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An interesting mind-puzzle/paradox for the 2D Ising model

+ 2 like - 0 dislike
183 views

Consider a 2D ferromagnetic Ising model below the critical temperature, on a rectangular strip with infinite height but finite width N. However, there is a twist: consider changing all of the horizontally-oriented bonds on one of the (finitely-many) columns to be antiferromagnetic. This special column is, say, 1/3 of the way across the strip (for our purposes, it could really be anywhere less than half-way across, but let's just say 1/3 for simplicity).

The partition function, using the column transfer matrix, is

$$Z=\sum_{i,j=1}^2\sum_{s_1,s_N}\langle s_1|GS_i\rangle\langle GS_i| T_{AF}|GS_j\rangle \langle GS_j| s_N\rangle$$

where $T_{AF}$ is the special antiferromagnetic transfer matrix for column $I=N/3$, and we are implicitly assuming that the width of the rectangle is much larger than the correlation length, and $|GS_{i=1,2}\rangle$ are the maximal eigenvectors of the bulk column transfer matrix $T_{FM}$. Now the hard question comes: How does the kink in the spontaneous magnetization manifest itself in this particular representation of the partition function?

In particular, consider applying a uniform external field $\epsilon\to 0^+$ to the entire lattice. Now, if the antiferromagnetic column were 1/2, instead of 1/3 of the way across the lattice, then (now thinking using the row transfer matrix) this external field will induce no splitting among degenerate transfer matrix eigenvectors, and the induced spontaneous magnetization will be zero. However, here, there is an asymmetry with the antiferromagnetic column 1/3 of the way across the lattice, and there will be a unique energy-favorable configuration: 1/3 of the spins misaligned with the external field, then 2/3 of the spins aligned. However, in our column representation of the transfer matrix, there is no information that distinguishes these (very) different scenarios!

asked Apr 14 in Theoretical Physics by David B Roberts (125 points) [ revision history ]
edited Apr 26 by David B Roberts

I am not sure I understand your question. You seem to be considering the model with free boundary condition, so the column of antiferromagnetic interactions does not affect the partition function: just change the sign of all the spins on the left of this column and replace the antiferromagnetic interactions by feromagnetic ones. This a bijection on the set of configuration, which does not change the energy, so you get the same partition function. It's similar to the classical mapping between the ferromagnet and the antiferromagnet on a bipartite lattice. Things are of course different when you introduce a magnetic field, but note that you haven't done so in your formula for $Z$, so it's not a surprise that it does not tell you anything about that case.

Thank you for your response. I agree that this question can be solved with a local gauge transformation. However, this was just a toy example - the real problem I am tackling has both frustration and nonuniform coupling strengths, which cannot be solved this way. I'm really trying to play around with external fields.  I have now responded to your feedback and written it out explicitly. Hopefully I have clarified the question - looking forward to your comments.

1 Answer

+ 2 like - 0 dislike

Yvan's comment helped me realize this: the trick is to factor out a global spin-flip operator $Q_F$ (equivalently the supercharge corresponding to fermion parity) from the antiferromagnetic transfer matrix:

$$T_{AF}=T_{FM} \cdot Q_F,~~~~~~~~[T_{FM}, Q_F]=0$$

Note the relation for the maximal eigenvectors of $T_{FM}$ (which follows directly as a corollary of the exactness of Yang's 1952 perturbative calculation of the spontaneous magnetization):

$$|GS_\pm\rangle = Q_F|GS_\mp\rangle$$

This is enough to see the kink in the magnetization in the column representation of the partition function:

$$Z=\sum_{s_0,s_N} \langle s_0 | T^{N/3-1}_{FM}T_{AF} T^{2N/3}_{FM}|s_N\rangle$$

$$~~=\sum_{s_0,s_N} \langle s_0 | T^{N/3}_{FM}Q_F T^{2N/3}_{FM}|s_N\rangle\underset{N\to \infty}{\sim}\sum_{s_0,s_N,\pm}\langle s_0|GS_\pm\rangle \langle GS_\pm|GS_{\mp}\rangle \langle GS_\mp|s_N\rangle$$

However, this still leaves open the question of how an improperly placed external field can still induce the correct magnetization in this column transfer matrix picture.

answered Apr 26 by David B Roberts (125 points) [ revision history ]
edited Apr 26 by David B Roberts

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