Well, terms with an odd number of operators are out immediately if the particle number is conserved. The two-operator term represents direct hopping, the lowest-order process, which presumably is the most important one. You can of course imagine an electron hopping from site 1 to site 2 and then...
You're right. I guess I should have said that the wave function must be square integrable and satisfy all other physical conditions of wave functions. (It certainly would be weird if the probability of finding the particle infinitely far away is non-zero.)
The idea is that the wave function must be square integrable, so it must fall off to zero faster than functions which aren't.
Or put another way, integrate \left( 1/\sqrt{|x|}\right)^2 and see what you get.
You don't need to calculate it directly, though it is a good exercise.
First, recall that the hopping term looks like c^\dagger_{i,\sigma} c_{i+1,\sigma} (specializing to 1D for simplicity) and assume that we start at half-filling. If we assume (like Altland and Simons do) that U\gg t, then...
I'm not familiar with this particular model, so I don't know the connection with the fractional quantum Hall effect or the eigenstates. However, it seems quite clear to me that while the spins themselves are at fixed positions, the dimerized bonds are free to move (as in changing which spins are...
The hydrogen eigenstates have specific angular momenta, so they must be eigenstates of the angular momentum operator. Does that operator commute with the position operator? Also check ChrisVer's point, does position really commute with the Hamiltonian in this case?
As far as the Zeeman effect and magnetism goes, I think people often tend to think of spin as a classical vector which points in some direction. It's not strictly right, but for many intents and purposes it works just fine as one tends to be more interested (at least in these applications) about...
Generally, one would treat the case of R=0 separetely, and solve the differential equation for this case too, which is obviously permitted. Your method is essentially the same, if one matches the solutions using the smoothness property (the derivative of the wave function is continuous too)...
It's quite tricky to find a simple and general criterion, but I remember seeing it be done for classes of equations, e.g. diffusion equations. If you are looking for a general answer, it probably will have something to do with symmetries and be rather distanced from how physicists would tend to...
Well, as has been pointed out earlier in the thread, it could be a classical field theory (such as classical electrodynamics) as well. The same ideas of an infinite number of degrees of freedom and a Lagrangian density apply, because it is still about a field. When you quantize the classical...
You can think of the Lagrangian density this way:
Probably, the first Lagrangian you wrote down was for a single particle,
L=\frac{m\dot{q}^2}{2} - V(q).
This L determines what the particle will do and where it will go for all times. For a system of particles, the Lagrangian is a sum of...
Haha, yeah, that's unfortunate.
Looking at your questions again, I think I can actually say something. Bear in mind though, that my answers might be influenced by my lack of knowledge of the problem at hand.
I think so. They were certainly meant to be interpreted that way in the original...
I'm still trying to make sense of that matrix in the context of contextuality vs. non-contextuality, and can't really answer your questions. The Wikipedia page seems rather inpenetrable to me as well, but it links to this blog post, which still seems a bit confused about the whole issue...
That formula seems to have an operator on one side, and a state on the other side. In general,
|k \rangle = \sum_r0 |r\rangle \langle r| k\rangle
which indeed follows directly from the resolution of identity. Taking |k\rangle and |r\rangle to be our one-particle basis states, i.e.
|k\rangle =...
We can do that. What book are you using? This is described quite well in Altland & Simon's book Condensed Matter Field Theory, for example, if you need a second resource.
Let's start by looking what these things actually means. a^\dagger adds a particle in some state, and a removes it, right...
If the operators represent different particles, then yes, they will commute. For creation and annihilation operators, this is apparent through the commutation relation
\left[ a_i, a_j^\dagger \right] = \delta_{ij}
The reason this works is that each particle has its own "Hilbert space" (a Fock...
While one can go as deeply into the formalism as one likes, the best non-technical explanation I've heard is that the Schrödinger equation is a wave equation. For all waves, the amplitude squared gives an intensity. In quantum mechanics the "intensity" is the probability of finding the particle...
On the ground that the operator can be written H=\alpha_i^\dagger H_{ij} \alpha_j. If H_{ij} is diagonal, so that
H=\sum_i \alpha_i^\dagger H_{ii} \alpha_i = \sum_i \epsilon_i \alpha_i^\dagger \alpha_i ,
it makes sense to call the operator diagonal. There is also the argument that a diagonal...
I think it is slightly off the topic Mandragonia eventually wanted, but you might find the Wigner function interesting. It is basically an attempt to provide a probability distribution in phase space, but in some regions associated with unphysical states, the function takes negative values.
If you deal with x and y, you only have two (real) field degrees of freedom. Hence, if you pass to some new fields L and R, you must somehow get the same number of field degrees of freedom. This could in principle be done by letting L and R be independent complex fields (4 field d.o.f.) with two...
You can think of a complete set of states as a basis in Hilbert space, much like an ordinary (geometrical) vector space. A state of the Hamiltonian H0 can be seen as a vector in this Hilbert space, which may be one of the basis states, or a linear combination of several of them. So if you change...
I guess that is the more common way of putting it, yeah.
I don't have an explicit example, but couldn't we have a local transformation y=f(x) that doesn't change the endpoints (i.e. limits) and J=\frac{\partial f}{\partial x} \not\equiv 1?
I think your statement should hold for linear...
I think you miss that, in general, \mathrm{d} \phi \neq \mathrm{d} \phi'.
For a simple example, consider S=\frac{1}{2}\phi^2. Obviously, this is invariant under the inversion \phi \rightarrow \phi'=-\phi, for which \mathrm{d}\phi'=-\mathrm{d}\phi and J=\frac{\partial \phi'}{\partial \phi}=-1.
It is not mentioned in the errata (for future reference, it is found at http://www.physique.usherbrooke.ca/pages/senechal/cft ), however I wonder if there really should be a negative sign in the exponent in eq. (2.69c). I don't have time to check it carefully now, but compare, for example...
Hello!
On p.424 in the second edition of Altland & Simons, they compute an average as part of a renormalization group analysis of a model for dissipative quantum tunneling. I'd like to use their result in another situation, but I would be much happier if I understood how they derived it. I have...