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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