# How to obtain a nonzero order parameter for a symmetry-breaking quantum phase transition?

If $$\hat{m_z}=\frac{1}{N}\sum_i \hat{\sigma^z_i}$$ is an order parameter for finite quantum system (transverse Ising model, say), then it will never break the $$\mathbb{Z}_2$$ symmetry since $$\langle\psi_0|\hat{m_z}|\psi_0\rangle=0$$ for the ground state $$|\psi_0\rangle$$ that satisfies the symmetry of system.

This is in fact not a unique problem to quantum systems. But for the classical systems, a quick fix that I'm aware of is replacing the scaling $$m_z\sim \tau^\beta$$ by $$m_z^2\sim \tau^{2\beta}$$, where $$\tau$$ is the control parameter and $$\beta$$ a critical exponent.

(a further possible step is the Binder cumulant/bimodality coefficient defined as the fourth moment scaling as $$\tau^{4\beta}$$ normalized by this second moment squared, so that no net scaling with the order parameter remains),

A naive, straightforward generalization to quantum systems is thus working with $$\langle \hat{m}_z^2\rangle\sim \tau^{2\beta}$$. This seems to be indeed what is done, for example in https://doi.org/10.1103/PhysRevB.87.174302 (eq. 54). But now I saw in a response to a different question here https://physics.stackexchange.com/a/460031/25292 that the scaling of the power of a variable should not be so trivial?

So, my questions are

• Is this procedure of squaring the operator of the order parameter and replacing $$\beta\rightarrow 2\beta$$ valid?
• In case not, what would be the alternative, for finite-size scaling purposes?

I just saw in an original paper on 1D quantum Ising, https://doi.org/10.1016/0003-4916(70)90270-8 that they used the square root of only the infinite-range correlation instead of from the full squared magnetisation. But it seems unclear what would be the result of this in a finite system.

• You cannot replaze m_z by m_z^2. These two operators don't even have the same symmetries. The standard procedure for thermodynamic phase transitions is to couple an external field h to m_z, and then consider the limit volume to infinity followed by h->0. – Thomas Jan 18 at 2:56
• @Thomas for the classical case, see e.g. journals.aps.org/prl/abstract/10.1103/PhysRevLett.47.693 (p.2, first paragraph) . Also for the quantum case, I think that the authors of the PRB paper above are smart people, so I'm wondering if there isn't something more to it. – Wouter Jan 18 at 4:19
• Binder looks at histograms. That is of course fine, – Thomas Jan 18 at 16:27

Up to some sloppy notations, what is written in the question is correct.

Being more rigorous with notations will help clarify the problem.

I will assume that the system is classical, since adding quantum fluctuations will only change the scaling by known scaling relations if the transition is at $$T=0$$ (which depends on the dynamical exponent $$z$$), or will not change anything if the transition is at finite temperature.

Let us call $$s(x)$$ the microscopic field, and $$M=\langle s(x)\rangle$$ the (natural) order parameter. Let us call $$s_0= L^{-d} \int d^dx\, s(x)$$, where $$L$$ is the linear size of the system, and $$d$$ its dimension. We have by invariance by translation that $$M=\langle s_0\rangle$$.

Scaling theory tells us that close to the transition (parametrized by the reduced temperature $$\tau$$), we have $$M\propto |\tau|^\beta.$$ This can be obtained by coupling $$s_0$$ to a source field $$j$$ (which is a volume times a magnetic field), and looking at the scaling behavior of the free energy (not the free energy density): $$F(\tau,j) = F(|\tau| s^{1/\nu},j s^{-\beta/\nu})= F_1( j |\tau|^{\beta}),$$ where the last line is obtained from $$s=|\tau|^{-\nu}$$. Because the theory is not critical in presence of a symmetry breaking source, we know that $$F_1$$ is analytic in its variable. Therefore, we immediately obtain that $$\langle s_0^k\rangle = |\tau|^{k\beta},$$ as can also be obtained using Binder's probability distribution of $$s_0$$.

For $$k=2$$, this is in agreement with the scaling of the susceptibility, since $$\chi = L^{-d} \int d^dx d^dy \,\langle s(x) s(y)\rangle=L^d \langle s_0^2\rangle,$$ using that $$\gamma=\nu d -2\beta$$.

In particular, this means that we can relate all the critical scaling of zero-momenta correlation functions to $$\beta$$ (up to some $$\nu d$$ terms). However, this does not mean that all correlation functions have scaling exponents that are simply related to $$\beta$$ and $$\nu$$. For instance, the specific heat exponent $$\alpha =2-\nu d$$ does not depend on $$\beta$$ (or equivalently to $$\eta$$). There is no contradiction because $$C_V = \int d^d x d^d y\, \langle s(x)^2s(y)^2\rangle$$ cannot be written in terms of $$s_0$$ alone.

• Not sure I understand. When you write $<s_0^k>\sim \tau^{k\beta}$ you seem to say that the susceptibility exponent $\gamma$ is twice the order parameter exponent $\beta$. This is not true. – Thomas Jan 18 at 16:35
• @Thomas: This is explicitly addressed in the answer already: $\langle s_0^2\rangle \propto \tau^{2\beta}$ is related to the fact that $\gamma=\nu d-2\beta$, which is true. – Adam Jan 18 at 18:36
• So you assume that $\nu=0$? That's also not true. (Even more obvious: Consider the disordered phase. Then $\beta=0$, but $\gamma\neq 0$.) – Thomas Jan 18 at 19:46
• @Thomas Of course I don't assume that $\nu=0$... $\chi$ is defined as I write (this is used e.g. in MC calculation, and has the same scaling than the more standard, $\chi=\int d^dx \langle s(x) s(0)\rangle$, but with the same scaling). Note that when computing the scaling of $\chi=L^d \langle s_0^2\rangle$, there is a scaling of $L$ that contributes to a b^{-d} which becomes $\tau^{\nu d}$. Multiplied by $\tau^{2\beta}$, one gets the correct scaling $\chi\propto \tau^{-\gamma}$ (using standard scaling relations, as written in the answer). – Adam Jan 18 at 19:57
• Sorry, should have read your answer more carefully. I thought $s_0$ was a local operator, not already integrated over space. Then what you say is of course true. – Thomas Jan 18 at 20:13