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This question already has an answer here:

The Boltzmann distribution function tells us what is the probability of a given particle with a given energy to be at a certain state.

Now, this is in contrast with the state function the Schrodinger equation gives us. I mean, the quantum state gives us a probability distribution in energy space, and the statistical mechanics also gives us such a distribution, but why the two distribution differ?

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marked as duplicate by knzhou, John Rennie quantum-mechanics Jun 10 at 10:53

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    $\begingroup$ ..... because they describe different situations? $\endgroup$ – Emilio Pisanty Jun 8 at 10:55
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This is actually a wonderful question and I would encourage you to think more about it.

As Emilio said in the comments. The two really describe two different scenarios. In the case of a wavefunction, we consider the system of interest as isolated and there is no notion of temperature (The notion of temperature most notably comes when our system is in contact with a heat reservoir). In such a case, the system has a fixed average energy i.e $\langle\psi| H |\psi\rangle$, that is the expectation value of the Hamiltonian with respect to the wavefunction $|\psi\rangle$.

This is different than the case when we attach our system to a heat reservoir of temperature T. Here, we consider an ensemble of wavefunctions $|\psi_i\rangle$ each of average energy $\epsilon_i$. Calculations of various properties of the system can be done by using the density operator (a.k.a density matrix) $ \sum_i p_i |\psi_i\rangle\langle\psi_i|$, here $p_i $ will be the probability as per the Boltzmann distribution of $|\psi_i\rangle$, the wavefunctions present in the system. (note: one thing you should think is that ask yourself is that wither this different from a superposition of different wavefunctions). This is in a sense how one goes from quantum mechanics to quantum statistical mechanics.

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