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 1d comment Are the path integral formalism and the operator formalism inequivalent? @AccidentalFourierTransform: I was not saying you cannot necessarily write a path integral, I was saying you might not ;-) See Qmechanic's links for some counter-examples. Though for anyons, I am not sure that you can define proper creation/annihilation operators (I might be wrong), as they are usually constructed as composite entities (charge plus flux for example). 2d comment Limits used to find non-rel limit of the Klein-Gordon equation Sure :) Tough I would still discuss pair production (I'm sure the OP has heard about that before). 2d comment Limits used to find non-rel limit of the Klein-Gordon equation You are not saying that in your answer: "otherwise the electromagnetic radiation would dominate compared to the electrostatic field" and " says nothing about the particles", whereas pair production has everything to do with the particles. And it is not because the OP ask about pre-QFT that the answer must be too... when the effect is clearly coming from QM + SR, which is QFT... 2d comment Limits used to find non-rel limit of the Klein-Gordon equation The second point is not about radiation, but particle-antiparticle pair production. If $\omega>2mc^2$, the EM field has enough energy to create pairs (the factor of 2 is important here) Apr 26 comment Are the path integral formalism and the operator formalism inequivalent? To complement Prahar comment, if you did not have the canonical commutation relation, the path integral would not be the one you have written (assuming you could write one, since you would not have the standard creation/annihilation operators and the coherent states and so on). Apr 25 comment why cannot fermions have non-zero vacuum expectation value? @TwoBs: no problem, it was useful to me too, and forced to organize my thoughts ! Though I don't think having non constant sources would change anything. The partition function will be analytical in all the $\eta(x)$ by the same argument I think. Apr 25 comment why cannot fermions have non-zero vacuum expectation value? @TwoBs: in fact, I'm not that happy with my argument, since one can find a density matrix that allows a non vanishing expectation value. But of course, it won't be a grassmann number, whereas a mean-field approximation of the action would generate that. But I think that your $\delta Z$ is not possible, because when you expand the exponential of the sources, there is only one term, $\bar\eta \psi$, so there is no way to generate a square root (which would do the job, indeed). Apr 24 comment why cannot fermions have non-zero vacuum expectation value? @innisfree: put it that way: a fermionic annihilation operator $\hat c$ is a standard QM operator. Therefore, the average of $\hat c$ with respect to any physical state should give a complex number. Thus having $\langle \hat c\rangle =\psi$ with $\psi$ a grassmann number is impossible. It is in fact trivial to check that $\langle \hat c\rangle=0$ for any physical state (that is, $a|0\rangle+b|1\rangle$ with $a,b$ complex number. Stated otherwise, fermionic coherent state are not physical states, just very useful to write down a path integral. Apr 24 comment why cannot fermions have non-zero vacuum expectation value? @innisfree: sorry, I meant the argument of TwoBs about bosons having a finite $\langle x\rangle$ with a potential $(x^2-v^2)^2$. This won't work for grassman variables. Apr 24 comment why cannot fermions have non-zero vacuum expectation value? @TwoBs: of course your argument works for bosons, never said otherwise. Your "example" is not one, since you haven't done any calculation showing it works. Because it won't. And your example with magnetization doesn't work either, since spins are quadratic in fermionic operators... which of course can have a finite vev. Apr 24 comment Matrix representation of the radial Laplace operator isn't symmetric (or hermition as a result) Might be easier to understand how to make this work by looking at the hermiticity of the momentum operator in 1D. Apr 24 comment why cannot fermions have non-zero vacuum expectation value? @innisfree: the argument works for bosonic variables, not for fermions. Apr 24 comment why cannot fermions have non-zero vacuum expectation value? @TwoBs: SSB won't change anything, since already for explicit symmetry breaking (for a condensate $\bar\psi\psi$) won't give you a finite $\langle\psi\rangle$, because Grassman variables are completely different from bosonic variables. Stated in a different way: I challenge you to find a reference talking about $\langle\psi\rangle\neq 0$ (without linear sources). Apr 21 comment Does the goldstone field really disappear? What have you tried ? What makes you think it won't work ? Apr 21 comment Does the goldstone field really disappear? If I recall correctly, you can rewrite the whole Lagrangian in terms of the Higgs mode interacting with a massive vector field, which means that effectively, the goldstone mode has disappear from the full Lagrangian. Apr 19 comment Pion-pion scattering amplitude with an effective Lagrangian Yes, they give the same contribution for the vertices 2 and 3 by symmetry. I think your confusion comes from a misunderstanding on how to count the diagrams. Depending on the factor in front of the interaction, you might have only "one diagram" because all equivalent diagrams add up to compensate a carefully chosen factor (1/24 for example). But one need to be careful, because using other convention will change the prefactors, but not the "real" number of different contributions. Apr 19 comment Pion-pion scattering amplitude with an effective Lagrangian If you use Wick theorem, there is no ambiguity. Take all terms, simplify everything, and you are good. Apr 18 comment Pion-pion scattering amplitude with an effective Lagrangian Then I don't understand the question. You have one (collective) interaction vertex, so you can use the standard rules to get the diagrams for it... A very basic way is to write down the 4-point correlation function and do first order perturbation theory with the help of the Wick theorem. Apr 18 comment Why are periodic boundary conditions used for the derivation of phonons? @ThomasElliot: in physics, infinity just means large enough. Typically, these finite size corrections will vanish as the inverse of the volume, i.e. very fast, and will usually not be the main error coming from your modelization (read, almost never be). There are very few cases (such as the BKT transition), where the finite size of the system really changes the physics. (In that case, it has been argued that one would need system of the size of Texas to see properly (vanishing magnetization) the BKT phase). But that's quite an advanced topic compare to plain vanilla phonons. Apr 18 comment Pion-pion scattering amplitude with an effective Lagrangian Schematically, you can rewrite the interaction term as $\sum_{p_1,p_2,p_3,p_4}\delta(p_1+p_2+p_3+p_4)\delta_{ab}\delta_{cd}(-p_2.p_3+p_‌​3.p_4+\frac{M^2_\pi}{4})\pi_a(p_1) \pi_b(p_2) \pi_c(p_3) \pi_d(p_4)$, where the term in the middle is juste the interaction vertex .