- #1
DeathbyGreen
- 83
- 15
\bf{Setup}
Hi! I am trying to derive the wavefunctions of the zero energy solutions of the Schrödinger 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 \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
$$
\begin{equation}
H\Psi = 0
\end{equation}
$$
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)
$$
\begin{equation}
\epsilon_k = -\mu - t\cos{k} \rightarrow \mu - t\cos{k_F} + t\sin{k_F}(k-k_F)
\\ \Delta = -i\Delta\sin{k} \rightarrow -i\Delta\sin{k_F} -i\Delta\cos{k_F}(k-k_F)
\end{equation}
$$
If I define Fermi velocity as \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
$$
\begin{equation}
\epsilon_k = -\mu - t\sin{k_F}(k-k_F)
\\ \Delta = -i\Delta\sin{k_F}
\end{equation}
$$
I assume \mu is 0 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 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
$$
\begin{equation}
H =
\left[\begin{array}{cc}
v_Fp & i\Delta_{k_F}\\
-i\Delta_{k_F} & -v_Fp
\end{array}
\right] = v_Fp\sigma_z - \Delta_{k_F}\sigma_y
\end{equation}
$$
(\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?
Thank you for your time!
Hi! I am trying to derive the wavefunctions of the zero energy solutions of the Schrödinger 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 \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
$$
\begin{equation}
H\Psi = 0
\end{equation}
$$
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)
$$
\begin{equation}
\epsilon_k = -\mu - t\cos{k} \rightarrow \mu - t\cos{k_F} + t\sin{k_F}(k-k_F)
\\ \Delta = -i\Delta\sin{k} \rightarrow -i\Delta\sin{k_F} -i\Delta\cos{k_F}(k-k_F)
\end{equation}
$$
If I define Fermi velocity as \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
$$
\begin{equation}
\epsilon_k = -\mu - t\sin{k_F}(k-k_F)
\\ \Delta = -i\Delta\sin{k_F}
\end{equation}
$$
I assume \mu is 0 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 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
$$
\begin{equation}
H =
\left[\begin{array}{cc}
v_Fp & i\Delta_{k_F}\\
-i\Delta_{k_F} & -v_Fp
\end{array}
\right] = v_Fp\sigma_z - \Delta_{k_F}\sigma_y
\end{equation}
$$
(\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?
Thank you for your time!