Anderson model with no hybridization

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  • Thread starter gonadas91
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Main Question or Discussion Point

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!!
 

Answers and Replies

  • #2
DrDu
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Perhaps you could try a Bogolyubov transformation to diagonalise the hamiltonian.
 

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