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In my earlier MO question, I was seeking for a proof for $\det A_{\infty}:=\det(I_{\infty}-M_{\infty}^2) =\sqrt[4]{1-x^2}$ where $M_n$ is the $n\times n$ matrix: $$M_n =\left[\frac{2i+1}{2(i+j+1)}\binom{i-1/2}i\binom{j-1/2}jx^{i+j+1}\right]_{i,j=0}^n.$$ Since then I learned of two different sources where this determinant appeared in Physics/Chemistry.

(1) L. Onsager and B. Kaufman in thier study of spontaneous magnetization of the Ising model. Here is a readable account.

(2) J. H. Schwarz and C. C. Wu in a paper entitled Evaluation of dual Fermion amplitudes, published in Physics Letters, Vol. 47 B, number 5, 10 December 1973. This can also be found in the survey by J. H. Schwarz, Superstrings: The first 15 years of superstring theory.

Incidentally, the matrices appear different but they have equal determinants. This prompted me to ask:

Question. Is it a mere mathematical coincidence that both determinants are the same? Or, is there a real (deeper) reason and connection between the two physical problems (Ising model and String Theory/free fermion theory) leading up to the matrices (hence determinants)?

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  • $\begingroup$ Have you any conjecture about $\operatorname{det}(I-M)$, $\operatorname{det}(I+M)$, or more generally $\operatorname{det}(f(I))$, say with an holomorphic $f$ with $f(0)=1$? PS: It is not clear to me if you are still wanting a computation for that determinant, or if you are ok with those references $\endgroup$ Commented Jan 11, 2017 at 22:22
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    $\begingroup$ The connection between the 2d Ising model and a theory of free Fermions is not a coincidence. See vol 1 of the book by Itzykson and Drouffe on statistical field theory for a thorough discussion. $\endgroup$ Commented Jan 11, 2017 at 22:27
  • $\begingroup$ @PietroMajer: I'm still interested in a proof, yes. On the other hand, the two split determinant you ask about do not seem to have a neat valuation, individually. $\endgroup$ Commented Jan 11, 2017 at 22:29
  • $\begingroup$ @AbdelmalekAbdesselam: Thanks. Is it possible to give a concrete transition between the two matrices in the present setting? $\endgroup$ Commented Jan 11, 2017 at 22:30
  • $\begingroup$ For $\det(I\pm M)$, see my comment on posting 259104. Please continue the discussion on the determinant problem there. $\endgroup$
    – Fred Hucht
    Commented Mar 21, 2019 at 12:47

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If the question is is there a real (deeper) reason and connection between the Ising model (in 2D) and free fermion (in 1D) the answer is yes. I don't know how deep. I believe this was shown for the first time neatly in H. Lieb's solution of the 2D classical Ising model.

The steps are the following

1) Write the partition function of the classical 2D Ising model (on a, say, $N\times N$ square lattice) as

$$ Z := \sum_{\sigma} e^{-\beta E[\sigma]} = \mathrm{Tr} (T^N), $$

where $\sigma$ is a classical configuration and $E[\sigma]$ is its energy. The above equation defines the transfer matrix $T$. The matrix $T$ turns out to be a product of Pauli operators defined on a line (i.e., one dimension less than 2D). In particular one obtains something like $T=T_1 T_2$

$$ T_1 \ \sim \ \exp{\left ( K_{\parallel} \sum_i \sigma_i^z \sigma_{i+1}^z \right ) }\\ T_2 \ \sim \ \exp{\left ( K^{\ast} \sum_i \sigma_i^x \right ) } $$

where $K_{||}, K^{\ast}$ are functions of the original parameters of the 2D Ising model ($J_{\parallel}, J_\perp$).

3) There is a mapping (named after Jordan-Wigner) that maps a system of $N$ spins to a system of $N$ fermions. This is not too surprising giving that both Hilbert spaces are of dimension $2^N$. So the next step is to write out $Z$ in terms of fermion operators $a_i, a_i^\dagger$.

4) It turns out that the resulting expression is integrable. What this mean is somehow involved but we can say it amounts to the fact that, essentially, it can be written as

$$ Z = \mathrm{Tr} e^{\Gamma[A_1]} \dots e^{\Gamma[A_n]} $$

where $\Gamma[A] :=\sum_{i,j} a^\dagger_i A_{i,j} a_j$.

5) Integrability is now expressed by the following identity

$$ \mathrm{Tr} \left( e^{\Gamma[A_1]} \dots e^{\Gamma[A_n]} \right) = \det \left(1 +e^{A_1} \dots e^{A_1} \right) $$

namely a trace over a $2^N$ dimensional space is reduced to a trace over an $N$ dimensional one (the one-particle space).

It should be pointed out that not every expression of $N$ Pauli operators is mapped into free fermions. This lucky coincidence is what is responsible for the integrability of problem.

EDIT

I just checked and it turns out that the original paper is by T. D. SCHULTZ, D. C. MATTIS, and E. H. LIEB and can be found here. Not surprisingly its title is Two-Dimensional Ising Model as a Soluble Problem of Many Fermions.

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