I've seen many people (inc. Xiao-Gang Wen) claim that grassmann numbers don't have any "physical meaning" - at least that they are aware of. What is the opinion of people on this board?
I want to take grassmann valued fields seriously as a physical concept, but I'm having a difficult time doing so. There are two main issues.
In the first place, Berezin integration has nothing to do with sums. Should we still think of a grassmann valued field as taking some stochastic sequence of configurations, each of which has some contribution to the observable we calculate? It seems not so, but then the entire notion of "path integration" seems only to apply to bosons - and most particles are not even bosons.
In the second place, it doesn't even seem like the concept of a "grassmann valued field" means anything in terms of quantum states. In the case of bosons, a field s(x) is normal (so [s(x), s*(x)] = 0; where s*(x) is the Hermitian conjugate of s(x)) and at a fixed time t = 0, [s(x),s(y)] = 0. Because of these two commutation rules, it is possible to simultaneously diagonalize the s(x)'s at each spacetime point and define a quantum state that is an eigenvector of each s(x). Because of this, field configurations are in one-to-one correspondence with a complete set of quantum states. For fermions, these commutation relations are replaced with anti-commutation relations. Similar results seem not to exist. The conclusion is that grassmann valued field configurations are NOT in one-to-one correspondence with fermionic quantum states. This makes interpretation even more difficult.
The grassman values are fudge factors to force the theory to work mathematically when it doesnt work otherwise. There is no physical meaning to it other than a physicist reaching into his ass and pulling it out.
Seriously, fermions only work in a classical sense. But the holy commandment that is quantum theory shall not be disobeyed so physicists made reality obey it.
>In the second place, it doesn't even seem like the concept of a "grassmann valued field" means anything in terms of quantum states
You are correct. Quantum physics is not physics based, its math based.
Oliver Morales
>fudge factors this I read something like this in one of Hawking's books
Nicholas Morris
What does "fermions only work in a classical sense" mean? The bosonic commutator obeys the axioms of a Lie algebra, and so a lot of classical things that follow for Poisson brackets follow for Bosons. They don't follow for fermionic fields. I am not sure in what sense fermions make sense in a classical context.
When you say "there is no physical meaning to it...", you must appreciate the parallels between the fermionic and boson path integrals, at least symbolically. To assume that all means nothing, and is just a fudge factor, doesn't make much sense to me. That is my main motivation to pursue this. To use your analogy, it's like if a physicist reached into his ass, but instead of finding shit, found a well-cut diamond. It works too well for me to ignore it.
Your final point that quantum physics is not physics based make no sense. Let's redefine "physics based" in such a way that it agrees with whatever definition you are using. Almost all of physics is not physics based, and fermionic physics is even less physics based. Why? Should I just accept this?
Oliver Gray
Fermions obey classic field theory. When physicists were calculating qft for fermions, they ran into a problem. It didnt work, so they derived grassman numbers to build qft for it. Essentially.
I dont think Im going to be able to provide you a satisfactory answer. Especially if you are asking in the context of learning for a class.
Fermion and Boson path integrals are fun to play around with. But neither have been confirmed experimentally. Neither were experimentally derived. The parallels between them do not satisfy me as a justification when we still haven't fully observed anything subatomic.
What I mean by not physics based is that the observations from which we derive quantum is incomplete even though quantum provides a "good enough" model from time to time.
Aaron Cruz
I'm still not sure what you mean by "Fermions obey classic[al] field theory". I could treat the Dirac equation as a PDE and solve it in a classical field-theoretic sense, but I would understand the result as a quantum probability distribution and not as a field. The funny thing is that if you forget about fields for a moment and introduce a finite number of quantum degrees of freedom (x1,x2,x2...) and treat them as you would in the usual quantum theory (sans field) with the exception that you add a factor of (-1) to the paths in one of the two homotopy classes for paths in R^{3}, you'd get the correct thing. So, the problem isn't quantum mechanics. It's the introduction of a field where the particle number is no longer fixed. I repeat: quantum mechanics is not at all the problem. Fermions can be easily described by quantum mechanics, as experiments in solid state physics show.
We've observed the anomalous magnetic moment of the muon. Is that not subatomic enough? I'm not sure what more justification you'd need. Quantum mechanics provides a "good enough" model every time. There is no exception.
