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14

Majorana fermions are fermions which are their own antiparticles. As a result, they only have half the degrees of freedom as a regular Dirac electron. One physical interpretation, at least for Majorana fermion quasiparticles in condensed matter systems, is that they can be thought of a superposition of an electron and hole state. Only Majorana bound states ...


13

Well, there is certainly not a Majorana representation, since any decomposition will have two terms which differ by a phase of -1, you can't find Majorana points. The singlet state is anti-symmetric, so there is no way as writing it in the form $\frac{e^{i\alpha}}{\sqrt{2}} \sum_{j=0}^1 | \phi_{1\oplus j} \rangle \otimes | \phi_{0\oplus j} \rangle$ since it ...


11

It seems to me that the stabilizer formalism provides an answer to your question (see section 3.1 of quant-ph/0603226 for an introduction to the formalism). Given the two states, you simply take the stabilizer group for that 2-dimensional subspace of the total Hilbert space, and they will give you a such a set of invariants. However, this of course has ...


10

This is an algorithm for the computation of the homogeneous polynomial invariants in the general case, however without any attempt to reduce the algorithm's complexity. The basic needed ingredient of the algorithm is the ability to average over the Haar measure of the group of the local transformations. In the qubit case it is just an integration over copies ...


9

Symmetry of the superconducting gap First of all, a bit of theory. Superconductivity appears due to the Cooper paring of two electrons, making non-trivial correlations between them in space. The correlation is widely known as the gap parameter $\Delta_{\alpha\beta}\left(\mathbf{k}\right)\propto\left\langle ...


9

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 ...


8

Majorana fermions as defined in http://en.wikipedia.org/wiki/Majorana_fermion are really fermions, as its name indicates. So Majorana fermion really have Fermi statistics. It is not proper to say Majorana fermions obey non-abelian statistics, since fermion always obey Fermi statistics by definition. The thing that people said to have non-Abelian statistics ...


7

It seems from Piotr's comments on my other answer, that he is looking for an invariant of the mathematical representation of the state, rather than an observable that remains unchanged. In this case the answer is very different, and hence I am posting a new answer, rather than replacing the old one (since it would mean entirely rewriting it, and the current ...


7

We know that we can describe a spin $1/2$ massless particle using only a single Weyl field (lets say left-handed $\psi_{L}$). To introduce a mass term we have to use two spinor fields (one left-handed and one right-handed) and this gives the Dirac mass term. The question is now that if we can describe a massive particle with a single Weyl field. Well yes, ...


6

There are many schemes to make topological superconductors. Some of these schemes have restrictions on the chemical potential $\mu$. You also need to know what type of topological superconductors you are dealing with. You can refer to the periodic table to determine this: In the paper from the link you provided the authors mention two types of ...


6

Majorana fermions are by definition fermions which are their own antiparticles, i.e. the do have spin and it's 1/2. An introduction to these fermions can be for example found here: http://arxiv.org/abs/0806.1690. In contrast bosons are their own antiparticles, e.g. photons, i.e. one does not need a "Majorana-boson" definition. Now, one has to say that these ...


6

In the general - the answer is no. Majorana representation's key point is to express a composite state of $n$ qubits as $n$ points is such way, that action of a collective rotation (i.e. $|\psi \rangle \mapsto U^{\otimes n} |\psi \rangle$ for $U\in \text{SU}(2)$) rotates each points in the same way (i.e. $| \eta_k\rangle \mapsto U | \eta_k\rangle $). In ...


5

This answer is incomplete, but it should provide an answer for almost all symmetric states (i.e. it suffices for all but a set of symmetric states having measure zero). The symmetric subspace is spanned by product states. We may then consider different ways in which a particular symmetric state decomposes into symmetric products; in particular, if any ...


3

Non-conservation of charge in Majorana terms The Dirac mass term is $m\bar\psi \psi$ where one field-factor $\bar\psi$ is complex conjugated (aside from other transpositions included in the Dirac conjugation) and the other is not. So one may assign a fermion number $1$ to $\psi$ which means that $\bar\psi$ automatically carries $-1$ and in the product, the ...


3

For a moment let's assume you subscribe to the Hamiltonian used by Fu-Kane to describe the TI surface with an induced superconducting proximity effect, \begin{equation} \mathcal{H}(\mathbf{k})=\left( \begin{array}{cc} H(\mathbf{k}) & i\sigma_y \Delta \\ -i\sigma_y \Delta^* & - H^*(-\mathbf{k}) \end{array} \right)=v_F(\sigma_x k_y-\tau_z\sigma_y ...


3

A Majorana fermion would be a fermion that is it's own anti-particle. This is a common trait for bosons - the photon, gluon, and Z bosons are all their own anti-particles. Obviously, a Majorana fermion must not interact through the electromagnetic force, that is, it must have zero charge. More technically, the wave equation that governs Majorana particles ...


