# Callen & Welton calculation of power of absorption

1. Mar 25, 2013

### Jano L.

In the paper

H.B. Callen, T.A. Welton, "Irreversibility and Generalized Noise", Phys. Rev. 83, 34, (1951)

the authors arrive at an interesting formula for average power continually absorbed by a quantum system ( = with discrete states of definite energy) under action of a harmonic perturbation $V_0 \sin \omega t$:

$$\text{Power} = \frac{1}{2}\pi V_0^2 \omega \sum_{n=1}^\infty \bigg\{|\langle E_n + \hbar \omega|Q|E_n \rangle|^2 \rho(E_n + \hbar \omega) - |\langle E_n - \hbar \omega|Q|E_n\rangle |^2 \rho(E_n - \hbar \omega) \bigg\} f(E_n)$$

In this formula, $\rho(E_n)$ is density of states at the energy level $E_n$, $f(E_n)$ is the Boltzmann distribution function $f(E_n) \propto e^{-\beta E_n}$ and $Q$'s is the dipole moment operator or something similar.

The derivation of the above formula is based on the idea that the system jumps from one Hamiltonian eigenstate to another with probability given by the so-called "Fermi golden rule". I assume that generally it gives non-zero positive absorbed power.

However, if we disregard the golden rule and calculate what happens to expansion coefficients or density matrix according to the time - dependent Schroedinger equation, we find out that there is no dissipation - the system just performs idle oscillations in phase with the external field (Rabi oscillations). Since the model does not include any action of other external forces back on the system which would damp its motion, it should be expected that the average absorbed power is zero.

So the two approaches give different answer. Of course, in reality there is dissipation everywhere so the non-zero average power is a welcome result, but it is not clear to me how it comes about.

It is quite common to think that dissipation is not due to the system itself (which is impossible if it is a Hamiltonian system), but due to action of external forces, e.g. due to bath or other bodies. However, the manipulation of the Fermi golden rule does not carry such idea of bodies external to the system. I think that this is because it is just a descriptive model "what roughly happens" and it should be supported or replaced by a more detailed theory.

I wonder, do you know of some way to derive the above Callen and Welton's formula for dissipated power using instead the approaches studied in non-equilibrium statistical physics, like stochastic Schroedinger equation, or via reduced density matrix calculations? Or alternatively, to derive the Fermi golden rule in this way?

2. Mar 27, 2013

### Jano L.

The question is probably too difficult. I'll try to ask differently. Do you think there is some derivation of the "Fermi golden rule" which does not just assume quantum jumps but shows how the quantum jumps arise effectively from Schroedinger's equation?

3. Mar 27, 2013

### psmt

Fermi's golden rule comes from the expansion of the time evolution operator to first order in the interaction Hamiltonian, in other words directly from the Schroedinger equation. "Quantum jumps" are not something put in by hand. What the rule tells you is the overlap (squared) of eigenstates of the free Hamiltonian with the state the system has evolved into, and this is turn tells you the probability of the system collapsing into a particular free eigenstate if you come along and measure the energy. Does that help?

4. Mar 27, 2013

### Jano L.

I am afraid this is not how the rule is often used. Callen and Welton do not have any measurement of energy in their derivation. They use the golden rule formula in such way that it suggests that the system jumps autonomously and randomly, and the rule serves just to find the rate of these jumps. I think this is how the rule is often used when some known external perturbation acts on the system.

5. Mar 27, 2013

### psmt

Good point. Could rapid decoherence in real systems be the resolution of this? (I don't mean a mathematically rigorous resolution, just that interactions with the enviroment can act like a measurement.)

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