# Showing existence of negative temperature for a quantum system

It may be shown that the partition function for a quantum system containing N distinguishable particles each of which has energy state $\epsilon_1$ and $\epsilon_2$ is given by

$$Z(\beta)=(e^{-\beta\epsilon_1}+e^{-\beta\epsilon_2})^N$$.

Show that the system exhibits negative temperature.

For a start I don't really understand why we don't just rescale the Kelvin scale if there are negative temperatures?

But for the purposes of the question I tried using the equation $\displaystyle E=-\frac{\partial \ln Z}{\partial \beta}$,

\begin{align} E=-\frac{\partial}{\partial \beta} \ln \left( e^{-\beta\epsilon_1}+e^{-\beta\epsilon_2} \right)^N &=-N\frac{1}{e^{-\beta\epsilon_1}+e^{-\beta\epsilon_2}}\frac{\partial}{\partial \beta} \left( e^{-\beta\epsilon_1}+e^{-\beta\epsilon_2} \right) \\ &=-N\frac{-\epsilon_1e^{-\beta\epsilon_1}+-\epsilon_2e^{-\beta\epsilon_2}}{e^{-\beta\epsilon_1}+e^{-\beta\epsilon_2}}\\ &=N\frac{\epsilon_1e^{-\beta\epsilon_1}+\epsilon_2e^{-\beta\epsilon_2}}{e^{-\beta\epsilon_1}+e^{-\beta\epsilon_2}}\\ \end{align}

which as $\beta=\frac{1}{k_BT}$,

$$N\frac{\epsilon_1e^{-\frac{1}{k_BT}\epsilon_1}+\epsilon_2e^{-\frac{1}{k_BT}\epsilon_2}}{e^{-\frac{1}{k_BT}\epsilon_1}+e^{-\frac{1}{k_BT}\epsilon_2}}$$

which doesnt seem to suggest negative energy as $e^{-x}>0$.

You should start by the definition of temperature $$\frac{1}{T} = \frac{\partial s}{\partial E }= k \frac{\partial \ln \Omega}{\partial E}$$

Also you have the partition function $$Z_1=\sum \exp(-bE_r) \text{ where } b=1/kT$$

Let's now assume a single magnetic dipole in a magnetic field inside a heat tank( we can assume the heat tank as the rest of the body). It is then $$Z_1=2 \cosh x \text{ where } x=μΒ/κΤ \text{ and } μ \text{ the magnetic moment.}$$

And $$p+_=\exp(\pm x)/Z_1$$ depending on the spin (we have a spin=1/2 or spin =-1/2)

Putting the partition function in the possibility relationship and solving to find b you get $$b=1/kT=[\ln(n/N-n)]/(2μB)$$ where n are, let's say, the spin up particles in the magnetic field , if we have not only one particle but a system of them(we can say this because of the additive property of the energy)

We see from here that T(absolute temperature) is positive or negative depending on $n$. If $n>N/2$ then $T<0$, that is the most of the dipoles are anti-parallel with the magnetic field.We can show that if n up is equal to n down then T is infinite. This means that a negative temperature is a ''hotter'' state than a positive temperature.

Intuitively you could say from the beginning through the T definition that negative temperature is a negative derivative($ΔΕ>0$ and $ΔΩ<0$ or the opposite). You can have this negative derivative if your system has an upper energy limit and a property of the system is independent of the other properties of the system( in our example the time needed for the spins to come in balance with each other is much smaller than the time needed to come in balance with the oscillations of the crystal grid(we can then say we have a spin temperature).

What negative temperature means, as i have understand, is that by raising the energy of the system the entropy is getting smaller(think of our example-the energy dispersal of the system- or what some say the disorder- is getting smaller, the weight function for n>N/2 are lesser as n grows).

PS: There are also applications for the pressure of the system if you want to search it. PS2:Some papers that may interest you in the subject, if you want:

http://arxiv.org/abs/1304.2066

http://www.physik.uni-augsburg.de/theo1/hanggi/Dunkel_Nature.pdf

http://arxiv.org/abs/1403.4299

• @sandstone Thanks for the edit. I'll try to learn better the math writing on the site. Commented Apr 17, 2015 at 12:41
• Commented Apr 17, 2015 at 14:32