# A Anderson model with no hybridization

1. Nov 4, 2017

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

$$H_{0} + \varepsilon_{0}d^{\dagger}d + \sum_{k} V_{k}c_{k}^{\dagger}d + \text{h.c}$$

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

$$\sum_{k} V_{k}c_{k}^{\dagger}d + \text{h.c}$$

is now changed to the term:

$$\sum_{k} V_{k}c_{k}^{\dagger}d^{\dagger} + \text{h.c}$$

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. Nov 6, 2017

### DrDu

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