@Relativisticcucumber for completeness, it is $a_0 + b_0\ln(\rho)$
is it safe to think of the propagator as the solution to the distribution equation $\partial_\mu \partial^\mu \Pi(x,x') = -\delta(x - x')$?
I am asking this question because of the following observation. 1) Schwartz heuristically writes $\Pi(x,x') \sim \frac{1}{\partial_\mu \partial^\mu}$ (usage of $\sim$ my own). 2) Taking this symbol literally immediately is problematic as it implies $\Pi(x,x')a_{\vec{0}}^\dagger \lvert 0 \rangle$ is not well-defined. But a particle with $0$ momentum is quite a usual object (as opposed to being exotic for which an excuse could more conceivably be made for the singularity).
@SillyGoose The function is not originally well-defined, which is why we had to use quite a bit of trickery to define it. e.g. $\mathrm i\varepsilon$ prescription. The intent is specifically to get the propagator to become the correct solution to that equation. I am not sure why you think that the particle with 0 momentum would be a problem; especially if the QFT has a mass gap, so that then the 0 momentum wave is still waving in time, there should be no real problems.
You already know that the 0 momentum wave is infinitely large and has basically no position dependence, right?
shouldn't we be able to consider "rest frames" of systems, though? or in this case because the system would not occupy a perfect momentum eigenstate i guess the problem could be avoided
But the goodness of 0 momentum waves means that we can, isn't it?
In quantum theory, the limit of an exact momentum eigenstate is also an assertion that it is everywhere in the universe at once.
The properties of "position eigenstates" and "momentum eigenstates" are pathological enough that it is perfectly fine that we never have access to them. The miracle that is Fourier analysis on Hilbert spaces allows us to work, mathematically, with these limiting forms and yet derive results we want.
It is thus quite amusing whenever bolbteppa rails about the issue.
is this the right idea for proving this way of writing the Feynman propagator? it is just stated in schwartz, so i thought i;d try to derive the expression.
I am particularly wondering if we can change the $\phi_-(x_1) \phi_+(x_2)$ into the commutator $[\phi_-(x_1), \phi_+(x_2)]$ because the second term of the commutator vanishes when sandwiched betwixt the vacuum
Because this same replacement is done in Schwartz's proof of Wick's theorem, even when we are not sandwiching them betwixt the vacuum. But maybe he is just saying because we will always sandwich these time ordered products between the vacuum it's okay?
I think I have a misunderstanding in Wick's theorem. For the case $\mathcal{T}\{\phi_1 \phi_2 \}$, we would have $ = : \phi_1 \phi_2 : + :D_F(x_1,x_2):$
But seemingly, there should not be a normal ordering on the feynman propagator? But Wick's theorem states that the contractions should be normal ordered as well.
I'm trying to formulate it in a way that makes it clearer...
Wick's theorem converts any product of operators into a combination of contraction and normal ordered product of operators. This applies also to the time ordered stuff. The thing that we are defining, and will end up being the propagator, is the time ordered product of operators; Wick's theorem applied to that, and it ends up being normal ordered, and a contraction. The normal ordered part disappears because of vacuum, and the contraction is the only thing left; that is the propagator
Contracted stuff are simple functions, and no longer operators.
But Wick’s theorem as stated on the wikipedia for it converts a product of operators into a combination of normal ordered contractions and the normal ordered product of operators.
I've read the stranger and enjoyed it. I thought about it through the lens of autism/anhedonia, but there was a nice essay I read about reading it from a postcolonial perspective too.
I'm sure the French will say it's not understandable if you don't read the French version. there's always discussion on the correct translation
@naturallyInconsistent he used sisyphus as a paradigmatic example of the absurd and camus’ response to this absurd is to imagine sisyphus happy; heuristically rebelling against the absurd by continuing to push
@naturallyInconsistent oh goodness. what is the $::$ supposed to mean in that sum?
@SillyGoose The contracted parts are scalar functions and are taken out of the normal ordering. It will be the same leaving them inside the normal ordering anyway
@Obliv study. There is no goodness that will come out of that mayhem. Be blissfully ignorant as much as you can. Let the adults shield ya as long as possible.
Yeah. But it kinda makes sense; the seasons are changing and especially in the Northern hemisphere where we are going from scorching hot to blizzard cold, bodies are weakened and pathogens ready to attack
@naturallyInconsistent I've been reading a lot recently, let me see if I can find it
I'm afraid I can't find a source except a few scribbles I was handed. Precisely, in that context is done for the third order anharmonic term of the phonon hamiltonian (which incidentally, I couldn't find on any MBT book in 2nd quantization. I check Bruus&Flensberg, Fetter&Walecka and a couple of other books)
@Ryder Rude. I have been thinking along the same lines. In the relativistic Doppler effect the Doppler shift is a function of relative velocity. But the photon's relative velocity is indeterminate until it hits a receiver - the photon can't possibly know anything about the velocity of a receiver it is yet to meet. To me this seems analogous to the collapse of the wave function.
@Ryder Rude. It has been suggested, somewhat dismissively, that this is no different to the idea that bodies don't have kinetic energy, only systems. But that doesn't seem to require any collapse.
@Ryder Rude. Perhaps I didn't explain it too well. I should have said 'the Doppler shift is a function of relative velocity between source and receiver'. We of course know the velocity of the photon.
