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If we want to calculate mean magnetisation of an equilibrium two-level-system, we know that we can resolve the identity $ \mathbf{1} = \sum_i | E_i \rangle \langle E_i |$ and giving us a uniform measure over the states of the system.

I then remember my classical stat mech, shove in some Boltzmann factors and obtain the gibbs state $$\rho = Z^{-1}\sum_i \mathrm{e}^{- \beta E_i}| E_i \rangle \langle E_i | $$ and can calculate the mean magnetisation, for a two level system with $E_0 = -B/2$ and $E_1=B/2$ we then get $\langle m \rangle = \mathrm{Tr}[\sigma_z \rho]/2 = \tanh \left(\beta B/2 \right)/2$. Alls seems very good.

However if I was ignorant to this result, I might also say I can write down a state $$ | \psi \rangle = \cos \frac{\theta}{2} | 0 \rangle + \mathrm e^{\mathrm i \phi}\sin \frac{\theta}{2} | 1 \rangle$$ which hase energy $\langle \psi | H | \psi \rangle = - B \cos \theta/2$.

So since I can write down a measure over my states $$ \mathbf{1} = \frac12\iint \sin \theta \, \mathrm d \phi \, \mathrm d \theta \, | \psi \rangle \langle \psi |$$ I can also contruct some sort of Gibbs state by weighting them all by a Boltzmann factor, this gives $$\rho = \frac1{2Z}\iint \sin \theta \, \mathrm d \phi \, \mathrm d \theta \, \mathrm e^{\beta B \cos \theta/2}| \psi \rangle \langle \psi |$$ However this yields $\langle m \rangle = \mathrm{Tr}[\sigma_z \rho]/2 = \coth (\beta B/2)/2-1/(\beta B)$, the classical result.

My question is: Is there a simple physical principal to which I can point to determine the correct process (one which is more satisfactory than simply observing that one of these approaches works and the other obtains a different answer)?

Or is it a matter of accepting $\rho = \exp (-\beta H)/Z$ as the definition of thermal equilibrium? (and hence the von Neumann entropy as the correct entropy?)

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Equilibrium can be defined only axiomatically. To be consistent with traditional thermodynamics, it has to be described by a density matrix of the form $e^{-\beta S}$ with an operator $S$ that consists of globally conserved quantities, in the simplest case just the Hamiltonian plus a constant (conventionally absorbed into $Z$). Thermal equilibrium is therefore always defined through a Gibbs state; for boundary conditions corresponding to a canonical ensemble by ρ=exp(−βH)/Z. In particular, a thermal state is always given by a density operator. It is impossible to refine the description - the textbook description of a density operator as a mixture of pure states is fictitious and very far from unique.

If a system is approximately in a pure state it is either very far from equilibrium, or it is approximately in the ground state at a temperature corresponding to an energy significantly smaller than the energy gap between the ground state and the first excited state (so that only the ground state contributes significantly to the canonical ensemble).

For example, the latter is the case for the electronic part of a molecule in cases where the Born-Oppenheimer approximation is applicable. (The latter fails when the energy gap gets too small.)

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  • $\begingroup$ Hi, thanks very much for your answer. However I have a question: Am I right in understanding we are taking an axiomatic approach to defining $\rho$, and taking this our starting point? This seems unsatisfying to me as I thought the usual open systems prescription was of (system)+(environment) and was to assume the global state is pure. This would imply that the equilibrium state of the system has to, at the very least, be consistent with Eigenstate Thermalisation Hypothesis, and not something we are at liberty to freely choose. $\endgroup$ – ComptonScattering Nov 23 '16 at 10:53
  • $\begingroup$ Your answer is consistent with ETH of course, but this seems to imply to me that it should be derived in some sense, rather than chosen. $\endgroup$ – ComptonScattering Nov 23 '16 at 11:02
  • $\begingroup$ @ComptonScattering: Equilibrium can be defined only axiomatically. To be consistent with traditional thermodynamics, it has to be described by a density matrix of the form $e^{-\beta S}$ with an operator $S$ that consists of globally conserved quantities, in the simplest case just the Hamiltonian. $\endgroup$ – Arnold Neumaier Nov 23 '16 at 11:26
  • $\begingroup$ Even when the global state is assumed as pure (which is often done although one can never check whether this is truly the case) the elimination of the environment produces a density matrix for the state of the subsystem considered. $\endgroup$ – Arnold Neumaier Nov 23 '16 at 11:28
  • $\begingroup$ For tiny systems people also look at nontraditional (Tsallis) thermodynamics, where the exponential is replaced by a different function. But even then equilibrium is described by a pure state. $\endgroup$ – Arnold Neumaier Nov 23 '16 at 11:29

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