# Why should the equal a priori states be eigenstates of the Hamiltonian?

When I have was taught quantum statistical mechanics, the course derived the canonical ensemble by assuming that eigenstates of the total system Hamiltonian have equal a priori probabilities, and then imposing Lagrangian constraints for the average energy and sum of probabilities being equal to one.

I want to know, why physically it is the eigenstates of the Hamiltonian that should have equal a priori probabilities and not some other complete set of states that span the Hilbert space?

One way to derive the canonical ensemble is by maximizing the entropy of the subsystem subject to given constraints (usually including a constraint on the average total energy). In this derivation, no a priori probabilites are assumed at all. On the contrary, maximizing the entropy amounts to finding the least presumptuous probability distribution compatible with the given constraints.

Another way to derive the canonical ensemble is to consider a subsystem of a larger system that is in the microcanonical ensemble. The microcanonical ensemble consists of a subspace (of the full Hilbert space) specified by some set of constraints, which usually includes a constraint on the total energy $$E$$, such as $$E for some given maximum $$E_0$$. The microcanonical ensemble assigns equal probabilities to all states in any orthogonal basis that spans the subspace. Here's the key: Within the given subspace, assuming equal probabilities to the vectors in one orthonormal basis is equivalent to assuming equal probabilities to the vectors in any other orthonormal basis. That's because when $$p_n=1/N$$ for all $$n$$, the density matrix $$\rho=\sum_{n=1}^N p_n|n\rangle\langle n|$$ is proportional to the identity matrix within the $$N$$-dimensional subspace spanned by the orthonormal state-vectors $$|n\rangle$$, and this is true no matter which orthonormal basis is used. Using a basis in which the basis vectors $$|n\rangle$$ are eigenstates of the Hamiltonian is often convenient, but it's not necessary; all predictions depend only on the density matrix and so are independent of which orthonormal basis is used.

• This seems like exactly what I was looking for, thank you. Could you flesh out or provide a reference for the next steps in how you arrive at the Canonical ensemble from the micro-canonical ensemble. I assume this involves tracing out the environment to find the reduced density matrix for the subsystem, but am not sure how to actually do this. Thanks
– J.L.
Apr 9, 2019 at 16:39
• @J.L. Explicitly tracing out the environment is not actually necessary. The purpose of the density-matrix argument is simply to say that we aren't required to use the energy basis (to answer the original question), but we still can use the energy basis, and once we've agreed to work just with a basis of energy eigenstates, we can use the same state-counting approaches that we would use in classical physics, like this one on Physics SE: "How is the distribution probability in the canonical ensemble derived?" (physics.stackexchange.com/q/70263) Apr 10, 2019 at 2:00
• I do not understand how the derivation you link to makes sense, they seem to just assume the final result, I think this maybe warrants a separate question and so have asked it in detail here <physics.stackexchange.com/questions/471968/…>
– J.L.
Apr 11, 2019 at 12:13
• @J.L. Good question! I posted an answer: physics.stackexchange.com/a/472183 Apr 12, 2019 at 3:54
• That is very helpful, thank you. I'm still not entirely sure how these 2 different perspectives (max-ent and this system bath perspective) end up being equivalent, but Im not sure yet how to pinpoint where my confusion lies.
– J.L.
Apr 12, 2019 at 11:28

The same ansatz is made in classical stat mech to derive the expression for probability of a configuration in terms of the partition function: configurations of equal energy are equally likely to be found.

If you want to ask why the classical ansatz: dynamically, systems tend to go to lower energy states. A ball rolls down a hill, charges attract/repel. Therefore there is a correlation between how likely a configuration is and its energy. It's natural to posit that equal energy configurations are equally likely to be found.