# What is the temperature of a pure quantum state?

I was wondering about temperatures and pure quantum states. I'm currently working on thermalization of isolated quantum systems, which can be described by pure quantum states (kets). How do we define a temperature for these?

Normally you work within the density matrix formalism so you can define an ensemble, which is a mixed state, for a certain temperature. In my scenario we prepare a system in a lab in a pure state, that is isolated, and let it thermalize to its equilibrium state. This must be (in very good approximation) still a pure state if the system is strongly isolated, because of the unitarity of evolution operator.

We can then probably assign a certain temperature to this equilibrium system, as we could bring this system in contact with a heath bath with a certain temperature and monitor if energy flows out of the system or into the system (aka it has a higher or lower temperature). But from a theoretical viewpoint I have no idea how to assign a temperature to a pure quantum state, so I'm quite dazzled by this.

I know the ground state is supposed to have $$T = 0 K$$, but that's all I could think of. The Internet is not giving me much more information.

Edit: Ofcourse I am talking about a certain many-body system with a corresponding Hamiltonian, which should be non-integrable and chaotic. This way the system will thermalize (in most cases) following the Eigenstate Thermalization Hypothesis. You could take the 1D spin chain non-integrable Ising model as example.

• what do you mean by "let it thermalize to its equilibrium state"? Do you just that you take the unitary time-evolution of the Hamiltonian of the local system and wait for a long enough time? or do you consider some external bath to which the system is weakly coupled? – user245141 Apr 1 '20 at 14:31
• No, no coupling to a bath. The point is that it thermalizes while isolated so that it stays a pure state because of unitary evolution. I must add that thermalizing of isolated quantum states is not completely understood yet from a theoretical viewpoint, but it certainly happens (experimental evidence), you might wanna google Eigenstate Thermalization Hypthesis. But anyways, this does not really change my question. – CFRedDemon Apr 1 '20 at 14:48
• I am talking about many-body systems ofcourse, like a long spin-chain or something. One particle systems are not going to thermalize ofcourse if prepared in a special initial pure state. – CFRedDemon Apr 1 '20 at 14:50
• Then I think that a central question here is whether your system is finite - and then you will remain in a pure state and one cannot ascribe temperature in the usual sense to the system as a whole (but you might be able to ascribe temperature locally), or if you can consider it to be infinite, with internal relaxation and dissipation mechanisms. Then, I think, assuming that your initial state has microscopic energy, you will evolve to the ground state (see for example Phys. Rev. B 73, 245326). Things might be different if you have disorder, though – user245141 Apr 1 '20 at 15:42
• Well that was my original question: What is the (or can you ascribe a) temperature of an isolated system in a pure state? You say you can't, which is weird to me because you can also isolate a gas, which starts off as a pure state, and describe it quantum mechanically . You know in the thermodynamic (and classical) limit it should have a certain temperature? It is a physical system, you don't even have to probe it, as I said you can put it in a heat bath you know the temperature of and see if heat flows out of the system. So your answer is probably not wrong, but not really satisfying. – CFRedDemon Apr 1 '20 at 16:15

The same question could be asked in classical physics: What is the temperature of a specific microstate of an ideal gas? Conceptually, the answer to the quantum version is essentially the same as the answer to the classical version.

If the question is how to actually calculate the temperature of a given pure state (at least in principle), then one way is to take a partial trace over half of the system. If the system is large enough so that the energy of interaction between the two halves is negligible, then the result should have the form $$\exp(-\beta H)$$ where $$H$$ is the self-Hamiltonian of the remaining half. This assumes that the whole-system pure state has attained "equilibrium", in the sense that it has evolved long enough to be in a "typical" state for the given macro-conditions.

This paper may be of interest:

From the abstract:

Currently there are two main approaches to describe how quantum statistical physics emerges from an isolated quantum many-body system in a pure state...

And here's an older review paper:

• That was the kind of answer that I was looking for. Thanks for providing more background! – CFRedDemon Apr 2 '20 at 12:28