Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Second Quantization-Kinetic operator

  1. Jan 26, 2014 #1


    User Avatar
    Gold Member

    I am feeling a little stupid tonight... So let me build the problem...
    For a single particle operator [itex]O[/itex], we have in the basis [itex] |i> [/itex] we have that:
    [itex] O= \sum_{ij} o_{ij} |i><j| [/itex] with [itex]o_{ij}=<i|O|j>[/itex]
    Then for N particles we have that:
    [itex] T=\sum_{a}O_{a}= \sum_{ij} o_{ij} \sum_{a} |i>_{a}<j|_{a} [/itex] with [itex]a=1,2...,N[/itex]

    How can we write this afterwards (for bosons) in respect to the creation and annihilation operators as:
    [itex]T= \sum_{ij} o_{ij} c_{i}^{t} c_{j} [/itex]

    From what I suspect, [itex]\sum_{a} |i>_{a}<j|_{a}[/itex] must be equal to the number operator... But for some reason I'm unable to see it...Also I tried taking this:
    [itex] O= \sum_{ij} o_{ij} |i><j| [/itex]
    [itex] O= \sum_{ij} o_{ij} c_{i}^{t}|0><0|c_{j} [/itex]
    And then summing over the particles:
    [itex] O= \sum_{ij} o_{ij} c_{i}^{t}c_{j} \sum_{a}|0>_{a}<0|_{a} [/itex]
    but I see no reason why the sum must be equal to unity [itex] \sum_{a}|0>_{a}<0|_{a} =1 [/itex]

    Please give hints, not answers
    Last edited: Jan 26, 2014
  2. jcsd
  3. Jan 29, 2014 #2


    User Avatar

    Staff: Mentor

    Consider a ##N##-particle operator ##O## as a sum of one-particle operators
    O = \sum_{i=1}^{N} O_i
    First assume that you can write it as
    O = \sum_{\alpha, \beta} \langle \alpha | O_1 | \beta \rangle c_\alpha^\dagger c_\beta
    If you change the basis, you can write
    O' = \sum_{\gamma, \delta} \langle \gamma | O_1 | \delta \rangle b_\gamma^\dagger b_\delta
    The creation and annihiliation operators satisfy the linear transformations
    b_\gamma^\dagger &= \sum_{\alpha} c_\alpha^\dagger \langle \alpha | \gamma \rangle \\
    b_\gamma &= \sum_{\alpha} \langle \gamma | \alpha \rangle c_\alpha
    O' &= \sum_{\gamma, \delta} \sum_{\alpha, \beta} c_\alpha^\dagger \langle \alpha | \gamma \rangle \langle \gamma | O_1 | \delta \rangle \langle \delta | \beta \rangle c_\beta \\
    &= \sum_{\alpha, \beta} \langle \alpha | O_1 | \beta \rangle c_\alpha^\dagger c_\beta \\
    &= O
    This means that the assumed representation of ##O## is independent of the basis. If you choose a basis ##| \gamma \rangle## that diagonalizes ##O_1##, then
    O = \sum_\gamma o_\gamma b_\gamma^\dagger b_\gamma
    where ##o_\gamma## is the eigenvalue associated with the ket ##| \gamma \rangle##, which is what we expected from the first formula above. Therefore, the assumption for the form of ##O## in any basis was correct.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook