@DanielSank They aren't minor! In many cases we don't even know whether putting the theory on a lattice and evaluating the quantum path integral on it, then taking a conitnuum limit gives the same results that we derive by other methods in the continuum! That's a big and in general unsolved problem in the field of doing lattice simulations in QFT
For the one-dimensional case you're looking at here you may be in luck because we actually can write down a mathematically well-defined continuum path integral and ask the mathematicians to verify it works out fine, though :P
Guys, I only care about the formal relations between stuff here and the only reason I'm discretizing anything is that I don't know how path integrals work yet.
An answer of the form "Here's how it works in the continuum, maybe it's the same in the discrete case" would be fine.
If you want to edit the question to make such an answer acceptable, please do.
@DanielSank are you hoping to start from a PDE and end up with the path integral directly from the equation, as in, you invent it starting from the equation directly?
I'm trying to understand how to deal with $J^T A^{-1} J$ without actually computing an inverse.
@bolbteppa I don't understand the question.
My basic understanding of the (Wiener) path integral is that we already know the Green's function for transition probabilties, and then we decide to think of it as a weighted sum over paths.
We then treat the "path integral" as a new object whose value we know because we know the Green's function, and we construct a calculus around it by discovering self-consistent algebraic manipulations.
We then evaluate nontrivial cases by referring them back to the known case.
@DanielSank the point of the $J$ terms is that differentiating with respect to $J$ lets you form correlation functions, but at least in physics/qft, $J = 0$
@Danu @DanielSank Ugh, it doesn't work so straightforward as I thought because this is not a relativistic theory, and you're not inverting the full operator that $W$ is a Green's function for, but merely the Laplacian. The Green's function/inverse operator for the Laplacian is well-known, but I'm afraid it doesn't directly relate to the diffusion propagator at all.
So I can't relate $A$ and $W$ even in the continuous case, since $A$ is just $\partial_x^2$ and not the $\partial_t - \partial_x^2$ that $W$ belongs to.
Would it be convenient to include an automatically generated comment when voting to close a question as homework-like? The way things are set up now, if I see a question that runs afoul the homework-like policy, I just cast a downvote. Some people less lazy than me put a comment that looks like
...
So I recently read the following statement that I've heard many times and internalized without ever understanding
If a "symmetry operator" (so an operator that commutes with the Hamiltonian) annihilates the vacuum, then one-particle states must furnish a representation of the symmetry." Do you know how to show it?
Is it correct to translate this statement to simply "symmetry operators do not change particle number"?
@Danu It's something Witten once said and everyone else is just mindlessly repeating :P (as can be seen by the odd use of "furnish" even if they never use the word "furnish" for any other kind of representation) I've once failed to prove that statement. David Bar Moshe has given a better possible answer here
@Danu Yes. And I think there is a way to resolve this: Particles are irreps of the Lorentz group. Any symmetry that commutes with the Hamiltonian must also commute with all Lorentz generators (this is Coleman-Mandula). But if the symmetry commutes with the Lorentz generators it clearly must carry all irreps to themselves.
@Danu Annihilation of the vacuum is also implied by C-M - since the vacuum forms a one-dimensional irrep, the generators of all symmetries acting on it are trivial.
If it doesn't annihilate the vacuum, it carries the vacuum out of its irrep.
Hi guys! Given a gauge group made up of two simple gauge groups, say G = G1 \times G2, is it generally possible to have situations in which just one of its subgroup confines? I would say it's possible. Indeed, the Standard Model with SU(3)c is an example... at order $\Lamba _{QCD}$ quarks confine and mesons appear in the spectrum which transform under the unbroken $U(1)_{em}$ but in this example the second subgroup (SU(2)XU(1)) is spontaneously broken.
Hmm, @Danu, now I don't know how to argue the image of the 1-particle irrep is, in fact, a 1-particle irrep. The multi-particle reps should decompose into sums of irreps after all - how does one know the 1-particle irrep is carried to another 1-particle irrep?
@DanielSank I guess you might be interested in "Path Integrals in Physics" by Chaichian & Demichev. The first half of the first book is just classical stuff; about Wiener path integrals and ways to compute them (mostly working out the discrete case and taking limits)
@Danu What works? So far I see that his answer is the same as mine, minus mentioning Coleman-Mandula.
He doesn't give any explanation why the 1-particle irrep needs to be carried to a 1-particle irrep and not an irrep that sits inside a many-particle state.
There's nothing about a representation that uniquely signifies it's "one-particle" or "many-particle" - after all, all information it carries is spin and mass.
So I understand that you can look at the $0$-momentum states, then $Q_i|\alpha\rangle=0$ for some $i$ implies it for all $i$ since $Q_i^2=H$ for all $i$ and the $Q_i$'s are self-adjoint. So he picks one of the $Q_i$'s, calls it $Q$ and writes $Q=Q_++Q_-$, where $Q_+$ maps bosons to fermions and $Q_-$ must be its adjoint.
Then, for the existence of a state such that $Q|\alpha\rangle=0$, it is sufficient to have nonzero index for $Q_+$. No problem there. Now, he says that this index can be expressed as $\operatorname{tr}(-1)^F$
Trace over what? Why do all fermion states contribute to this trace? Or do only the ones that are annihilated contribute somehow?
I vaguely recall some statement from another book that says that for the non-vacuum states there's always a boson-fermion symmetry so those don't contribute. But I don't recall precisely what the idea is.
Right, it's because for $E>0$, $Q$ yields an isomorphism between the bosonic and fermionic Hilbert spaces at energy $E$ (page 189 of Hori et al's Mirror Symmetry book).
in my lecture notes in a passage we basically have these equality: $\pi^i \pi^j = [\pi^i, \pi^j] $ . Now I think that to justify this passage the operator $\pi$ has to be antisymmetric and therefore the anticommutator between $\pi^i and \pi^j$ has to be 0
@Runlikehell Write $\pi^i\pi^j = \frac{1}{2}(\{\pi^i,\pi^j\} + [\pi^i,\pi^j])$, which is true just by definition, then use that the sum of something anti-symmetric times something symmetric is 0.
[If] one observer is a timelike, future pointing worldline ...
There are indeed different interpretations of the notion "observer" in use; as concisely contrasted by this Wikipedia page as
(a) "observer referring to an (inertial) reference frame", or
(b) "observer referring to an individu...
[If] one observer is a timelike, future pointing worldline ...
There are indeed different interpretations of the notion "observer" in use; as concisely contrasted by this Wikipedia page as
(a) "observer referring to an (inertial) reference frame", or
(b) "observer referring to an individu...
