Recent content by Sir Beaver

I think the answer is in the definition of 'powerful'. If we would read the word as 'general', then I agree that DFT is more general than Fermi liquid theory, since DFT does not make any assumption on the strength of the interactions involved. I would still have an objection to your example. You...

Dear bjnartowt, Very interesting question! There are several approaches to show if a considered many-body approximation is conserving. Personally, I find the easiest-to-understand method is to consider the time-dependent case, and specialize to the steady-state case afterwards. In this...

As you say, if the integral converges for ##x \to 0^+##, the result is always zero. However, even if the integral diverges, it is still possible that one can get a finite value. This depends on the speed of divergence. Try it out with ##f(t) = 1## and ##f(t) = 1/t##. What happens for ##f(t) =...

I agree with the point that it is not one-to-one in general, and this solves the issue with a string of operators of two or more. However, is it really what I use here? Considering the equality above (which I think we agree on then) $$ T [ \hat{A}_H (t) ] = T [ \hat{A}_S (t)], $$ is it not...

I mean, that if we have bosonic operators, ## A ## and ## B ##, the equality $$ T[ \hat{A} (t) \hat{B} (t') ] = T [ \hat{B} (t') \hat{A} (t) ] $$ holds, even if ## A ## and ## B ## themselves do not commute. Thus, if the operators are under the time-ordering operator, they can be treated...

Hi all, I have a severe confusion about the time-ordering operator. It is the best thing ever, I think, since it simplifies many proofs, due to the fact that operators commute (or anti-commute, but let's take bosonic operators for simplicity) under the time-ordering. However, sometimes I...

A tip which has helped me is to always extract as much as possible before doing the integral. Doing that, you can see that it is actually the same integral as before, e^-9t. The error is simply that you multiplied with 81,while you should have divided.

Well, I am a little confused. The free electrons in a metal makes up the Fermi sea, and one can immediately use them as an electron source by shining light on the metal by photoemission. Was this the question?

I would suggest by starting with the probability, write the probability as \psi^* \psi, and take the time derivative. Make connection with the Schrödinger equation after that.

I think this is really good, both as a reference to integration methods and as theory. One thing I could think of is to be careful about \int _a ^x f(x) dx , to have a different symbol for the dummy variable.

Also, the program will give the same sequence of numbers every time, since one has to initialize the 'seed', which can be done according to CALL SEED(ii), where i is an integer. Different integers will give different sequences of random numbers.

Just to clarify, what hiyok says is that a Hamiltonian which is spherically symmetric will commute with L^2 and Lz. This is not true for any system. Also, in physics, the statements 'Hamiltonian has this symmetry' and 'system has this symmetry' is the same thing. Finally, just because the...

Hello, This program shows the principle of diagonalizing a matrix in fortran, using a 2D geometry. To compile, write gfortran -llapack 2D.f90 then write ./a.out to run the program Or ifort -llapack 2D.f90 module shared !every function and routine which implicit none !includes the...

The homogeneous solution looks correct, although you do not have to differentiate it. You merely have to add the solutions together in the end, y=y_H+y_{P}, so you are done with y_H. In the particular solution you have assumed that the polynomial is p(x)=Ax, which is not the general case. In...

Well, the general solution is the sum of the homogeneous solution and a particular solution. The homogeneous solution (gotten from putting the right hand side = 0) is obtained using the characteristic equation. To find a particular solution, one can make a clever guess how the solution...