Recent content by arojo

Hi DrDu, Actually I started by doing precisely that, but I got a messy result. Which certainly is analytical but hard "to read", at least from the point of view of getting an idea of what is going on without doing the numerics. Actually I should rephrase my question as is there any elegant...

Hello all, I am developing a model of multiple gaps in a square lattice. I simplified the associated Hamiltonian to make it quadratic. In this approximation it is given by, H = \begin{pmatrix} \xi_\mathbf{k} & -\sigma U_1 & -U_2 & -U_2\\ -\sigma U_1 & \xi_{\mathbf{k}+(\pi,\pi)} & 0 &...

Hi, You should think this phenomena not in the classical picture but as a quantum phenomena where this effect is due to a Macroscopic Wave Function, just as an example you can think as an electron "turns" around the nuclei, it does not loss any energy. Finally remember that in the Josephson...

Actually I am not a supporter of the High Energy Physics (HEP) but when comment results we have to be a little bit more objective and to comment the technicalities of the problem we should be more formal about the physics and mathematics behind it. I am not going to explain in full detail how...

Hi, I do not know the general answer, but if we limit ourselves to the [p,q] = ih case, as it was remarked earlier in the discussion you must know what does it mean \langle x|i\hslash|x\rangle? Actually if \langle x| x\rangle is the probability for the particle of being at x, then the...

Hi, I think your question does not have an easy answer, the raison why bosons and fermions are different is their statistic. From this we get the fermi and bose statistic, which is related to their spin. The Pauli exclusion principle tells us that the wave function of a set of fermions must...

Hi, You do not hold x fixed, it is just a change of variables, if you calculate explicitly the Hessian you will see. Actually you must do this using the 5 variables, x, y1, y2, y3 and y4.

Thanks

Hi, There is no problem in the equation, the density matrix \rho is a diagonal matrix (with the probabilities of each state in the diagonal), that is why you have just a sum over one index. Actually you could see it otherwise, an Observable with physical must be a real quantity, then what...

Hi, I do not know the answer to your question, but I think there is a misunderstanding, the fact that two particles in a system does not mean that the ensemble will behave as a new particle (either like boson or a fermion). This only happens when the mentioned particles are bounded to each...

I agree with you that there is a problem, but the definitions I am using are the most standard, actually for confirmation I took them from Bruus "Many body quantum theory in condensed matter". So in principle the definitions are right, besides error of my part recopying them. Then if we change...

Actually the problem is little bit different. It will be more like: \gamma_{-k,-\sigma}^{\dagger} = \gamma_{k,\sigma} Which could have some physical meaning as: Annihilating a particle with a momentum k and an spin \sigma is equivalent to create a particle with a momentum -k and...

Maybe I was not very clear, my question is not related to the fact if it is allowed or not to follow the procedure I propose. My question is why we do not get equation (2) from (1) if we do the proposed change of variables?

I am not sure of your comment, because if I follow the definitions for α and β, I can do the following \epsilon_{k}^- = \frac{\epsilon_{k} -\epsilon_{-k}}{2} So if we replace k → -k we obtain: \epsilon_{-k}^-= - \epsilon_k^- For \epsilon_{-k}^{+}= \epsilon_k^+ , therefore is we...

Hey, I received some comments asking for a more clear latex expression. So equations (1) and (2) become: \gamma_{k, σ}=\alpha_{k}^{*} c_{k, σ}+\beta_{k} c_{-k,-σ}^{\dagger} (1) \gamma_{-k, -σ}^{\dagger}=-\beta_{k}^{*} c_{k, σ}+\alpha_{k} c_{-k,-σ}^{\dagger} (2) Where c_{k, σ}...