# Partition function as a description of the system

by phys_student1
Tags: description, function, partition
 P: 96 Hi, 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?
 Sci Advisor P: 2,108 Partition function as a description of the system Your computation of $e^{-\beta H}$ isn't correct, I don't think. An easy way to compute $e^{-\beta H}$ is to first figure out how to diagonalize $H$. That is, you find some matrix $U$ such that $U H U^{-1} = D$ where $D$ is a diagonal matrix, with entries $\lambda_i$. (Those $\lambda_i$ are just the eigenvalues of $H$) Then: $e^{-\beta H} = U^{-1} (e^{-\beta D}) U$ Now, one of the magical properties of the trace is that for any matrices $A, B, C$, $tr(A B C) = tr(B C A)$ So $tr(e^{-\beta H}) = tr(U^{-1} (e^{-\beta D}) U) = tr(e^{-\beta D} U U^{-1}) = tr(e^{-\beta D})$ Computing $e^{-\beta D}$ is easy: it's just a diagonal matrix with entries $e^{- \beta \lambda_i}$. The trace of a matrix is just the sum of the diagonal entries, so So $tr(e^{-\beta H}) = tr(e^{-\beta D}) = \sum_i e^{-\beta \lambda_i}$ For the particular $H$ that you're talking about, I believe it's true that the two eigenvalues are: $\lambda_1 = \sqrt{a^2 + b^2}$ and $\lambda_2 = - \sqrt{a^2 + b^2}$ So $tr(e^{-\beta H}) = e^{-\beta \sqrt{a^2 + b^2}} + e^{+\beta \sqrt{a^2 + b^2}} = 2 cosh(-\beta \sqrt{a^2 + b^2})$
 Quote by stevendaryl An easy way to compute $e^{-\beta H}$ is to first figure out how to diagonalize $H$.