1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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]
    2)[itex]\hat{J}[/itex][itex]_{z}[/itex][itex]|[/itex][itex]jm\rangle[/itex]=mℏ[itex]|[/itex][itex]jm\rangle[/itex]
    3)[itex]\hat{J}[/itex][itex]_{\pm}[/itex][itex]|[/itex][itex]jm\rangle[/itex]=[itex]\sqrt{(j∓m)(j±(m+1))}ℏ[/itex][itex]|[/itex][itex]j(m±1)\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-m^{2})}{4}[/itex]
    P(m-1)=[itex]\frac{ε^{2}(2+m-m^{2})}{4}[/itex]
    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
    Pm=1

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

    Thanks.
     
  7. Apr 4, 2014 #6
    anyone?
     
  8. Apr 4, 2014 #7

    DrClaude

    User Avatar

    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
    anyone?
     
  11. Apr 22, 2014 #10
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Quantum Mechanics , bra-ket , angular momentum eigenkets, eigenvalues
Loading...