Conformal invariance in the FK-representation of the quantum Ising model and convergence of the interface to the $\mathrm{SLE}_{16/3}$

Abstract

We study the interface in the FK-representation of the 1D quantum Ising model and show that in the limit, it converges to the $\mathrm{SLE}_{16/3}$ curve.

Appendix: Computation of residues

For a non-negative integer k and an integer m, define the following meromorphic function on $\mathbb {C}$,

$$\begin{aligned} g_{k, m} (z) := \frac{1}{z} \left( \frac{1}{z+1} + \frac{1}{z-1} \right) ^k \left( \frac{z+1}{z-1} \right) ^{2m} = \frac{2^k z^{k-1}}{(z-1)^{k+2m} (z+1)^{k-2m}}. \end{aligned}$$

We notice that the possible singularities are at 0, 1 and $-1$.

Lemma A.1

When $k=0$, we have ${{\mathrm{\mathrm{Res}}}}(g_{k, m}, 1) = {{\mathrm{\mathrm{Res}}}}(g_{k, m}, -1) = 0$ for all integer m.

Proof

We fix $k=0$. When $m = 0$, the result is trivial. For $m \geqslant 1$, the function $g_{k, m}$ does not have any pole at $-1$, so it is clear that ${{\mathrm{\mathrm{Res}}}}(g_{k, m}, -1) = 0$. The residue of $g_{k, m}(z)$ at $z = 1$ is the residue of $g_{k, m}(y+1)$ at $y = 0$. We have

$$\begin{aligned} g_{k, m} (y+1)&= \frac{1}{y+1} \left( 1 + \frac{2}{y} \right) ^{2m} \\&= \left[ \sum _{k \geqslant 0} (-y)^k \right] \left[ \sum _{l=0}^{2m} {2m \atopwithdelims ()l} \left( \frac{2}{y} \right) ^l \right] , \end{aligned}$$

where the coefficient of $\frac{1}{y}$ is given by

$$\begin{aligned} \sum _{l=1}^{2m} {2m \atopwithdelims ()l} 2^l (-1)^{l-1} = - [(1-2)^{2m} - 1] = 0. \end{aligned}$$

When m is negative, the proof is similar. $\square$

Lemma A.2

For $k \geqslant 1$, we have ${{\mathrm{\mathrm{Res}}}}(g_{k, m}, 1) + {{\mathrm{\mathrm{Res}}}}(g_{k, m}, -1) = 0$.

Proof

When $k \geqslant 1$, the singularity at 0 is removable. We observe that $|g_{k, m} (z)|$ behaves like $|z|^{-k-1} \leqslant |z|^{-2}$ when |z| is large, giving

$$\begin{aligned} \lim _{R \rightarrow \infty } \frac{1}{2 \pi {{\mathrm{\mathrm{i}}}}} \int _{\partial B(0, R)} g_{k, m} (z) \mathrm{d}z = 0. \end{aligned}$$

Moreover, when $R > 1$, we have

$$\begin{aligned} \frac{1}{2 \pi {{\mathrm{\mathrm{i}}}}} \int _{\partial B(0, R)} g_{k, m} (z) \mathrm{d}z = {{\mathrm{\mathrm{Res}}}}(g_{k, m}, -1) + {{\mathrm{\mathrm{Res}}}}(g_{k, m}, 1). \end{aligned}$$

Thus, we get the desired result. $\square$

Lemma A.3

For $k \geqslant 1$ and $k \leqslant 2 |m|$, ${{\mathrm{\mathrm{Res}}}}(g_{k, m}, 1) = {{\mathrm{\mathrm{Res}}}}(g_{k, m}, -1) = 0$.

