Closed JohanCarlsen closed 1 year ago
Hi @JohanCarlsen!
I don't think your expression is correct. Keep in mind that what we define $\Delta E$ as in the project description is the energy change resulting from flipping a single spin, not the energy change from flipping $N$ spins.
You may find it useful to think about all the possible configurations of neighbouring spins that a single spin can be surrounded by.
Hm. Okay. I have done this: $$\Delta E = -J\sum_{kl}^Ns_k^1sl^1+J\sum{kl}^Ns_k^0s_l^0$$ If only one spin (eg. $s_l$) flips, then $s_k^1=s_k^0$, and $s_l^1=\pm1$ and $s_l^0=\pm1$. This means that if $s_l^1=1$, then $sl^0=-1$ and vice versa. This led me to $$\Delta E=-J\sum{kl}^Ns_k(s_l^1-s_l^0)=2Js_l^1\sum_k^Ns_k$$
But you are saying that this is not correct, if so why? I thought I was on to something :'D
Oh wait, is $\Delta E$ the energy difference of a single particle due to its flipping???
The "energy shift due to flipping a single spin", as the project description says. :) The energy change $\Delta E$ is still an energy change in the total system energy $E$ (we don't discuss single-particle energies), but it's the energy change caused by flipping one single spin in the lattice.
Then I'm back to not understanding why my first expression is not correct. The energy is given as a sum over all neighboring spin pairs, so the difference has to be a sum over all spins as well.
If I understand your notation correctly, I think you have mathematically described the situation where we flip one spin for each pair of interacting spins. But the situation you should be describing is flip one spin in the entire lattice. That is, all but four of the spin pairs in the lattice will be completely unaffected by the flip.
Aha! Thank you!
You're welcome -- good luck with the project! :)
Hi! I've been trying to find the five values that $\Delta E$ can take, and ended up with the expression $$\Delta E=\pm2J\sum_k^Ns_k$$ I assume that the sum is the same as $M$(?), but $M$ can take many values, no?