Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A Calculating Einstein's coefficients in QFT vs equilibrium

  1. Jul 27, 2017 #1
    Einstein predicted constraints on the coefficients of stimulated emission and absorption of radiation by atoms. He did that by assuming that the gas of atoms had to reach thermal equilibrium. For the gas to reach thermal equilibrium the coefficients had to be related in a certain way, otherwise it would reach thermal equilibrium.

    Now here's my thought. Since these coefficients are are a consequence of QFT and not statistical mechanics, they have nothing to do with equilibrium in a causal way. In other words, it's almost a coincidence that QFT turned out to be consistent with reaching an equilibrium state.

    Question: Isn't that weird? is QFT somehow built to achieve equilibrium? I realize that it would be unphysical if QFT didn't turn out like that, but is there a weird backward link from macroscopic equilibrium to what the theory is? or am I just paranoid?

    Thanks for reading to the end!
    Contact me:
    Youtube: https://www.youtube.com/thephysicsconnection
    Tumblr: https://www.tumblr.com/blog/the-physics-connection
    Instagram: https://www.instagram.com/the_physics_connection/?hl=en
  2. jcsd
  3. Jul 27, 2017 #2
    Macroscopic equilibrium is related to what happens at the microscopic scale. The ensamble tends to equilibrium because the equilibrium configuration is the most probable state for the system. It doesn't matter if these are billiard balls or photons of a field interacting with atoms.
  4. Jul 27, 2017 #3
    I agree. But if it did not depend on the details of the microscopic theory, then how can you derive a constraint on the microscopic theory (historically, before QFT was invented) from macroscopic equilibrium?
  5. Jul 27, 2017 #4
    I think the constraint you talk about is called detailed balance, is that what you mean? I have studied this years ago, and I'm not sure I actually understand your question. Can you elaborate a little bit more? thermodynamics was developed long before quantum mechanics, and the principles it relies on are empirical. The microscopic theory was understood after that, and the transition from microscopic theory to macroscopic theory is given by statistical mechanics.

    Perhaps this can help: https://en.wikipedia.org/wiki/Einstein_coefficients#Detailed_balancing
  6. Jul 27, 2017 #5


    User Avatar
    2016 Award

    Staff: Mentor

    You don't have to assume that everything is in thermal equilibrium. In particular, lasers only work in nonequilibrium systems. The assumption that there exists a thermal equilibrium everywhere is true for all systems, with or without QFT. It is a mathematical result.
  7. Jul 27, 2017 #6
    Yes the question is about this statement from the wikipedia article:
    On the one hand the probabilities are associated with each atom and does not depend on the presence of other atoms. On the other hand you can derive relationships between these coefficients just by assuming thermal equilibrium and Planck's law.
    I'm not saying there is anything wrong with that, it may be common practice. But there is something interesting and weird about that. Because from the point of view of building a microscopic theory of how atoms behave you have to make sure that whatever theory you propose satisfies these relations. And it's not obvious to me that any self-consistent theory (in the sense of being solvable, and producing predictions) will satisfy the Einstein relations.

    I hope that makes more sense.
  8. Jul 27, 2017 #7
    That's probably true. but I'm sure there are reasonable assumptions on the physical theory when this is proved (I haven't looked at proof, if you have a link, I would appreciate it if you shared it). I'm not sure that ergodicity is easy to prove generically.
  9. Jul 27, 2017 #8
    The quote answers your question, I think. It is considered the case for thermal equilibrium because at thermal equilibrium you have the same probability for transitions in one direction and its opposite. But as it says in the quote, the results are universal, the only assumption is that there is an equilibrium state, and as mfb said, this can be proven mathematically. The coefficients are related only to the transition probabilities between one state and another. At thermal equilibrium you will have the same probability of emission of a photon corresponding to a decaying from one atomic state to another of lower energy than the promotion of one electron from a lower energy to another of higher energy by absorption of a photon from the field. The Einstein coefficients are related to the probabilities of this transitions.
  10. Jul 27, 2017 #9
    Thank you for the response! Does that mean that you can derive the relations mathematically? without reference to any specific physical theory?
  11. Jul 27, 2017 #10
    No. You have a physical theory behind, and your calculation relies on it. The atomic transition probabilities could be calculated using quantum mechanics. To calculate how much emission and absorption you will have for a determinate species in a gas you have to introduce statistical mechanics. Note that there are many processes possible. Take a look to the full article in wikipedia: https://en.wikipedia.org/wiki/Einstein_coefficients

    This is not limited to the thermal equilibrium situation. Mathematics tells you that there is an equilibrium state, and that allows you to calculate the Einstein coefficients using that particular state and statistical mechanics.
    Last edited: Jul 27, 2017
  12. Jul 27, 2017 #11

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

  13. Jul 28, 2017 #12


    User Avatar
    Science Advisor
    2016 Award

    It's the other way around! The tendency of macroscopic systems to evolve into thermal equilibrium can be derived from the underlying microscopic dynamics. That's the great achievement of the pioneers of "statistical mechanics" (nowadays I'd rather call the subject "statistical physics" since it's very general, including all phenomena from classical mechanics and field theory to quantum (field) theory).

    Roughly the idea is the following: You start from the general QFT equations of motion of a general state, i.e., with an initially given statistical operator, describing an arbitrary state. This leads to the socalled Kadanoff-Baym equations which are in principle an infinite tower of equations (related and pretty similar to the Dyson-Schwinger equations of vacuum QFT but more general in involving an arbitrary state, not only the vacuum). Then you apply some coarse-graining procedure like the gradient expansion to arrive at some quantum transport equation (in the non-Markovian limit it's basically the Boltzmann-Uehling-Uhlenbeck equation).

    The collision term of the Boltzmann equation is basically the macroscopic description of the dynamics, and it involves the (in-medium) transition-matrix elements (S-matrix elements). Now the very general property of the unitarity of the S-matrix leads to the Boltzmann H-theorem, i.e., that the entropy (defined as the Shannon-Jaynes-von-Neumann information-theoretical expression) is never decreasing. Equilibrium, i.e., the stationary solutions of the Boltzmann equation are thus characterized as the states of maximum entropy.

    Of course, the thermal equilibrium is much simpler to formulate than the general off-equilibrium case, and the maximum-entropy principle provides the corresponding Statistical operators. Most convenient is the grand-canonical ensemble, where the average total energy (and perhaps various conserved charges) are determined by the initial state. The maximization of the entropy under these constraints then leads to the grand-canonical equilibrium stat. op.
    $$\hat{\rho}_{\text{GC}} = \frac{1}{Z} \exp[-\beta(\hat{H}-\sum_i \mu_i \hat{Q}_i)], \quad Z=\mathrm{Tr} \exp(\ldots).$$

    For the analogous derivation of the relativstic Boltzmann equation in the classical realm, see


    I've also an unfinished manuscript of the QFT-many-body case:

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Discussions: Calculating Einstein's coefficients in QFT vs equilibrium
  1. QFT vs QM 101 (Replies: 67)

  2. QFT vs Wave Function (Replies: 31)