Proof

The previous lemma tells us that it is enough to show that the residue is zero at either 1 or $-1$. If m is positive, we notice that $g_{k, m}(z)$ does not have any pole at $-1$, thus the residue at $-1$ is zero. If m is negative, the residue at 1 is zero. $\square$

Lemma A.4

More generally, for all $k \in 2 \mathbb {N}$,

$$\begin{aligned} {{\mathrm{\mathrm{Res}}}}(g_{k, m}, 1) = {{\mathrm{\mathrm{Res}}}}(g_{k, m}, -1) = 0. \end{aligned}$$

Proof

Assume that $k = 2l > m \geqslant 0$ for a positive integer l. Look at the residue of $g_{k, m} (z)$ around $z = 1$ is equivalent to looking at the residue of $g_{k, m} (y+1)$ around $y=0$,

$$\begin{aligned} {{\mathrm{\mathrm{Res}}}}(g_{k, m}(z), z=1) = {{\mathrm{\mathrm{Res}}}}(g_{k, m}(y+1), y=0). \end{aligned}$$

We have the following equivalent relations

$$\begin{aligned}&{{\mathrm{\mathrm{Res}}}}(g_{2l, m}(y+1), y=0) = 0 \nonumber \\&\quad \Leftrightarrow {{\mathrm{\mathrm{Res}}}}\left( \frac{(y+1)^{2l-1}}{y^{2l+2m}(y+2)^{2l-2m}}, y=0 \right) = 0 \nonumber \\&\quad \Leftrightarrow \frac{(1+y)^{2l-1}}{(1+\frac{y}{2})^{2l-2m}} [y^{2l+2m-1}]= 0 \end{aligned}$$

(A.1)

where the notation $R(y) [y^k]$ gives the coefficient of $y^k$ in the Laurent series of R(y).

We expand the rational fraction to evaluate this coefficient by applying the three following identities,

$$\begin{aligned} {{2m-2l \atopwithdelims ()2m+p}}&= (-1)^{2m+p} {{2l+p-1 \atopwithdelims ()2m+p}} \end{aligned}$$

(A.2)

$$\begin{aligned} {{2l+p-1 \atopwithdelims ()2m+p}}&= \sum _{q=0}^{2l-1} {{p \atopwithdelims ()q}} {{2l-1 \atopwithdelims ()2m+p-q}}, \end{aligned}$$

(A.3)

$$\begin{aligned} \sum _{k=0}^n {{n \atopwithdelims ()k}} {{ k \atopwithdelims ()r}} (-x)^k&= (-x)^r (1-x)^{n-r} {{n \atopwithdelims ()r}}. \end{aligned}$$

(A.4)

Equation (A.2) comes from the general definition of binomial coefficients, and in our case, we have $2m-2l < 0$. Equation (A.3) is an easy combinatorial identity and Eq. (A.4) a simple expansion. Then, the left-hand side of Eq. (A.1) equals

$$\begin{aligned}&\sum _{p=0}^{2l-1} {{2l-1 \atopwithdelims ()p}} \left( \frac{1}{2} \right) ^{2l+2m-1-p} {{2m-2l \atopwithdelims ()2l+2m-1-p}} \\ (p \leadsto 2l-1-p)&\quad = \sum _{p=0}^{2l-1} {{2l-1 \atopwithdelims ()p}} \left( \frac{1}{2} \right) ^{2m+p} {{2m-2l \atopwithdelims ()2m+p}} \\&\quad = \sum _{p=0}^{2l-1} {{2l-1 \atopwithdelims ()p}} \left( -\frac{1}{2} \right) ^{2m+p} {{2l+p-1 \atopwithdelims ()2m+p}} \\ (\text{ Equation } ~(A.3))&\quad = \sum _{p=0}^{2l-1} {{2l-1 \atopwithdelims ()p}} \left( -\frac{1}{2} \right) ^{2m+p} \sum _{q=0}^{2l-1} {{p \atopwithdelims ()q}} {{2l-1 \atopwithdelims ()2m+p-q}} \\ (q \leadsto p-q)&\quad = \sum _{p=0}^{2l-1} {{2l-1 \atopwithdelims ()p}} \left( -\frac{1}{2} \right) ^{2m+p} \sum _{q=0}^{2l-1} {{p \atopwithdelims ()q}} {{2l-1 \atopwithdelims ()2m+q}} \\&\quad = \left( \frac{1}{2} \right) ^{2m} \sum _{q=0}^{2l-1} {{2l-1 \atopwithdelims ()2m+q}} \sum _{p=0}^{2l-1} {{2l-1 \atopwithdelims ()p}} {{p \atopwithdelims ()q}} \left( -\frac{1}{2} \right) ^p \\ (\text{ Equation } ~(A.4))&\quad = \left( \frac{1}{2} \right) ^{2m} \sum _{q=0}^{2l-1} {{2l-1 \atopwithdelims ()2m+q}} \left( -\frac{1}{2} \right) ^q \left( \frac{1}{2} \right) ^{2l-1-q} {{2l-1 \atopwithdelims ()q}} \\&\quad = \left( \frac{1}{2} \right) ^{2m+2l-1} \sum _{q=0}^{2l-1} (-1)^q {{2l-1 \atopwithdelims ()2m+q}} {{2l-1 \atopwithdelims ()2l-1-q}} =0 \end{aligned}$$

