2
$\begingroup$

So I have the question where a system of two QM oscillators both have energies $\left(n+\frac12\right)\hbar\omega_0$, and the total energy of the system is given to be $E'=\left(n'+\frac12\right)\hbar\omega_0$ where both $n$ and $n'$ are positive integers. I am asked to find the number of microstates $\Omega$.

Now my first guess was that $\Omega=\frac{\left(n'+1\right)!}{n'!}=n'+1$ . I got this by taking the "stars and bars" approach.

However on the given solution it is given that $\Omega=\frac{(n')!}{\left(n'-1\right)!}=n'$.

Can anyone explain to me why my approach is wrong and the reasoning behind the approach given?

As a side note, I know that at very high energies the extra "$+1$" becomes negligible this is just to get my reasoning correct.

$\endgroup$
1
$\begingroup$

yes so the microcanonical partition function is

$\Omega = \frac{\mathrm{Number~of~Macrostates}}{\mathrm{Number~of~Microstates}}$

So the $n'!$ is your number of macrostates (which is always the case for all systems normally it is n!).

The $(n'-1)!$ comes by avoiding double counting.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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