As far as I know by inserting a harmonic potential $V(x) = \frac{1}{2}m \omega x^2$ into the time-independent schrödinger equation I can obtain the wave-functions eigenstates and eigenvalues (energies).

So let us assume we have a quantum harmonic oscillator. Depending on the set-up this harm. osci. can be in a pure state $\mid \psi \rangle = \mid 0\rangle$ with density matrix $\rho = \mathrm{diag}\{1,0,0,..\}$ or in a thermal state with a density matrix $\tau(\beta) = \mathrm{diag}\{P_1,P_2,P_3,...\}$ where $P_1\leq P_2 \leq P_3 \leq ...$

So both states (pure vs. thermal) relate to the same Hamiltonian with the same eigenvalues it's just that the probability distribution is different. Is that correct?

Also I find it somewhat confusing when we are talking about 'states the system is in' vs. 'eigenstates of the Hamiltonian' which are 2 completely different things as far as I understand this. Or am I missing something?

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    $\begingroup$ Indeed, an eigenstate of the Hamiltonian is a pure state, which is distinct from a mixed state. A thermal state is a mixed state. $\endgroup$ – knzhou Jan 12 '19 at 14:26
  • $\begingroup$ Hm, when you say 'eigenstate of the Hamiltonian is a pure state' you mean not any specific pure state $\mid n \rangle$ but just that in general the property that any eigenstate of the Hamiltonian is a state that has only one 1 on the diagonal of the density matrix and rest is $0$, correct? I.e. all these $\mid 0\rangle ,\mid 1\rangle , \mid 2\rangle ....$ are eigenstates of H. $\endgroup$ – CatoMaths Jan 12 '19 at 14:40
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    $\begingroup$ All eigenstates of the Hamiltonian are pure states. But not all pure states are eigenstates of the Hamiltonian. For example, $|n \rangle + |m \rangle$ is a pure state. $\endgroup$ – knzhou Jan 12 '19 at 14:50
  • $\begingroup$ The form of the density matrix depends on the basis you're using. Yes, if you use the energy basis, then the density matrix of an energy eigenstate is mostly zeroes, with a single $1$ on the diagonal. $\endgroup$ – knzhou Jan 12 '19 at 14:51
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    $\begingroup$ You've asked a lot of questions about the basics of quantum mechanics, and while you might get a few answers, this really won't be efficient. It would be much better to sit down for a few hours with a good book. Almost any book on quantum mechanics will cover all of the questions you've asked. $\endgroup$ – knzhou Jan 12 '19 at 14:51

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