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Operator in second quantization

  1. Jan 26, 2010 #1

    I'm struggling with the second quantization formalism. I'd like to derive the hamiltonian of a system with non-interacting particles
    [tex]\hat{H}=\int dx\,a(x)^\dagger \left[\frac{\hat{P}}{2m}+V(x)\right]a(x),[/tex]
    where [tex]a(x) = \hat{\Psi}(x)[/tex].

    I know the second quantized representation of a single-particle operator [tex]\hat{O}[/tex] which is diagonal in the basis [tex]\{|\alpha\rangle\}[/tex]:

    [tex]\hat{O}=\sum_i o_{\alpha_i} a_{\alpha_i}^\dagger a_{\alpha_i}[/tex]

    My idea was, as a first step, to derive the expression of the linear momentum operator:

    [tex]\hat{P}=\sum_i p_{p_i} a_{p_i}^\dagger a_{p_i} =\sum_i \int dx\, \langle x|p_i\rangle a^\dagger(x) \int dx\, \langle p_i|x\rangle a(x) ??[/tex]

    The above is probably wrong (and I don't know how to proceed in order to derive an expression which is of similar form as [tex]\hat{H}[/tex])

    Any help is much appreciated.
  2. jcsd
  3. Jan 26, 2010 #2
    It is wrong, but not very wrong. I guess you are trying to use the change of basis:
    a_p = \int dx\, \langle p_i|x\rangle a(x)
    a_p^\dagger = \int dx\, \langle x|p_i\rangle a^\dagger(x)

    to expand the diagonal representation
    [tex]\hat{P}=\sum_p p a_{p}^\dagger a_{p}[/tex]

    Your biggest mistake is failing to use different labels for dummy variables of integration. You should instead have written:
    [tex]\hat{P}=\sum_p p \int dx_2\, \langle x_2|p_i\rangle a^\dagger(x_2) \int dx_1\, \langle p_i|x_1\rangle a(x_1)[/tex]
    [tex]\hat{P}=\iint dx_2dx_1a^\dagger(x_2) a(x_1)\sum_p p \langle x_2|p_i\rangle \langle p_i|x_1\rangle [/tex]

    This takes you closer. Can you manipulate it further to derive what you want?
  4. Jan 26, 2010 #3
    Thanks, that makes sense to me. As far as I know, the momentum operator in 3 dimensions should be
    [tex]\hat{P}=-i\hbar \int d^3 x a^\dagger (x) \nabla a(x)[/tex]

    Your equation slightly modified gives
    [tex]\int \int dx_1\, dx_2\, a^\dagger (x_2)\langle x_2|\left(\sum \hat{P}_i |p_i\rangle \langle p_i|\right)|x_1\rangle a(x_1)[/tex]

    The inner term simplifies to [tex]\langle x_2|\hat{P}|x_1\rangle = -i\hbar \nabla \langle x_2|x_1\rangle[/tex]?! Because [tex]\langle x_2|x_1\rangle = \delta(x_1-x_2)[/tex] the above double integration indeed gives the desired result.
  5. Jan 26, 2010 #4
    I'd have a preference for saying
    [tex]\langle x_2|\hat{P}|x_1\rangle = -i\hbar\delta(x_2-x_1)\nabla_{x_1}[/tex]
    since derivatives of delta functions tend to scare me.

    As an aside, I'm also a fan of keeping the final answer like:
    H = \int dxdx' a(x')^\dagger\delta(x-x')\left[\frac{-\hbar^2}{2m}\nabla^2 + V(x)\right]a(x)
    since that reminds me that we are really looking at a quadratic form.

    FWIW, I mentioned some references on second quantization in this post in this thread:
  6. Jan 26, 2010 #5
    Thanks a lot.

    I'm trying now to derive the second quantized expression for a general two-body operator [tex]\hat{V}[/tex].

    Diagonal representation: [tex]\hat{V}=\frac{1}{2}\sum_{ij} v_{ij}\hat{P}_{ij}[/tex]
    where [tex]\hat{P}_{ij} = a_i^\dagger a_j^\dagger a_j a_i[/tex] counts the number of pairs of particles in states [tex]|i\rangle, |j\rangle[/tex].

    When following the same procedure as before I get an ingegral over [tex] dx_1 \ldots dx_4 [/tex] with the terms

    [tex]\langle x_1|v_i\rangle \langle x_2 | v_j \rangle \langle v_j | x_3 \rangle \langle v_i |x_4\rangle [/tex].

    Is it correct to rearrange these terms like

    [tex]\langle x_1|\otimes \langle x_2| |v_i\rangle \otimes |v_j \rangle \langle v_j | \otimes \langle v_i | |x_3\rangle \otimes |x_4 \rangle [/tex]

    so as to plug in [tex]\hat{V} = \sum_{ij} v_{ij} |v_i\rangle \otimes |v_j\rangle \langle v_j|\otimes \langle v_i|[/tex] ?
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