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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
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@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
    
"It is easy to change the model above to have the probability instead be a rate write down a time evolution." Something broke wrt grammar :) –  DanielSank Jul 1 at 7:00
    
@RyanThorngren: I'd like to see this question answered. Is there some way I could improve my answer to make it acceptable? –  DanielSank Sep 16 at 8:35

1 Answer 1

Can one intuit (or calculate!) what this rate is from the electromagnetic field?

Sure. For a given problem you'll have a system you care about $S$, and an environment $E$, which in this case is the electromagnetic field. The Hamiltonian for the system is something like

$$H = H_S + H_E + H_I$$

where $H_S$ is the Hamiltonian for the system $S$, $H_E$ is the Hamiltonian for the electromagnetic field, and $H_I$ is an interaction. The exact form of the interaction depends on the problem at hand. Suppose you've got an atom in an electromagnetic resonator (ie two mirrors pointed at one another). Then the interaction between the atom and the light field might be well approximated by a dipole term:

$$H_I = \vec{d} \cdot \vec{E}$$

where $\vec{d}$ is the dipole moment of a relevant electron transition in the atom, and $\vec{E}$ is the electric field.

Now you want to get a decoherence rate. If one of the modes in the optical resonant cavity is on resonance with the electron transition, the interaction Hamiltonian can be reduced to a simpler form called the Jaynes-Cummings form:

$$H_I=g\left( \sigma_+a + \sigma_- a^{\dagger} \right)$$

where $\sigma_-$($\sigma_+$) is the lowering (raising) operator for the electron transition and $a$($a^{\dagger}$) is the lowering operator for the mode of the electromagnetic field which is on resonance with the electron transition. If you just flat out compute the time dependence of this thing you find that excitations of either the atom of the electromagnetic field oscillate back and forth between the atom and light field with frequency $\omega=g$ $^{[1]}$.

Now, that might not sound like decoherence at first sight, but it actually is. Suppose we start the system with the atom in state $\left(|0\rangle+|1\rangle\right)/\sqrt{2}$ and then allow it to interact with the light field. The time dependent state of the system is then

$$|\Psi(t)\rangle = \cos(\omega t)|10\rangle + i \sin(\omega t)|01\rangle$$

where $|10\rangle$ means that the atom is excited and the light field is in the ground state, and $|01\rangle$ means that the atom is in the ground state but the light field is excited. Now let's ask what is the state of the atom if we ignore the light field. To answer that, you have to compute the reduced density matrix for the atom. Abbreviate $c(t)\equiv \cos(\omega t)$ and $s(t)\equiv \sin(\omega t)$ to simplify notation. The full density matrix is

$$\rho(t) = c(t)^2 |10\rangle\langle10| + s(t)^2|01\rangle\langle01| - i c(t)s(t)|10\rangle\langle01| + ic(t)s(t)|01\rangle\langle10|.$$

If you trace over the light field to find the reduced density matrix for the atom, you get

$$\rho_{\text{atom}} = c^2(t)|1\rangle\langle1| + s(t)^2|0\rangle\langle0|.$$

This is, in general, a mixed state. A system in a mixed state will show some degree of reduction in its quantum coherence $^{[2]}$. This state goes from being pure to mixed, and back again with an oscillation frequency of $\omega$. So there's your calculation of the decoherence rate.

Now of course, in this problem the coherence actually comes back because the interaction of the atom and light field is oscillatory. There isn't a true "rate" here. However, if you have enough environmental degrees of freedom interacting with $S$, the recurrence time for coherence to be regained can be astronomically long. In these cases, the system $S$ appears to have irrevocably decohered, and in many cases the coherence follows an exponential decay, so you can define a true rate.

Note that I mentioned large numbers of degrees of freedom. In the case of an extended photon field the number of photon modes goes to a continuum and there are really a huge number of environmental degrees of freedom which can interact with the atom. In this case, if you were to compute the decoherence rate of the atom, you would re-derive Fermi's golden rule for decay rates.

If you want to see how to compute decay rates using path integrals look up papers by Caldeira and Leggett.

[1] Or maybe it's $\omega = g/2$, I forget which it is.

[2] If that statement needs clarification, please ask.

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Thanks for taking the time to answer an old question of mine and for providing another interesting perspective! –  Ryan Thorngren Jul 3 at 22:14
    
@RyanThorngren: Glad to help. Is there some way in which I can improve this answer? –  DanielSank Jul 4 at 0:13

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