I put an extra answer, since I believe the first Jeremy's question is still unanswered. The previous answer is clear, pedagogical and correct. The discussion is really interesting, too. Thanks to Nanophys and Heidar for this.

To answer directly Jeremy's question: you can ALWAYS construct a representation of your favorite fermions modes in term of Majorana's modes. I'm using the convention "modes" since I'm a condensed matter physicist. I never work with particles, only with quasi-particles. Perhaps better to talk about mode.

So the unitary transformation from fermion modes created by $c^{\dagger}$ and destroyed by the operator $c$ to Majorana modes is
$$
c=\dfrac{\gamma_{1}+\mathbf{i}\gamma_{2}}{\sqrt{2}}\;\text{and}\;c{}^{\dagger}=\dfrac{\gamma_{1}-\mathbf{i}\gamma_{2}}{\sqrt{2}}
$$
or equivalently
$$
\gamma_{1}=\dfrac{c+c{}^{\dagger}}{\sqrt{2}}\;\text{and}\;\gamma_{2}=\dfrac{c-c{}^{\dagger}}{\mathbf{i}\sqrt{2}}
$$
and this transformation is always allowed, being unitary. Having doing
this, you just changed the basis of your Hamiltonian. The quasi-particles
associated with the $\gamma_{i}$'s modes verify $\gamma{}_{i}^{\dagger}=\gamma_{i}$,
a fermionic anticommutation relation $\left\{ \gamma_{i},\gamma_{j}\right\} =\delta_{ij}$,
but they are not particle at all. A simple way to see this is to try
to construct a number operator with them (if we can not count the
particles, are they particles ? I guess no.). We would guess $\gamma{}^{\dagger}\gamma$
is a good one. This is not true, since $\gamma{}^{\dagger}\gamma=\gamma^{2}=1$
is always $1$... The only correct number operator is $c{}^{\dagger}c=\left(1-\mathbf{i}\gamma_{1}\gamma_{2}\right)$.
To verify that the Majorana modes are anyons, you should braid them
(know their exchange statistic) -- I do not want to say much about
that, Heidar made all the interesting remarks about this point. I
will come back later to the fact that there are always $2$ Majorana
modes associated to $1$ fermionic ($c{}^{\dagger}c$) one. Most has
been already said by Nanophys, except an important point I will discuss
later, when discussing the delocalization of the Majorana mode. I
would like to finnish this paragraph saying that the Majorana construction
is no more than the usual construction for boson: $x=\left(a+a{}^{\dagger}\right)/\sqrt{2}$
and $p=\left(a-a{}^{\dagger}\right)/\mathbf{i}\sqrt{2}$: only $x^{2}+p^{2} \propto a^{\dagger} a$
(with proper dimension constants) is an excitation number. Majorana
modes share a lot of properties with the $p$ and $x$ representation
of quantum mechanics (simplectic structure among other).

The next question is the following: are there some situations when
the $\gamma_{1}$ and $\gamma_{2}$ are the natural excitations of
the system ? Well, the answer is complicated, both yes and no.

- Yes, because Majorana operators describe the correct excitations of
some topological condensed matter realisation, like the $p$-wave
superconductivity (among a lot of others, but let me concentrate on this specific one, that I know better).
- No, because these modes are not excitation at all ! They are zero
energy modes, which is not the definition of an excitation. Indeed,
they describe the different possible vacuum realisations of an emergent
vacuum (emergent in the sense that superconductivity is not a natural
situation, it's a condensate of interacting electrons (say)).

As pointed out in the discussion associated to the previous answer,
the normal terminology for these pseudo-excitations are zero-energy-mode.
That's what their are: energy mode at zero-energy, in the middle of
the (superconducting) gap. Note also that in condensed matter, the
gap provides the entire protection of the Majorana-mode, there is
no other protection in a sense. Some people believe there is a kind
of delocalization of the Majorana, which is true (I will come to that
in a moment). But the delocalization comes along with the gap in fact:
there is not allowed propagation below the gap energy. So the Majorana
mode are necessarilly localized because they lie at zero energy, in
the middle of the gap.

More words about the delocalization now -- as I promised. Because
one needs two Majorana modes $\gamma_{1}$ and $\gamma_{2}$ to each
regular fermionic $c{}^{\dagger}c$ one, any two associated Majorana
modes combine to create a regular fermion. So the most important challenge
is to find **delocalized** Majorana modes ! That's the famous
Kitaev proposal arXiv:cond-mat/0010440 -- he said unpaired Majorana instead of delocalised, since delocalization comes for free once again. At the
end of a topological wire (for me, a $p$-wave superconducting wire)
there will be two zero-energy modes, exponentially decaying in space
since they lie at the middle of the gap. These zero-energy modes can
be written as $\gamma_{1}$ and $\gamma_{2}$ and they verify $\gamma{}_{i}^{\dagger}=\gamma_{i}$
each !

To conclude, an actual vivid question, still open: there are a lot
of pseudo-excitations at zero-energy (in the middle of the gap). The
only difference between Majorana modes and the other pseudo-excitations
is the definition of the Majorana $\gamma^{\dagger}=\gamma$, the
other ones are regular fermions. How to detect for sure the Majorana
pseudo-excitation (zero-energy mode) in the jungle of the other ones
?

This post imported from StackExchange Physics at 2014-04-04 16:38 (UCT), posted by SE-user FraSchelle