8
$\begingroup$

I was recently studying Blackbody Radiation and the principle behind it (as far as used in the Plancks original paper) is to find the energy distribution which maximizes the number of ways in which energy quanta can be distributed.

Since the radiation formula is derived due to quantization + maximum entropy reasons, can it also be derived using a principle of least action?

$\endgroup$
5
  • 2
    $\begingroup$ Well, what would you propose would be the "action" to be analyzed? $\endgroup$ Nov 16 '20 at 18:26
  • $\begingroup$ Well, I have tried to work on it. And I try to pursue this thought further so whatever I have thought is really vague. Here goes nothing: $\endgroup$
    – Lost
    Nov 17 '20 at 7:12
  • $\begingroup$ 1) I propose a simple dynamical path as in the normal lagrangian but in entropy space. A space where each point corresponds to an entropy value. Then trying to proceed from there. I could be totally wrong, its a vague thought thats why I put it here so that more ideas could be known. $\endgroup$
    – Lost
    Nov 17 '20 at 7:14
  • 1
    $\begingroup$ You can write down an expression for the entropy (subject to constraints depending on the ensemble you are working in), and find the distribution that maximizes the entropy using variational methods. I think this is done in Griffifths' Quantum Mechanics book. If no one else fills in the details I can write this up in a few days. $\endgroup$
    – Andrew
    Nov 21 '20 at 7:24
  • $\begingroup$ Okay, thank you I'll look up Griffiths. Though what I am looking for is a functional from which the distribution can be derived maybe without assuming quantas as a priori. What I have in mind is that the choice of any continuous energy distribution should always give a non-stationary action and thus there exclusion should become natural. $\endgroup$
    – Lost
    Nov 21 '20 at 7:35
2
+50
$\begingroup$

In so far as I'm aware, there is not direct connection between the two as you might like, however there are a few things that might be worth mentioning. I will take this back a few steps from the question itself because I feel there's some important context to the point I would like to make in the end.

Since the problem at hand is looking for a steady-state distribution, the Liouville theorem (or its equivalent in a quantum system) implies that the probability distribution we seek must (Poisson) commute with the Hamiltonian, and hence can only be a function on the conserved quantities of the system. A key fact that's often overlooked is that there's more than one way to do this. One would be to assert that the distribution is simply a constant. This would give the distribution associated to the microcanonical ensemble. Perhaps more common is the assumption that the distribution depends only on the energy (Hamiltonian).

Now, another important fact to understanding things lies in probability theory itself. Entropy maximization (perhaps the best understanding of what this should really mean to us actually comes, I think, from information theory and signal processing -- the application Shannon originally invented the notion of entropy to deal with. See for example Cover and Thomas' book), is a fairly good principle all things considered, but alone this doesn't do much for us, we need constraints as well.

In general, we might say that we know something about a system, and it might be convenient to assert that this something takes the form of an expectation value, which we assume to be a fixed, known quantity. For a nice discussion of how entropy maximization works in the presence of constraints, see here. Roughly the same arguments work out in the quantum case as well. If you were to consider the specific question: "What is the maximum entropy distribution such that the expectation of the energy is held fixed," you would find the unnormalized answer to be the Boltzmann factor, $e^{-\beta H}$ where $\beta$ is the Lagrange multiplier enforcing the constraint $E=\langle H\rangle$ which we identify with the inverse temperature. In the quantum case, we find the density matrix to be the Boltzmann factor, but with the Hamiltonian operator rather than function.

Now, in the quantum case, the expectation of any operator $\mathcal{O}$ is given by $\langle \mathcal{O}\rangle=\text{tr}(\mathcal{O}\rho)$ where $\rho=Ne^{-\beta H}$ is the density matrix ($N$ being some normalization). This trace can be identified (under certain technical conditions) with a path integral in which "time" has been replaced by a complex time $\tau$ which is periodic with period $i\beta$ (up to minus signs). I believe I recall a nice discussion of this appearing in the conformal field theory book by Di Francesco, Mathieu, and Senechal. Under certain other technical conditions, the path integral may be expressed in terms of the classical action as $$ \langle\mathcal{O}\rangle=\int\mathcal{D}[\phi]\mathcal{O}e^{-\beta S}. $$

Finally now we can come to the principle of least action. Integrals of this form can often be approximated by asymptotic series. For example, see here. As an aside, this is the approximation working in the background any time you see Feynman diagrams floating about.

The lowest order term in this sort of approximation scheme is proportional to $e^{-\beta S[\phi_c]}$ where $\phi_c$ is the classical solution. That is, the solution which extremizes the action.

This may feel like an unsatisfactory connection between stationary action principles and thermal physics, but it is nonetheless the only connection I can think of which is technically accurate.

$\endgroup$
2
  • $\begingroup$ Ok. So what I got from this is: when we consider the question "What is the maximum entropy distribution such that the expectation of the energy is held fixed?" And add the constraint of discrete energy quantas we should get back the Planck Distribution Formula. In QM then this would be like indetifying a path integral (since its connected to trace which gives expectation value of energy) + (again) subject to the constraint that energy values are discrete. This should give back the distribution. But is possible to have a functional that automatically rejects the non-discrete energy cases. $\endgroup$
    – Lost
    Nov 22 '20 at 12:59
  • $\begingroup$ @Lost No, I don't believe so. The calculus of variations is necessarily bound by the limitations of calculus, namely we need everything we deal with to be smooth (more or less). This is equivalent to the statement that differential equations produce (at least locally) smooth solutions. Discretization in differential equations comes only from the imposition of boundary conditions. Though related, boundary conditions can't be implemented nicely as a variational problem. The only other way around this is to explicitly work in quantum mechanics. $\endgroup$ Nov 22 '20 at 23:03

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.