Recent content by vkroom

The condensation of Cooper pairs is a property of their statistics (Bose-Einstein ). It has nothing to do with interactions between them. In reality I think one requires some sort of repulsive interactions for condensation to occur. But at an elementary level one may think of condensation of...

all electrons in a lattice cannot have their momenta go to zero due to Pauli exclusion principle. the lowest energy electrons are actually the one that are most easily excited. ironically they have a very large velocity called Fermi velocity. For most metals its around 10^6 m/s. Cooper pairing...

Cooper pairs are *dynamical* pairing of fermions. Cooper pairs being bosons ultimately condense in momentum space. There is not much use in considering pairs of bosons with \pm \vec{k}{} since most of them will end up having the same momentum due to condensation.

In general Fermi liquid description is valid when 1. the interaction is short ranged. 2. the interaction energy scale is smaller than the kinetic energy of the electrons. Although for short range interactions one can still get fermi liquids at relatively high (interaction energy) /...

You don't want it to vanish. All I meant is, by that particular choice of operator \hat{O} your Hamiltonian becomes H = H_U + \mathcal{O}(t^2), which is what you deduce below. When written in this way one can see that the \mathcal{O}(t^2) are perturbations about the Hubbard U term, i.e. your...

upto first order in t: H_U + H_t - t[O,H_U] - t[O,H_t] ... (Eq 1) Now choosing H_t - t[O,H_U] =0 \Rightarrow H_t = t[O,H_U] Put this back to Eq. 1, and you'll have no order t term left.

as one approaches a critical point the correlation length starts increasing, i.e. units farther and farther away becomes aware of each others dynamics. in math terms the correlation functions \sim e^{-r/\chi}, where \chi is the correlation length, and r is the distance from your origin. this...

I think we are messing up something here. \gamma \rightarrow e^- + e^+ is mass creation from energy: E=m_0 c^2 + p^2c^2. If the photons have less energy than twice the rest mass of electrons this creation process will not occur except maybe as vacuum fluctuations.

I've been thinking on similar lines for sometime. I was trying to use some differential geometric ideas to construct local field theories. Do you think there's some reference out there that I may be able to look into regarding these matters or, would it be possible to mention some people who...

Your doubt is valid. The 'J' infact destroys the Kondo screening. In the dilute impurity limit one can devise the Kondo argument using lattice version of the Anderson Model. As 'J' is tuned up the Kondo screening becomes poorer and is speculated to either suppress it completely or at least...

X. G. Wen's book might help you in this regard. Look into the chapter on 'Spin Liquids'.

Are you sure? Fermi liquid theory works so well because of its stability to perturbations. For generic momenta even four fermion interactions are irrelevant. Only the Cooper channel makes four fermion interactions marginal and hence we get superconductors. This is one of the very few...

Consider the infinite U limit. Here the electrons are frozen at the lattice sites because the U prohibits double occupancy, even as a virtual state. Now if you tune U to a finite value then electrons start hopping in the process getting delocalized, thus lowering its energy. The sign of t...

because most lattices are translation invariant, momentum is a "good" quantum number, i.e. it is conserved. this leads to a diagonal representation of all other conserved quantities in the momentum space, which is especially simple to work with.

Yes! Theoretically one cannot observe beyond the Planck`s length because the uncertainty principle will imply energies large enough to form black holes. Consequently one may not obtain information beyond such black holes. In this sense with our current understanding of Nature one may assume...