Recent content by The_Duck

It doesn't matter. But if your potential goes to some finite value at ##x = \pm \infty##, it's conventional and convenient to set that value to zero. That's what we do for the finite square well. For the infinite square well this is impossible, because the potential is infinite outside of a...

He is using the relation \ln(1 + x) = x + O(x^2) Here ##x## has an ##\alpha## in it, so the ##O(x^2)## term is absorbed into the ##O(\alpha^2)##.

I don't think so. A spin zero state is completely rotationally symmetric, so can't have a moment pointing in any particular direction.

Write down a general ##2 \times 2## matrix as \left(\begin{array}{cc} a+bi & c+di \\ e + fi & g + hi \end{array} \right) Now require the matrix to be self-adjoint and traceless. What constraints does this put on ##a,b,\ldots,h##? Try to see how the resulting matrix can be written as a linear...

Maybe you can be more specific. What field theory are you studying? What quantity disagrees between the lattice and the continuum? If there is disagreement between lattice and continuum results, that just means you have failed to construct a lattice version of your field theory.

Any values for ##A## and ##B## that satisfy ##A^2 + B^2 = 0## give a solution. For each ##k## there is an infinite family of solutions. You can think of them all as linear combinations of the "basic" solutions ##\sin(kx)## and ##\cos(kx)##.

No, there's no reason to set B to zero. Assuming ##x## here is the angle around the circle, your constraints are: * ##\psi(0) = \psi(2 \pi)## * ##\frac{d}{dx}\psi(0) = \frac{d}{dx}\psi(2 \pi)## * ##\psi(x)## should solve the Schrödinger equation If you impose these constraints you will find...

The point is that the total mass of a composite system, such as a composite particle, is exactly the total energy of all its constituents, via ##E = mc^2##. If quarks have constituents, those constituents must have kinetic energies of order ##10^6## MeV or greater (I'm actually not sure what the...

I have read that page but unfortunately none of the critical exponents listed seem to be related to the heat of vaporization (or if they are it isn't obvious to me). Just now I have found some papers that say that empirically it is indeed a power law: {\rm heat}\,{\rm of}\,{\rm vaporization}...

The Wikipedia page on the heat of vaporization has a nice graph showing the heat of vaporization going to zero at the critical points of various substances. Is there a known form for the heat of vaporization as a function of temperature near the critical point? I imagine it is probably a power...

Sure. An overall shift in the energy density *does* affect gravity, by effectively shifting the cosmological constant, as I described above. So gravity is sensitive to the absolute energy density. One way of thinking about it is the following. Consider electromagnetism. In electromagnetism the...

Right. Have a look at the action for classical gravity. Here ##R## is the Ricci scalar, ##\Lambda## is the cosmological constant, and ##\mathcal{L}_M## is the Lagrangian density for matter and fields (excluding the gravitational field). If we change the vacuum energy of some field, this is...

Yes. Since we have the freedom to shift the Hamiltonian without affecting the dynamics, we can choose to shift it such that the energy of the vacuum state is zero, or any other desired value. So the absolute vacuum energy is arbitrary and it's fair to call it unphysical.

The Schrödinger equation is ##i \hbar \frac{d}{dt} | \psi(t) \rangle = \hat H |\psi(t) \rangle## Suppose we have some solution ##|\psi(t)\rangle## to this equation. Now let's add an overall constant ##\delta## to the Hamiltonian: ##i \hbar \frac{d}{dt}|\psi'(t)\rangle = (\hat H + \delta) |...

Let's write down a possible field history in which a soliton at ##t=0## goes to the vacuum at ##t=1##: \phi(x,t) = (1-t)v\tanh(x/\xi) + tv The starting configurationi at ##t=0## and the ending configuration at ##t=1## both have finite total energy. But If you calculate the energy of the field...