Operator which is written in k space

[aaaa202]

I have an operator which is written in k space as something like:

H = Ʃ_kc_ka_k
where a_k and c_k are operators. So it is a sum of operators of different k but there are no crossterms as you can see. Being no crossterms does this mean that the operator is diagonalized in the language of linear algebra. That a matrix is diagonalized means that it is represented in a basis of eigenstates but this is an expansion of an operator so I am not sure.
Edit; any operator is of course an expansion in operators related to the basis of eigenstates {e_k} by the equation:
Ʃ_i,jle_i><e_jl a_ij
Now to be diagonalized would mean the vanishing of all matrix elements where i≠j. Can this be said about the given Hamiltonian above. How do I know how the c_k and a_k are related to my eigenbasis.

 

[Mark44]

I have an operator which is written in k space as something like:

H = Ʃ_kc_ka_k
where a_k and c_k are operators. So it is a sum of operators of different k but there are no crossterms as you can see. Being no crossterms does this mean that the operator is diagonalized in the language of linear algebra. That a matrix is diagonalized means that it is represented in a basis of eigenstates but this is an expansion of an operator so I am not sure.

What's the actual question? Your sum looks like the product of two diagonal matrices C and A, where c_i is an entry on the diagonal of matrix C and a_i is an entry on the diagonal of matrix A. As it seems to me, C and A are the operators, but the entries in the matrices aren't.

Without knowing what the entries in the matrices are, it's hard to say what H is. Otherwise, your sum looks like some inner product to me. More information would be helpful.

 

[aaaa202]

The a_k and c_k are operators as stated? So its a sum of operators. The a_k is the creation operator for the momentum state lk> and c_k is the anihillation operator for the same state.

 

[aaaa202]

Wait I know now. It is of course because the product a_k c_k is just the number operator.

 

[Mark44]

This thread is probably more appropriate in the Adv. Physics section, so I'm moving it.