# Tag Info

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 ...

10

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 ... 9 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 ... 8 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, ... 7 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 ... 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 ... 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 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 $). In ... 6 It seems that this question is addressed in this paper: http://arxiv.org/abs/1011.5229 (Edit: I just noticed this is seems to be restricted to qubits, so it probably does not answer your question ... ) 5 This answer is incomplete, but it should provide an answer for almost all symmetric states (i.e. it suffices for all but a set of symmetric states having measure zero). The symmetric subspace is spanned by product states. We may then consider different ways in which a particular symmetric state decomposes into symmetric products; in particular, if any ... 5 Your statement is wrong in one point: Regular$\beta\beta$decay happens in nature, even with dirac neutrinos. However, this shows the continuous electron (or positron) energy distribution one expects from just two (largely independent)$\beta$decays (see image below). What's interesting is that, if neturinos are marjorana particles indeed, a different ... 4 It seems that you are asking if the fundamental theorem of algebra is true. A symmetric$n$qubit state can be seen as a homogeneous polynomial of degree$n$in two variables. The factorization corresponds to the roots of the dehomogenized version of this polynomial. 3 Non-conservation of charge in Majorana terms The Dirac mass term is$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, \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 ... 3 You simply have to include the Majorana mass term (into the most general Lagrangian compatible with the given symmetries) at a tree level because the masslessness of$\chi_{++}$isn't guaranteed by any symmetry that is valid at the quantum level. In fact, as I will argue in a minute, the would-be symmetry is violated even at the classical level. Then the ... 2 When you apply the expression on a state, you can clearly see what it does: The new state created by applying the Majorana operators must have the same energy as the original one. So when applied on a zero-energy state, this creates a degeneracy and the so-called Majorana zero mode. 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 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 ... 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{array}{c} ... 2 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$(A')), so the parity of the ground state is changed. b) Every (original) hamiltonian ... 2 Neutrino mass is not in conflict with electroweak theory. One can introduce neutrino masses by modifying the Higgs or lepton sector of the Standard Model. The simplest method, that which you propose, is introducing a right handed neutrino with a Yukawa coupling with the left handed neutrino (extending the lepton sector). The right handed neutrino, however, ... 2 Yes, they can via a dimension 5 operator that contains two Higgs doublets and two lepton doublets. This is sometimes called Weinberg operator, it violates lepton number conservation and it was introduced in this work http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.43.1566 . Since the mass comes from an higher dimensional operator (i.e. an ... 2 In this context, the right-handed neutrino is a singlet under the Standard Model gauge groups. Only the right-handed neutrino is allowed a Majorana mass. The left-handed term is not gauge invariant. If the SM and right-handed neutrino fields were embedded into a Grand Unified gauge group, the right-hand term would break that symmetry. It is expected ... 2 You can compare neutrinoless double-beta decay to the problem of \begin{align} \pi^+ &\to e^+ + \nu_e, & \text{with branching ratio} &\sim 10^{-4} \\ \pi^+ &\to \mu^+ + \nu_\mu, & \text{ branching ratio} &\sim 1 \end{align} Here the pion has spin zero and is at rest in some reference frame; in that reference frame the neutrino and ... 1 The transformation rule for\hat{C}$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 = ...

Only top voted, non community-wiki answers of a minimum length are eligible