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Is this Dirac Ket Bra correct?

  1. Aug 1, 2010 #1
    A B are two-body and one-body operators respectively.
    Is the following equation correct? If so, Would you give me the proof in real space?
    [tex]\sum\limits_{ijklm}\langle ij|A|km \rangle \langle m |B |l\rangle= \sum\limits_{ijkl}\langle ij|A B |k l\rangle[/tex]
  2. jcsd
  3. Aug 1, 2010 #2


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    I assume |km> is a tensor product state. [itex]|km\rangle=|k\rangle\otimes|m\rangle[/itex].

    If [itex]A=A_1\otimes A_2[/itex], then [itex]\langle ij|A|km\rangle=\langle i|A_1|k\rangle\langle j|A_2|m\rangle[/itex], so when you do the sum over m, you get

    [tex]\sum_{ijkl}\langle i|A_1|k\rangle\langle j|A_2B|l\rangle[/tex]

    If A2 is the identity operator, this simplifies to

    [tex]\sum_{ijkl}\langle i|A_1|k\rangle\langle j|B|l\rangle=\sum_{ijkl}(\langle i|\otimes\langle j|)(A_1|k\rangle\otimes B|l\rangle)=\sum_{ijkl}\langle ij|(A_1\otimes B)|kl\rangle[/tex]

    Notational abuse is common when dealing with tensor products. If you write [itex]A_1=A[/itex], i.e. [itex]A=A\otimes I[/itex], and [itex]AB=A\otimes B[/itex], even though this doesn't really make sense, you get an expression that looks the way you want it to. But is it really the same? You need to think about how your A and B are defined, and in particular if they're defined the same way at different places in your equation.

    I don't know what you mean by "real space".
    Last edited: Aug 1, 2010
  4. Aug 1, 2010 #3
    Thank you for your reply.
    Yes, the state can be represented by the tensor product, but the operator A is not so.
    Here , A is coulomb interaction and is not separable, B is the kinetic term.
    Is the above equation correct?
  5. Aug 1, 2010 #4


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    This would be a good time for you to post your own attempt to prove that it is. (It's sort of the custom around here. If you want help with something, show us what you've done so far).
  6. Aug 2, 2010 #5
    what's the rule for matrix multiplication for a 4x4 matrix and a 2x2 matrix?
    as operators they act on completely different spaces! there is no way to
    compose them (if AB is "do B then A").

    what you have written is wrong.

  7. Aug 2, 2010 #6


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    Every operator is a limit of linear combinations of separable operators.
  8. Aug 29, 2010 #7
    Thank You all!
    I can not prove, then I ask for help.
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