# Probabilities in statistical mechanics

I am reviewing some concepts in statistical mechanics and am becoming confused with how to calculate probabilities when a system has $N$ non-interacting particles.

For instance, let's say we have $N$ electrons with magnetic moment $\vec{\mu} = (g e/2 m)\vec{S}$. If we apply a strong magnetic field parallel to $\vec{S}$, then

$$E = - \vec{\mu} \cdot \vec{B} = \pm \frac{g e \hbar}{4 m} = E_{\pm}$$

depending on the orientation of the spin of the electron. Therefore, the partition function for one electron is simply

$$\xi = 2 \cosh \left(\frac{g e \hbar B}{4 m k T}\right)$$

And the probability to find the electron with spin parallel to the magnetic field is simply $e^{-\beta E_{+}}/Z$. So far, so good.

However, what happens when we have $N$ such electrons? Statistical mechanics says that the partition function for the system is now

$$Z = \frac{\xi^N}{N!}$$

if we assume that the electrons do not interact with each other.

This is where I get confused. Now, if we want to find the probability that 75% of the electrons have energy $E_+$, then the Boltzmann argument doesn't hold anymore:

$$\frac{0.75 N }{N} \neq \frac{e^{-\beta E_{+}}}{Z}$$

If the Boltzmann ratio doesn't hold, how can one proceed to calculate the aforementioned probability?

• I don't get it, if you imagine your $N$ electrons to be independent, then the probability you are looking for is independent on $N$. You just want that $e^{-\beta E_+}/\xi = 0.75$ don't you? – gatsu Aug 23 '13 at 11:15
• My intuition also tells me it should be independent of $N$. However, if all the electrons are independent, the probability should follow a binomial distribution, so that the probability would be $N!/((0.75N)!(0.25N)!) \; p^{0.75N} (1-p)^{0.25N}$, where $p$ is the probability of one electron to have energy $E_+$. This does depend on $N$ though... – physguy Aug 23 '13 at 17:28

It seems like you want to consider $N$, distinguishable, non-interacting spins with spin $1/2$, so let's assume we don't have to consider subtleties arising because of identical particles. When there are $N$ distinguishable, the Hilbert space $\mathcal H_N$ of the system has dimension $2^N$, and when they don't interact, the energy eigenbasis can be labeled by sequences \begin{align} (+\tfrac{1}{2},+\tfrac{1}{2},+\tfrac{1}{2},-\tfrac{1}{2},-\tfrac{1}{2},\dots, -\tfrac{1}{2},+\tfrac{1}{2},+\tfrac{1}{2},-\tfrac{1}{2}) \end{align} which specify that particles $1$ through $3$ are spin up, particles $4$ and $5$ are spin down, etc. We denote a general such state as $|\mathbf m\rangle =|m_1, m_2, \dots m_N\rangle$ with $m_i=\pm1/2$. The probability that the system is in a given state $|\mathbf m\rangle$ is \begin{align} \mathrm{prob}(\mathbf m) = \frac{e^{-\beta E_\mathbf{m}}}{Z} \end{align} where the partition function is \begin{align} Z = \sum_{\mathbf m}e^{-\beta E_\mathbf{m}} \end{align} and $E_{\mathbf m} = E_{m_1} + E_{m_2} +\cdots+ E_{m_N}$ is the total energy of the state $|\mathbf m\rangle$. Now, to determine the probability that the system is in any state for which $3/4$ of them are up, we simply determine all sequences $\mathbf m$ for which this condition holds, and we sum up the corresponding probabilites. Since $3/4$ of the spins are up if and only if the the numbers $m_i$ add up to $(\frac{3}{4}N\cdot \frac{1}{2} + \frac{1}{4}N\cdot(-\frac{1}{2}) = \frac{1}{4}N$, the probability we're after is \begin{align} \mathrm{prob}\left(m_1+\cdots+m_N = \tfrac{1}{4}N\right) = \sum_{m_1+\cdots+m_N = \frac{1}{4}N} \frac{e^{-\beta E_\mathbf{m}}}{Z} \end{align} where the right hand side represents a sum over all sequences $\mathbf m$ whose elements sum to $N/4$, and now you just have to figure out how to perform the sum.

• That does make sense. However, I'm trying to figure out what field strength $B$ I should apply, given a temperature $T$, to polarize 75% of the electrons. It seems like finding a way to perform that sum would be cumbersome, and then finding a way to isolate $B$ would be even harder. The question I'm looking at apparently requires at most 20 minutes of thinking and calculating (it's a question for quals preparation). – physguy Aug 23 '13 at 17:32
• @physguy I see ok. Yes if it's a qual problem then there is probably a clever way to do this; I'll try to think about it today if I have some time. – joshphysics Aug 23 '13 at 17:46

Ok I see where is your problem. I think that there is subtlety in stat. mech. that you are rightfully pointing out.

My first guess/answer was to think that since the electrons are assumed independent then if we ask say what is the temperature at which 75% of electrons are up then we simply need to solve:

$$0.75 = \frac{e^{\beta E_+}}{\xi}$$

You pointed out that in fact, since we have $N$ particles the probability, instead of being the simple stuff written above, should follow a binomial distribution that would involve $N$.

What happens is that we are both right. The answer I gave replies to the question "if I have a system of electrons at equilibrium and I measure the spin of one of the electron, what should be the temperature so that the probability to be up is 75%?".

As usual in statistical mechanics, this particular answer imagines that you have an infinite number of replica corresponding to each possible states in your system. To answer the question, you measure the spin of one particle in each replica and observe that the frequency within the ensemble with which you find a spin up satisfies what I have written.

Your answer replies to a slightly different question which is "let's consider a single system of N electrons at equilibrium. Out of these N electrons, what is the probability that 75% of them have spin pointing upward?"

As you say, the answer involves a binomial distribution BUT if $N$ is big enough, this binomial distribution will become a gaussian distribution for $N_{+}/N$ centered around $\frac{e^{\beta E_+}}{\xi}$ and in the thermodynamic limit will eventually become a delta function.

This is exactly what would happen if you tried to actually measure the probability for a tossed coin to be head or tail.

This underlines a big issue from the beginning of the XXth century and that is kind of still ongoing which is about the definition of probability in mathematics: you have the frequentist approach which is the one you use, the ensemble approach which is the one I used and the a priori approach which would yield the same result as mine as well.