In the theory of phase transitions, an order parameter is usually defined as some quantity which distinguishes the two phases of the system by being zero in one phase, and non-zero in the other (see e.g. this and this question). This definition has always confused me since it seems like a very broad definition. The same phase transition may be studied through many distinct quantities acting as the order parameter. This question also discusses this non-uniqueness, however, without any conclusive answers.

Now even though this confused me a bit, I never thought there would be an actual problem with having this broad definition. But then I saw people talking about the order parameter critical exponent. At first I thought that maybe the critical exponent of the order parameter is unique, even though the order parameter itself is not. However, this is clearly not the case.

Let $M(t)$ be a valid order parameter of some phase transition, where $t\equiv \frac{T-T_c}{T_c}$ is the reduced temperature. Since it's an order parameter, $M$ is identically zero in one phase (e.g. $t\le 0$), and non-zero in the other. By definition, for any function $f(M)$ which has only one root at $M=0$, the quantity: $$M^*(t):= f(M(t))$$ also completely satisfies the definition of an order parameter. At $t\leq0$, $M^*=f(0)=0$, and at $t>0$, $M^* \neq 0$; since $M\neq 0$, and $f$ doesn't have any roots other than the origin. Now near the transition $t \to 0$, $M \propto t^\beta$, where $\beta$ is the order parameter critical exponent. Using a Taylor expansion, the behavior of $M^*$ near the transition is: $$M^*(t)=f(M(t))=f(A t^\beta)=0+f'(0)At^\beta+\frac12 f''(0)At^{2\beta}+O(t^{3\beta})$$ Now if $f'(0)\neq 0$, everything works fine and the critical exponent for $M^*$ is the same of that of $M$. However, when $f'(0)=0$, the critical exponent of the two order parameters will be different.

For example, for the mean field theory result of the Ising model, if $M \sim t^{1/2}$, then another order parameter defined by the square goes like $M^*:=M^2 \sim t^{1}$. In my opinion, there can only be two possible resolutions to this problem:

  1. The order parameter, and indeed its critical exponent are both not uniquely defined, and when people talk about the order parameter and the critical exponent, they implicitly assume that everyone they're talking to has the same particular quantity in mind.
  2. There is a more specific and restrictive definition for what an order parameter is, that me (and apparently many others) don't know about.

It would be great if someone could tell me which one is correct!

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – David Z
    Commented Feb 9, 2019 at 7:08
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    $\begingroup$ @DavidZ Extended discussions? There were four comments! $\endgroup$ Commented Feb 9, 2019 at 17:36
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    $\begingroup$ I allowed myself to replace the notation $M'$ with $M^*$, since you use $'$ to denote the derivative. I think it's less confusing like this. Very good question anyway, I am also wondering, could the choice of a different order parameter change the order of the transition? Now that would be a big problem! $\endgroup$
    – valerio
    Commented Feb 10, 2019 at 11:02
  • $\begingroup$ That makes sense. Yeah I really hope there's a more restrictive definition for an order parameter. Norbert Schush suggested in his comments that the order parameter needs to be the expectation value of an observable which anticommutes with the Hamiltonian's symmetry, so that it can effectively describe the spontaneous symmetry breaking. I don't know whether there should be any other restrictions on the definition yet. $\endgroup$ Commented Feb 10, 2019 at 17:16

3 Answers 3


I think the conventional wisdom is correct, the choice of order parameter does not matter (as long as the symmetries are preserved, and we redefine conjugate variables).

I think the main mistake in the post is that you assume that if $M$ scales as $M\sim t^\beta$ then $M^a$ scales as $M^a\sim t^{a\beta}$. But this is not the case, $M$ is a stochastic variable, and the scaling behavior is $\langle M\rangle \sim t^\beta$. There is no simple relation between the scaling of $\langle M\rangle$ and $\langle M^a\rangle$. Also note that using $M^2$ is not a good example, because $M^2$ has different symmetries (and is indeed not an order parameter in the Ising case).

The way to investigate this is to start from the Landau-Ginzburg functional $$ {\cal F} = \int d^3x \left(\gamma(\nabla M)^2 +\alpha M^2 +\beta M^4 + \ldots +Mh\right) $$ and perform a change of variables. As long as the new variable has the same symmetries as the old variable the form of the LG functional is not changed, and the universal predictions remain the same.

