Recent content by hmdkdl

## |n_c,n_d> ## is the fock state with ## n_c (n_d) ## particle in state c (d), which can be 0 or 1. I think my question has a very simple the answer: if we consider multiparticle states we have three eigenvalues, but if we are looking for state of one paticle, we have just two. Thanks a lot...

Your method is more easy and more straightforward. Thanks. Now, is it true that we have three eigenvalues: \epsilon = \pm |\phi|,0 ? with eigenstates: |1,0> , |0,1> and a linear combiantion of |0,0> and |1,1> That's true. But why shouldn't I? In other words, when can I mix them? The...

It is from a paper about graphene ( PRL 101, 026805 (2008) ). This hamiltonian with sum over wave vector k and \phi = \phi(k) , is the hamiltonian of graphene in the tight bonding approximation, and I think it is a very standard form of hamiltonian for graphene. if \phi = |\phi| e^{i...

Yes, you're right. {a,b}=0 gives nothing but {a,b*}=0 gives two other equations that can be used to find the final answer. By the way, my final answer is: H = \pm |\phi| (c^{\dagger}c + d^{\dagger}d - 1 ) This hamiltonian has three eigenvalues: \epsilon=\pm |\phi|, 0 However, my...

My try I supposed that the transformations are like this (stars on u,v,w,z means complex conjugate and on a,b,c,d means dagger): a = uc + vd* --> a*=u*a + v*d b = wc + zd* --> b*=w*c + z*d we have fermions, so: {a,a*}=1 --> |u|^2 + |v|^2=1 {b,b*}=1 --> |w|^2 + |z|^2=1 Now the...

Homework Statement Hi, I want to diagonalize the Hamiltonian: Homework Equations H=\phi a^{\dagger}b + \phi^{*} b^{\dagger}a a and b are fermionic annihilation operators and \phi is some complex number. The Attempt at a Solution Should I use bogoliubov tranformations? I...