# If we cool a topologically ordered system down to zero temperature, will it end up in a pure or mixed ground state?

If a quantum system with degenerate ground states is fully ergodic at zero temperature, then it is maximally mixed over the ground-state (GS) manifold; i.e. its density matrix $\rho$ is the projection operator onto the Hamiltonian's lowest-energy eigenspace.

But if the GS degeneracy results from a global symmetry, then we always find experimentally that the system is not in the ergodic mixed state, but instead in a symmetry-breaking pure state. For example, for the transverse quantum Ising model with degenerate "all up" $|\uparrow\rangle$ and "all down" $|\downarrow\rangle$ ground states, we never find an experimental system to be in the ergodic and symmetric mixed state $\rho = \frac{1}{2}|\uparrow\rangle \langle \uparrow| + \frac{1}{2}|\downarrow\rangle \langle \downarrow|$, but instead in one of two asymmetric pure states $|\uparrow\rangle$ or $|\downarrow\rangle$.

(Exactly how to interpret this statement is an extremely subtle issue - suffice it to say than no experimentalist ever sees a meter reading "$S_i^z = 0$", but instead either "$S_i^z = +\frac{1}{2}$" or "$S_i^z = -\frac{1}{2}$". Some would argue that from the many-worlds perspective, the system is in such in incoherent mixture, but is entangled with the experimentalist in such a way that she can't see both branches simultaneously.)

Now consider a system where the GS degeneracy results not from a global symmetry but instead from topological order - for example, a physical realization of the toric code with periodic boundary conditions, whose GS manifold has a robust (topologically protected) fourfold degeneracy. If we were to cool such a physical system down to zero temperature, without making any special efforts to maintain quantum coherence, what state would it end up in?

I could imagine three plausible possibilities:

1. The ergodic, maximally mixed state $$\rho = \frac{1}{4} | 1 \rangle\langle 1 | + \frac{1}{4} | 2 \rangle\langle 2 | +\frac{1}{4} | 3 \rangle\langle 3 | +\frac{1}{4} | 4 \rangle\langle 4 |.$$
2. A pure minimally entangled state (MES).
3. A random pure ground state (i.e. a coherent superposition of the four MESs with random coefficients).

The argument for possibility #2 is by analogy with global symmetry breaking, in which the ergodic mixture over the whole GS manifold is broken down to one of a few "special, physically natural" pure states (in the case of global symmetry breaking, states like $|\uparrow\rangle$ and $|\downarrow\rangle$ that respect the cluster decomposition property). The argument for possibility #1 is that for the toric code, all the pure states in the GS manifold respect cluster decomposition, so there's no reason for decoherence to break ergodicity. The argument for possibility #3 is that decoherence would break ergodicity down to a pure state as in global SSB, but since all the states in the GS manifold respect cluster decomposition, the non-MESs are just as physically natural as the MESs.

Is there any consensus as to which of these possibilities would be observed in a real experiment?

• #1 and #3 are actually the same -- or at least, there is no observable difference between them. An ensemble of random pure states with uniform distribution is, by definition, the maximally mixed state. Apr 10, 2018 at 2:26
• Actually, #2 is also the same as #1 and #3. Since there's no reason why a single MES ought to be favored (and no way of knowing which one is obtained, without measuring the highly non-local Wilson loops), our state of knowledge about the system should be represented by an incoherent mixture of all the MES. And that is again the maximally mixed state. The situation for spontaneous symmetry-breaking is different because the order parameter is locally measurable. Apr 10, 2018 at 2:36
• @DominicElse That second comment should really be an answer. Apr 24, 2018 at 16:30
• @NorbertSchuch I turned it into an answer. Apr 24, 2018 at 20:37

Let me try to expand the comment I wrote into an answer.

