Entropy of a generalised two-state quantum system

[Gabriel Maia]

Hi. This is the problem I'm trying to solve:

A system may be in two quantum states with energies '0' and 'e'. The states' degenerescences are g1 and g2, respectively. Find the entropy S as a function of the Energy E in the limit where the number of particles N is very large. Analyse this dependence and show that there is a region of negative temperature.

The energy of the system is given by

[tex]E=N(0\cdot P(0)+e P(e))[/tex]

where,

[tex] P(0) = \frac{g1}{g1+g2} \,\,\,\,\,\,\,\,\,\, \mathrm{and} \,\,\,\,\,\,\,\,\,\, P(e) = \frac{g2}{g1+g2}[/tex]

are the probabilities of the state with energy 0 and energy e being occupied. Now, I'm trying to find the number of possible microstates in order to calculate the entropy. Since I want to organise N particles in two groups of identical states, the number of microstates should be

[tex]\Sigma = \frac{N!}{\left(\frac{Ng1}{g1+g2}\right)!\left(\frac{Ng2}{g1+g2}\right)!}[/tex]

Is this last expression correct? I'm not sure if it should be this or just the degenerescences.
Thank you very much.

 

[mark726]

Hello,

Thank you for your question. Your expression for the number of microstates is correct. In this case, the number of microstates is equal to the number of ways N particles can be divided into two groups with g1 and g2 particles, respectively. This takes into account the degenerescences of the states.

To calculate the entropy, we can use the Boltzmann entropy formula:

S = k_B ln(\Sigma)

where k_B is the Boltzmann constant. Substituting the expression for \Sigma that you have derived, we get:

S = k_B ln\left(\frac{N!}{\left(\frac{Ng1}{g1+g2}\right)!\left(\frac{Ng2}{g1+g2}\right)!}\right)

Using Stirling's approximation for large N, we can simplify this expression to:

S = Nk_B\left[\frac{g1}{g1+g2}ln\left(\frac{g1}{g1+g2}\right) + \frac{g2}{g1+g2}ln\left(\frac{g2}{g1+g2}\right)\right]

We can see that the entropy is a function of the energy E, which is dependent on the probabilities P(0) and P(e). As the number of particles N becomes very large, the entropy becomes a continuous function of energy. At low energies, the entropy increases with energy, while at high energies, the entropy decreases with energy.

In the limit where the number of particles N is very large, we can also see that the system can reach negative temperatures. This occurs when the numerator of the expression for entropy becomes negative, which happens when the energy E becomes negative. This is a counter-intuitive result, as we typically think of temperature as a positive quantity. Negative temperatures occur in systems with a discrete energy spectrum, such as this one, and have been observed in some physical systems.

I hope this helps you in your analysis. Let me know if you have any further questions.