# General counting of microstates for a specific energy. Indistinguishable vs distinguishable

I have the following problem: I have two systems $$A$$ and $$B$$, each one with 5 distinguishable particles, $$N_A=N_B=5$$. The systems have infinite energy levels with energy $$E_m = m \cdot \epsilon$$ with $$m=0,1,2,3...$$ and I want to know the number of micro-states that are compatible with the macro-states $$E_A=10 \epsilon$$ (for system A) and $$E_b=5 \epsilon$$ (for system B) for each isolated system

Since all energies are positive (or zero), I start to do it by hand doing all the possible combinations for $$A$$ using only the first eleven levels ($$m=0,...,10$$) because it's impossible to find a macro-state $$E_A=10 \epsilon$$ in this system with some particle in $$m>10$$. For example I start for $$A$$ system with $$N_0=4,N_1=0,N_2=0,...,N_9=0,N_{10}=1$$. Maybe as you guess, this make me loss a lot of time, but finally I found the number of micro-states $$\Omega_A (E_A) =841$$ (I'm almost sure that this number is wrong). Also I repeat this for $$B$$. I wonder if there is an easy way to calculate the number of micro-states with this energy for each isolated system

Also I want to put both system in thermal contact and equilibrate, then calculate the probability that the system $$A$$ have an energy $$E$$, $$P_A(E)$$. I'm not sure how can I do that in a general way. And how can I do it if particles are indistinguishable?

I wrote a quick python script (see below), which finds that there are $$\Omega_{\text{dist.}}(10) = 1001$$ inequivalent ways of writing $$E_A = 10$$ as a sum of 5 non-negative integers. Similarly, for $$E_B = 5$$ the script gives $$\Omega_{\text{dist.}}(5) = 126$$.

For indistinguishable particles, we do the same but insist that $$n_1 \le n_2 \le \ldots \le n_5$$ to prevent over-counting. The script then gives $$\Omega_{\text{indist.}}(10) = 30$$ and $$\Omega_{\text{indist.}}(5)=7$$. These last ones are easy to verify by hand.

This brute-force approach is fairly unsatisfactory, since it does not really give any insight. But I hope it will help you cross-check any formula you might come up with.

Now here's the weird part. It feels like there should be a clear formula that calculates this, but I could not find anything in my statistical mechanics lecture notes. By chance I found that dropping the condition (if 0 <= n5 and n5 <= n4:) in the second script seems to give exactly the answer of the first script (i.e. $$1001$$ if $$E = 10$$ and $$126$$ if $$E=5$$).

Intrigued by this, I did a bit of algebra to find that the sum that is being calculated can be represented as a polynomial as follows: $$\Omega_{\text{dist.}}(E) \stackrel{?}{=}\sum_{i=0}^{E}\sum_{j=0}^{i}\sum_{k=0}^{j}\sum_{l=0}^{k} 1 = \frac{1}{24} ( E^4 + 10 E^3 + 35 E^2 + 50 E + 24).$$ Super random I know, but I checked it against the first script for $$0 \le E \le 100$$ and it seems to work. It might be that this formula ceases to hold for higher values of $$E$$, and I cannot really explain why it should be correct in the first place, but it is certainly interesting.

Python scripts:

# Distinguishable Particles:
E = 10  # system A
count = 0
for n1 in range(E+1):
for n2 in range(E+1-n1):
for n3 in range(E+1-(n1+n2)):
for n4 in range(E+1-(n3+n2+n1)):
print(n1, n2, n3, n4, E-(n1+n2+n3+n4))
count += 1
print(count)


For system B, just change E = 10 to E = 5.

# Indistinguishable Particles
E = 10  # System A
count = 0
for n1 in range(E+1):
for n2 in range(n1+1):
for n3 in range(n2+1):
for n4 in range(n3+1):
n5 = E-(n1+n2+n3+n4)
if 0 <= n5 and n5 <= n4:
print(n1, n2, n3, n4, E-(n1+n2+n3+n4))
count += 1
print(count)

• It's not the answer that I looking for, but it is incredibly helpful! I also thought about the possibility to make some code to compute it, and it makes more clear. Thank you Do you know something about the second part? Also I want to put both system in thermal contact and equilibrium, then I presume that $\Omega_{A+B}=\Omega_A (E_A') \cdot \Omega_B (E-E_A')$ with $E=E_A+E_B$ so with you code it should be more or less solved, right? – user239504 Feb 28 at 18:24
• Putting the systems into contact allows energy to be exchanged freely, meaning that now we have to find the number of possible configurations that a total of $N_A + N_B = 10$ particles can be in, with a total energy of $E=E_A + E_B = 15\epsilon$ available. This will not agree with my formula for $\Omega_{\text{dist.}}(E)$ above, because the formula assumes $5$ particles in the system. Why do you think the formula $\Omega_{A+B} = \Omega_A(E_A') \Omega_B(E - E_A')$ holds, and what are the primed energies $E_A'$ and $E_B'$? – Umut Feb 28 at 18:35
• I assume each system is in equilibrium before coming into contact with the energy $E_A$ and $E_B$ (A and B respectively), but once you put them together they will change the energy until they equilibrate and then the new equilibrium values should be $E_A '$ and $E_B'$, then $E = E_A + E_B = E_A '+ E_B'$, right? And I thought that when I put two systems in contact, in equilibrium the total number of microstates is $\Omega_{A + B} (E) = \Omega_A (E_A ') \cdot \Omega_B (E_B')$ or it will not always be true? – user239504 Feb 28 at 18:50