Heidar

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Apr 19	awarded	Nice Answer
Apr 17	comment	Why does in string theory the amount of supersymmetry have to be $\cal{N} \leq 2 $? I'm glad it helped a little. I hope somebody will pose a slightly deeper answer.
Apr 17	comment	Why does in string theory the amount of supersymmetry have to be $\cal{N} \leq 2 $? Correction: "can't do it with only one spinor" should say "can't do it with only two spinors".
Apr 17	comment	Why does in string theory the amount of supersymmetry have to be $\cal{N} \leq 2 $? The same as $\mathcal N=2$ in D=10. All I'm saying here is correct morally, but there might be some details which aren't completely correct. However, more details can be found in most SUSY notes.
Apr 17	comment	Why does in string theory the amount of supersymmetry have to be $\cal{N} \leq 2 $? more components. As you see in the above link, type I string theory has 16 supercharges ($\mathcal N=1$) and type II has 32 ($\mathcal N=2$), this is related to spinors in 10 dimensions (which I think have 16 components/spinor dimension). In lower dimensions you can still have, say, 32 supercharges but you can't do it with only one spinor since the spinors have fewer components there, thus you must have $\mathcal N\geq 2$. When people talk about $\mathcal N=8$ SUGRA, it's in $D=4$ I think. In four dimensions Dirac spinors have 4 components, so $8\times 4 = 32$ super charges. (cont.)
Apr 17	comment	Why does in string theory the amount of supersymmetry have to be $\cal{N} \leq 2 $? My point was that those two statements are not necessarily inconsistent with each other, one has to specify the space-time dimension. The fact that $\mathcal N\leq 2$ (understood appropriately) comes from analyzing the spectrum of superstring theory (see here for the results en.wikipedia.org/wiki/…). However, this is a statement in 10 dimensions. If $\mathcal N=1$ that means you have a spinor $Q_\alpha$, but the number of components $\alpha$ is given by representation theory of Clifford algebra. In higher dimensions, spinors have (cont.)
Apr 16	comment	Why does in string theory the amount of supersymmetry have to be $\cal{N} \leq 2 $? I think its important to know which dimension you are in, knowing only say $\mathcal N=2$ is not enough to determine how much supersymmetry there is and therefore hard to compare to $\mathcal N=8$.
Apr 7	awarded	Taxonomist
Mar 25	comment	Name of fermionic CFT theory Are the $\phi_\alpha$'s weight $\frac 12$ chiral primaries? Naively it seems that the second term does not make sense unless $D=4$. Am I misunderstanding something? By the way, I assume this is 2-dimensional CFT right?
Mar 23	comment	Basic questions in Majorana fermions @Jeremy Yes precisely, they obey very different particle statistics (as in behavior of the wave function under particle exchange). This is actually why they are interesting in the first place. To think about them in terms of statistical physics (Fermi-Dirac distribution etc) seems, however, to be very subtle. Even at zero temperature.
Mar 23	comment	Basic questions in Majorana fermions @NanoPhys Yes exactly. I generally think that people should make the distinction clear, because the physics of the HEP and CMP Majorana is very different and this often leads to misunderstandings. Worse, I always get frustrated when the truly exotic features of the Majorana zero modes are neglected in favor of a wrong analogy to HEP. Very interesting comments regarding the Kouwenhoven experiment! Indeed, closure of the gap is crucial from a theoretical POW and the absence of it seem rather puzzling. I hope new experiments will soon clarify the situation. I know several groups are working hard.
Mar 17	comment	Basic questions in Majorana fermions Just noticed that this distinction is made on Wikipedia , which is nice. Besides "Majorana bound state", they also use "Majorana zero mode", which is also a good and accurate name. en.wikipedia.org/wiki/Majorana_fermion#Majorana_bound_states
Mar 17	revised	Basic questions in Majorana fermions edited tags
Mar 17	comment	Basic questions in Majorana fermions Sorry for this way too long comment. To summarize: I think this is a really nice and accurate answer, my only problem was just a minor detail in the first sentence which is not correct and is caused by the misleading terminology. The fact that they are NOT fermions is why we are interested in them in the first place.
Mar 17	comment	Basic questions in Majorana fermions Regarding what a better terminology would be. Karsten Flensberg (one of the authors of the review cited above), often call these particles for "Majorana bound states" (but for some reason not in this review). I personally prefer this term since it's less misleading. (continued)
Mar 17	comment	Basic questions in Majorana fermions I think there is a way to think of the condensed matter Majorana as a true fermion, but then one has to couple it to a BF-gauge theory (which effectively makes the particle an anyon). This is however not how people usually think about Majorana's in Condensed matter physics. If I remember correctly, something like that was done by T. H. Hanssons group recently. (continued)
Mar 17	comment	Basic questions in Majorana fermions 2) The truly amazing features of the condensed matter particles are not mentioned (non-abelian statistics, topological order,TQC, ...) in favor of a wrong-ish analogy to dark matter (I have even seen a talk by Leo Kouwenhoven (who found them experimentally) doing this exact thing, not mentioning the importance of this discovery at the expense of a mostly wrong and less interesting connection to particle physics). (continued)
Mar 17	comment	Basic questions in Majorana fermions The term "Majorana fermion" often lead to other misunderstandings too. I often see popular articles saying the Majorana is a candidate for dark matter and it has been found in condensed matter systems. This is problematic since: 1) two different objects are confused with each other which only share a name and a few mathematical details, but are otherwise physically VERY different. (continued)
Mar 17	comment	Basic questions in Majorana fermions It also seems that it's a little misleading to say they are their own anti-particles. This is true for the high-energy Majorana, but in the condensed matter case it just means that $c^{\dagger}=c$ (which I don't think is the same thing). Moreover the identity $c^{\dagger}=c$ makes sure the Majoranas don't satisfy fermi statistics, which is not the case for the particle physics Majorana. (continued)
Mar 17	comment	Basic questions in Majorana fermions Well, "Majorana fermion" is the correct terminology in the sense that this is the term most often used in the field. My problem is that this is often very misleading, since these objects are NOT fermions at all but actually non-abelian anyons. You explain this very correctly and clearly above, my problem was just that you say they are fermions in the first sentence (which is not really correct). On the contrary, the high-energy Majorana fermion IS actually a fermions and therefore in a certain sense less exotic than the condensed matter version. (continued)