There is a lovely set of two-parameter spin chains that can be mapped to quadratic fermions and studied quite exactly:

$$H = -\sum_{i} \frac{1+\gamma}{2}\sigma^x_i\sigma^x_{i+1}+\frac{1-\gamma}{2}\sigma^y_i\sigma^y_{i+1} + h \sigma^z_i$$

I'll consider $\gamma, h \geq 0$ in the following. These models show phase transitions with infinite correlation length on crossing the vertical line $\gamma=0, h<1$ and on crossing the horizontal line $h=1, \gamma>0$. In fact, these phase transitions flow to conformal field theories with a central charge that can be determined quite simply: on mapping to fermions, the central charge is half the number of points in the single particle spectrum with a linear dispersion relation.

Below, I plot $\gamma, h \geq 0$ with the critical lines where phase transitions occur. I also draw what the single-particle dispersion relation is after mapping to quadratic fermions. It's easy to see, using the rule above, that $c=1/2$ for the $h=1$ line and $c=1$ for the $\gamma=0, h<1$ line.

enter image description here

However, exactly at $\gamma=0, h=1$, there's a quadratic dispersion relation. My questions are: Does this critical point flow to a conformal field theory? If not, is there a general theory of what happens when one has two lines of fixed points intersect, with the lines each described by different central charges?

Using exact diagonalization on small system sizes, and barring any coding mistakes, I find this point's entanglement entropy to be well-described by a central charge of $1$, where I have fit $S(l)$ to the function $\frac{c}{3}\ln \left(\frac{L}{\pi} \sin{\frac{\pi l}{L} }\right) +C$

enter image description here

Numerically, the fit gives $1.04$, a good sign that despite the quadratic dispersion relation, properties of this model are determined by the $c=1$ fixed point of the rest of the $\gamma=0, h<1$ line.

EDIT: I accidentally dropped the factor of $1/2$ in $(1\pm\gamma)/2$ in the Hamiltonian above when making my above plot. When I correctly include the factor, I instead get an entanglement entropy of about $10^{-3}$ everywhere in the chain, which is really just zero, in line with both the excellent answers to this question noting product state ground states. It turns out that the vicinity of this multicritical point also has very surprising entanglement properties, as elaborated in comments to the answers.

  • $\begingroup$ Interesting. If I'm not mistaken, every point on the horizontal line should be the same fixed point because the Ising model is the only unitary CFT with $c = 1/2$. $\endgroup$ Dec 10, 2021 at 0:36
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    $\begingroup$ As reviewed in the introduction of arxiv.org/abs/2111.01106 , the critical O(n) model leads to two different CFTs: a dilute model when the coupling takes a critical value $K=K_c(n)$, and a dense model for $K>K_c(n)$. But these two CFTs have different central charges. So you should not be too suprised if the central charge is a discontinuous function of the microscopic model's parameters. $\endgroup$ Dec 10, 2021 at 8:44
  • $\begingroup$ I think you are making a mistake: at the multicritical point, the ground state is a product state (i.e. zero entanglement entropy) for any system size. $\endgroup$ Dec 11, 2021 at 19:21

2 Answers 2


Let us consider the multicritical point that you point out at $\gamma=0$ and $h=1$: $$ \boxed{ H = - \sum_n \left( \frac{1}{2}\left[\sigma^x_n \sigma^x_{n+1} + \sigma^y_n \sigma^y_{n+1} \right] + \sigma^z_n \right) } \;. $$

Let us use the following two facts of Pauli matrices: $$ \frac{1}{2}\left(\sigma^x_n \sigma^x_{n+1} + \sigma^y_n \sigma^y_{n+1} \right) = \sigma^+_n \sigma^-_{n+1}+\sigma^-_n \sigma^+_{n+1} \quad \textrm{and} \quad \sigma^z_n = \sigma^+_n \sigma^-_n - \sigma^-_n \sigma^+_n = 1-2 \sigma^-_n \sigma^+_n.$$

Hence, by including an irrelevant global offset of the energy, and presuming periodic boundary conditions for simplicity, we can rewrite our Hamiltonian as: $$H = -\sum_n \left( \sigma^+_n \sigma^-_{n+1}+\sigma^-_n \sigma^+_{n+1} - \sigma^-_n \sigma^+_n - \sigma^-_{n+1} \sigma^+_{n+1} \right) = \sum_n \left( \sigma^-_n - \sigma^-_{n+1} \right)\left(\sigma^+_n - \sigma^+_{n+1} \right).$$

Magically, this gives us a frustration-free system. More precisely, if we define $\boxed{\Gamma_n := \sigma^+_n - \sigma^+_{n+1}}$, then we have $$ \boxed{ H = \sum_n \Gamma_n^\dagger \Gamma_{n}^{\vphantom \dagger} } \;. $$ Note that written this way, the spectrum of $H$ is manifestly non-negative, since for any operator $A$, the combination $A^\dagger A$ has real and non-negative eigenvalues. This means that if there exists a state with zero energy, it must be the ground state. Having zero energy means the state must be annihilated by $\Gamma_n$ (for every $n$). One such state is clearly $|\psi\rangle := |\uparrow\rangle^{\otimes N}$, which is killed by the raising operator; moreover, it is easy to see that this is the only state with this property!

