Physics Asked by Limechime on March 7, 2021
I am trying to understand some calculations to get the excitation energy $Delta E_text{M} = E_text{M} – E_0$ (M is the number of domain walls) in the 1d Ising model in the absence of a magnetic field:
$$ H = – Jsum_{i=1}^N s_i s_{i+1} $$
As far as I know, you can take two approaches — open boundary conditions and periodic boundary conditions. Open boundary conditions means that the Hamiltonian becomes
$$ H_text{OBC} = – J ( s_1 s_2 + s_2 s_3 + dots + s_{N-1} s_N )$$
while the periodic boundary conditions lead to
$$ H_text{PBC} = – J ( s_1 s_2 + s_2 s_3 + dots + s_{N-1} s_N + s_N s_1 ) .$$
This leads to different ground state energies
$$ E_{0, OBC} = – J ( N – 1 )$$
and
$$ E_{0, PBC} = – J N .$$
If I understood everything correctly, every domain will contribute an energy $- (L_i – 1) J$ (where $L_i$ are just the number of spins in the domain $i$) plus an additional $+J$ per domain wall in the OPC leading to
$$ E_{M, OPC} = – J sum_{i=1}^M (L_i – 1) + J M = – JN + 2JM = E_{0, OPC} + Delta E_{M, OPC} $$
but in the PBC, I would get an energy $- J L_i$ per domain plus $0$ per domain wall since the energy contribution is $propto – JN$. This leads to
$$ E_{M, PBC} = – J sum_{i=1}^M L_i = – JN = E_{0, PBC} $$
I am really confused by this and I feel like I’ve mixed things up quite a lot. I was expecting both results to match (or at least to match in thermodynamic limit).
Simplify your life by rewriting your Hamiltonian as $$H_{rm OBC} = -J (N-1) -J sum_{i=2}^N (s_{i-1}s_i -1) $$ and $$ H_{rm PBC} = -J N -J sum_{i=1}^N (s_{i-1}s_i -1), $$ where in the latter sum I have set $s_0equiv s_N$. In the ground states, $s_{i-1}s_i = 1$ for all $i$ and you recover your formulas for $E_{0,{rm OBC}}$ and $E_{0,{rm PBC}}$.
Now, if you have $M$ domain walls (that is, $M$ pairs of neighbors with disagreeing spins), then the energy clearly increases by $2MJ$, so $E_{rm M, OBC} = E_{0,{rm OBC}} + 2MJ$ and $E_{{rm M, PBC}} = E_{0,{rm PBC}} + 2MJ$.
Answered by Yvan Velenik on March 7, 2021
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