# Density of states in $k$ space for the BCS Hamiltonian

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?

• It is a Fourier mode, because $k$ labels plane wave states. – Thomas Mar 5 '17 at 15:47
• but the eigenstate $|\phi_k^+\rangle$ is not a plane wave. – StarBucK Mar 5 '17 at 15:48
• Well you can consider more complicated cases where $a_k$ does not destroy plane wave states, but the standard text book problem is that the reference state is just a Fermi sphere of plane wave states. – Thomas Mar 5 '17 at 15:51
• In fact, the number of $|\phi_k^+\rangle$ in $dk$ will be the same as the number of plane waves in $dk$ as in both case we use the same quantum number to differentiate them ? – StarBucK Mar 5 '17 at 16:49
• This is indeed the case: For any set of Bloch waves labeled by $k$, the same counting in $k$-space applies. For the density of states in energy space, you need to know the dispersion relation and the geometry of the Fermi surface. – Thomas Mar 6 '17 at 3:44

So let us introduce the quantum field theory approach to calculate the density of state systematically, which applies to both normal metals and BCS superconductors. The key observation is that the electron band (dispersion relation) is not well defined in BCS superconductors anymore, it must be replaced by the concept of electron spectral weight in the momentum-frequency space. The spectral weight (or the spectral function) $A(\omega,\boldsymbol{k})$ describes the probability density of observing an electron mode at frequency $\omega$ and momentum $\boldsymbol{k}$. Then the density of state (DOS) is just the integral of the spectral weight over the momentum space or over the Brillouin zone (BZ) $$\text{DOS}(\omega)=\int_\text{BZ} A(\omega,\boldsymbol{k})\mathrm{d}^d\boldsymbol{k}.$$ The spectral function can be calculated as the imaginary part of the electron Green's function $A(\omega,\boldsymbol{k})=-2\text{Im}G(\omega+\text{i}0_+,\boldsymbol{k})$, where the Green's function is defined as $G(t-t',\boldsymbol{x}-\boldsymbol{x}')=-\langle c(t,\boldsymbol{x})c^\dagger(t',\boldsymbol{x}')\rangle$ in the space-time, and given by $G(\omega,\boldsymbol{k})=(\omega-h(\boldsymbol{k}))^{-1}$ after Fourier transforming to the momentum-frequency space.
Now we apply this approach to the BCS superconductor described by the following Hamiltonian $$H_\text{BCS}=\sum_{\boldsymbol{k}}c_{\boldsymbol{k}\sigma}^\dagger \epsilon_{\boldsymbol{k}} c_{\boldsymbol{k}\sigma}+\Delta(c_{\boldsymbol{k}\uparrow}c_{-\boldsymbol{k}\downarrow}+\text{h.c.}),$$ with $\epsilon_{\boldsymbol{k}}=\boldsymbol{k}^2/(2m)-\mu$ and $\Delta\in\mathbb{R}$ for example. In the Nambu basis $\psi_{\boldsymbol{k}}=(c_{\boldsymbol{k}\uparrow},c_{-\boldsymbol{k}\downarrow}^\dagger)$, we have $$\begin{split}H_\text{BCS}&=\sum_{\boldsymbol{k}}\psi_{\boldsymbol{k}}^\dagger h(\boldsymbol{k})\psi_{\boldsymbol{k}},\\ h(\boldsymbol{k})&=\epsilon_{\boldsymbol{k}}\sigma^3+\Delta\sigma^1,\end{split}$$ where $\sigma^3$ and $\sigma^1$ denote the Pauli matrices acting on the Nambu spinor $\psi_{\boldsymbol{k}}$. The Green's function of the Nambu spinor is given by $$G(\omega,\boldsymbol{k})=(\omega-h(\boldsymbol{k}))^{-1}=\frac{\omega\sigma^0+\epsilon_{\boldsymbol{k}}\sigma^3+\Delta\sigma^1}{\omega^2-\epsilon_{\boldsymbol{k}}^2-\Delta^2}.$$ Note that this $G(\omega,\boldsymbol{k})$ is a $2\times 2$ matrix at each momentum-frequency point. The (1,1)-component $G(\omega,\boldsymbol{k})_{11}$ corrsponds to the correlation $-\langle c_{\boldsymbol{k}\uparrow}c_{\boldsymbol{k}\uparrow}^\dagger\rangle$ and is therefore the Green's function of the up-spin electron. We expect the Green's function of the down-spin electron to be the same as the up-spin electron due to the time-reversal symmetry. With this, we can compute the electron spectral function $$A(\omega,\boldsymbol{k})_{11}=-2\text{Im}G(\omega+\text{i}0_+,\boldsymbol{k})_{11}.$$ The result is shown in the following figure. Darker color means larger spectral weight and higher probability of observing the electron at this momentum-frequency point.
We can observe the BCS gap of $2\Delta$ across the Fermi energy ($\omega=0$). Two superconducting coherence peaks appear on the gap edges, as electrons are depleted in the gap and piled up to the coherence peaks. In this example, the gap has a U-shape because the Fermi surface is fully gapped. In $d$-wave superconductors like cuprates where the spectrum is nodal, there will be a V-shape gap instead in the density of state. This is what can be seen in the STM $\text{d}I/\text{d}V$ (differential conductance) spectrum.