Isaiah Smith
>In the first place, Berezin integration has nothing to do with sums. Why should they? >Should we still think of a grassmann valued field as taking some stochastic sequence of configurations, each of which has some contribution to the observable we calculate? Yes? Berezin integration is a linear functional from the algebra of Grassmann numbers to [math]\mathbb{C}[/math] so I don't see why not? >It seems not so, Why? >but then the entire notion of "path integration" seems only to apply to bosons No? It applies to any particle with any spin statistics given sufficient regularity conditions (i.e. Swieca-Strozzi type or when there's some non-trivial topology/geometry to exploit, like in TQFTs or CFTs). >and most particles are not even bosons. In dimension larger than two there are only bosons and fermions. >In the second place, it doesn't even seem like the concept of a "grassmann valued field" means anything in terms of quantum states. Except it does? They're representations of the generators of an infinite dimensional Clifford algebra. >In the case of bosons, a field s(x) is normal (so [s(x), s*(x)] = 0; where s*(x) is the Hermitian conjugate of s(x)) and at a fixed time t = 0, [s(x),s(y)] = 0. That's not what a boson is at all. Bosons satisfy the Heisenberg algebra [math][\psi^\dagger(x),\psi(y)]_{x^0 = y^0} = \delta({\bf x}-\bf{y}) \neq 0[/math]. In general bosons are operator-valued distributions, and so are fermions. >Because of these two commutation rules, it is possible to simultaneously diagonalize the s(x)'s at each spacetime point That's not true at all. There'd be no off-diagonal S-matrix elements otherwise.
Sebastian Cook
>and define a quantum state that is an eigenvector of each s(x). No, the Fock states are eigenstates of the position and momentum operators, not of the fields themselves. >Because of this, field configurations are in one-to-one correspondence with a complete set of quantum states. No they're not. Do you know what instantons are? >For fermions, these commutation relations are replaced with anti-commutation relations. Similar results seem not to exist. Why would you want to get similar results as bosons for fermions? >The conclusion is that grassmann valued field configurations are NOT in one-to-one correspondence with fermionic quantum states. That doesn't follow at all. >This makes interpretation even more difficult. By your own misunderstanding of the entire matter it appears.
Levi Sullivan
You might have been able to be helpful. Unfortunately, I think an anxiety-induced existential crisis may have led you to be too thirsty in your desire to call someone other than yourself a moron.
In many placed, you've completely missed my point and given some irrelevant definition. In particular, where you mention the Heisenberg algebra, I was referring to the field strength operator, not the creation/annihilation operator. Please see any introductory QFT text to see that I am correct. Here is a helpful reference (which you check for yourself, since I'm sure you are not busy with any important research): theory.caltech.edu/~kapustin/Ph205/2013/fall2.pdf
See the top of page 2. A lot of your misunderstanding stems from this. It's such a amateurish mistake. I feel some second-hand embarrassment.
I might be able to get something out of this, though. What is the definition of path integration you use in axiomatic field theory?
Ryder Murphy
>Unfortunately, I think an anxiety-induced existential crisis may have led you to be too thirsty in your desire to call someone other than yourself a moron. Good psychological projection, though I wouldn't be surprised if you don't know the definition of a projection in the mathematical sense either at this point. >field strength operator None of those things are "field strength" operators, they're the canonical field coordinates; no where does the notes you've linked mention any "field strengths" at all. Conventionally what people in the know (i.e. not you) call the field strength is the curvature of the connection 1-form of a gauge potential, which has nothing to do with the quantization of the canonical field coordinates [math]\{\phi_a,\pi_a\}[/math] for a free non-gauge scalar field (which is what you've linked). >not the creation/annihilation operator In the free field theory, the canonical field coordinates can be decomposed into plane-wave modes the coefficients of which generate the Heisenberg algebra. In fact Wick's theorem as the entirety of perturbative QFT relies on this assumption (i.e. the existence of an interaction picture). >A lot of your misunderstanding stems from this. Another projection. You sure like pulling shit out of your ass don't you? >What is the definition of path integration you use in axiomatic field theory? Undergrad explanation: it's the sum over all paths after inserting multiple resolutions of identities made of position/momentum eigenstates into the Green function (propagator) [math]\langle x;t_0| e^{-iH (t-t_0)}|y,t\rangle[/math]. Graduate explanation: it's the integration over the moduli space of connections. Please go read Weinberg before you embarrass yourself again. The Dunning-Kruger is overwhelming.
