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I am interested in whether there is a field theoretic description (there is, so what is it?) of the tensor product (aka density matrix) model of open quantum systems. In particular, I am interested in how QFT might express the model of decoherence where the environment has some probability per time of "measuring" the state.

For example, a qubit has classical states 0 and 1, which form a favored basis of the quantum Hilbert space. Let $A$ be an observable with this as eigenbasis and distinct eigenvalues. A photon interacting with the qubit causing it to decohere with probability $p$ can be modeled by saying the photon measures $A$ on the qubit with probability $p$. In other words, we have the unitary evolution

$|0\rangle_S \otimes |un\rangle_E\rightarrow\sqrt{1-p}|0\rangle_S\otimes |un\rangle_E+\sqrt{p}|0\rangle_S\otimes |0\rangle_E$

$|1\rangle_S \otimes |un\rangle_E\rightarrow\sqrt{1-p}|1\rangle_S\otimes |un\rangle_E+\sqrt{p}|1\rangle_S\otimes |1\rangle_E$,

where $|un\rangle_E$ is the state of the environment $E$ which nows nothing about the qubit $S$, and $|0\rangle_E$ is the state that knows the qubit is in state 0, and so on ($|un\rangle_E, |0\rangle_E,$ and $|1\rangle_E$ are all orthogonal).

This whole model seems to conceptually rest on the environment's photon being some discrete quantity.

The thing is, if the photon is really just an excitation in the all-permeating electromagnetic field, it is more like there is a little bit of interaction between the environment and system all the time, not just at Poisson-random times.

It is easy to change the model above to have the probability instead be a rate write down a time evolution. It seems from the field theory perspective that this is more conceptually accurate. Can one intuit (or calculate!) what this rate is from the electromagnetic field?

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Sorry, this question must be based on some misunderstanding. For low enough speeds and particle production switched off, quantum field theory as a quantum mechanical theory is exactly equivalent to non-relativistic quantum mechanics or other models (in the limit). One may be used to different notation on both sides but there is an exact isomorphism. So whatever derivation of decoherence (or any other physical situation) you may have seen in non-relativistic QM may be of course translated to QFT and it's the same thing. –  Luboš Motl Sep 15 '12 at 10:14
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Much of the standard derivations of decoherence do assume things like production of electromagnetic radiation i.e. photons so they implicitly use a quantum field theory. –  Luboš Motl Sep 15 '12 at 10:15
    
@LubošMotl Most questions are! I believe you that the descriptions from QM and QFT are equivalent. I am asking for the translation. –  Ryan Thorngren Sep 15 '12 at 18:50
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