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Ladder operators and matrix elements..

  1. Sep 26, 2015 #1
    Please I need your help in such problems..
    in terms of ladder operators to simplify the calculation of matrix elements... calculate those
    i) <u+2|P2|u>
    ii) <u+1| X3|u>
    If u is different in both sides, then the value is 0? is it right it is 0 fir both i and ii?
    when exactly equals 0, please explain slowly, my background is chemistry and not physics..
    Thank you.
     
  2. jcsd
  3. Sep 27, 2015 #2

    strangerep

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    You'll need to supply a bit more context...

    I guess P and X are momentum and position operators respectively? But what is your "u"?
     
  4. Sep 27, 2015 #3
    Well, in terms of ladder operators, X= (h/2mw)1/2(a+a+) you can say = constant (a+a+) it is a real
    P=i(hwm/2)1/2 (a-a+) or constant (a-a+)
    about u and u+1 are functions; some books write them like this and others write PSI or PHI of subscript u and u+1;I couldn't type that way as there is no symbols here.
    a+a=n occupation number operator , [a,a+]=1

    If needed I can give an example I know..
     
  5. Sep 27, 2015 #4

    stevendaryl

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    I'm assuming that [itex]|u\rangle[/itex] means the state such that the number operator [itex]N = a^\dagger a[/itex] has value [itex]u[/itex]? In that case, it's completely straight-forward, if tedious, to compute expressions such as

    [itex]\langle u+2 | P^2 | u \rangle[/itex]

    Just write [itex]P[/itex] in terms of [itex]a[/itex] and [itex]a^\dagger[/itex], and apply the rules:

    [itex]a |u\rangle = \sqrt{u} |u - 1\rangle[/itex]
    [itex]a^\dagger |u\rangle = \sqrt{u+1} |u+1\rangle[/itex]
    [itex] \langle n | m \rangle = 0[/itex] if [itex]n \neq m[/itex]
    [itex]\langle n | n \rangle = 1[/itex]

    So [itex]P^2 = A (a^2 - a a^\dagger - a^\dagger a + (a^\dagger)^2)[/itex] (for some constant [itex]A[/itex])
    [itex]P^2 | u \rangle = A (a^2 |u \rangle - a a^\dagger |u\rangle - a^\dagger a |u\rangle + (a^\dagger)^2 |u\rangle)[/itex]

    Then you just work out what [itex]a^2 |u\rangle[/itex] is, etc.
     
  6. Sep 27, 2015 #5
    ok great. I know till this step, my question is to separate these operations and get the final out put like this
    <u+2|aa|u> - <u+2|aa+|u> - <u+1|a+a|u>.... so on
    and solve each one on its own, when I get any different matrix output, the term equals 0; like <u+1|u> or <u+2|u-1>... right? Cuz my confusion is about this part.

    the other part is how to work a3,, if aaa, the 1st one from left operates with the left , and the 1st a from right operates with the right function of the matrix, the middle a operator works with which function.. as it gives different values..
    I am sorry for my silly questions, cuz this what I miss and can not find in books; it seems like a bases that I do not have.
     
  7. Sep 27, 2015 #6

    stevendaryl

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    Well, just look at the term

    [itex]\langle u+2|a\ a|u \rangle[/itex]

    Use the rule: [itex]a |u\rangle = \sqrt{u} |u - 1\rangle[/itex]. So that term simplifies(?) to:

    [itex]\langle u+2|a\ a|u \rangle = \sqrt{u} \langle u + 2 | a | u-1\rangle[/itex]

    Use the same rule again: [itex]a |u-1\rangle = \sqrt{u-1} |u - 2\rangle[/itex]. So we have:

    [itex]\langle u+2|a\ a|u \rangle = \sqrt{u(u-1)}| \langle u + 2 | u-2\rangle[/itex]

    which is zero.

    The only term you're going to get a nonzero result for is:

    [itex]\langle u+2|a^\dagger\ a^\dagger|u \rangle[/itex]
     
  8. Sep 27, 2015 #7

    stevendaryl

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    By the way, this should probably be in the Advanced Physics Homework section.
     
  9. Sep 27, 2015 #8
    Thank you so much for this clarification.. it is very helpful.
    Next time I will follow the right section.. new here.:smile:
     
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