Pure and mixed thermal states

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Hey guys,
I was reading about thermal states and now I have a doubt: is a thermal state always a mixed state with density matrix ρ=exp(-βH)/Tr(exp(-βH)), or is there also a pure thermal state?
Thank you
 

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  • #2
vanhees71
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Think about what happens for [itex]T=1/\beta \rightarrow 0[/itex]!
 
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ρ→1 and we get a pure state, is that correct?
So there are pure thermal state, they are just thermal state at an (ideal) zero temperature?
If I am right, do such pure thermal states exist in nature?
Thanks a lot for your help!
 
  • #4
Physics Monkey
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As long as the ground state is not degenerate, the zero temperature limit does give a pure state. However, one can never really reach zero temperature by cooling in the physical world. On the other hand, there are other ways to prepare a nearly isolated pure state.

In fact, there is a sense in which pure states may display thermal properties. Imagine a closed system begun in some pure initial state which evolves unitarily. If the initial state is, on average, highly excited (e.g. has a finite energy density above the ground state), then one would expect on general grounds that the system should "thermalize" in some sense. Yet the state of the whole system must remain pure. However, as long as we look at small pieces of the whole system, we may imagine that the rest of the system acts as an effective bath, and the state of the small subsystem may look thermal.
 
  • #5
vanhees71
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A pure state is by definition described by a statistical operator that is a projection operator
[tex]\hat{\rho}=|\psi \rangle \langle \psi|[/tex]
with a normalized state vector [itex]|\psi \rangle[/itex], i.e., with
[tex]\langle \psi|\psi \rangle.[/tex]
Indeed if the ground state is not degenerate, then
[tex]\lim_{T \rightarrow 0} \hat{\rho}_{\text{can}} = |\Omega \rangle \langle \Omega|,[/tex]
where [itex]|\Omega \rangle[/itex] is the energy-eigenvector for the lowest energy-eigenvalue, and this is uniquely defined (up to a phase, which cancels in the statistical operator).

To stress it again: The identity operator generally cannot be a proper statistical operator, because for usual physical systems the state space is a true infinitely-dimensional Hilbert space, and a statistical operator must have trace 1. The identity operator in a proper Hilbert space has no finite trace and thus cannot represent a mixed state.
 

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