Wavefunction is essentially a single particle concept. It is easily extended to multiparticle system as follows- if one has say five electrons the wavefunction of this five electron state is any completely antisymmetric function of five coordinates which is square integrable in the five dimensional space. Given a five electron ket in fock space $|K\rangle$, its wavefunction is denoted as $\langle x_1 x_2 ...x_5|K\rangle$. But for a superconductor its effective Hamiltonian doesn't conserve the particle number. Then can one come up with any reasonable defination of a wavefunction for a single quasiparticle excitation of the superconductor over its ground state denoted by $\gamma_i^{\dagger} |G\rangle$ where $|G\rangle$ is superconducting groundstate composed of Cooper pairs and $\gamma_i^{\dagger}=\sum_k u_i^kc_k+v_i^kc_k^{\dagger}$ is Bogoliubov quasiparticle creation operator and $c's$ being electron operators and $u's$ and $v's$ being some complex numbers.

In Kitaev chain and it's solid state realisation one usually talks about Majorana fermion(Bogoliubov excitation) being localised at the two ends. How can one do that without a reasonable definition of wavefunction for superconducting states? The papers usually interpret eigenvectors of $H_{BdG}$ in coordinate space as representative wavefunctions. Is it justified?


2 Answers 2


The localization of Majorana zero modes has a well-defined meaning: consider a Kitaev chain with two ends. Because of the zero modes, there are two nearly degenerate ground states, let us call them $|0\rangle$ and $|1\rangle$, which have opposite fermion parities. They are localized as "single-particle wavefunctions" in the following sense: if one computes the matrix element $\langle 1|c^\dagger(x)|0\rangle$ where $c^\dagger(x)$ is the creation operator for fermions, the result is an exponentially decaying function of $x$ away from the edge. This definition works even when the system is interacting. Intuitively it means that the weight for creating a single fermion excitation is localized near the edge, and in the bulk there is a finite gap to the single particle excitations.

  • $\begingroup$ Yes what you have described is what papers call wavefunction and I have issue with that. The matrix element you wrote is the tunneling amplitude into the state at site x at zero energy- not the amplitude of finding a set of particles at site x in a closed system. It seems this nomenclature has become standard maybe partly because it's hard to misunderstand its meaning which is the way you and most papers define. $\endgroup$
    – Noob Rev B
    Apr 27, 2015 at 7:43
  • $\begingroup$ This is why it is called "single-particle wavefunction". Of course this is a many-body state, but since the superconductor is essentially non-interacting (at least at the level the mean-field Hamiltonian) if you want you can build any many-particle state by successively creating single-particle states $\endgroup$
    – Meng Cheng
    Apr 27, 2015 at 15:27

In addition to what Meng Cheng said, remark that the mean-field Bogoliubov-Gennes Hamiltonian, say

$$H=\sum_{k}\left(c_{k}^{\dagger}H_{0}c_{k}+\Delta_{k}c_{k}c_{-k}+\Delta_{k}^{\ast}c_{-k}^{\dagger}c_{k}^{\dagger}\right)$$ can be diagonalised by the Bogoliubov transformation you mentioned :


which simply signifies that $$H=\sum_{q}\epsilon_{q}\gamma_{q}^{\dagger}\gamma_{q}$$ (all details in the book by P.G. de Gennes: Superconductivity of metals and alloys.

We interpret the above expression as the total energy of the system, provided we identify $\epsilon_{q}$ as the energy of the mode $q$ which is created by the operator $\gamma_{q}^{\dagger}$. Then, provided you know the vacuum, say $\left|\emptyset\right\rangle $, what forbids you to define the wave function of the mode $q$ as $\Psi_{q}\left(x\right)\sim\left\langle x\right|\gamma_{q}^{\dagger}\left|\emptyset\right\rangle $ ? It has all the properties of a wave-function (normalisation, significance of $\left|\Psi_{q}\left(x\right)\right|^{2}$, ...), and in fact it is the wave-function of an excitation quasi-particle above the superconducting ground-state.

For Majorana modes, there is (at least) one wave-function which decays exponentially: $\Psi_{0}\left(x\right)\sim e^{-qx}$ and thus it has to be located at the edge with an other system.

From the definition of the $\gamma_{q}^{\dagger}$ above, you can see that choosing convenient form of the functions $u_{q}^{k}$ and $v_{q}^{k}$ can lead to the condition $\gamma_{q}^{\dagger}=\gamma_{q}$, which is the definition of a Majorana fermion. In superconductors one should never forget that the operator $c_{k}^{\dagger}$ are not really electronic ones: they represent the creation of an excitation in a normal metal / semi-conductor, ... So we should avoid discussing Majorana fermion, and keep the more accurate Majorana modes.

  • $\begingroup$ If you want to see a simple calculation of the Majorana wave-function in $p$-wave superconducting wire, you might be interested in the Appendix of arxiv.org/abs/1306.2519 $\endgroup$
    – FraSchelle
    Feb 18, 2015 at 10:35

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