Given the BdG Hamiltonian of a 1D p-wave superconductor we can obtain the zero-energy excitation solution as Eq. 16.24 from Topological Insulators and Topological Superconductors (Bernevig & Hughes):

$|\Psi(x)\rangle = \frac{1}{\sqrt{2}}exp\left(-\frac{1}{|\Delta|}\int_{0}^{x}\mu(x')dx'\right)\left(\begin{array}{c} 1\\ -i \end{array}\right)$

associated with this wave function one can define a field operator that destroys this zero-mode $\gamma_{0}$ such as $\gamma_{0}^{\dagger} = \gamma_{0}$ and $\gamma_{0}^{2}=1$ so it's a Majorana 'fermion' and it's occupation number is not well defined. My question is: how can I interpret $|\Psi(x)|^2$ as the probability to have a Majorana in position $x$ if it's occupation is not well defined? Does it make sense?

In general we can construct a regular fermion from two Majoranas by $f=\frac{1}{\sqrt{2}}(\gamma_{A}+i\gamma_{B})$, this way I can define $\hat{n}=f^{\dagger}f$ and talk about occupying this state. In a finite p-wave wire we obtain two zero-energy solutions. Doesn't it make more sense to construct a regular fermion wave function from the Majorana pair, let say perhaps $|\Psi_{pair}|^2 = |\Psi_{A} + \Psi_{B}|^2$ ?


1 Answer 1


For the first part, although it does not quite make sense to ask "the probability to have a Majorana in position x" (there are no "Majorana" in the system; the system only has electrons), the wavefunction $|u|^2$ ($|v|^2$) does have a physical meaning as the weight for a single electron (hole) excitation, see the answers in a closely related question:

Can one define wavefunction for Bogoliubov quasiparticle excitation in a superconductor?

Of course, in this case $|u|^2=|v|^2$ because it is a Majorana zero mode.

For your second question, if the zero modes are exactly degenerate, any combination is an eigenstate. However, the wire has a finite length, so the energy of zero modes are generally splitted away from zero by a small amount, and the correct combination to approximate the splitted zero modes (in the spirit of WKB approximation) should be determined by particle-hole symmetry of the BdG Hamiltonian. See http://arxiv.org/pdf/0905.0035v3.pdf for more details.

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    $\begingroup$ A small mistake: The wave function is clearly the one of a Bogoliubov quasi-particle, not an electron (excitation or not) :). To the OP: It's kind of an accident that some Bogoliubov quasi-particles have unpaired Majorana statistics. Since the observables are invariant under unitary transformation, you can define the combination you want in principle. If it's an observable, you can observe it in an experiment, by definition. Only observables matter of course in quantum mechanics. $\endgroup$
    – FraSchelle
    Oct 31, 2015 at 11:43
  • $\begingroup$ Sure, the wavefunction is of a Bogoliubov quasiparticle, but what I mean is that the "electron" component of this guy is also given by $|\Psi(x)|^2$, as well as the hole component, because it is Majorana. $\endgroup$
    – Meng Cheng
    Oct 31, 2015 at 16:10
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    $\begingroup$ For the linear combination, you can form arbitrary linear combination as you want, but generally they will not correspond to an energy eigenstate of the Hamiltonian. If we want to find an approximate eigenstate of the BdG Hamiltonian using the two Majorana edge states, in the spirit of WKB approximation, one has to respect the particle-hole symmetry of the BdG Hamiltonian, which fixes the relative phase. $\endgroup$
    – Meng Cheng
    Oct 31, 2015 at 16:12
  • $\begingroup$ First of all thank you both. But things are getting a bit confusing for me. First: why $|\Psi(x)|^2$ is the electron as well as the hole component if its given by $|\Psi(x)|^2= |u|^{2} + |v|^{2}$? Second: the zero-mode solutions are degenerate eigenstates, so in principle any linear combination of them will also be an eigenstate (if they are orthonormal) with zero energy, right? $\endgroup$ Nov 1, 2015 at 1:10
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    $\begingroup$ You should, definitely. But it is also important to keep in mind that this fermionic excitation has some unusual feature, i.e. a superposition of two localized states (which can be far from each other). And in many situations, one is really looking at one of them, and that's perfectly fine. $\endgroup$
    – Meng Cheng
    Nov 2, 2015 at 23:59

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