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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 ... 9 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, ... 8 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 ... 7 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 ... 7 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 ... 7 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 ... 6 Jane, \partial_\tau is clearly a derivative with respect to a bosonic time \tau, so it commutes with everything else (except for functions of \tau itself, with which it has a nonzero commutator), rather than anticommutes. Only if both objects have a fermionic character (if both of them are Grassmann-odd), they anticommute with one another (or they have ... 6 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) ... 6 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 ... 5 It's correct that you only replace the denominators 1/(p^2-m^2+i\epsilon) by -2\pi i \delta(p^2-m^2) in the propagators to compute the discontinuities. The fermionic propagators must first be rewritten so that they contain the denominator I just mentioned. You're right that the numerator isn't affected in the Cutkosky rules. In some formal sense, you ... 5 No that is not how it works. A 360 rotation multiplies the wave function by a factor -1 which by itself is not observable. It does not switch up and down spins. An experiment which demonstrated the effect would have to involve forming interference patterns between streams of electrons where one stream was being rotated through 360 degrees (e.g. using ... 5 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: ... 5 They're more complicated cousins of the Fierz identities, http://en.wikipedia.org/wiki/Fierz_identity The article above also recommends you Okun's book for the general recipe to prove similar identities. Note that all the identities you wrote except for the third one are just normal Fierz identities because the first factor may be cancelled as it ... 5 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 ... 5 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 states 10 |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 ... 5 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. ... 5 If by unification, one means that the bosons' and fermions' properties are linked to each other by a principle, the answer according to everything we know is Yes because the only principle able to link properties is a symmetry and a symmetry mapping bosons to fermions and vice versa is clearly a Grassmann-odd generator which has to carry a half-integral spin ... 4 I will just answer the first part of question: is phonon attraction stronger than Coulomb? Short answer: No. Longer answer: Nothing (in condensed matter) is ever stronger than the Coulomb force. Longest answer: There are two aspects to consider. First is the self-screening of the electrons, which will add a mass term to the photons, giving a Yukawa-esque ... 4 Here's my attempt at answering some of your multitude of points... I know you're often looking at N=2 susy. D_{++} type covariant derivatives often appear in harmonic superspace, where the ++ indicates the harmonic charge. The spinor components of your gauge field are Lie algebra valued fermionic spin-1/2 fields. That means they take a spacetime point ... 4 If the integral I:=\int d\theta on the algebra {\cal A}  of superfunctions f(\theta)=\theta a + b should be 1) a (graded) linear operation, 2) translation invariant, i.e., \int d\theta ~f(\theta+\theta') =\int d\theta~f(\theta), 3) and if the output \int d\theta~ f(\theta) should not depend on the integration variable \theta, then it is ... 4 The solution to this problem comes from the sneaky fact (Kugo, 1978) that while the FP ghost field is hermitian c^\dagger (x) = c(x) while the anti-ghost field is anti-hermitian \bar c^\dagger (x)=-\bar c (x) . As a result, the plane wave expansion for the ghost/anti-ghost fields (Becchi, 2008), Scholarpedia are:$$ c^a(x)={1 \over(2 \pi)^{3/2}} ...

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Like an ideal gas, a Fermi gas is composed of non-interacting particles. This is typically an idealization--few gases are composed of entirely non-interacting particles--but it's no more of an idealization than the classical ideal gas is. The differences between Maxwell-Boltzmann statistics (which yields the classical ideal gas), Bose-Einstein statistics ...

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The canonical momenta don't change if you add a total derivative to the Lagrangian. The particular total derivative you wanted to add to the Lagrangian as well as the Lagrangian itself has free $i,j$ indices. You surely meant something else because the Lagrangian should have no free indices like that. Let me assume that you meant both expressions to be ...

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For one, Brian Cox was probably talking about fermions and, more specifically, about identical fermions. Bosons, such as light, can share the same energy level and even the same quantum state. For example, in Bose-Einstein condensates, all the particles (i.e. atoms) share the same ground state energy level. The simplification that you mention probably ...

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A composite particle consisting of an even (odd) number of fermions behaves like a boson (fermion). Swapping $\require{mhchem}\ce{^3He}$ atoms involves swapping an odd number of fermions (3 nucleons and 2 electrons): an odd number of sign changes of the wave function corresponds to no sign change or bosonic symmetry. Swapping $\ce{^4He}$ atoms involves ...

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It can. This is exactly what happens when Helium-3 becomes superfluid. It's also what happens in superconductivity, which you mention in your question, when electrons combine into Cooper pairs. Well, it's not exactly what you ask since neither liquid Helium-3 nor electrons are a gas. It's very unlikely a gas of fermions could pair up to form a gas of bosons ...

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Be careful that there are some integrals, which implies that $S_I^n\neq 0$ ! For instance $S_I^2=\int_{x_1,x_2,x_3,x_4} (\psi_1^\dagger\psi_1)V_{12}(\psi_2^\dagger\psi_2)(\psi_3^\dagger\psi_3)V_{34}(\psi_4^\dagger\psi_4)$, where $\psi_i=\psi(x_i)$, $V_{ij}=V(x_i,x_j)$ and I assumed that the interaction is of the form ...

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This is not an answer just some thoughts from playing with the expression. I've read the question before you included the phase, so for now let $\phi = 0$ (sorry it this makes my response useless for you). I'll simply write $Z$ instead of your $Q(\beta, \beta_c)$ and also drop the arguments where obvious. Denote by $Z_{abc\dots}$ the partition function ...

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