I was just repeating the Bose-Einstein and Fermi-Dirac distributions for the Sommerfeld model, and I noticed something I don’t quite understand. In our statistical physics course, the Bose-Einstein distribution is given by:
n_{be} = \frac{1}{e^{(\varepsilon-\mu)/k_{B}T}-1}
In the lecture notes on the Einstein model, however, it is given by:
n_{be} = \frac{1}{e^{\hbar\omega/k_{B}T}-1}
The energy that would be added to the harmonic oscillator if n increases by 1 is \hbar\omega, so I think this means \varepsilon = \hbar\omega. If this is the case, then in the lecture notes we somehow have that the chemical potential \mu=0.
Why is this the case? What does the chemical potential even mean if we are not considering particles, but rather excitations of an oscillator?
In the Einstein model the atoms are seen as independent atoms. Therefore I think they do not experience any potential of other atoms of whatever on the place where the atoms are. In this model atoms are seen like they are all in their own infinite well. Therefor the chemical potential is in this model zero.
What does the chemical potential even mean if we are not considering particles, but rather excitations of an oscillator?
In a harmonic oscillator a potential over the whole oscillator decreases the probability to measure higher energy states if I am right. This because the wave function \psi(r) is influenced by E-V.
Besides that, we are considering atoms in the Einstein model. In some sense that is a particle, isn’t it? Or is this not correct physically speaking?
Note, I am no expert on quantum mechanics etc. I could be wrong. Please correct me.
I agree that it makes sense for there to be no other potential. The weird thing is that n normally stands for a number of particles that have a certain chemical potential \mu which determines the change in entropy when a particle is added or removed. In the Bose-Einstein distribution, however, n counts the number of excitations in a certain harmonic oscillator (so it is not counting atoms). This leaves me kind of confused about the meaning of \mu.
n can be used with different meanings in different formula’s and different books. Sometimes confusing, but the world of physics would be complicated when every book would give the same meaning to every symbol.
The amount of atoms thus far in this course has got the symbol N. In materials with only one type of atoms, this equals the amount of particles which have a certain chemical potential \mu.
Ps. My earlier explanation why the chemical potential \mu does not matter for the Einstein model gives no explanation why \mu is not used in the Debije model. Actually I know no physical reason why it is not used in the Debije model. I guess it is not mentioned because it is not very interesting for this course and only makes stuff more complicated.
Chemical potential is the energy at which adding a boson costs no extra energy. We set the zero of energy as the state where the solid doesn’t oscillate (since the laws of physics always allow to add a constant energy, this is a choice, although a very reasonable one). Therefore for us \mu = 0.
That makes sense. Is it true that in this problem “adding a boson” means moving a boson from the reservoir (of all the other states) to the system (the particular state we are considering). In that case you are just moving \hbar\omega_0 energy around and not adding or subtracting energy. Is that what it means or am I still looking at it in the wrong way?