}[/itex]-4)<1m'l1m>+[itex]\frac{1}{2ℏ^{2}}[/itex]<1m'l J[itex]^{2}_{+} + J^{2}_{-} [/itex] l1m> ) and not just [itex]\frac{ε^{2}}{4}[/itex][itex]\frac{1}{2ℏ^{2}}[/itex]<1m'l J[itex]^{2}_{+} + J^{2}_{-} [/itex] l1m> *

    So the next line in my original post would be:

    [itex]\delta_{m'm}[/itex](1-[itex]\frac{ε^{2}(2-m^{2})}{4}[/itex]- [itex]\frac{ε}{2}(\sqrt{(1-m)(2+m)}[/itex][itex]\delta_{m'(m+1)}[/itex]+[itex]\sqrt{(1+m)(-m)}[/itex][itex]\delta_{m'(m-1)} + [/itex] [itex]\frac{ε^{2}}{4}[/itex]<1m'l J[itex]^{2}_{+} + J^{2}_{-} [/itex] l1m>

    I.e from this the solution concludes:(where P represents the probability)

    P(m)=1 - [itex]\frac{ε^{2}(2-m^{2})}{2}[/itex]

    Whereas * I would also neglect ε[itex]^{2}[/itex] term proportional to <1m'l1m> to attain

    (which obviously does not make sense , but in terms of neglecting the J[itex]^{2}_{+} + J^{2}_{-} term for the same reason)

  7. Apr 4, 2014 #6
  8. Apr 4, 2014 #7


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    Staff: Mentor

    Could you please post the original question? It might help clear up the possible presence of ##O(\varepsilon^4)## terms.
  9. Apr 11, 2014 #8
    An angular eigenstate l jm > is rotated by an infinitesimal angle ε about its y-axis. Without using the explicit form of the matrix element

    d[itex]^{j=1}_{mm'} = [/itex] < jm' l exp([itex]\frac{-i\hat{J_{y}\phi}}{ℏ}) [/itex] l jm >

    calculate the probabilities up to O(ε[itex]^{2}[/itex]) to find the system in other l j m' > states after the rotation.
  10. Apr 17, 2014 #9
  11. Apr 22, 2014 #10
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