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A Zero Energy Wavefunctions in 1D superconductor

  1. May 22, 2017 #1
    [itex]\bf{Setup}[/itex]
    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!
     
  2. jcsd
  3. May 23, 2017 #2

    DrDu

    User Avatar
    Science Advisor

    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?
     
  4. May 23, 2017 #3
    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 [itex] \epsilon_k \rightarrow \epsilon_{k_F} + v_Fp[/itex], could I drop the [itex] \epsilon_{k_F} [/itex] term?
     
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