0
$\begingroup$

The partition function is defined as a sum over all microstates $j$ as:

$Z=\sum_{j}exp(-\beta E_j)$

or

$Z=\int_{-\infty}^{\infty} exp(-\beta E)dE$

if the states are continuous.

We can use $Z$ to get the probability $p_i$ that a system has microstate $i$ via:

$p_i=\frac{exp(-\beta E_i)}{Z} $

Now, does this work for finding the probability that a system is in multiple states?

For example, does

$p_{1,2}=\frac{exp(-\beta E_1)+exp(-\beta E_2)}{Z} $

define the probability of the system being in states $1$ and $2$?

If we have continuous states between $E_1$ and $E_2$, does

$p_{1,2}=\frac{\int_1^2 exp(-\beta E)dE}{Z} $

define the probability that the system has a microstate between states $1$ and $2$?

$\endgroup$
  • 1
    $\begingroup$ Yes, that's how you find the probability for being in state 1 or state 2. $\endgroup$ – knzhou Apr 3 '18 at 22:53
  • $\begingroup$ @knzhou What about the continuous case? Can we do the integral between states 1 and 2? $\endgroup$ – Drew Apr 4 '18 at 0:49
  • $\begingroup$ Your equations for the continuous case are not correct: they are missing the density of states (the part that makes the partition function dependent on the system you are describing). $\endgroup$ – user8153 Apr 8 '18 at 10:30
1
$\begingroup$
  • Discrete case The probability of canonical microstate $i$ is $$ p_i = \frac{e^{-\beta E_i}}{Z} $$ The probability to find the system in a region $\mathcal R$ of microstates is the sum over all these probabilities: $$ p_{i\in\mathcal R} = \sum_ {i\in\mathcal R} \frac{e^{-\beta E_i}}{Z} $$

$~$

  • Continuous case The probability to find the canonical microstate in the region $(\Gamma,\Gamma+d\Gamma)$ of phase space is $$ P(\Gamma) d\Gamma = \frac{e^{-\beta E(\Gamma)}}{Z} d\Gamma $$ where $\Gamma = (\mathbf r_1,\cdots; \mathbf q_1\cdots)$ is the vector with the positions and momenta of all particles. The probability to find the system in a region $\mathcal R$ of microstates is the sum over all these probabilities: $$ P(\Gamma \in \mathcal R) = \frac{1}{Z} \int_{\Gamma\in\mathbf R} e^{-\beta E(\Gamma)} d\Gamma. $$

In both cases the result can be expressed in the form $$ p(\mathcal R) = \frac{Z_{\mathcal R}}{Z} $$ with the denominator calculated by adding/integrating the factor $e^{-\beta E}$ over all phase space and the numerator is the summation/integral of the same factor in region $\mathcal R$.

$\endgroup$

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.