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.

  • 1
    $\begingroup$ 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. $\endgroup$
    – Thomas
    Jan 18, 2021 at 2:56
  • $\begingroup$ @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. $\endgroup$
    – Wouter
    Jan 18, 2021 at 4:19
  • $\begingroup$ Binder looks at histograms. That is of course fine, $\endgroup$
    – Thomas
    Jan 18, 2021 at 16:27

1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$
    – Thomas
    Jan 18, 2021 at 16:35
  • $\begingroup$ @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. $\endgroup$
    – Adam
    Jan 18, 2021 at 18:36
  • $\begingroup$ So you assume that $\nu=0$? That's also not true. (Even more obvious: Consider the disordered phase. Then $\beta=0$, but $\gamma\neq 0$.) $\endgroup$
    – Thomas
    Jan 18, 2021 at 19:46
  • $\begingroup$ @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). $\endgroup$
    – Adam
    Jan 18, 2021 at 19:57
  • $\begingroup$ 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. $\endgroup$
    – Thomas
    Jan 18, 2021 at 20:13

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