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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 $$\text{Grassmann-odd/fermionic/an ... 9 The fermion doubling is manifested through the existence of extra poles in the Dirac propagator on the lattice. These poles cannot be made to disappear at the continuum limit. (The number of doublers can be reduced by different discretizations but not eliminated at all, this is essentially the Nielsen-Ninomiya theorem). The reason for the fermion doubling ... 9 Actually a paper recently came out, and highlighted in Popular Science, discussing using fermionic field concepts to model crowd avoidance at Netflix. You can imagine that the same concept could be used to consider in any situation where there are large numbers of people competing for limited preferred items. Update Now that we have a few minutes, ... 9 Antiparticles naturally arise when studying the Dirac equation within quantum field theory. Recall that we may expand a Dirac spinor field as a plane wave, namely,$$\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 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 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 This is a nice puzzle--- but the answer is simple: the composite bosons can occupy the same state when the state is spatially delocalized on a scale larger than the scale of the wavefunction of the fermions inside, but they feel a repulsive force which prevents them from being at the same spatial point, so that they cannot sit at the same point at the same ... 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 ... 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 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 – ... 7 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 ... 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 ... 6 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 ... 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 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 ... 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 The most elementary reason is that the Dirac field Hamiltonian is bounded below only when you use anticommutation relations on the creation/annihilation operators instead of commutators. A free quantum field theory with energy unbounded below has no stable vacuum. It is easiest to demonstrate this in two dimensions, where there are no polarization issues. ... 6 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$$for some convenient trace orthogonal basis. To find the k_i for the specific basis and space that you're working in, you multiply by some M_j and take the trace: ... 6 Keeping it simple, let's asume that \psi(a) creates a particle in the state a (i.e., characterized by some collection of quantum numbers that we call a),$$ \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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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 ...

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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 ...

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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 ...

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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 ...

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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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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. ...

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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 ...

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I think the answer is it depends on distance (relative to the size of your system). Another well known example of a boson which is comprised of fermionic components is the helium-4 atom, which has integer spin (both the nucleus and the neutral atom itself). Fermionic or bosonic behavior of a composite particle (or system) is only seen at large (compared ...

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This is because the path integral ${\cal Z}$ is an infinite-dimensional version of a Grassmann-odd Gaussian integral $$\int \!\mathrm{d}^n \bar{\theta} ~\mathrm{d}^n\theta ~e^{\sum_{i,j=1}^n\bar{\theta}_i ~M^i{}_j ~\theta^j}~\propto~\det(M),$$ where the indices $i,j$ can be interpreted as DeWitt's condensed notation.

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First, watch out for the factors of 2 and $\sin(k)$s in your line 3 (after doing the fourier transform). Second, you do not want to set those terms to zero. Instead, remember that $k$ is just a dummy index. I could consider each term as a separate sum, and for some of them, I'll set $k \rightarrow -k$. Then -e^{ik}c_{k}^{\dagger}c_{-k}^{\dagger} ...

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