What is a Doppler shift or general relativity supposed to have to do with the "collapse of the wavefunction", which doesn't even exist in all interpretations?
You can't just claim random things are connected because they're both mysterious to you.
@JohnHobson i think ur idea is entirely classical phenomena. There is no wavefunction collapse in them
@JohnHobson "photon's relative velocity is indeterminate". Are u referring to the Heisenberg uncertainty principle?
The relativistic Doppler effect is not really about photons. It is best understood using classical light. So there is no Heisenberg uncertainty involved @JohnHobson
@JohnHobson but here, it seems like u r not talking about Heisenberg uncertainty cuz u say "We of course know the velocity of the photon"
How to better represent mass? Is is considered as quantity of matter? Any phenomenon which directly related to quantity of matter in better and better way can be considered as mass. Like inertial mass, gravitational mass etc..
"the photon can't possibly know anything about the velocity of a receiver it is yet to meet". When viewed from the frame of reference of the receiver, i think the frequency of the light wave is already time-dilated from the moment it is emitted
You can find some interesting physics and math for pre-modern people tho
Old school astronomy
One fun trick they did for comparing angles in astronomy was to fill some tube with water, angle it so as to point toward a star, and then mark the water level to indicate the angle at which that star is
I guess i am just wondering what the origin of haag’s theorem can be thought of as. Seemingly the big difference between QM and QFT is the commutation relation that is put front and center
@SillyGoose wiki says that Haag's theorem also uses Poincaire invariance as an assumption. So at least Haag's proof does not apply to other QFTs, even tho the equal time commutation relations would.be the same
but maybe a more general proof can apply to other QFTs
but I'm not aware of such results
but Stone Von Neumann does suggest that something like Haag's theorem may arise whenever here is an infinite number of CCR
Haag himself often called the crucial aspect the "vacuum polarization" - each pair of c/a operators "shifts the vacuum" by a finite amount, so that the overlap $\langle 0\vert \Omega\rangle$ diverges in the case of infinitely many pairs, meaning the two vectors can no longer be considered part of the same Hilbert space
@ACuriousMind this would mean that something like Haag's theorem may arise in non rel QFTs. But also, some non rel QFTs are equivalent to non rel QM, which seems to mean that Haag's theorem shouldn't apply
or maybe it does always apply. I don't think any non rel QFT is completely equivalent to non rel QM
as a modern summary of the literature on the topic, I enjoyed reading this PhD thesis by Klaczynski, though I'm not sure about the significance of the "resurgent transseries" part at the end
Streater and Wightman (PCT, Spin and Statistics, and all that) is also a good older read on this and similar topics in axiomatic QFT
suppose we take a free system and an interacting system in ordinary QM (e.g. free particle and Harmonic oscillator). The vacua of these two lie in the same Hilbert space
then if we re-write these two as non rel QFTs, the vacua should still live in the same Hilbert space
because the energy eigenvalues and the inner products would be the same as in the QM formulation
@ACuriousMind does this mean that something like Haag's theorem need not hold in all QFTs?
@RyderRude Haag's theorem does rely on Poincaré invariance, cluster decomposition or any of the other related assumptions of relativistic QFT. There are Euclidean QFTs that don't exhibit vacuum polarization (see e.g. the reference to Lévy-Leblond in the PhD thesis I linked).
However, these Eucldean theories are extremely special - see the discussion in section 1.4., where it is shown they correspond to superrenormalizable theories
I don't see an example anywhere, just vague claims about "lattice models", which I have repeatedly told you are much more complicated than you pretend.
i think my above example does not make sense. it was a misunderstanding
@ACuriousMind do we know if Haag's theorem can apply to lattice QFTs? e.g. we take a free lattice QFT and an interacting QFT on the same lattice. Can they not have the same Hilbert space? The lattice size is infinite to ensure Stone Von Neumann doesn't apply
@RyderRude Again, Haag's theorem is for relativistic QFTs. A lattice model obviously does not have Poincaré invariance (and its status as a "QFT" in the same mathematical sense as the Euclidean or Minkowski theories is debatable), so it is outside of the scope of the theorem.
@RyderRude Why do you insist on talking about lattice theories when all of our past discussions about them have resulted in me telling you they're much more subtle than you seem to think, and now you ask me again with the same naive ideas? If you want to keep being so obsessed with lattice theories you need to actually learn how people construct them and what they're used for.
i am thinking that : suppose we take the theory of free gravitons at the lattice spacing =Planck scale. And then we take the theory of interacting gravitons with the same lattice spacing. Since the latter theory has infinities, it means the two are unitarily inequivalent
Most practical usages of "lattice theory" simply use it to compute a discretized path integral, there's no full-blown physical theory with operators and Hilbert spaces etc. there
What i think as we have defined mass as a quantity of matter. Any phenomenon which directly related to quantity of matter by any process, we name it mass. e.g. object in acceleration in response of known force is directly related to quantity of matter, that's why we name this property as mass. Am i correct?
I have doubt about idea of mass. Because previously we explained mass as a quantity of material. Then we have changed it to resistance against change in motion or gravitational charge. If we don't directly explain mass why we used this idea?
> The Kochen-Specker theorem is equivalent to the statement that the spectral presheaf Σ of the algebra of bounded operators has no global elements if the dimension of the Hilbert space is greater than 2.