In conclusion, this proves that the ground state of the multicritical point is a product state, independent of system size. In particular, the entanglement entropy is zero for all system sizes, implying that your plot had some kind of coding error (maybe you accidentally took $h =1/2$ instead of $h=1$).

Fun fact: the above argument can be extended to show that the above product state is the ground state for all $h\geq 1$ (with $\gamma=0$).

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    $\begingroup$ Hi Ruben, isn't $h=1$ the band touching point for a free fermion chain (like chemical potential $\mu=2t$ or $-2t$? Is this just the state with no fermions at all in the system? $\endgroup$
    – Meng Cheng
    Dec 11, 2021 at 20:31
  • $\begingroup$ @MengCheng That is correct! I just wanted to give the proof in the spin language to avoid any confusion with having to do a nonlocal JW transformation. (Also, the above proof generalizes to arbitrary lattices in arbitrary dimensions, which is why I like iit :) ) $\endgroup$ Dec 11, 2021 at 20:48
  • $\begingroup$ Besides a coding error, it could also be that he/she actually uses U(1) symmetry in exact diagonalization, and the ground state is different when you have more than one fermions? $\endgroup$
    – Meng Cheng
    Dec 11, 2021 at 21:00
  • $\begingroup$ @RubenVerresen Argh, I accidentally used the wrong value of $h$ which put me on the $c=1$ line below the multicritical point in my entanglement entropy plot. This answers my question, and I'm accepting it. Could you elaborate on or give a reference for your comment that there is an essential singularity in the entanglement entropy? $\endgroup$
    – user196574
    Dec 11, 2021 at 22:18
  • $\begingroup$ @MengCheng Agreed, but fixing $U(1)$ charge means having $h$ in the Hamiltonian has no effect at all, which I would still count as a 'coding error' (given the way the question was phrased). $\endgroup$ Dec 11, 2021 at 22:23

This is a very well-understood model (as might be expected, since it maps to free fermions!). The reference Damle and Sachdev, PRL, 76, 4412 (1996) discusses the phase diagram, critical properties, and exact finite-temperature correlation functions for this model, see their Figure 2 for a phase diagram equivalent to yours. (To map to your notation, take $J=1/2$ and $\Delta = h$. They allow negative $\gamma$ and $h$, so your phase diagram is a single quadrant of theirs, but the phase diagram is symmetric anyways so there's no new information there.)

Your description of the CFTs away from the multicritical points is correct and consistent with the reference. One has $c=1/2$ at the $h = \pm 1$, $\gamma \neq 0$ lines, describing the onset of Ising order of either the $\sigma^y$ or $\sigma^z$ magnetization. Along the $\gamma=0$, $|h|<1$ line, the Mermin-Wagner theorem prevents any symmetry breaking, but one has a $c=1$ "Luttinger liquid" state which Damle and Sachdev call the "(Ising)$^2$" universality class since it may be written as two Ising models coupled by an exactly marginal operator.

However, as you noted, the multicritical points at $\gamma=0$, $h=\pm1$ have a quadratic dispersion. Therefore, the low-energy theory does not even have Lorentz invariance, so clearly it cannot have conformal invariance (the conformal group contains Lorentz symmetry as a subgroup) [*]. So one cannot define a central charge at this point, and one shouldn't use results which follow from conformal invariance. This universality class is sometimes called the "U(1) dilute Bose gas" due to its appearance in the 1d Bose-Hubbard model (see Chapter 16 of Sachdev's Quantum Phase Transitions, which spends quite a bit of time on the $\gamma=0$ line of your model). The two lines of CFTs are obtained by relevant perturbations of this multicritical point, which is how Damle and Sachdev approach computing the full universal crossovers of the theory.

I do not know the general form of the entanglement entropy for a scale-invariant (1+1)d QFT without conformal invariance, though I strongly suspect it is also logarithmic with the size of the cut. However, the entanglement entropy for this particular model has been computed analytically for a finite (but large) cut in an infinite system size, see https://arxiv.org/abs/quant-ph/0304108 and https://arxiv.org/abs/quant-ph/0409027. It seems they find that the entanglement entropy approaches zero at the multicritical point (after all it's just a product state of fermions, right?). Perhaps your numerics are using systems which are too small?

[*]: The often-quoted proof from field theory that scale invariance implies conformal invariance in (1+1) dimensions assumes Lorentz invariance. See also a previous answer of mine; the 2D classical Pakrovsky-Talapov critical point I discuss there is the same universality class as the quantum multicritical point we are discussing.

  • $\begingroup$ It turns out that the multicritical point is an essential singularity in entanglement entropy: you can obtain any limiting value by approaching the multicritical point in a particular way. However, if one is exactly at the multicritical point, its entanglement is exactly zero for any system size. $\endgroup$ Dec 11, 2021 at 19:23
  • $\begingroup$ @SethWhitsitt Your linked answer is very nice to see! I accidentally made the plot at a value of $h$ that was less than $1$ (namely, I forget the factor of $1/2$ in $(1\pm\gamma)/2$), which put me on the critical line below this multicritical point, which is a little embarrassing on my part. $\endgroup$
    – user196574
    Dec 11, 2021 at 22:23

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