In most continuous phase transitions, there is a well-defined order parameter $\langle \psi \rangle$ of some observable that is zero above the transition temperature, and continously grows below the transition. In the cases that I am familiar with, this observable is usually the expectation value of some physical quantity (spin, density, momentum, position, number) that is generally a local observable. On the other hand, it has become increasingly clear that there is a lot of non-local physics captured in quantities like entanglement, so perhaps more complex types of non-local order parameters may exist.

My question is, can entanglement entropy, or a similar entanglement witness, ever be an order parameter in a phase transition? If yes, what class of phase transitions do they appear in? If not, can you explain why?

I am interested to hear about both the trivial answer, where an entanglement measure and conventional observable can both be the order parameter, and the non-trivial case where only entanglement can be treated as the order parameter. I suppose at some level this whole discussion must involve quantum phase transitions, not just classical phase transitions.


3 Answers 3


Entanglement over an ensemble can be used to tell measurement induced phase transitions. The idea arises in a quantum circuit where measurements and unitary gates take place randomly over time. The ratio between measurements and gates determines whether the system is driven to a "more entangled" or "more disentangled" phase.

An example can be found here. You can see in figure 3 that the entanglement entropy is obtained for chains of different sizes, and the phase transition is obtained from a scaling law. The entanglement is, however, defined over an ensemble of quantum circuits run under the same ratios of gates/measurements.

This is a pretty recent field of research. More information can be found here.


I don't think I can answer your question, but I can at least wave my hands in the general direction of an answer.

In two-dimensional gapped systems, it is believed that the entanglement entropy of a subregion follows an "area law" in that it is proportional to the length of its boundary, \begin{align} S \sim \alpha L \end{align} for some non-universal $\alpha$ independent of the system size. Kitaev and Preskill proposed that for topologically ordered phases, there is an universal constant term $\gamma$, i.e. \begin{align} S \sim \alpha L - \gamma. \end{align} $\gamma$ is called the topological entanglement entropy. For example, different Laughlin phases of a fractional quantum Hall system have different values of $\gamma$, so we could perhaps think of $\gamma$ as an order parameter for these phases. (There is no gauge-invariant local order parameter that detects FQH order.)

More generally, one can consider the "quantum conditional mutual information" (see Chapter 5 of Zeng et al.), which detects topological order as well as other kinds of quantum order. For example, it can distinguish between the two gapped phases of the transverse field Ising model (one with trivial order and the other with symmetry-breaking/GHZ order) and between topological and trivial phases of the toric code in a magnetic field.

As another example, the so-called "entanglement spectrum" (the spectrum of the logarithm of the reduced density operator) can be used as a signature of quantum Hall order (see Li and Haldane).

  • $\begingroup$ Interesting, is it possible to measure $\gamma$ without having to reconstruct the full reduced density matrix? I am thinking of something analogous to heat capacity for normal phase transitions. $\endgroup$
    – KF Gauss
    Commented Oct 27, 2021 at 9:29
  • $\begingroup$ @KFGauss There are some proposals for measuring entanglement entropies or entanglement spectra: arxiv.org/abs/1204.2819 arxiv.org/abs/1605.08624 arxiv.org/abs/1712.08581 The last of these includes an experimental implementation of the measurement protocol, but on a two-site system rather than a many-body system. All of these proposals require microscopic control of the system, and the ability to prepare multiple copies of the state being measured. As far as I can tell they are only practical for cold atom systems, not for solid state systems... $\endgroup$
    – d_b
    Commented Oct 27, 2021 at 23:45
  • $\begingroup$ ... They are certainly not thermodynamic measurements in the same way that a measurement of the heat capacity is. $\endgroup$
    – d_b
    Commented Oct 27, 2021 at 23:47

It will be hard, though I don't think impossible.

Generally speaking, when we consider phase transitions, we talk about states that minimize the Free energy functional: i.e. an interplay between energy and (total) entropy. If we want to observe something quantum though, we are in the realm of quantum phase transitions where the entropy contribution vanishes and the 'phase' refers to the ground state of the Hamiltonian, such that the transition arises from the interplay of different terms in it.

The hard part is, if the Hamiltonian is gapped, it is known that the entanglement in the ground state will be following an 'area' law, meaning it is short ranged... and this would be true on both sides of the transition.

The gap in the Hamiltonian has to close however precisely at the critical point of the transition. And indeed, there is generally long-range entanglement (volume-law) around the transition point itself.

I could imagine that there are more exotic phases, where this gapping of the spectrum is circumvented in some way (or alternatively, one looks at exited states), though I can't think of a good example now. It is actually a nontrivial issue. For a normal quantum Ising-ferromagnet, there are strictly two almost degenerate ground states (so-called GHZ-states, macroscopic superpositions between all-up and all-down) below the gap-but as it turns out, in the thermodynamic limit (infinite system size, where phases are truly defined), this becomes equivalent to a classical mixture.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.