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A Anderson model with no hybridization

  1. Nov 4, 2017 #1
    Hi guys, I wass recently thinking about the Anderson model in its non-interacting limit U=0. I ommit spin in the following and then the hamiltonian is

    [tex]H_{0} + \varepsilon_{0}d^{\dagger}d + \sum_{k} V_{k}c_{k}^{\dagger}d + \text{h.c} [/tex]

    It is well know that due to the hybridization term V, a resonance of width ~ V^2 happens in the density of states of the impurity site. But suppose this term

    [tex] \sum_{k} V_{k}c_{k}^{\dagger}d + \text{h.c} [/tex]

    is now changed to the term:

    [tex] \sum_{k} V_{k}c_{k}^{\dagger}d^{\dagger} + \text{h.c} [/tex]

    i.e., particle number is NOT conserved in this modified Anderson hamiltonian at U=0. Would we still observe a resonance peak at the impurity density of states? The physics of such term are a bit unclear to me since it does not represent an hybridization term properly speaking, Thanks!!
  2. jcsd
  3. Nov 6, 2017 #2


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    Science Advisor

    Perhaps you could try a Bogolyubov transformation to diagonalise the hamiltonian.
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