3. Consider the simplest finite 2D lattice, namely 4 spins arranged in a square.
ID: 1769404 • Letter: 3
Question
3. Consider the simplest finite 2D lattice, namely 4 spins arranged in a square. (The spins can point up or down perpendicular to the plane of the square, and B is in the up-spin direction.) Explicitly enumerate all possible energy states, and use these to work out the partition function and the net magnetization M as a function of temperature. Comment on whether this shows a ferromagnetic phase transition. When you enumerate the energy states, keep in mind that only interactions between adjacent or nearest-neighbor spins should be counted (no interactions across the diagonals).Explanation / Answer
Momentum conservation
On the lattice, a momentum state |k? can mix off the potential with momentum states |k?? only when |k?? has the same eigenvalue for the operators Tx and Ty, where
Tx=eiaPx
Ty=eiaPy
Since Tx and Ty commute with the full Hamiltonian. This means that a momentum state only mixes with other states which acquire the same phase for x and y translations by a, and this justifies Bloch's form for the eigenstates--- they are periodic in x and y up to a phase.
The mixing in general will split the energy of the states, which gives rise to the bands. The bandgaps occur at the exact point where the phase acquired when you translate one of x,y by a becomes equal to (an integer multiple of) 2?.
To see why, it is easiest to start with a tight-binding model. Consider a potential which is a deep square well with very high walls, but periodic in x and y. The bound states are only mixed by tunneling, so that the bands are very narrow, their energy is nearly the same as the bound state energy, and these states are mixed with a phase which varies in a linear way from well to well.
If you give a periodic boundary condition on x which says that ?(x+a,y)=ei??(x) where ? is close to 2?, but slightly less, then ? in each well is very close to an energy eigenstate, say the ground state, but with a little twist in the tunneling region. The moment this phase ? equals 2?, however, there is the requirement that the result must be orthogonal to 0 phase shift, and this condition was absent just before.
It is impossible to satisfy this condition when ?=2? by using a state which is close to the ground state. You must use the first excited state to continue past this value of k. This gives rise to the gap discontinuity.
The discontinuity appears immediately when you add a periodic potential. It always occurs at the first point where you move from the solution being the lowest energy state with the given lattice phase condition to the next lowest energy state with this phase condition. This happens exactly at the boundary of the k-space of the lattice.
Separable potentials
In this case, there is a zero-counting theorem for arranging the eigenstates. The n-th eigenstate has n zeros. In the Bloch case, the parameter "n" is the number of 2? twists of the phase as you scan from left to right. The first band has no twists, the second band has 1 twist, etc.
The additivity of the energy means that the 2-d bands are sums of the bands of each motion separately. I don't know what more there is to say.
General theorem
it should be possible to prove that the energy as a function of k is continuous except at the boundary of the lattice k-space region It is obvious physically, but I didn't give a proof.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.