where the sum in the last line is the coefficient in front of $x^{2m+2l-1}$ in $(1-x)^{2l-1}(1+x)^{2l-1} = (1-x^2)^{2l-1}$, which is zero because the polynomial only contains monomials of even degree. $\square$

Proposition A.5

For $\zeta = m + {{\mathrm{\mathrm{i}}}}t \in \mathbb {L}_1$ with $m \in \frac{1}{2} \mathbb {Z}$ and $t \in \mathbb {R}$, let $f_\zeta$ as defined in (3.22). Consider C the path defined in Proposition 3.18. Then,

$$\begin{aligned} \int _C f_\zeta (z) \mathrm{d}z = 0. \end{aligned}$$

Proof

We will expand the exponential into series and show that this integral is zero for all terms. We can do this because the series converges uniformly on all compacts to the exponential function. Therefore, it makes sense to exchange the integral and the series. After expanding, we get

$$\begin{aligned} f_\zeta (z) = \sum _{k \geqslant 0} \frac{(2 {{\mathrm{\mathrm{i}}}}t)^k}{k!} g_{k, m} (z) \end{aligned}$$

and the residue theorem along with Lemmas A.1 and A.2 allows us to conclude. $\square$

The previous proposition is important because it shows that the Green’s function as defined by (3.23) does not depend on the lift of the logarithm.

Proposition A.6

Around $\zeta = m \in \mathbb {Z}\backslash \{ 0 \}$, the function G is $\mathcal {C}^{2|m|}$.

Proof

To show this, we need to check that for all $k \leqslant 2 |m|$, we have

$$\begin{aligned} \int _C g_{k, m}(z) \ln _1(z) \mathrm{d}z = \int _C g_{k, m}(z) \ln _2(z) \mathrm{d}z \end{aligned}$$

where $\ln _i$ is chosen such that $\ln _i(1) - \ln _i(-1) = (-1)^i {{\mathrm{\mathrm{i}}}}\pi$. Here, $\ln _1$ corresponds to the logarithm chosen in the upper half-plan and $\ln _2$ in the lower half-plane. From Proposition A.5, we can fix a lift of the logarithm such that $\ln _1 - \ln _2$ is non zero around 1 and $-1$, for example $\ln _1 (1) = 2 {{\mathrm{\mathrm{i}}}}\pi$, $\ln _1 (-1) = 3 {{\mathrm{\mathrm{i}}}}\pi$, $\ln _2 (1) = 0$ and $\ln _2(-1) = -{{\mathrm{\mathrm{i}}}}\pi$.

Let $I_1$ be the integral on the left-hand side and $I_2$ be the one on the right-hand side. Since $\ln _1 - \ln _2$ is non zero at 1 and $-1$, we can write

$$\begin{aligned} I_1 - I_2&= 2 {{\mathrm{\mathrm{i}}}}\pi [ 2 {{\mathrm{\mathrm{i}}}}\pi {{\mathrm{\mathrm{Res}}}}(g_{k, m} (z), z = 1) + 4 {{\mathrm{\mathrm{i}}}}\pi {{\mathrm{\mathrm{Res}}}}(g_{k, m} (z), z = -1) ] = 0 \end{aligned}$$

due to Lemma A.3. $\square$

Proposition A.6 provides us with the regularity of Green’s function; in particular, the second condition giving in Sect. 3.4.1 is satisfied. This comes to complete the proof of Proposition 3.18.