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my professor in the lecture defined a general thermal 'state' (which actually is a density matrix) in the following way

$$\tau(\beta) = \frac{e^{-\beta H}}{Z}\tag{1}$$ where Z is the partition function defined as $Z = \mathrm{Tr}(e^{-\beta H})$.

Given a harmonic potential, we consider a harmonic oscillator to be initially in such a thermal state.

If we insert the for a harmonic oscillator corresponding Hamiltonian $H = \sum \omega a^{\dagger}a$, where $a^{\dagger},a$ are creation/anihiliation operators, into (1) we obtain

$$\tau(\beta) = (1-e^{-\beta \omega})\sum_n e^{n\beta \omega} \mid n\rangle\langle n \mid$$

My question here is, what exactly is such a thermal 'state' (density matrix) in this context?

Expressed in the energy-eigenbasis of the systems Hamiltonian it is diagonal with its entries decreasing$\tau(\beta) = \mathrm{diag}\{P_1,P_2,P_3,...\}$ with $P_1 \leq P_2 \leq P_3 \leq...$

So from this I can say that it is NOT a pures state. But is there anything else I might be missing? I find it hard to think in terms of 'Our harmonic oscillator is in the thermal state'.

Can someone give me contextual understanding to this? Thanks(:

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  • $\begingroup$ I'm not really sure what you're asking. A harmonic oscillator is a physical system. Systems have states. What you have is the thermal state, which is the state that the system takes at temperature $T$. What do you want to know besides that? $\endgroup$ – knzhou Jan 12 at 12:43
  • $\begingroup$ As far as I know if we insert the potential V(x) our system is in, into the time-independent schrödinger equation we obtain the systems eigenstates and corresponding eigenvalues. But this approach doesn't take temperature T into consideration at all. So....are you trying to tell me that if we have a system in a quadratic potential with temperature T and somehow try to look for the eigenstates we'll find that the eigenstates are the thermal state? $\endgroup$ – CatoMaths Jan 12 at 12:52

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