# A Zero Energy Wavefunctions in 1D superconductor

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1. May 22, 2017

### DeathbyGreen

$\bf{Setup}$
Hi! I am trying to derive the wavefunctions of the zero energy solutions of the Schrodinger equation in a 1D p-wave superconductor (Kitaev model). I am starting with the Hamiltonian
$$H = \left[\begin{array}{cc} \epsilon_k & \Delta^{\ast}_k\\ \Delta_k & -\epsilon_k \end{array} \right]$$
where $\epsilon_k = -\mu - t\cos{k}$ ($\mu$ is the chemical potential, $t$ the hopping parameter, and $k$ the crystal momentum) and $\Delta = -i\Delta\sin{k}$ ($\Delta$ is a pairing parameter from the superconductivity). Zero energy modes are thought to exist near the gap closing at the Fermi energy.
$\bf{Solution\quad Attempt}$
We can define
$$H\Psi = 0$$
It was suggested that I assume $\epsilon_k$ is linear and approximate it in terms of Fermi velocity, $v_F$. I had some difficulty determining how to define the Fermi velocity without taking the partial derivative of the full dispersion relation, $E(k) = \sqrt{\epsilon_k^2 + \Delta_k^2}$. So I decided to Taylor expand around the Fermi momentum (the zero modes are thought to occur near this point)
$$\epsilon_k = -\mu - t\cos{k} \rightarrow \mu - t\cos{k_F} + t\sin{k_F}(k-k_F) \\ \Delta = -i\Delta\sin{k} \rightarrow -i\Delta\sin{k_F} -i\Delta\cos{k_F}(k-k_F)$$
If I define Fermi velocity as $\frac{\partial\epsilon_k}{\partial k}$ then $v_F = t\sin{k_F}$, and also that $\frac{dv_F}{dk} = 0$ (my justification being that the "acceleration" at the Fermi level would be 0 since it is the maximum momentum value), these become
$$\epsilon_k = -\mu - t\sin{k_F}(k-k_F) \\ \Delta = -i\Delta\sin{k_F}$$
I assume $\mu$ is $0$ because we are looking at the Fermi level. So
$$\epsilon_k = t\sin{k_F}(k-k_F) \rightarrow v_Fp\\ \Delta = -i\Delta_{k_F}$$
where $p=(k-k_F)$ is the momentum with respect to the Fermi level, which I replace by $p \rightarrow -i\partial_x$ (justification that the momentum near the Fermi level follows the same form everywhere) and $\Delta_{k_F} = \frac{v_F\Delta}{t}$. So my simplified Hamiltonian becomes
$$H = \left[\begin{array}{cc} v_Fp & i\Delta_{k_F}\\ -i\Delta_{k_F} & -v_Fp \end{array} \right] = v_Fp\sigma_z - \Delta_{k_F}\sigma_y$$
($\sigma$'s are the Pauli Matrices) I then multiply from the left by $\sigma_z$, assume a wavefunction form of $\Psi = \chi_{\nu}\phi$, with $\chi_{\nu}$ an eigenvector of $\sigma_x$ (the first term in the simplified Hamiltonian is the identity matrix after multiplying by [/itex]\sigma_z[/itex]) and $\phi \propto e^{\lambda x}$.
$\bf{Question}$
Are my assumptions for the simplifications of the Hamiltonian matrix elements correct? Is my understanding of Fermi velocity and its derivative okay?

2. May 23, 2017

### DrDu

Sounds mostly right to me. However $\mu$ isn't 0 but rather $\epsilon_{k_F}$, at least at T=0. Furthermore, I don't believe that acceleration vanishes at the Fermi energy. This is rather an artifact of you breaking the Fourier sum after the linear term.
Btw. I thought there is no superconductivity in one-dimensional systems?

3. May 23, 2017

### DeathbyGreen

Thank you! The system is theoretical regarding the SC in 1D question. There are attempted experimental realizations of this using a 1D quantum wire placed on an s wave SC to gain a pairing term from the proximity effect. So if I instead have $\epsilon_k \rightarrow \epsilon_{k_F} + v_Fp$, could I drop the $\epsilon_{k_F}$ term?