2

Majorana fermions obey non-abelian statistics and it will be anyonic if your Majorana mode is confined to two dimensions. In $3\text{D}$, you still have the possibility of non-abelian statistics but it is no longer anyonic as the braid group is trivial. Here are some useful references: Majorana Fermions and Non-Abelian Statistics in Three Dimensions, J. ...


2

Majorana and Dirac equations are usually considered as two different and mutually exclusive equations. However, both of them can be considered as a special cases of the more general equation. Let's start with Dirac equation written in terms of the "left" ($\xi$) and "right" ($\dot\eta$) spinor components: \begin{equation} \begin{array}{c} ...


2

In this context, the right-handed neutrino is a singlet under the Standard Model gauge groups. Only the right-handed neutrino is allowed a Majorana mass. The left-handed term is not gauge invariant. If the SM and right-handed neutrino fields were embedded into a Grand Unified gauge group, the right-hand term would break that symmetry. It is expected ...


2

Yes, they can via a dimension 5 operator that contains two Higgs doublets and two lepton doublets. This is sometimes called Weinberg operator, it violates lepton number conservation and it was introduced in this work http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.43.1566 . Since the mass comes from an higher dimensional operator (i.e. an ...


2

Neutrino mass is not in conflict with electroweak theory. One can introduce neutrino masses by modifying the Higgs or lepton sector of the Standard Model. The simplest method, that which you propose, is introducing a right handed neutrino with a Yukawa coupling with the left handed neutrino (extending the lepton sector). The right handed neutrino, however, ...


2

Because of the Majorana condition $\psi=\psi^C$, Majorana fermions are singlets with respect to gauge symmetries, including $SU(2)$. No chiral symmetry forbids a Majorana mass. The natural mass scale for Majorana fermions is the Planck mass. The right-handed neutrino could be Majorana, with interesting implications. See the see-saw mechanism; the ...


1

The transformation rule for $\hat{C}$ or that of $\hat{C}\gamma_{0}^{T}$ is different from the usual one. The whole charge conjugation operation is given by $VK$, where $V\equiv\hat{C}\gamma_{0}^{T}$ is unitary and $K$ is complex conjugation. (i.e., $VK$ is antiunitary.) Then, under a basis transformation, $$ VK \ \rightarrow \ U^{\dagger}VK U = ...


1

We have no reason to expect, in general, that mass eigenstates and interaction eigenstates should be the same. In your example, the Lagrangian written the mass-diagonal basis $\Psi_L,\Psi_R$ will contain additional mixing terms: $$ \bar\Psi_R \hat A_{RL} \Psi_L + \bar\Psi_L \hat A_{LR} \Psi_R $$ If $\hat A$ is a gauge field, physically, these terms ...


1

You misunderstood the classification I believe. Let's take an example. In class D and 1D, the classification tells you there are two possible vacua (you understood this apparently). This is the famous $\mathbb{Z}_{2}$ ensemble in the classification. Next the classification tells you also that: at the boundary between the two gapped vacua, a Majorana mode ...


1

Simple, combine both real- and $\mathbf{k}$-space pictures! The basic idea is to split up your $n$-dimensional system into multiple $(n-1)$-dimensional systems. For example, say you have a 2D square lattice and you define your edges along the $x$-direction. Then you need to break the 2D lattice into 1D lattices pointing in the $x$-direction. In other words, ...


1

It seems to me that what you need is much more an introduction to superconductivity than to Majorana modes physics. I suggest you to open any book called superconductivity to have more details than the ones I give below. A standard description of a superconductor consists in saying it is a perfect metal with an electron-electron attractive potential. A ...


1

a) The parity of the ground state is the sign of the pfaffian of $A$. With the transformation (12), $A'= WA W^t$, you get pfaffian ($A'$) = det $W$ pfaffian ($A$) So, if det $W=-1$, then pfaffian $(A')$ = $-$ pfaffian $(A)$, so sgn(pfaffian $(A')$) $= -$sgn(pfaffian $(A')$), so the parity of the ground state is changed. b) Every (original) hamiltonian ...


1

The answer is yes, a single electron can tunnel between two topological superconductors hosting unpaired Majorana modes. Actually, since the Majorana pair flips (changes sign) by the injection of an electron inside the junction, a new term appears in the Josephson relation between two Majorana wires, which would be absent for a Josephson junction between ...


1

The paper you link to contains a model which is simply the Standard Model coupled to a singlet scalar. Its hierarchy problem is just as severe as that of the Standard Model, and as such it is a highly unnatural theory. (To be clear: what I mean by "unnatural" here is that the theory has quadratically divergent corrections to the Higgs mass, and as such the ...



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