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Let, $$\hat{H} = a\hat{S_x} + b\hat{S_z}$$where Sx,Sz are the spin operators, a,b constants. Assume the system is coupled to a reservoir.

For clarity, Let $$\hbar=\beta=1$$ The density matrix is

$$ρ=\frac{e^{-β\hat{H}}}{Z}=

\frac{1}{Z} \left(\begin{array}{cc}0&e^{-a/2}\\e^{-a/2}&0\end{array}\right) +

\frac{1}{Z} \left(\begin{array}{cc}e^{-b/2}&0\\0&e^{b/2}\end{array}\right) =

\frac{1}{Z} \left(\begin{array}{cc}e^{-b/2}&e^{-a/2}\\e^{-a/2}&e^{b/2}\end{array}\right)$$

And $$Z=tr(e^{-β\hat{H}})=2\cosh{\frac{b}{2}}$$

The partition function is a description of the statistical properties of the system. But here it is ignorant to the fact that H also contains Sx part. The constant 'a' does not come into play at all. How, then, can we depend on Z to derive our statistical functions for the system?

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# Partition function as a description of the system

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