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Question about Braket notation

  1. Apr 23, 2009 #1
    Ok, here is my question.

    When you have < r | i >, this equals Sri. So logically if that is that case, if you had SriSaj this would equal < r | j >< a | j >, right?

    If so, then what does < r | j ><a | j > equal? I'm working a problem where I am trying to get a final answer of < r | h | a >. Is that even possible from what I have here?

    By the way, S = Kronecker delta.
  2. jcsd
  3. Apr 23, 2009 #2


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    In an orthonormal basis, yes, [tex]\left\langle {r}
    \mathrel{\left | {\vphantom {r i}}
    \right. \kern-\nulldelimiterspace}
    {i} \right\rangle = \delta _{ri} [/tex].

    However, what you wrote, [tex]\delta _{ri} \delta _{aj} = \left\langle {r}
    \mathrel{\left | {\vphantom {r i}}
    \right. \kern-\nulldelimiterspace}
    {i} \right\rangle \left\langle {a}
    \mathrel{\left | {\vphantom {a j}}
    \right. \kern-\nulldelimiterspace}
    {j} \right\rangle [/tex] is not true. I suspect a typo? Did you mean to put 'i' instead of the first 'j'?

    EDIT: Ok until latex is back up, the first equation is <r|i> = S_ri. The second is S_ri*S_aj = <r|i><a|j>.
  4. Apr 23, 2009 #3
    Yes sorry, that was a typo on my part. That first j was supposed to be an i.
  5. Apr 23, 2009 #4


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    Ok well, to answer your question, what you have with <r|j><a|j> is S_rj*S_aj.
  6. Apr 23, 2009 #5
    Ok, now, is it possible for <r|i><a|j> = <r|h|a> ?
  7. Apr 24, 2009 #6


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    H is your Hamiltonian i assume? I can't imagine you'll get that manipulation ever if it is.
    Last edited: Apr 24, 2009
  8. Apr 24, 2009 #7

    Vanadium 50

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    Those don't have the same units. The left hand side is a pure number, and the right hand side has dimensions of energy.
  9. Apr 24, 2009 #8
    Ok, I messed up in my problem. I have my problem worked down to this:

    (I have to make E = Capital sigma (summation) because the LaTeX isn't working for some reason.)

    E<i|h|j>[<r|i><a|j> - <r|j><a|i>]

    (Sigma is summed over ij, I just don't know how to make it show on here)

    Now is it possible for this to equal <r|h|a> ? In other words:

    E<i|h|j>[<r|i><a|j> - <r|j><a|i>] = <r|h|a> ?
  10. Apr 24, 2009 #9

    Well, if |i> and |j> form a complete set of basis vectors (and i believe they do in this case), then the above expression reduces to this,
    E<i|h|j>[<r|i><a|j> - <r|j><a|i>] = <r|h|a> - <a|h|r> = 2*iota*Im( <r|h|a> )
    where Im() represents the imaginary part.
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