# Tag Info

54

Great, important question. Here's the basic logic: We start with Wigner's Theorem which tells us that a symmetry transformation on a quantum system can be written, up to phase, as either a unitary or anti-unitary operator on the Hilbert space $\mathcal H$ of the system. It follows that if we want to represent a Lie group $G$ of symmetries of a system via ...

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To a reasonable approximation the protons and neutrons in a nucleus occupy nuclear orbitals in the same way that electrons occupy atomic orbitals. This description of the nucleus is known as the shell model. The exclusion principle applies to all fermions, including protons and neutrons, so the protons and neutrons pair up two per orbital, just as electrons ...

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After the answers by joshphysics and user37496, it seems to me that a last remark remains. The quantum relevance of the universal covering Lie group in my opinion is (also) due to a fundamental theorem by Nelson. That theorem relates Lie algebras of symmetric operators with unitary representations of a certain Lie group generated by those operators. The ...

15

I'd like to add to Josh's answer, because he didn't really explain what a universal covering group is. Essentially, a space $T$ is a covering space of another space $U$ if, for an open subset of $U$, there's a function $f$ that maps a union disjoint open subsets of $T$ to the subset of $U$. Or, more simply worded, pick a piece of your space $U$, and I'll ...

15

This is a legitimate question but one for which you probably won't get any real, satisfying answer rather than just "because that's how nature works". You can "derive" the impossibility for two fermions to have the same quantum numbers from the requirement for many-fermion states to be antisymmetric with respect to the exchange of any two particles, that is,...

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Actually a paper recently came out, and highlighted in Popular Science, discussing using fermionic field concepts to model crowd avoidance at Netflix. You can imagine that the same concept could be used to consider in any situation where there are large numbers of people competing for limited preferred items. Update Now that we have a few minutes, rather ...

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Just in view of the double universal covering provided by $SU(2)$, $SO(3)$ must a quotient of $SU(2)$ with respect to a central discrete normal subgroup with two elements. This is consequence of a general property of universal covering Lie groups: If $\pi: \tilde{G} \to G$ is the universal covering Lie-group homomorphism, the kernel $H$ of $\pi$ is a ...

