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Representation of second quantization

  1. Oct 3, 2009 #1
    In coordinate representation in QM probality density is:


    in RSQ representation operator of density of particles is


    Is this some relation between this operator and density matrix?

    Operator of number of particles is

    [tex]\hat{N}=\int d^3\vec{r}\hat{\psi}^{\dagger}(\vec{r})\hat{\psi}(\vec{r})[/tex]

    Why I can now use

    [tex]\hat{\psi}^{\dagger}(\vec{r})=\sum_k\hat{a}_k^{\dagger}\varphi^*_k(\vec{r})\qquad \hat{\psi}(\vec{r})=\sum_k\hat{a}_k\varphi_k(\vec{r})[/tex] ?

    where [tex]\{\varphi_k\}[/tex] is complete ortonormal set.

  2. jcsd
  3. Oct 4, 2009 #2


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    Science Advisor

    Which operator? What's [tex]\hat{\psi}^{\dagger}[/tex]? Do you mean [tex]a^\dagger|\psi>[/tex]?

    Anyway, the matrix elements for creation/annhilation operators are:
    [tex]<m|a^\dagger|n> = \sqrt{n+1}\delta_{m,n+1}[/tex]
    [tex]<m|a|n> = \sqrt{n}\delta_{m,n-1}[/tex]

    They don't form a complete set; the creation/annihilation operators aren't self-adjoint.
  4. Oct 4, 2009 #3

    I defined

    \hat{\psi}^{\dagger}(\vec{r})=\sum_k\hat{a}_k^{\dagger}\varphi^*_k(\vec{r})\qquad \hat{\psi}(\vec{r})=\sum_k\hat{a}_k\varphi_k(\vec{ r})

    [tex]\hat{a}_k^{\dagger},\hat{a}_k[/tex] are operators you talking about and [tex]
    [/tex] form complete set. You did not read my post.
  5. Oct 4, 2009 #4
    I usually think of the density matrix as something like
    \hat\rho(r,r') = \hat\psi^\dagger(r) \hat\psi(r')
    so your density operator would be the diagonal element.

    I think your question about the transform relies on the assumption that you can write
    \delta(r) = \sum_k \phi_k(r)
    which certainly works for [tex]\phi_k[/tex] as plane waves but I am not sure about other basis sets.

    It's late and I'm tired so I apologize if this doesn't make much sense.
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