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## Hot answers tagged fermions

17

I don't have an answer to the question "why would one want to consider such crazy stuff in physics?" since I don't know much physics, but as a mathematics student I do have an answer to the question "why would one want to consider such crazy stuff in mathematics?" What physicists call Grassmann numbers are what mathematicians call elements of the exterior ...

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A fermion is any particle, elementary or composite, that obeys Fermi-Dirac (as opposed to Bose-Einstein) statistics relating to how identical particles behave when you swap two of them. Due to an important but complicated result, this is taken to amount to having half-integer spin. A lepton is one type of elementary particle with spin 1/2. The only leptons ...

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The deuterium nucleus is a boson, with spin $\hbar$ (and positive parity). Unlike other stable nuclei, deuterium doesn't have any bound excited states; however if it did they would also have integer spin. The deuterium atom is a fermion, which may have spin $\frac12\hbar$ or $\frac32\hbar$, to be combined with the orbital angular momentum (which is zero in ...

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A supernumber $z=z_B+z_S$ consists of a body $z_B$ (which always belongs to $\mathbb{C}$) and a soul $z_S$ (which only belongs to $\mathbb{C}$ if it is zero), cf. Refs. 1 and 2. A supernumber can carry definite Grassmann parity. In that case, it is either $$\text{Grassmann-even/bosonic/a c-number},$$ or $$\text{Grassmann-odd/fermionic/an ... 14 I don't have a very satisfactory description of the microscopic picture, but let me share my thoughts. The Pauli exclusion doesn't quite say that fermions can't be squeezed together in space. It says that two fermions can't share the same quantum state (spin included). A black hole has an enormous amount of entropy (proportional to its area, from the ... 12 Actually a paper recently came out, and highlighted in Popular Science, discussing using fermionic field concepts to model crowd avoidance at Netflix. You can imagine that the same concept could be used to consider in any situation where there are large numbers of people competing for limited preferred items. Update Now that we have a few minutes, ... 11 Anomalies (not anamolies) are a whole subject whose basics are covered by one or several chapters of almost any good enough quantum field theory textbook so it's counterproductive to retype this whole chapter here. But generally, in quantum field theory, anomalies are quantum mechanical effects breaking symmetries that exist in the classical theory – ... 10 Yes, there is a very good reason why leptons and quarks mix. It would be shocking if they didn't mix. Just to avoid confusions, we're not saying that leptons and quarks mix with each other: they don't because leptons are eigenstates of color or baryon charge with different eigenvalues than quarks which are also eigenstates: this difference prevents mixing ... 9 I always thought of Fierz identities as a kind of completeness relation (*) for products of spinors. To use the bra-ket notation:$$|a\rangle\langle b| = \sum k_i \langle b|M_i|a\rangle M_i$$for some convenient trace orthogonal basis. To find the k_i for the specific basis and space that you're working in, you multiply by some M_j and take the trace: ... 9 The fermion doubling is manifested through the existence of extra poles in the Dirac propagator on the lattice. These poles cannot be made to disappear at the continuum limit. (The number of doublers can be reduced by different discretizations but not eliminated at all, this is essentially the Nielsen-Ninomiya theorem). The reason for the fermion doubling ... 9 A fermion is any particle characterized by Fermi–Dirac statistics and obeying the Pauli exclusion principle. So for example quarks are fermions, as are Helium-3 atoms. A fermion does not have to be an elementary particle. I'm not even sure that it has to be spin \tfrac{1}{2}, though I can't think of any fermions that aren't. A lepton is a spin ... 9 Antiparticles naturally arise when studying the Dirac equation within quantum field theory. Recall that we may expand a Dirac spinor field as a plane wave, namely,$$\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 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 So what people mean by 'non-local' varies from context to context and person to person. Wen has a very particular meaning to this. 1) In fermionization in D=1+1 the Jordan-Wigner fermions are, in the bosonic language, operators supported over many sites. The emergent (mutual)-fermions in the toric code are also supported at the ends of strings. 