10

Let me quote Phys. Rev. B 83, 115132 (2011) The one-dimensional representations of $S_n$ correspond to bosons and fermions. One might have hoped that higher-dimensional representations of $S_n$ would give rise to interesting 3D analogues of non-Abelian anyons. However, this is not the case, as shown in Ref. 18,19: any higher dimensional representation of $... 10 On the deepest level, particles are indistinguishable if and only if they have the same quantum numbers (mass, spin, and charges). However, in statistical mechanics one often studies effective theories where there are additional means of distinguishing particles. Two important examples: In modeling molecular fluids, two atoms on the same molecule are ... 10 For the partition sum, you have so sum$e^{-E}$($T=1$) over all possible eigenstates of the system where$E$is the energy of the corresponding state. Two bosons can be in the 10 states$|kl\rangle$, with$1\leq k \leq l \leq 4$where we accounted for the degeneracy by introducing an additional state with$E_4 =2E$. The corresponding partition sum reads (... 10 I write below the statement of the aforementioned theorem which assumes, as hypotheses, the validity of so-called "Wightman axioms" in the four-dimensional Minkowski spacetime. You see that there is nothing imposed by hand. It is actually a no-go theorem. Quantizing free fields, it establishes in particular that the standard choice is the only possible. ... 10 It's a force like no other. It is fundamentally a quantum property and there is no classical way to think of it (at least to my knowledge). That's just how the universe is, and we haven't understood any deep reason "why" it should be that way. Mathematical consistency seems to dictate it. It comes down to the observation that there can be some objects such ... 9 How to obtain this braiding matrix from Non-Abelian Chern-Simon theory? To obtain braiding matrix$U^{ab}$for particle$a$and$b$, we first need to know the dimension of the matrix. However, the dimension of the matrix for Non-Abelian Chern-Simon theory is NOT determined by$a$and$b$alone. Say if we put four particles$a,b,c,d$on a sphere, the ... 9 Antiparticles naturally arise when studying the Dirac equation within quantum field theory. Recall that we may expand a Dirac spinor field as a plane wave, namely, $$\psi= \sum_{s=1}^2 \int \frac{\mathrm{d}^3 p}{(2\pi)^3} \frac{1}{\sqrt{2E_{p}}} \left[ b^s_p u^s(p)e^{ipx}+c^{s\dagger}_p v^s(p)e^{-ipx}\right]$$ and similarly for the conjugate field. Notice ... 8 You are right. In a space-time with one time dimension and$D$spatial dimensions, finding possible different statistics is equilalent to look at the fundamental group (first homotopy group) of$SO(D)$For$D=1$, the fundamental group is trivial. For$D=2$, the fundamental group is$\mathbb{Z}$. For$D>=3$, the fundamental group is$\mathbb{Z}_{2}$So,... 8 The "spin" tells us how the wavefunction changes when we rotate space (or spacetime). Just because I double all charges by convention, the behaviour of the wavefunction will not be any different. What will happen is that the "doubling" or charges will lead to the "halving" of your definition of angles such that the physical results (which depends on angle ... 8 You can find the right statistics using QFT. In QFT you can write down the fields for spin-$0$and spin-$\frac{1}{2}$, then you can show that fermions have to be antisymmetric otherwise there will be an infinity number of negative energy states (if they are finite you can always shift your definition of ground state with the lower state). In this sense you ... 8 When formulating a physical theory, one usually begins with a set of axioms. The theory itself will be just as useful as its axioms are accurate. In particular, when dealing with a supposedly fundamental theory, the following set of axioms seems natural and well-motivated by experiments: There is a set of fundamental entities, referred to as particles, ... 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 The classic place to start would be the book "PCT, Spin & Statistics, and All That", by R.F.Streater and A.S.Wightman. The spin statistics theorem can be proved as rigorously as you likely can want in the context of the Wightman axioms. The difficulty with this statement relative to your question is that we cannot prove that interacting fields satisfy ... 7 Initially Pauli exclusion principle imposed for fermions was purely phenomenological conception introduced in order to explain experimental facts. The reason is that there is no well-defined theoretical description of the relation between the spin and the statistics in non-relativistic quantum mechanics (within which Pauli has formulated his principle ... 6 Both fractional/non-Abelian statistics and fractional charges come from the same origin: long-range entanglements. This is why fractional/non-Abelian statistics common for fractional charges. One way to realize long-range entanglements is through the string-net liquid phase of a pure bosonic model. The ends of strings in string-net liquid are non-local and ... 6 There are two possible answers to why$T^2=-1$: a) Why not. The total phase of a quantum state is unphysical. So a symmetry may be realized as a projective representation. Here T may be viewed as a projective representation of time reversal$T_{phy}$which satisfy$T^2_{phy}=1$. b) If we define the time reversal symmetry to be realized as a regular ... 6 Please, let me first refer you to the original paper by Leinaas and Myrheim where the existence of anyonic statistics was first predicted before its actual discovery. All the ingredients of the understanding of the special properties of the two dimensional case already exist in this old paper, however, I'll try to cast it in a more modern terminology: By ... 6 The only direct experimental evidence for anyons which the scientific community broadly accepts is the evidence described here for$\nu = 1/3$Laughlin states in the fractional quantum Hall effect. Experimental evidence for nonabelian anyons (which are necessary for universal topological quantum computing) is much weaker. The only promising lead seems to ... 5 The spin-statistics thing isn't a problem, it is a theorem (a demonstrably valid proposition), and it shouldn't be addressed, it should be understood and celebrated. The Higgs field gives us interactions between chiral fermions and the Higgs,$yh\cdot \chi_\alpha\eta^\alpha$which produces mass terms$m \chi_\alpha\eta^\alpha$if the Higgs field has a ... 5 I will firstly point out some apparent misconceptions in the question and subsequently I will explain what goes wrong when quantizing a theory of integer spin fields or particles with anticommutators, and vice versa. First, if we quantize a real Klein-Gordon field using anticommutators, the Hamiltonian vanishes (or is a field-independent constant). At the ... 5 The way Shankar addresses the problem (pg. 278) is by introducing an "Exchange Operator"$P_{1,2}$, which would swap your two particles as follows:$P_{1,2} |\xi_1, \xi_2 \rangle = |\xi_2, \xi_1 \rangle$I like the operator notation because it makes it clear (to me, at least) that applying the operator twice is just the identity operator, since swapping ... 5 The (unitary) "phase" factor for non-Abelian anyons satisfies the (non-Abelian) Knizhnik-Zamolodchikov equation: $$\big (\frac{\partial}{\partial z_{\alpha}} + \frac{1}{2\pi k} \sum_{\beta \neq \alpha} \frac{Q^a_{\alpha}Q^a_{\beta}}{z_{\alpha} - z_{\beta}}\big )U(z_1, ....,z_N) = 0$$ Where$z_{\alpha}$is the complex plane coordinate of the particle$\...

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