What is the relation between chemical potential and the number of particles?

[Luqman Saleem]

Chemical potential is defined as the change in energy due to change in the number of particles in a system. Let we have a system which is defined by the following Hamiltonian:
$$H = -t \sum_i^L c_i^\dagger c_{i+1} + V\sum_i^L n_i n_{i+1} -\mu \sum_i^L n_i$$
where $c^\dagger (c)$ are creation (annihilation) operators, $n$ is number operator, $t$ is hopping parameter, $V$ is nearest-neighbor interaction, $L$ is the total number of sites and $\mu$ is chemical potential.
What I understand by chemical potential is, if we set μ=some constant, then no matter how many sites (L) we add to the system, the number of particles will always be conserved. (Please correct me if I am wrong)

QUESTION:
What is the relation between chemical potential and the number of particles? i.e. if I set μ = 10 then how many particles are allowed in the system?

 

[stevendaryl]

I think it's worth your while to try to understand the classical (non-quantum) significance of thermodynamics quantities first.

Let's consider what's called a "Grand Canonical Ensemble". You imagine that you have some system, maybe a box with particles enclosed. This system is in contact with a much larger system, and can exchange energy and particles with this larger system.

What will happen as you let the two systems come into equilibrium is that energy will flow from the hotter of the two systems to the colder of the two systems, until they reach the same temperature. There is a similar kind of equilibrium for number of particles: particles will flow from one system to the other until they reach a particle equilibrium. The equilibrium conditions are given by:

$\frac{\partial S_1}{\partial E} = \frac{\partial S_2}{\partial E}$
$\frac{\partial S_1}{\partial N} = \frac{\partial S_2}{\partial N}$

where $S_1$ is the entropy of system 1, and $S_2$ is the entropy of system 2, $E$ is energy and $N$ is the number of particles. $S_1$ depends on the energy and number of particles in the first system, and $S_2$ depends on the energy and number of particles of the second system.

The first quantity, $\frac{\partial S_i}{\partial E}$ is a way of defining the temperature of system $i$ (where $i$ is 1 or 2): $\frac{1}{T_i} \equiv \frac{\partial S_i}{\partial E}$. So the first equilibrium condition says that the temperatures are equal. The second quantity, $\frac{\partial S_i}{\partial N}$ is a way of defining the chemical potential: $- \frac{\mu}{T_i} \equiv \frac{\partial S_i}{\partial N}$.

Now, in actuality, if energy and particles can flow back and forth between the two systems, then that means that the energy and number of particles available to the small system, system 1, is not constant. Instead, there is an associated probability distribution. We can understand that probability distribution via a quantity known as the grand canonical partition function, $Z$:

$Z = \sum_E \sum_N e^{- (E - \mu N - TS)/(kT)}$

where $k$ is Boltzmann's constant.

Once we have that quantity, we can compute various quantities:

$\langle N \rangle = k T \frac{\partial ln Z}{\partial \mu}$
$\langle E \rangle = k T^2 \frac{\partial ln Z}{\partial T} + \mu \langle N \rangle$

where $\langle N \rangle$ is the average number of particles in a system with partition function $Z$ when it is free to exchange heat and particles with a much larger system in equilibrium, and $\langle E \rangle$ is the average energy. ($ln Z$ means the natural logarithm of $Z$).

Now, to make this more amenable to quantum mechanics, let's rewrite $Z$:

It turns out that thermodynamically, the entropy $S$ is $k ln W$ where $W$ is the number of possible states with a given energy and number of particles. Let's (to first approximation) ignore particle interactions, and assume that the total energy of a system is simply given by:

$E = \sum_i n_i E_i$

where $E_i$ is the single-particle energy of state $i$, and $n_i$ is the number of particles in that state. So under that assumption, we can rewrite $Z$ as follows (I'm going to skip the proof that this is equivalent to the previous form):

$Z = \sum_i \sum_{n_i} e^{-(E_i - \mu) n_i/(k T) }$

Now that we've written $Z$ in this form, we can easily make the transition to quantum mechanics.

That was pretty long-winded, but the point is that the connection between the chemical potential and the number of particles is probabilistic:

$P(n_i) = e^{-(E_i - \mu) n_i/(k T) }/Z$

is the probability that there are $n_i$ particles in state $i$. You can't figure out the number of particles without knowing the temperature and the chemical potential.

 

[stevendaryl]

So under that assumption, we can rewrite $Z$ as follows (I'm going to skip the proof that this is equivalent to the previous form):

$Z = \sum_i \sum_{n_i} e^{-(E_i - \mu) n_i/(k T) }$

Actually, that's not quite correct. The actual formula is this:

$Z = \Pi_i (\sum_{n} e^{-(E_i - \mu) n/(k T) })$

where $\Pi_i$ means the product over all possible single-particle states $i$. The sum over $n$ is the sum over all possible number of particles that can be in state $i$. For Bosons, that's any number. For Fermions, it's 0 or 1.

 

[Luqman Saleem]

Thank you so much. It was really very help.