# Statistical mechanics: energy variance of ensemble

1. Jan 19, 2010

### Dawei

I posted this once already without seeing the rule that HW questions must go here--sorry

So, the problem: I'm really lost on where to get started here. It's a two state system, one with energy 0 and the other with energy ε. I already have ensemble average, <E>, found to be:

ε / (e^βε + 1) , where β is thermodynamic beta, 1/KbT.

How do I convert this to an expression for the variance of the energy?

A more complete description of the problem can be found http://books.google.com/books?id=z6...in+thermal+physics&cd=1#v=onepage&q=&f=false", on the top half of page 41:

I've already found http://en.wikipedia.org/wiki/Partit...chanics)#Relation_to_thermodynamic_variables" explanation, but I'm not connecting how it could be applied to this example:

Please, can anyone point me in the right direction?

Last edited by a moderator: Apr 24, 2017
2. Jan 19, 2010

### Mapes

Can you identify the partition function Z in this system?

3. Jan 19, 2010

### Dawei

The partition function would be the denominator of the probability function:

Z = (1 + e-βε) ?

No?

4. Jan 19, 2010

### Mapes

Great, so have you calculated the variance $\partial^2 \ln Z/\partial\beta^2$?

5. Jan 19, 2010

### jambaugh

From general probability the variance of a variable X is <X^2>-<X>^2.
So you can directly calculate <E^2> from the probabilities and subtract out <E>^2.

But the partition function gives you a short-cut. See the wikipedia article:http://en.wikipedia.org/wiki/Partition_function_(statistical_mechanics)" [Broken]

$$\langle \Delta E^2\rangle = \frac{\partial^2}{\partial\beta^2} \ln(Z)$$
where
$$\beta = 1/kT$$
and Z is your partition function.

To understand where the partition function comes from and what it is one must go back to the original problem. That problem is to determine the probability distribution (quantum density matrix, or classical distribution) which maximizes the entropy but which also preserves a given set of expectation values.

The method for solving constrained optimization problems is the Method of Lagrange "[URL [Broken] Method[/URL]and the beta's are the lagrange multipliers.

In the Canonical Ensemble the only expectation value of interest is the Energy. In Grand Canonical Ensemble, particle number is also variable and thus included.

Given the Boltzmann, or VonNeumann entropy has the form:
$$S=-\sum_k p_k \ln(p_k)$$
which has derivatives with respect to probabilities of $$\ln(p_k) + 1$$...

when we solve for the optimizing probabilities, due to the logarithmic form of the entropy's derivative, we get them as exponentials of the lagrange multipliers times the fixed values of the quantities to have fixed expectation value.

There is one more constraint and that is the sum of probabilities (trace of the density matrix) must add up to 1. This extra constraint really has its own lagrange multiplier and that is what really defines the partition function.

A definition of the partition function is then that it is the exponential of the lagrange multiplier corresponding to the normalization constraint which we can view as the expectation value of the trivial random-variable (observable) 1 being 1. (When we vary the probabilities to optimize entropy they are just numbers and not yet probabilities, so the "expectation value" is just a formula.)

$$p_k = e^{-z\mathbf{1} - \beta E_k} =\frac{ e^{-\beta E_k} }{e^{z}} = \frac{1}{Z} e^{-\beta E_k}$$
where
$$Z=e^{z}$$

If all this doesn't make much sense... don't worry about it. Tuck it away for future reference. But when you have the time you'll find it enlightening to go through the derivation of the partition function.

Last edited by a moderator: May 4, 2017
6. Jan 19, 2010

### Dawei

Embarrassing it may be, it's been a while since I've done a partial derivative. Here is my attempt:

∂ ln(1 + e-βε)/∂β

= (-εe-βε) / (1 + e-βε)

Then differentiating again with quotient rule, I get this mess:

[ (1 + e-βε)(ε2e-βε) - (-εe-βε)(-εe-βε) ] / [(1 + e-βε)2 ]

Could that be close to correct?

EDIT: Thank you for your added comments jambaugh. A lot of it did go over my head, but I'll take some time and try to digest it.

Last edited: Jan 19, 2010
7. Jan 20, 2010

### jambaugh

I see no errors at first glance. You can simplify before taking that second derivative, multiply numerator and denominator by eβε. Should give you a much simpler form.