[itex]\bf{Setup}[/itex](adsbygoogle = window.adsbygoogle || []).push({});

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

$$

\begin{equation}

H =

\left[\begin{array}{cc}

\epsilon_k & \Delta^{\ast}_k\\

\Delta_k & -\epsilon_k

\end{array}

\right]

\end{equation}

$$

where [itex]\epsilon_k = -\mu - t\cos{k}[/itex] ([itex]\mu[/itex] is the chemical potential, [itex]t[/itex] the hopping parameter, and [itex]k[/itex] the crystal momentum) and [itex]\Delta = -i\Delta\sin{k}[/itex] ([itex]\Delta[/itex] is a pairing parameter from the superconductivity). Zero energy modes are thought to exist near the gap closing at the Fermi energy.

[itex]\bf{Solution\quad Attempt}[/itex]

We can define

$$

\begin{equation}

H\Psi = 0

\end{equation}

$$

It was suggested that I assume [itex]\epsilon_k[/itex] is linear and approximate it in terms of Fermi velocity, [itex]v_F[/itex]. I had some difficulty determining how to define the Fermi velocity without taking the partial derivative of the full dispersion relation, [itex]E(k) = \sqrt{\epsilon_k^2 + \Delta_k^2}[/itex]. So I decided to Taylor expand around the Fermi momentum (the zero modes are thought to occur near this point)

$$

\begin{equation}

\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)

\end{equation}

$$

If I define Fermi velocity as [itex]\frac{\partial\epsilon_k}{\partial k}[/itex] then [itex]v_F = t\sin{k_F}[/itex], and also that [itex]\frac{dv_F}{dk} = 0[/itex] (my justification being that the "acceleration" at the Fermi level would be 0 since it is the maximum momentum value), these become

$$

\begin{equation}

\epsilon_k = -\mu - t\sin{k_F}(k-k_F)

\\ \Delta = -i\Delta\sin{k_F}

\end{equation}

$$

I assume [itex]\mu[/itex] is [itex]0[/itex] because we are looking at the Fermi level. So

$$

\begin{equation}

\epsilon_k = t\sin{k_F}(k-k_F) \rightarrow v_Fp\\

\Delta = -i\Delta_{k_F}

\end{equation}

$$

where [itex]p=(k-k_F)[/itex] is the momentum with respect to the Fermi level, which I replace by [itex]p \rightarrow -i\partial_x[/itex] (justification that the momentum near the Fermi level follows the same form everywhere) and [itex]\Delta_{k_F} = \frac{v_F\Delta}{t}[/itex]. So my simplified Hamiltonian becomes

$$

\begin{equation}

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

\end{equation}

$$

([itex]\sigma[/itex]'s are the Pauli Matrices) I then multiply from the left by [itex]\sigma_z[/itex], assume a wavefunction form of [itex]\Psi = \chi_{\nu}\phi[/itex], with [itex]\chi_{\nu}[/itex] an eigenvector of [itex]\sigma_x[/itex] (the first term in the simplified Hamiltonian is the identity matrix after multiplying by [/itex]\sigma_z[/itex]) and [itex]\phi \propto e^{\lambda x}[/itex].

[itex]\bf{Question}[/itex]

Are my assumptions for the simplifications of the Hamiltonian matrix elements correct? Is my understanding of Fermi velocity and its derivative okay?

Thank you for your time!

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# A Zero Energy Wavefunctions in 1D superconductor

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