Partition function of a particle with two harmonic oscillators

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SUMMARY

The discussion focuses on the calculation of the partition function for a system consisting of two harmonic oscillators, as outlined in Daniel V. Schroeder's "Introduction to Thermal Physics." The partition function is derived by summing the individual partition functions of each oscillator, represented as \( z = Z_A + Z_B \). The confusion arises from the interpretation of the energy levels, where the total energy is not simply the sum of the two oscillators but includes all possible states of both oscillators. The correct formulation emphasizes the need to consider the eigenvalues of the combined Hamiltonian.

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mjmnr3
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Homework Statement
Consider now a particle, which can sit in two harmonic oscillators. Oscillator A has
energy levels ε=nhω, and oscillator B has energy levels ε= ε_0 + nhω, where ε_0 > 0 and
n = 0, 1, 2, . . .. The particle can hop freely between the two oscillators, and the available
states for the particle are therefore the harmonic oscillator levels in A and B.

Question b:
Give an expression for the partition function for the particle, when the temperature is T.
What is the probability P that the particle sits in oscillator A?
Relevant Equations
$$
z=\sum_{s} e^{-E(s) / k T}
$$
Here is the solution I have been given:
deleteme7.PNG


But I really don't understand this solution. Why can I just add these two exponential factors (adding two individual partition functions?).-------------------------------------------------------------------------------------------------------------------------------------

if ##E_{A}=n \hbar \omega## and ##E_{B}=n \hbar \omega+\epsilon_{0},## Why is
$$
z=\sum_{n=0}^{\infty} e^{-E_{A} / k T}+\sum_{n=0}^{\infty} e^{-E_{B} / K T}=Z_{A}+Z_{B}
$$
Because I would solve it differently. The way Daniel Schroeder (Introduction to thermal physics) teaches you to calculate the partition function is:
##z=\sum_{s}^{\infty} e^{-E(s) / k T}##, where ##s## is the different quantum states of the system. So how does the two exponential factors appear?

I interpret the system's hamiltonian consisting of a sum of the two harmonic oscillators $$\mathrm{H}=\mathrm{H}_{\mathrm{A}}+\mathrm{H}_{\mathrm{B}}$$, So what I should really find is the eigenvalues to this operator:
$$\hat{H}|s\rangle=E(s)|s\rangle$$
if ##H_{A}## and ##H_{B}## commutes, then they should have simultaneous eigenfunctions and we should expect the energy to be the sum of each hamiltons eigenvalue: ##E(s)= E_{A}+E_{B}##
So in that case, the partition function would be:
$$
z=\sum_{n=0}^{\infty} e^{-\left(E_{A}+E_{B}\right) / \kappa T}=\sum_{n=0}^{\infty} e^{-E_{A} / k T} e^{-E_{B / K T}}=\sum_{n=0}^{\infty}\left(e^{-n \hbar \omega /(k T)} e^{-n \hbar \omega /(k T)} e^{-\epsilon_{0} /(k T)}\right)
$$
So the issue, is probably that I really don't understand the setup of this system, and what they mean by one particle and two harmonic oscillators.
 
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Here is the full problem text:
deleteme6.PNG

I use "An introduction to thermal physics" by Daniel V. Schroeder, If you prefer to refer directly to the litterature.
 
For the system in question b, the possible states of the particle are the states of oscillator A as well as the states of oscillator B. So, the possible energies of the particle in question b are the energy levels of A and the energy levels of B.

1609638385245.png


Note that the energy levels for part b are not the sum of the energies of oscillator A and oscillator B. The energy levels in part b are all the energy levels of A and all the energy levels of B.
 
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