I would like to understand how we can find the density of states in $k$ space for the BCS Hamiltonian.
First, let's talk about free electrons. When we deal with free electrons, the Hamiltonian is $\frac{p^2}{2m}$ and it is more simple to write it in $k$ space: $$H=\frac{h^2k^2}{2m}.$$
An eigenvector is then $\psi_k=\delta(k-k_0)$ and we know how many modes we have in an interval $dk$ because of Fourier series. If we have a function confined in a box of size $L$, we can periodise the function with period $L$ on all the space, and two Fourier modes will be separated by $\frac{2 \pi}{L}$.
So, we have $\rho(k)dk= \frac{Ldk}{2 \pi}$ modes in $dk$.
To summarize: the $k$ is really a Fourier mode here. And because of Fourier theory I know how many modes I have in $dk$.
Now, lets talk about the BCS Hamiltonian,
$$ H=\sum_{k,\sigma} E_k \hat{a}^{\dagger}_{k, \sigma}\hat{a}_{k, \sigma} -|g_{eff}|^2 \sum_{k_1, k_2, \sigma_1, \sigma_2}\hat{a}^{\dagger}_{k1+q, \sigma_1}\hat{a}^{\dagger}_{k2-q, \sigma_2}\hat{a}_{k2, \sigma_2}\hat{a}_{k1, \sigma_1}$$
and after diagonalisation, we have
$$ H= \sum_k E_k \gamma^{\dagger}_{k+}\gamma_{k+} -E_k \gamma^{\dagger}_{k-}\gamma_{k-}.$$
I note $$|\phi^+_k\rangle=\gamma^{\dagger}_{k+}|0\rangle_{BCS}.$$
$k$ is now just a quantum number used to "note" an eigenvector, it is not a Fourier mode. How can one determine the density of states in $k$ space now?
No comments:
Post a Comment