Statistical physics - does total energy matter, or only differences?

[laser1]

Homework Statement

Desc

Relevant Equations

Z

In statistical physics, in a two level system, I'll give an example to show what I am talking about:

situation 1) energy 0 and energy E
situation 2) energy E/2 and energy -E/2

Are these two situations equivalent? Computing the partition function for both of them it seems they are different, but I am not sure. Because from my experience when dealing with energies in mechanics only the energy difference is important. Cheers

 

[pines-demon]

Can you write the two expressions? Why do you think it matters?

 

[laser1]

Can you write the two expressions? Why do you think it matters?

The two expressions for $Z$:

1) $e^{-\beta \epsilon}+1$
2) $2\cosh\left(\frac{\beta \epsilon}{2}\right)$

I think it matters because they are two different expressions!

 

[pines-demon]

The two expressions for $Z$:

1) $e^{-\beta \epsilon}+1$
2) $2\cosh\left(\frac{\beta \epsilon}{2}\right)$

I think it matters because they are two different expressions!

Well the partition functions cannot be exactly the same because energies are slightly shifted so it matter for energy-level dependent quantities, for example if you calculate the mean energy $\overline{E}=-\partial \ln Z/\partial \beta$ you get:

1. $\epsilon/(e^{\beta \epsilon}+1)$
2. $\epsilon \tanh(\beta \epsilon/2)/2$

As $\beta \to 0$, Eq.1 goes to $\epsilon/2$ and Eq.2 goes to $0$ (the average value) as you would expect because of the choice of energy reference. However if you calculate a quantity that does not depend on the energy reference, for example the energy fluctuations around the mean $\Delta E^2=\overline {E^2}-\overline{E}^2=\partial^2 \ln Z/\partial \beta^2$ you get
$$\Delta E^2=\frac{\epsilon^2 e^{\beta \epsilon}}{(e^{\beta \epsilon}+1)^2}$$
for both. You can try with other energy-independent quantities.

 

[pines-demon]

Here is another example, imagine that the higher energy state has a magnetic moment $m$ associated to it, and the lower energy state has a moment $-m$. The average magnetic moment is
$$\overline{m}=m\frac{\exp(-\beta \epsilon)-1}{\exp(-\beta \epsilon)+1}=m\frac{\exp(-\beta \epsilon/2)-\exp(\beta \epsilon/2)}{\exp(-\beta \epsilon/2)+\exp(\beta \epsilon/2)}$$
which is the same value for both.

 

[pines-demon]

Another way to check all of this is to write a system with energies $-E_0$ and $E-E_0$. If you calculate the mean energy it will depend on $E_0$, but if you write the other two quantities above, $E_0$ will vanish.

 

[vela]

Well the partition functions cannot be exactly the same because energies are slightly shifted so it matter for energy-level dependent quantities, for example if you calculate the mean energy $\overline{E}=-\partial \ln Z/\partial \beta$ you get:

1. $\epsilon/(e^{\beta \epsilon}+1)$
2. $\epsilon \tanh(\beta \epsilon/2)/2$

As $\beta \to 0$, Eq.1 goes to $\epsilon/2$ and Eq.2 goes to $0$ (the average value) as you would expect because of the choice of energy reference.

You can see the shift a little more easily if you rewrite the first partition function $Z_1$ as
$$Z_1 = e^{\beta \epsilon}+1 = e^{\beta \epsilon/2}[\underbrace{2\cosh (\beta \epsilon/2)}_{Z_2}].$$ When you calculate the mean energy, the exponential factor contributes a shift of $\epsilon/2$ to the mean energy you get from the second partition function $Z_2$.