Thanks, François. That's insightful.
In any case, it's easy to see where the SU(N) comes from, but not terribly illuminating. If you work in the holomorphic representation, then the Hamiltonian is H = a^\dagger a which clearly has an SU(N) symmetry. This is also a very elaborate way of...
If you did some quantum mechanics, then you might recall that the angular momentum states group themselves into multiplets labeled by j. Same thing is going on, except that for groups other than SU(2), you get more labels.
Blegh. It's a helluva long road. Quickest thing to do is to fix a gauge like the Lorenz gauge, and then find out what the Hamiltonian is. The factor of -1/4 will give you a canonical Hamiltonian that is something like \int\!dx\, \left(\frac{1}{2} \mathbf{E}^2 + \frac{1}{2}|\boldsymbol\nabla...
The point is that you were never performing an infinite sum to begin with (the theory isn't defined to arbitrarily short wavelengths). If you had a theory defined to arbitrarily small wavelengths, then the actual expression would look like your sum at low orders, and look like something else at...
I think you're better off using [A,BC] = [A,B]C + B[A,C] and then integration by parts. Also, it's a good time in your life to realize that a commutator behaves very much like a functional derivative (recall QM 101).
A neutral gold atom has 79 electrons and mass of 3.27x10^-25 kg. As a metal, gold has a density of 19300 kg/m^3. So there are 5.90x10^28 atoms and 4.66x10^30 electrons in a cubic meter of gold. This is the "first year" answer. Hope that helps.
My suspicion is yes, but it will be hard. Smells like something that you'd do with Young tableaux.
Edit: I think the CG theorem involves both symmetric and antisymmetric tensors. Do you know how to prove the theorem in the usual way (highest weight procedure)?
You can also ask what is the operator conjugate to the number operator of the harmonic oscillator. Supposedly this is how Dirac found the creation and annihilation operators (though I'm not sure). In any event, you might want to look at the Weyl representation of the canonical commutation relations.
Perhaps best to say that virtual photons are the ones that you don't detect. The distinction is more of a practical one since photons themselves are "mathematical concepts/entities when you do perturbative expansions in quantum field theory". Forgetting this leads to problems like IR divergences...
I have made notes from various sources on this subject if you're interested
http://www.mathematics.thetangentbundle.net/wiki/Differential_geometry/spin_connection [Broken]
http://www.physics.thetangentbundle.net/wiki/Gravitational_physics/fermions_in_curved_space [Broken]
I had to study this...
When Q acts on a the most general function on superspace (superfield) Y = \phi + \bar\theta \psi + \bar\theta\theta F then it interchanges \phi and \psi (I'm ignoring the auxiliary field now), which is a SUSY transformation if you build your action out of superfields. Since you want non-trivial...
You can measure it in Newtons, if you want. But that's not very useful.
The proton's, mass-energy comes form the binding energy of the quarks (negative), plus the masses of the individual quarks (positive), plus their relativistic kinetic energy (positive). The last two quantities are hard to...
Well, it's the other way around. Anyway, you normally specify the incoming momenta, and integrate over the outgoing momenta. The momentum dependence of the 4pt function shows up in it's Fourier transform as well as from the LSZ formalism
I disagree. Diagrams at a certain order have meaning and correspond to something physical, but many diagrams contribute *at a given order in the perturbation series*, and singling out one of them and asking "what does this diagram represent" is not always a good question. For example, IR...
Relativistic quantum mechanics demands a quantized field approach, due to properties of the Lorentz group. So, in QFT, the path integral approach is still used, but it is a sum over field configurations and not particle configurations.
Incidentally, you can incorporate spin in the path integral...
Yes, the Schrodinger equation is invariant under coordinate changes x \to x + b.
You'll also find that if you do the transformation on your eigenstates, you end up with cosines, which are solutions to the SE with the new boundary condition.
LSZ only chops off external propagators, so is no big deal (although the derivation of it is quite awful).
The n-point function, after you throw LSZ at it, represents an element of the S-matrix with given number of in- and out-states. What do single Feynman diagrams correspond to, then...
It's called r because when Schwarzschild wrote the metric, he used spherical coordinates to cover space. But coordinates merely assign numbers to points, and don't talk about distances between them. For instance, I can use polar coordinates on a cone, but circles won't have circumference 2pi r...
Mmm, I would lump de Broglie's under old quantum physics also... but maybe that's just me. Copenhagen was probably the definitive point, where "everything was understood". I don't know if the probablistic interpretation was known to Schrodinger originally.
In any event, Eisberg and Resnick is a...
This is great and all, but it isn't actually correct. I mean, an electron is spread out all over space. It's not like a runner on a race track confined to a single lane. I wouldn't spend too much time trying to visualize Bohr orbits or de Broglie waves since these are hold-overs from when...
Like I said, in the Bohr model, which is based on de Broglie's ideas, things are very sketchy and very classical-mechanics-ish. There it doesn't make sense for angular momentum to be zero, so that orbital isn't even considered. Schrodinger's equation only has harmonic solutions (by that, we...
The whole standing wave argument for the atom is very heuristic and not really the way to go (solving the Schrodinger equation is). For example, the lowest state of hydrogen is spherically symmetrical, and there is no "wave". However, if you're interested, look at the n=2, m_l = +/- 1 solutions...
Kaluza's ideas are pretty neat and are actually used in some string theory contexts. However, there have been many attempts to extend it to include known particles and such attempts have failed. For instance, at the time I don't think he knew about spin or nuclear forces.
Forces are carried by particles, and as a first approximation, the potential experienced by two objects charged under said interaction is given by the Fourier transform of something like \frac{1}{k^2 - m^2}, where m is the mass of the force carrier. For the electromagnetic and gravitational...
I think the point of this discussion is that the latter is not well defined. R that I calculated is simply the geometrical or arclength distance at constant t between the two points. A person hovering above the horizon would slice spacetime differently than Schwarzschild coordinates. For an...
Oops. Thanks George. I was a bit quick to respond. The circumference is always 2\pi r. What I should've said is that the distance between two radial points is
R = \int_a^b \frac{1}{1-\frac{2M}{r}} dr,
which diverges as you take either point towards the horizon. You don't want the points...