  • $\begingroup$ This doesn’t sound right to me. The OP’s reasoning should be perfectly fine if mean field theory holds. $\endgroup$
    – knzhou
    Commented Feb 10, 2019 at 22:32
  • $\begingroup$ A change of variables might preserve the symmetries, but it need not preserve the form of the kinetic term, plus it may render kinetic terms nonanalytic. I suspect that’s why the universality argument fails. $\endgroup$
    – knzhou
    Commented Feb 10, 2019 at 22:34
  • $\begingroup$ @knzhou The failure of mean field theory is the whole point in the study of phase transitions. At the very least you have to include Gaussian fluctuations, so that if $<M>$ vanishes, $<M^2>$ does not. $\endgroup$
    – Thomas
    Commented Feb 10, 2019 at 22:41
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    $\begingroup$ It seems to me that the OP is using $M$ to denote the true order parameter (ie the magnetization), and not the field which is integrated over (call it $\phi$). That is $M=\langle \phi\rangle$. (This problem of course arises because people often use the term "order parameter" for the field, which is true only at a mean-field level.) Thus $M^2\propto t^{2\beta}$ is fine. $\endgroup$
    – Adam
    Commented Feb 14, 2019 at 16:33
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    $\begingroup$ @Adam When we say the order parameter is the chiral condensate we mean that $\phi=\bar\psi\psi$ and ChiSB is signalled by $\langle\phi\rangle\neq 0$. We don mean $\phi=\psi$ and $|\langle \phi\rangle|^2\neq 0$. When we consider alternative order parameters in QCD we consider things like $\phi'=(\bar\psi\psi)^2$ so that $\langle\phi\rangle=0$ but (possibly, not actually realized in nature) $\langle\phi'\rangle=0$. To summarize: The interesting operation is to change the power of the field in the operator, not to change the power to which the VEV is raised (that's a trivial operation). $\endgroup$
    – Thomas
    Commented Feb 15, 2019 at 14:00

The viewpoint of classical statistical mechanics is indeed that the order parameter is non-unique. For example, in his excellent Lectures On Phase Transitions And The Renormalization Group, Goldenfeld writes

The order parameter for a given system is not unique; any thermodynamic variable that is zero in the un-ordered phase and non-zero in an adjacent (on the phase diagram), usually ordered phase, is a possible choice for an order parameter. Trivially, we could perfectly well choose $M^3$ as the order parameter in a ferromagnet.

Unfortunately, I don't think he really dwells on the consequences of such a choice. As you've noted in the question, such a functional redefinition of the order parameter generically leads to changed critical exponents.

In the Landau-Ginzburg formalism, the Landau functional is given in powers of the order parameter and its gradients. I believe this should restrict the allowable functional redefinitions to analytical functions. It's helpful to further restrict this definition by requiring that the order parameter be a function of observables. This is still not unique, and is one reason for the debate about what the right order parameter is (if any) for e.g. the Mott transition.

Next we can consider the coupling to some external parameter (e.g. a magnetic field, which would give the term $\mathbf{M}\cdot\mathbf{H}$). The first derivative of the Landau functional with respect to this external field is precisely the order parameter1. This is kind of circular, because we could also redefine the field. However, the model is obviously most useful when the external parameter is chosen as a physical parameter or field. And if the coupling between the field and observables is known, this fact can be used to fix2 the nature of the $\mathbf{M}$ in the $\mathbf{M}\cdot\mathbf{H}$ term. In practice, this typically leads to an order parameter that's a linear observable, as Nortbert Shuch pointed out in a comment.

So, yes, when people talk about the order parameter or the critical exponent for a certain transition, they do tend to assume that everyone's talking about the same quantity. However, it should also generally be a physically motivated order parameter that couples to an appropriate field.

1: The first derivative can also include higher-order terms in the order parameter, but we can still isolate the lowest-order term.

2: Keep in mind that the Landau functional is, at least in principle, possible to derive as a mean-field theory of the underlying Hamiltonian.


Weinberg makes the following suggestion (on page 225 of the Quantum Theory of Fields, Vol II):

enter image description here

In your example, the magnetization $M$ is an order parameter in this sense: in the normal state it transforms in the vector representation of the unbroken rotation group, it becomes massless right at the transition and in the symmetry broken state it creates Goldstone bosons. A general function $f(M)$ won't have this property.

  • $\begingroup$ I think the OP was just thinking of a discrete symmetry. Also, even if you think of a vector field $\vec{M}$ you can always change to any function $\vec{M}f(\vec{M}^2)$, which also transforms as a vector. $\endgroup$
    – Thomas
    Commented Feb 15, 2019 at 3:01
  • $\begingroup$ Then this would be a field redefinition, not affecting weinbergs reasoning. $\endgroup$ Commented Feb 15, 2019 at 16:02

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