A Can decoherence be associated with heat generation?

[Prathyush]

This question is inspired by a comment that @thephystudent made where he said that

"The dephasing between the Bragg pulses is not unitary, I believe it can be explicitly written in Lindblad form and generates heat. I believe this Point of view is the same as (among others) the papers of Allahverdyan and Van Nieuwenhuizen cited, but I don't know how far it applies to your work."

The context for this comment is that a density matrix that is not diagonal in the energy basis undergoes decoherence and the off diagonal terms go to zero. Can this type of decoherence be associated with the generation of heat? How does one define heat in such a context?

Any references will be also be much appreciated.

 

[Demystifier]

No, the production of heat is associated with dissipation, which is independent from decoherence. See Schlosshauer, Decoherence and the Quantum-to-Classical Transition, Sec. 2.11.

 

[vanhees71]

But isn't there some entropy production (supposed the Lindblad operators are choosen appropriately)? I know, entropy is not heat, and you need a temperature measure to define heat via $T \mathrm{d} S$ though.

 

[A. Neumaier]

But isn't there some entropy production

Yes, since Lindblad equations don't preserve all pure states.

 

[Demystifier]

But isn't there some entropy production (supposed the Lindblad operators are choosen appropriately)? I know, entropy is not heat, and you need a temperature measure to define heat via $T \mathrm{d} S$ though.

Heat is associated with thermal entropy, which is a maximal possible entropy under given constraints. The entropy produced by decoherence (Lindblad) is usually not maximal entropy.

 

[f95toli]

No, the production of heat is associated with dissipation, which is independent from decoherence. See Schlosshauer, Decoherence and the Quantum-to-Classical Transition, Sec. 2.11.

I don't think that is correct; at least not in the way the terminology is normally used. If you take a regular two-level system and assume no pure dephasing you have T2=2*T1.
Now, T1 will of course depend on the amount of dissipation in the system; and if the TLS is just a resonator you have that this is directly proportional to Q; i.e. the dissipation. I realize that you know this which is why I suspect there is an issue with the terminology here.

Note that when the goal is to maximise the coherence time of a real system (e.g. a qubit) you are in reality very rarely in this limes since T2 is typically limited by things like spectral diffusion etc . However, it is nevertheless true that T2 will ultimately be limited by "classical" dissipation; and in some systems (e.g. superconducting qubits) we are now at a point where we are frequently limited by plain old dissipation.

 

[Prathyush]

@f95toli I don't understand the terminology you are using? What are T1 and T2 ?