Calculating Einstein's coefficients in QFT vs equilibrium

In summary, Einstein predicted constraints on the coefficients of stimulated emission and absorption of radiation by atoms by assuming the gas of atoms had to reach thermal equilibrium. These coefficients are consequences of QFT and are not related to equilibrium in a causal way. However, it is a coincidence that QFT turned out to be consistent with reaching an equilibrium state. There is a link between macroscopic equilibrium and the microscopic theory, known as detailed balance, which is a universal mathematical result. This can be seen in the relationship between the coefficients at thermal equilibrium derived from statistical mechanics. However, it is not obvious that any self-consistent theory will satisfy the Einstein relations, making it a unique and interesting aspect to consider.
  • #1
alemsalem
175
5
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
 
Physics news on Phys.org
  • #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.
 
  • #3
Telemachus said:
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.
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?
 
  • #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
 
  • #5
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.
 
  • Like
Likes Telemachus
  • #6
Telemachus said:
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
Yes the question is about this statement from the wikipedia article:
The Einstein coefficients are fixed probabilities per time associated with each atom, and do not depend on the state of the gas of which the atoms are a part. Therefore, any relationship that we can derive between the coefficients at, say, thermodynamic equilibrium will be valid universally.
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.
 
  • Like
Likes Telemachus
  • #7
mfb said:
The assumption that there exists a thermal equilibrium everywhere is true for all systems, with or without QFT. It is a mathematical result.
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.
 
  • #8
alemsalem said:
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.
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.
 
  • #9
Telemachus said:
the only assumption is that there is an equilibrium state, and as mfb said, this can be proven mathematically.
Thank you for the response! Does that mean that you can derive the relations mathematically? without reference to any specific physical theory?
 
  • Like
Likes Telemachus
  • #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:
  • #12
alemsalem said:
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?
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

http://th.physik.uni-frankfurt.de/~hees/publ/kolkata.pdf

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

http://th.physik.uni-frankfurt.de/~hees/publ/off-eq-qft.pdf
 
  • Like
Likes Telemachus and Demystifier

1. What is the difference between calculating Einstein's coefficients in QFT and in equilibrium?

The main difference between calculating Einstein's coefficients in quantum field theory (QFT) and in equilibrium is the approach and assumptions used. In QFT, the calculations are based on the principles of quantum mechanics and involve the interactions between particles and fields. In equilibrium, the calculations are based on the laws of thermodynamics and assume a system in thermal equilibrium.

2. How do you calculate Einstein's coefficients in QFT?

In QFT, Einstein's coefficients can be calculated using Feynman diagrams, which represent the different interactions between particles and fields. These diagrams involve calculating the amplitudes of each interaction and using them to determine the probabilities of each process occurring.

3. What is the significance of Einstein's coefficients in QFT?

Einstein's coefficients play a crucial role in determining the rates of different processes in QFT, such as emission and absorption of particles. They also provide insight into the dynamics of the system and can be used to make predictions about future behavior.

4. How do the calculations of Einstein's coefficients in QFT differ from those in classical physics?

In classical physics, Einstein's coefficients are calculated using the laws of classical mechanics and thermodynamics. These calculations do not take into account the quantum nature of particles and fields, and therefore cannot accurately predict the behavior of systems at the quantum level.

5. Can Einstein's coefficients be calculated for all systems in QFT?

While Einstein's coefficients can be calculated for many systems in QFT, there may be cases where the complexity of the interactions makes it impossible to determine them accurately. In these cases, approximations and simplifications may be used to estimate the coefficients.

Similar threads

  • Quantum Interpretations and Foundations
Replies
21
Views
2K
Replies
13
Views
2K
Back
Top