This is something probably very basic but I was led back to this issue while listening to a recent seminar by Allan Adams on holographic superconductors. He seemed very worried to have a theory at hand where the chemical potential is negative. (why?)
- For fermions, isn't the sign of the chemical potential a matter of definition?
The way we normally write our equations for the Fermi-Dirac distribution the chemical potential happens to that value of energy at which the corresponding state has a occupation probability of half. And within this definition the holes in a semiconductor have a negative chemical potential.
- It would be helpful if someone can help make a statement about the chemical potential which is independent of any convention.
{Like one argues that negative temperature is a sign of instability of the system.}
Also isn't it possible for fermions in an interacting theory to have a negative chemical potential?
Also if there is a "physical argument" as to why bosons can't have a positive chemical potential? (Again, can an interacting theory of bosons make a difference to the scenario?)
And how do these issues change when thinking in the framework of QFT? (No one draws the QCD phase diagram with the chemical potential on the negative X-axis!)
In QFT does the chemical potential get some intrinsic meaning since relativistically there is a finite lower bound of the energy of any particle given by its rest mass?
Answer
consider the grand canonical ensemble, $$ \rho \sim \exp[-\beta (E-\mu N)] $$ In the exponent, the inverse temperature $\beta = 1/kT$ is the coefficient in front of one conserved quantity, the (minus) energy, while another coefficient, $\beta\mu$, is in front of the number of particles $N$. The chemical potential is therefore the coefficient in front of the number of particles - except for the extra $\beta$. The number of particles has to be conserved as well if $\mu$ is non-zero.
As long as the sign of $N$ is well-defined, the sign of the chemical potential is well-defined, too.
Now, for bosons, $\mu$ can't be positive because the distribution would be an exponentially increasing function of $N$. Note that in the grand canonical ensemble - which is really the ensemble in which $\mu$ is sharply defined - the dual variable to $\mu$, namely $N$, is not sharply defined. However, the probability for ever larger - infinite $N$ would be larger, so the distribution would be peaked at $N=\infty$. Such a distribution couldn't be well-defined. We want, in the thermodynamic limit, the grand canonical ensemble, while assuming a fixed $\mu$, also generate a finite and almost well-defined $N$, within an error margin that goes to zero in the thermodynamic limit. That couldn't happen for bosons and a positive $\mu$.
This catastrophe would be possible because $N=\sum_i n_i$, a summation over microstates $i$, and every $n_i$ can be an arbitrarily high integer for bosons. For fermions, the problem doesn't occur because $n_i=0$ or $1$ for each state $i$. So for fermions, we can't argue that $\mu$ has to be positive. Note that $E-\mu N$ in the exponent is the sum $\sum_i N_i (e_i-\mu)$. For bosons, the problem occurred for states for which $e_i-\mu$ was negative i.e. $e_i$ was low enough. For fermions, however, the number of such states - and therefore the maximum number of fermions in them - is finite, so the divergence doesn't occur if the chemical potential is positive. For fermions, positive $\mu$ is OK.
In fact, for fermions, both positive and negative $\mu$ is OK. Also, it is easy to see that if both particles and antiparticles exist, $\mu$ of the antiparticle has to be minus $\mu$ of the particle because only the difference $N_{\rm particles} - N_{\rm antiparticles}$ is conserved; this is true both for bosons and fermions.
So if the potential for electrons is positive, the potential for positrons or holes (which play the very same role) has to be negative, and vice versa.
Nothing changes about the meaning of the chemical potential when one switches from classical physics to quantum physics: in fact, above, I was assuming that there are "discrete states" for the particles, just like in quantum physics - otherwise we wouldn't be talking about bosons and fermions which are only relevant in the quantum setup. Classical physics is a limit of quantum physics in which the number of states is infinite because $\hbar$ goes to zero, so a finite number of particles never ends up in "exactly the same state". In some sense, Ludwig Boltzmann, while working in the context of classical statistical physics, was inherently using the thinking and intuition of quantum statistical physics - he was a truly ingenious "forefather" of quantum physics.
In relativity, one has to be careful how we define the energy of a state. Note that the physically meaningful combination that appears in the exponent is $e_i-\mu$, so if one shifts $e_i$ e.g. by $mc^2$, the latent energy, one has to shift $\mu$ in the same direction by the same amount. The notions of chemical potential obviously work in relativity, too. Relativistic physics is not a "completely new type of physics". It is just a type of the old physics that happens to respect a symmetry - the Lorentz symmetry.
Again, in quantum field theory which combines both quantum mechanics and relativity, statistical physics including the notion of the chemical potential also works but one must be careful that particle-antiparticle pairs may be created with enough energy. That implies $\mu_{p}=-\mu_{np}$, as I said.
There can't be any general ban on a negative chemical potential of fermions: fermionic $\mu$ can have both signs. However, in the particular theory that Allan wanted to describe, he could have had more detailed reasons why $\mu$ should have been positive for his fermions. I am afraid that this would be an entirely different, more specific question - one about superconductors. As stated, your question above was one about statistical physics and I tend to believe that the text above exhausts all universal facts about the sign of the chemical potential in statistical physics.
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