2) ... 8 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 ... 8 There are answers in the note by Polchinski linked by Matt, and an article by Shankar in Review of Modern Physics: Renormalization-group approach to interacting fermions. Just to flesh out was it meant by "stability" and "Fermi surface". The Fermi-liquid can be thought of as a phase characterized by several properties: arbitrarily long-lived, gapless ... 8 Great question that exposes some really confusing terminology. This is a rather long answer, and the punchline is basically in the second-to-last paragraph, but I think (hope) it's worthwhile to read the whole answer because I tried to give a somewhat systematic description of fermionic states using a specific, simple example along the way. Firstly, let's ... 8 In contrast with the previous incorrect answers that I hadn't noticed, there isn't any ambiguity or confusion about the Bose-Einstein or Fermi-Dirac statistics for composite systems such as atoms. A particle – elementary or composite particles – that contains an even number of elementary (or other) fermions is a boson; if it contains an odd number, it is a ... 8 Our current best experimentally verified theory, quantum field theory, isn't based on matter being particles or waves - all matter consists of excitations in quantum fields. The interactions of the quantum fields may appear particle like or wave like, so the wave-particle duality is a duality in the way the fields interact not a duality in the matter itself. ... 8 I think that you are confused. When you rotate something by 360 degrees, you won't change the direction in space of anything. You will only change the wave function to minus itself - if there is an odd number of fermions in the object (which is usually hard to count for large objects). If you have electrons with spins pointing up and you rotate them around ... 8 When physicists say that a quantum field \phi(x) is real-valued, they are usually referring to Feynman's path integral formulation of quantum field theory, which is equivalent to Schwinger's operator formulation. The values of a field \phi(x) in the path integral formulations are numbers. E.g.: If the numbers are real, we say that the field \phi(x) ... 8 In the actual local quantum field theories, theories of point-like particles, the mass correction due to the renormalization effects from (2) is divergent. It has a short-distance divergence so it is infinite. One needs to cancel the "infinite part" so that there's a finite leftover. What is the separation of the physical observed mass to (1) and (2) depends ... 7 Neutron degenerate matter can undergo a phase transition to a superfluid state. The process is thought to be analogous to Cooper-pairing, but the coupling interaction due to the long-range nuclear force is of order 1 MeV, so can occur at temperatures below about 10^{9} K in neutron star interiors. The neutrons (fermions) form bosonic pairs in an analogous ... 6 The terminology of a mode of a free quantum field \phi(x) comes from writing it as a Fourier transform, often also called mode expansion:$$ \phi(\vec x) = \int \frac{\mathrm{d}^3 p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}}\left(a(\vec p)\mathrm{e}^{\mathrm{i}\vec x\cdot\vec p} + b(\vec p)^\dagger\mathrm{e}^{-\mathrm{i}\vec x\cdot\vec p}\right)$$where for a ... 6 The fundamental representation of a Lie group G, as commonly used in this context, is the smallest faithful (i.e. injective) representation of the group. We do not require fermions to belong to the fundamental rep, it is just the case that, in the Standard model, they always either belong to the fundamental or the trivial representation (as that is ... 6 All the known laws of physics, including the exclusion principle, are believed to be valid at all times during the collapse, up until the matter that you're talking is just about to hit the singularity. ("Just about to hit" may mean when the density reaches the Planck density, so that quantum gravity effects become important, or it may be a little earlier, ... 6 It depends on your definition of force. Force means a change in momentum, ~dp/dt , so any change in momentum in a Feynman diagram is a force. For example this diagram for compton scattering says yes. If one is talking of gauge theories and exchanged bosons , because those are the ones that build up the three, electromagnetic, weak, strong ( maybe ... 6 Keeping it simple, let's asume that \psi(a) creates a particle in the state a (i.e., characterized by some collection of quantum numbers that we call a),$$ \psi(a)|0\rangle=|a\rangle .$$and \psi(b) does the same for b. We can create a state with two particles:$$ \psi(b)\psi(a)|0\rangle = \psi(b)|a\rangle = |a;b\rangle  ...

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