@StanShunpike That's why I set the "equation of motion" in quotes - in ordinary QM, there is no equation of motion in the classical sense because there are no trajectories to be determined - there is no motion.
@StanShunpike Well, I'd be interested in anyone's opinion.
@StanShunpike Yep. And now we have these states evolving in the Schrödinger picture or these operators evolving in the Heisenberg picture. You might call the time evolution equations "equations of motion", but they are not the classical e.o.m.
And, in QFT, the classical e.o.m. of the field gain again a certain relevance, so you have to be careful what people mean when they say "equation of motion".
It is well-known that the Earth's magnetic field may flip (geomagnetic reversal), and some data suggests it is weakening or distorting. The following shows the Earth's magnetic field at its surface in June 2014 (by ESA) which shows some distortion.
My question is, can such changes in the Earth...
@infinitesimal. Super introductory means I have a physics degree, but have never studied gauge theory, and know math up to calc 3, diff eq, and linear algebra
@Sean Well, I learned it "on the way" when doing quantum field theory. And that I mostly pieced together, I know of not many resources that are explicitly about gauge, and none introductory
@ACuriousMind Tonic water traditionally has some sweetner in it, to cover the bitterness of the quinine. Now, I don't like mine very sweet, but there should be a hint of sweetness in there.
@ACuriousMind I think asking about a paper is off topic. Based on my experience, a good question should be (1) relatively self-contained (in case the link breaks) and (2) show some effort to understand the derivation by specifying the particular problem they are having. That link you gave doesn't do either of those.
@ACuriousMind Therefore, I think really the person should give more thought. Asking SE people to write an entire derivation is a lot of work and I don't think is in the spirit of answering as a means to teach.
@ACuriousMind So the Schrodinger equation tells us how the wave function evolves. And from this, we can generate expectation values for the classical quantities like position and momentum. So the Heisenberg Uncertainty principle doesn't prevent us from making an average. But the gist of the idea is that those quantities don't actually exist as definable information.
@StanShunpike Yes. For states that are not eigenstates of the respective operators, the quantities the operators measure do not exist as well-defined numbers for the states as such.
Yes, or your positions and momenta are not actually meaningful ;)
But, in the case of gauge symmetry, you often have the case that your position and momenta contain redundant information (this is again this mess with the constraints). One can go to a so-called "reduced phase space" where every position and momentum is invariant, and meaningful
@ACuriousMind The running out of close votes reminds me of the NFL Salary cap. The salary cap fluctuates with the total revenue of the league. I feel like close votes should be the same way, if we observe many questions needing to be closed, the total number of votes should increase during an influx and the go down once the influx is over.
Because that question clearly is off topic.
Not that I really care, but it does seem like a waste of effort if you read it and can't close it.
I fear I will become like that dude on math.SE that uses his 40 votes every day to downvote bad closed questions so the roomba (automatic deletion routine) eats them
@ACuriousMind I'm still confused about gauge symmetry. Like the simplest one I gather is electric potential. obviously, the idea of differences in electric potential comes from the Gradient theorem. So there is a symmetry under V --> V + C yield the same electric field?
A global symmetry is just given by applying an element $g$ of the symmetry group to everything: $\phi(x) \mapsto g\phi(x)$, while a local symmetry is given by choosing an element $g(x)$ of the symmetry group for every point and then applying it: $\phi(x) \mapsto g(x)\phi(x)$.
Yes, it is one way of writing it. You will often consider the infinitesimal version $\phi(x) \mapsto \phi(x) + \epsilon(x)\phi(x)$ instead, where $\epsilon$ now goes into the symmetry algebra instead of the group
(Also, people like to introduce various factors of $\mathrm{i}$ and $g$ and $\hbar$ here, which doesn't make it easier to follow anyone)
@StanShunpike that depends on the $\phi$ ;P
Whenever you give a field, you also have to specify in which representation of the group it transforms
Often, you will have your symmetry group given as a matrix group, say $\mathrm{SU}(N)$
And stuff in your theory then either is an $N$-vector that transforms by multiplication with the matrices, or just a number, in which case the symmetry group does nothing to it.
@0celo7 Strictly speaking, nothing, I think, and that's why the mathematical approaches first construct the Fock spaces and then define the fields out of the creation/annihilation operators
Or rather, you say "The KG holds as an operator equation".
Classical fields, like the electrical field must vanish at infinity, because otherwise their energy would be infinite. This can be used in computations to exclude certain solutions.
In quantum mechanics, the wave function must be normalized, because of the probalistic interpretation. (The proba...
CFT question: (Ignoring the atomic scale) does water look the same at all length scales at 200'F the way it does at boiling point 212? If it's in equilibrium, why isn't it invariant under a scale (conformal) transformation!? :(
@ACuriousMind (1) why isn't something more specific specified? that seems a bit vague. any real valued function? and (2) what is the $\phi(x)$ then? That's the field correct? And we are applying a local gauge transformation $g(x)$ yielding $\phi(x) \mapsto g(x)\phi(x)$.
for (1) i meant why isn't something more specified for $\epsilon$
I think Stokes, Green etc... do not hold on unbounded domains, and physically it doesn't make sense for a physical field to be anything other than 0 at infinity
@0celo7 Careful, the gauge transformation group is not the gauge group - the gauge transformations are the smooth functions into the gauge group, and the group of gauge transformations is always infinte-dimensional
@bolbteppa Yeah, that's basically my third argument - the problem is that "is 0 at infinity" is not a rigorous statement - you need to give the limit behaviour.
@ACuriousMind I think it's larger in the sense that strings have infinite spins, whereas one Yang-Mills theory in QFT has one spin.
Oh you meant the functions.
@ACuriousMind But what if we have the functions into group A and the functions into group B. If A is bigger than B, then are the functions into A bigger than than the functions into B?
@0celo7 ST has gauge groups like $E_8 \times E_8$, which is indeed bigger than anything you usually get in QFTs. I'm not sure what they'd have to do with spin, because one usually doesn't count the spacetime (reparametrization/Lorentz) symmetries as "gauge symmetries", though you technically probably could
@0celo7 You have one state for every momentum, so you've got infinitely many states for one particle, too. But probably only countably infinitely many normalizable states...
@ACuriousMind Does the Heisenberg uncertainty principle mean that we cannot have both position and momentum as observables for a particle ever or just for a given moment in time?
Yes, infinities are silly. If you ever are tempted to look into cardinalities, the continuum hypothesis and all that, it's a rabbit hole you'll never emerge from again
@StanShunpike You have them both as observables, always. The operators don't care what states they act on. The HUP just tells you there are no simultaneous eigenstates (it's just a quantification of the statement that non-commuting operators can't have a basis that is an eigenbasis for both, essentially)
It is well-known that the Earth's magnetic field may flip (geomagnetic reversal), and some data suggests it is weakening or distorting. The following shows the Earth's magnetic field at its surface in June 2014 (by ESA) which shows some distortion.
My question is, can such changes in the Earth...
Classical fields, like the electrical field must vanish at infinity, because otherwise their energy would be infinite. This can be used in computations to exclude certain solutions.
In quantum mechanics, the wave function must be normalized, because of the probalistic interpretation. (The proba...
