# Fermi's Golden Rule

Consider a system with countable quantum states. One can define $J_{ij}$ as the rate of transition of probability from i-th to j-th quantum state. In H-theorem, if one assumes both $$H:=\sum_{i} p_{i}\log p_{i}$$ $$J_{ij}=J_{ji}$$ then they can prove the H always decrease. The latter is Fermi's Golden Rule states that the transition rate's matrix is symmetric.

I have seen in Federick Reif's book Fundamentals of Statistical and Thermal Physics he has proven Fermi's rule. Briefly, consider a quantum system which obeys Schrödinger's equation:$$\mathrm{i}\hbar\frac{\mathrm{d}\psi}{\mathrm{d}t}=H\psi$$ where $H$ is Hermitian. Then one can use these relations to prove Fermi's Golden Rule in this specific case: (I show i-th eigenvector with $\psi_{i}$.) $$J_{ij}\propto |\langle{\psi_{j},H\psi_{i}}\rangle|^2=\langle{\psi_{j},H\psi_{i}} \rangle\overline{\langle{\psi_{j},H\psi_{i}}\rangle}$$ and H is Hermitian, so: $$J_{ij}\propto |\langle{\psi_{j},H\psi_{i}}\rangle|^2=\overline{\langle{H\psi_{j},\psi_{i}} \rangle}\langle{H\psi_{j},\psi_{i}}\rangle=\langle{\psi_{i},H\psi_{j}} \rangle\overline{\langle{\psi_{i},H\psi_{j}}\rangle}$$ Hence: $$J_{ij}=J_{ji}$$

As a result, we can prove that entropy for an isolated system always increases at least for some special cases with these assumptions:

I. If our quantum states are countable.

II. If our system can be described with a Hamiltonian that is Hermitian.

I have a question: do you have an example of a system does not obey these two assumptions? If so, is Fermi's Golden Rule a principle? How can we prove it using quantum mechanics? Do you know some articles about it?

• What about the most trivial of all closed systems? A single spin? How do I interpret its dynamics (a simple oscillation) as an increase in entropy? That's not a criticism of the idea but a straight question. – CuriousOne Dec 25 '15 at 8:24
• @CuriousOne single spin and a simple oscillation have Hamiltonians that are Hermitian; and their states are countable. This argument states their entropy cannot decrease. Is it false? – Kiarash Dec 25 '15 at 8:27
• I am trying to understand what the argument says. For closed systems with a finite number of states I intuitively expect a quantum version of the Poincare recurrence theorem, though, so the statement can only be correct in the short term limit (even though that limit is even much, much longer than it would be for a classical system with the same number of degrees of freedom). Am I missing something? – CuriousOne Dec 25 '15 at 8:37
• @CuriousOne You are right. Even for a system with infinite countable states, Poincare recurrence theorem is true for wave function but I am not sure whether it can be a counter-example for the prove or not. If as you said it is true only for short term limit, what is false in H-theorem? – Kiarash Dec 25 '15 at 10:30
• I don't think there is anything wrong with it or the generally accepted statistical mechanics/thermodynamics connection per Gibbs, per se, if these are applied judiciously and they are obviously transferable to quantum mechanics (probably even more easily, since we can formulate even large problems as collections of harmonic oscillators). For most physical systems and for most practical time scales one can work with this... and then there are the few cases when one can't. – CuriousOne Dec 25 '15 at 10:39

Something important is missing from your presentation of the Fermi rule. The Schroedinger eq. you mention, $i\hbar(d\psi/dt) = H\psi$, is never going to produce any transitions between eigenstates of $H$ itself: by definition eigenstates are stationary states.

What you probably refer to is something like $$i\hbar\frac{\partial \psi_I}{\partial t} = V_I(t) \psi_I$$ which is the interaction picture form of the Schroedinger eq. in the presence of a perturbation $V$, $i\hbar(d\psi/dt) = (H + V)\psi$. Here $\psi_I(t) = e^{(i/\hbar)Ht}\psi(t)$ and $V_I(t) = e^{(i/\hbar) H t} V e^{-(i/\hbar) H t}$. If $\psi_{i(j)}$ are eigenstates of $H$, then Fermi's golden rule indeed gives the transition rate between them to 2nd order in the (small) perturbation $V$: $$J_{i\rightarrow j} \sim |\langle \psi_j|V|\psi_i\rangle|^2$$ In other words, Fermi's rule concerns an open system undergoing a weak interaction with its environment, usually represented by an electromagnetic field, or more generally, by a an external thermodynamic "bath".

This being said, Fermi's rule is known to be equivalent to the Markov approximation for open systems, see R.Alicki, "The Markov master equations and the Fermi golden rule", Int.J.Theor.Phys.Vol.16(5), 351-355(1977). A very important consequence of the Markov approximation is that the dynamics is no longer time reversible: while under the original Hamiltonian dynamics entropy is conserved, under the Markov approximation it is not. In fact, as you point out, it is possible to justify the H-theorem, and under some additional conditions it can be shown that the dynamics is governed by a dynamical semigroup with a Lindblad-type generator (see for instance Sec.IIC on the "Secular Approximation" here).

Now to the actual question: Fermi's rule is definitely not a principle. The Markov approximation, and therefore Fermi's rule, holds provided the relaxation time $\tau$ it describes for the system is much longer than the bath relaxation time, $\tau >> \tau_{bath}$. When the system-bath interaction is too strong and/or the time scale for the system's relaxation becomes comparable to that of the bath, both the Markov approximation and Fermi's rule cease to apply. What this means is that the dynamics is no-longer memoryless, but depends on the past history of the system. The transition from Markovian to non-Markovian dynamics can be seen even in such a simple system as a qubit in a dissipative environment, which makes it important for entanglement and decoherence problems. For instance, a qubit undergoing revival of coherence driven by a dissipative bath no longer follows a Markovian dynamics. See for instance this recent review on "Non-Markovian dynamics in open quantum systems".

• Thank you very much for your clear answer. I am sophomore and I think I need to study some preliminaries before regarding this question. I guess one should study density matrix rather than eigenstates when they are talking about quantum statistical mechanics. – Kiarash Dec 25 '15 at 15:55
• I would be happy if you look at my another question. physics.stackexchange.com/q/225428 – Kiarash Dec 25 '15 at 15:58
• About density matrices - absolutely. If you are just starting out in quantum statistics: build a solid background in equilibrium concepts, but pay attention to density matrix techniques (master equations) for a good start in non-equilibrium issues. Will be taking a look at the other question in a couple of hours. – udrv Dec 25 '15 at 17:48
• Sorry about getting back so late. On the 2nd question: The major points on discrete $\Omega$ vs. limit of large no. of particles/degrees of freedom have already been made in the other comments. There is also the law of large numbers invoked for large factorials (Stirling's approx.), and sometimes peak integration; both approx. smooth out the discrete nature of $\Omega$. Perhaps just to add that approximating $\Omega$ is not so much about fitting a continuous function to discrete values as about approximating the sum the function represents by an integral that converges to the same limit. – udrv Dec 26 '15 at 23:56
• Also keep in mind that what is needed is the variation of $\ln\Omega$, which approaches the continuous limit much faster than that of $\Omega$ itself ($\partial \ln\Omega = \partial \Omega/\Omega$). – udrv Dec 26 '15 at 23:56