Firstly, I will make a philosophical point. Consider a spin-1/2 particle. We can consider two different bases $|\uparrow\rangle, |\downarrow\rangle$ (the $S_z$ basis) or $|+\rangle, |-\rangle$ (the $S_x$ basis). There is no observable difference between the statement that the system is in the state $|\uparrow\rangle$ or $|\downarrow\rangle$ with equal probability, and the statement that the system is in the state $|+\rangle$ or $|-\rangle$ with equal probability, or indeed the statement the statement that the system is some random state $|\psi\rangle$ sampled from the uniform measure on $\mathbb{C}^2$ (the one which is invariant under any unitary). They all give the maximally mixed state $\rho = \frac{1}{2} \mathbb{I}$, from which all measurement statistics can be computed. Depending on the interpretation of QM you subscribe to, there may or may not be an ontological difference. But this answer will only consider observable differences.

As such, returning to the specific question, there is no observable difference between the statements that the system is in one of the 4 MES with equal probability (#2) and the statement is in a uniformly random state in the whole 4-dimensional ground state subspace (#3). They are both equivalent to the statement that the system is in a maximally mixed state (#1).

However, there remains the following possibility: what if all the MES do not have equal probability? Here it's instructive to consider the differences between topological order and spontaneous symmetry breaking. In an Ising model, for example, which spontaneously breaks spin-flip symmetry at low temperatures, there is a two-fold degenerate subspace corresponding to the two possible spin alignments: call them $|\uparrow\rangle$ and $|\downarrow\rangle$. You can imagine an experiment where you start the system at infinite temperature and then allow it to cool in a box that prevents you from easily seeing the magnetization inside (of course, the box can't be completely isolated, because heat has to pass through it to cool the system). Then the experimenter has no information whatsoever about whether the system has ended up in a $|\uparrow\rangle$ state or a $|\downarrow\rangle$ state. So before opening the box, the only rational probability distribution for the experimenter to assign is to say they have equal probability, so the mixed state they assign to the system is the maximally mixed state over the ground-state subspace. However, as soon as they open the box it will be very easy for them to observe the magnetization to find out which the way the spins were aligned; indeed, short of closing their eyes it might be very hard for them not to observe this information. This "collapses" the probability distribution and they will see either $\uparrow$ or $\downarrow$ (whether this corresponds to a QM wavefunction collapse or just a classical update of probabilities based on new information depends on which basis you wrote the maximally mixed state in, which as mentioned above is unobservable data).

By contrast, for a topologically ordered system the different ground states are locally indistinguishable. So just glancing at the system gives the experimenter no information whatsoever about which ground state the system is in, and they are forced to continue using the maximally mixed state to describe their state of knowledge. Of course, if the experimenter has access to a quantum computer (for instance), then they might be able to do some (necessarily highly non-local) measurement that does give them information. Then the nature of the resulting state will obviously depend on which observable they choose to measure.

Random side note: this kind of thing is why I am skeptical of interpretations of QM that claim to assign some notion of "reality" to the QM wavefunction. Mixed states seem a lot more physical than pure states, and mixed states are clearly a generalization of classical probability distributions, which (doesn't matter whether you're a Bayesian or a frequentist) are never considered "real" degrees of freedom.

• How do you allow something to cool in complete isolation? You need someplace for the energy to go, which necessitates interaction with the environment. Apr 24, 2018 at 20:41
• @probably_someone Good point. I should have said that the box doesn't allow any information about magnetization to pass through in any way that is easily observable outside the box. Obviously you need some thermal exchange. I will update my answer. Apr 24, 2018 at 20:46
• Are you sure that thermal exchange without any information exchange is even possible, though? The equivalence between thermodynamic entropy and information entropy seems to suggest otherwise. Apr 24, 2018 at 20:49
• @probably_someone That's why I said "easily" -- the information will always leak, but it might be too scrambled to observe in practice. Apr 24, 2018 at 20:50
• I think the key point that I didn't fully appreciate is that if you only have one copy of a system, then there's no way to measure its thermal expectation value for any operator. If such a measurement device existed, then we could use it to experimentally distinguish between a qubit which is "either in the $|\uparrow\rangle$ or the $|\downarrow\rangle$ pure state, but we don't know which" (which would return $\langle S^z \rangle = \pm 1/2$) from a qubit which is "in the unique mixed state $\frac{1}{2} |\uparrow \rangle \langle \uparrow| + \frac{1}{2} |\downarrow \rangle \langle \downarrow|$" Apr 24, 2018 at 22:01