I am currently studying Statistical Mechanics and already have a background in probability and statistics. However, there are still things that remain unclear to me. So far I understand that time averages may be substituted by ensemble averages according to the hypotheses of Ergodicity, which we assume is true. Hence, to evaluate $\langle O(q,p)\rangle$ we just need to compute

$$\int \textrm{d}q\,\textrm{d}p\,\rho(q,p)O(q,p)$$

where $\rho(q,p)$ is the ensemble density. My first question is:

  1. What does this density represent? That is, if $F(q,p)$ where its cummulative distribution function, what would it represent? Normally $F(x)=P(X\leq x)$ for some random variable $X$. What is the random variable in this case?

My second question is how to calculate observable densities. For instance, if I want to know $P(H(q,p)\leq \epsilon)$, I might want to know the density associated to the (random variable) $H(q,p)$. This also makes me wonder:

  1. Why $H(q,p)$ is a random variable?

  2. I have read that the density of an observable $\Omega$ is simply computed as $\langle \delta (\Omega(q,p)-\omega)\rangle$, i.e.

$$\int \textrm{d}q\,\textrm{d}p\,\rho(q,p)\,\delta (\Omega(q,p)-\omega)$$

What is this based on? I have never seen anything similar in my probability lessons. Note that I study mathematics, and we are not used to the Dirac Delta. But in any case, I still do not understand the meaning of this.

  1. In a similar manner, could one just compute $P(\Omega(q,p)\leq \omega)$ as

$$\int \textrm{d}q\,\textrm{d}p\,\rho(q,p)\,\theta (\omega-\Omega(q,p))$$

where $\theta(\cdot)$ is Heaviside step function, or it makes no sense?

Thanks to anyone who takes the time to answer these questions. I deeply appreciate your dedication.

  • $\begingroup$ If you've a mathematician's background, then I would recommend that you first read an introduction to statistical mechanics with such readers in mind. These lecture notes, for instance. Otherwise, there is our book, although it is restricted to lattice models. $\endgroup$ – Yvan Velenik Apr 28 at 7:48
  • $\begingroup$ You may also read the first two chapters of Ruelle's famous book. $\endgroup$ – Yvan Velenik Apr 28 at 8:17
  • $\begingroup$ Rough answers: 1. $(p,q)$ is a random vector with (joint) distribution ρ. 2. $H(p,q)$ is a random variable because it is a function of the random vector $(p,q)$. 3. and 4. Look at the answers to this question. $\endgroup$ – Yvan Velenik Apr 28 at 8:25

1) one good way to think of an ensemble is that we have many different systems that were prepared in an identical manner, and then we sample with equal probability one of them and measure it. As each system will have different $p$ and $q$, then $\rho(p,q)$ is the probability-density to sample such values. In the canonical ensemble, for example, this is given by $\rho(p,q) = \exp[-\beta H(p,q)]/Z$ where $Z=\int\! dp dq \exp[-\beta H(p,q)]$ (I assume here that $p$ and $q$ are dimensionsless, otherwise we have to divide by some factor that will not play any physical role). In this interpretation, $p$ and $q$ give the state of the system, and are a random variable.

2) As $H$ is a function of $p$ and $q$, and they are a random variable, $H$ (and any other observable that depends on $p$ and $q$) is also a random variable.

3) Formally, you can convince yourself that this is a proper density by seeing that $$\int d\omega\rho_\Omega(\omega) = \int dq dp \rho(p,q) = 1$$ and also $\rho_\Omega(\omega) \geq 0$.

Intuitively, this expression places weights of $\rho_\Omega(\omega)$ on values of $\Omega(p,q)$ that match the weights of $p$ and $q$ using $\rho(p,q)$.

4) Yes. This is the explicit expression that you get when you integrate the probability density of $\Omega$

$$ P(\Omega \leq \omega) = \int_{-\infty}^{\omega}d\omega' \rho_{\Omega}(\omega ') = \int dp dq \rho(p,q)\int_{-\infty}^{\omega}d\omega' \delta(\Omega(p,q)-\omega') = \int dp dq \rho(p,q) \theta(\omega-\Omega(p,q))$$

| cite | improve this answer | |

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.