# Homework Help: Statistical mechanics: Sums of exponentials with sums.

1. Sep 24, 2011

### Beer-monster

1. The problem statement, all variables and given/known data

I'm working through an example from class and the textbook, but I'm confused about how the steps progress mathematically.

The example involves the Gibb's partition for a paramagnet.

$$\sum_{s} exp(\beta \mu B \sum_{i}^{N} s)$$

Where s = -a,-a+1...a for each spin.

Am I right in thinking the sum in the exponent is a sum over this particles in the sample (N). Since a sum in an exponent is a product of the exponentials we get:

$$[\sum_{s} exp(\beta \mu B s)]^{N}$$

However, I'm not sure what do with the second sum. I can't see a neat way proceed. Can anyone point me in the right direction?

2. Sep 26, 2011

### G01

The second sum is a sum over spin states. If you're dealing with quantum systems with a discrete number of spin states, then it's probably best to write the sum explicitly and re-express the result in a more simple form. HINT: use some hyperbolic trig identities.

If you are treating the spin classically and the value of s can take on any value $s_o cos( \theta )$ (where theta is the angle from the z axis), then your best best will be to convert the sum to an integral in theta.

3. Sep 27, 2011

### Beer-monster

Thanks for the response. I'm treating the spins as quantum. I managed to solve the taking out the factor of a and treating the rest as a geometric series i.e.

$$[e^{-\beta\mu B a} \sum_{s=1}^{(2a+1)} exp(\beta \mu B s)]^{N}$$

Which, using the formula for the sum of a finite geometric series gives:

$$[e^{-\beta\mu B a} \frac{1-e^{(2a+1)\beta\mu B}}{1-e^{\beta \mu B}}]^{N}$$

Problem, is I can't seem to resolve that down to a hyperbolic function and I'm sure it should. Any ideas?

4. Sep 27, 2011

### G01

Oh ok. So, your dealing with a quantum spin system of arbitrary integer or half integer spin, not just spin 1/2. That makes it slightly more complicated but, not by too much.

The sum of the finite geometric series is the way to go.

I don't think the partition function can be simplified directly to a single hyperbolic trig function as it stands, but for large values of a, your expectation values for the Energy, Magnetization, Specific heat, etc. definitely can be.

Try this: Keep the partition function in exponential form, but simplify and multiply through the exponential,$e^{-\beta \mu B a}$. Take $\frac{\partial Z}{\partial \beta}$ with everything in exponential form. Then you can use that, along with Z to find all the state variable expectation values. Then for large a, these quantities will be expressible in terms of only hyperbolic trig functions.