- #1
arojo
- 16
- 0
Hello everybody,
I am having some trouble to understand the significance behind the Bogoliubov transformation in the case of the BCS theory in the mean field approximation (MFA). Without going into all the details of the calculation the final result is a Bogoliubov transformation like:
(1) $γ_{k, σ}=α_{k}^{*} c_{k, σ}+β_k c_{-k,-σ}^{\dag}$
(2) $γ_{-k, -σ}^{\dag}=-β_{k}^{*} c_{k, σ}+α_k c_{-k,-σ}^{\dag}$
where $γ_{k, σ}$ in the annihilation operator in the new basis, α and β are the Bogoliubov coeffs.
My question is the next, once we get the previous result why can not we get equation (2) from (1) by making the change of variables k→-k, σ→-σ and the applying $^{\dag}$ in both sides?
I know that the vector in (1) and (2) are eigenvectors of the BCS hamiltonian, therefore, linearly independent from each other. But I still do not get why we can not do the transformation I mentioned before.
Thanks
I am having some trouble to understand the significance behind the Bogoliubov transformation in the case of the BCS theory in the mean field approximation (MFA). Without going into all the details of the calculation the final result is a Bogoliubov transformation like:
(1) $γ_{k, σ}=α_{k}^{*} c_{k, σ}+β_k c_{-k,-σ}^{\dag}$
(2) $γ_{-k, -σ}^{\dag}=-β_{k}^{*} c_{k, σ}+α_k c_{-k,-σ}^{\dag}$
where $γ_{k, σ}$ in the annihilation operator in the new basis, α and β are the Bogoliubov coeffs.
My question is the next, once we get the previous result why can not we get equation (2) from (1) by making the change of variables k→-k, σ→-σ and the applying $^{\dag}$ in both sides?
I know that the vector in (1) and (2) are eigenvectors of the BCS hamiltonian, therefore, linearly independent from each other. But I still do not get why we can not do the transformation I mentioned before.
Thanks