Recent content by saaskis

The equation \langle \psi|AB|\psi\rangle = \langle \psi|A|\psi\rangle \langle \psi|B|\psi\rangle \ \forall \psi also leads to [A,B]=0. But assuming that [A,B]=0 and that both are usual Hermitian observables does not seem to imply the above equation for general states, even though any state...

I don't have Datta now available, but if I remember correctly, he showed how to derive the "Bloch" Hamiltonian for general systems. If you understand what he does, you should be able to reproduce the same calculation for graphene. You know how to derive the bandstructure for a unit cell of two...

I thought that increasing the size of the unit cell simply folds the bandstructure? So I cannot see how this folding would lead to an opening of a gap, that does not sound very physical to me... But for a unit cell of 8 atoms, you have to construct the effective Hamiltonian Heff(k), which is...

If you are worried about the standard phrase "let us study spinless fermions" that you can find in many books on QM and condensed matter, then my answer would be no, there is no contradiction in imposing anticommutation relations on spinless fermions. Very often in non-interacting problems the...

This may confuse someone, but here's my way to think about the prefactors. There are 3*3=9 ways to connect the vertices together. After this, there are 4! ways to connect the external points to the vertices (since there are 2+2 free fields left in the vertices). This means that there are...

I don't quite understand what you mean by "twisting" the diagrams to get the three groups, but afaik, the three groups roughly correspond to the "s", "t" and "u" channels familiar from QED. And to see to which group a given connected diagram belongs, you simply have to look at the four-momentum...

Whoops, if you are not that familiar with the second quantization, we of course have to solve the problem without it first :) Earlier you asked about using the Pauli paramagnetism formula, and yes, you are correct, the formula is applicable. But you should really try to understand the...

Okay, so the relevant quantum numbers are the lattice momentum k and the spin s. The free Hamiltonian is H=\sum_{k,s} \epsilon_k c_{ks}^{\dagger} c_{ks}, and now you want to add a magnetic field h to the system, say along the z-axis. What is hS^z_{tot} written in terms of c_{ks}^{\dagger}...

Hi! The simple way: calculate the magnetization <m> for the Hamiltonian H'=H-hm (where H is your original Hamiltonian and m is the magnetization operator, usually sum over spins), differentiate the result with respect to h and set h equal to zero. You could also determine the...

You achieved positive MR only after Fe was added, so it's not a weak antilocalization effect in carbon? EDIT: Probably not due to room temperature...

I guess this is true (I don't know anything about quantum gravity), but one has to notice that non-renormalizable theories are not useless as effective field theories. Since non-renormalizable vertices have a coupling constant with a negative mass dimension -d, the coupling constant is something...

Classical statistical field theory is already good enough. The field Hamiltonian is something like \beta H = \int dx \left[(\nabla m)^2+M^2 m^2\right] \sim \sum_p m(-p) (p^2+M^2)m(p), which is the simplest rotationally invariant Hamiltonian available. The partition function is Z = \int Dm(p)...

The previous answer may be more rigorous, but this here is my way to understand it. Since the number of photons can fluctuate freely, the average number of photons is obtained by minimizing the free energy with respect to N, i.e. by setting dF/dN=0. But by statistical mechanics, dF/dN at...

The actual diagrams are the same in every basis, only the mathematical expressions representing the diagrams are different. As for your question about the internal electron line: if two diagrams only differ by some sort of rotation of propagator lines, the diagrams are the same (topologically...

The main reason is that in a translationally invariant system, Green's functions are diagonal in momentum space. This simplifies all calculations and turns matrix equations (e.g. Dyson equation) into algebraic equations that can be easily solved. The diagonality of correlation functions is one...