An ensemble of non-interacting spin particles is in contact with a heat bath at temperature $T$, and is subjected to an external magnetic field. Each particle can be in one of the two quantum states of energies $ \pm \epsilon_0$. If the mean energy per particle is $-\frac{\epsilon_0}{2}$ , then the free energy per particle is:

  1. $-2\epsilon_0 \frac{\ln{\frac{4}{\sqrt{3}}}}{\ln{3}}$

  2. $-\epsilon_0 \ln(\frac{3}{2})$

  3. $-\epsilon_0 \frac{\ln2}{\ln3}$

  4. $-2\epsilon_0 \ln2$

I have tried in the following methods:

  1. The mean energy per particle is $-\epsilon_0/2$.
    So the total energy of the system of $N$ particles is $-N\epsilon_0/2$. And total number of micro-states is $2^N$.
    Then the free energy $F=E-TS=-N\epsilon_0/2-Nk_BT\ln2$. Hence free energy per particle is $f=F/N=-\epsilon_0/2-k_BT\ln2$.

  2. The partition function of the system is $Z=[e^{\beta \epsilon_0}+e^{-\beta \epsilon_0}]^N$.
    So the free energy per particle $f=F/N=-k_BT \ln[2\cosh(\beta \epsilon_0)]$.

But none of that yields the answer. Could you please explain where is the fault?


1 Answer 1


Here's what's wrong with your two methods.

  1. The $2^N$ microstates do not all have the same energy. Since you are thinking in terms of a microcanonical distribution for this method, you want to only count the states with total energy $E$. But really since this is in contact with a heat bath you should be thinking in terms of a canonical distribution like your second method (although counting the states with energy equal to the average total energy should be pretty close to this).

  2. Are you sure this isn't one of the answers? The temperature is not a free parameter if you have fixed the average energy. In terms of the partition function the average energy is equal to $$-N\frac{\epsilon}{2}=-\frac{\partial}{\partial\beta}\ln Z.$$ This fixes $\beta$ in terms of $\epsilon$.

  • $\begingroup$ Actually I was curious and tried it, and the "pretty close" I said for method 1 is equivalent to dropping the $O(\ln N)$ corrections to Stirling's approximation. So both of these methods done correctly (using $\ln N! = N \ln N - N$ in the 1st) give the right answer. $\endgroup$
    – octonion
    Jun 12, 2017 at 0:44
  • $\begingroup$ It is a question from Joint CSIR UGC NET 2015 December Exam (Physics paper, question no: 60) and the correct answer is given to be the option 1. @octonion $\endgroup$ Jun 12, 2017 at 6:44
  • $\begingroup$ @SwarnadeepSeth If you read carefully what octonion wrote, you will obtain the correct answer (which is 1). $\endgroup$
    – valerio
    Jun 12, 2017 at 6:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.