# Partition function as a description of the system

1. Dec 13, 2013

### phys_student1

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?

Last edited: Dec 13, 2013
2. Dec 14, 2013

### BruceW

I don't understand how $\exp(-\beta \hat{H} )$ is equal the thing you have written out on the right hand side. When there is an exponential of a matrix, you need to write it out as the Taylor series of an exponential, where each term will contain the matrix, to some power.

This will give an infinite series, which is not something that is usually 'nice'. But you can use the properties of the Pauli matrices to get a 'nice' answer. This is not obvious if you have not been shown how to do this problem before.

3. Dec 14, 2013

### dauto

Also, the calculation of the trace doesn't look right. The trace is supposed to be taken over the space of eigenfunctions of your solutions. That has nothing to do with the trace of spin matrix. For instance, a harmonic oscillator doesn't have spin matrix at all but it has an infinite number of eigenstates. the trace is taken over that infinite space.

4. Dec 14, 2013

### stevendaryl

Staff Emeritus
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 $tr(e^{-\beta H}) = e^{-\beta \sqrt{a^2 + b^2}} + e^{+\beta \sqrt{a^2 + b^2}} = 2 cosh(-\beta \sqrt{a^2 + b^2})$

5. Dec 14, 2013

### phys_student1

Yes I think I missed up things while trying to use Latex.

6. Dec 14, 2013

### BruceW

oh wow, that's a much easier way than I was suggesting. and much more general, than just relying on the nice properties of the spin matrices. I'll keep that in mind. the sad part is that I probably learned that not so long ago, but had forgotten about it already!