# "pure state" of the EM field?

1. Sep 7, 2015

### pellman

In Ballentine's Quantum Mechanics book, as part of a discussion of pure states vs non-pure (mixed) states, he says

Polarized monochromatic light produced by a laser can approximate a pure
state of the electromagnetic field. Unpolarized monochromatic radiation and
black body radiation are examples of nonpure states of the electromagnetic
field.
I think that he is referring to the classical EM field and using "state" in a more broad sense than quantum states, though I could be wrong about that.

So in what sense is unpolarized light not a pure state? I would think that, just as superpositions of pure state quantum wave functions give another pure state, that superpositions of the EM field would also give "pure states". That is, any solution of Maxwell's equations would be considered a pure state.

But really I am muddled on this concept.

2. Sep 7, 2015

### sophiecentaur

I think that restricting the view of lasers to QM makes life very hard. If you regard 'ordinary light' as coming from a large number of individual atoms, each of with produced its photon at a random time then this is similar to a large number of radio transmitters with their antennae, all with slightly different frequencies and with the antennae in random directions. That is your impure state and it's down to bandwidth (line width?) When the radiation is stimulated, magically all the outputs from all the atoms become so close in phase that it's as if they'd all been produced from one continuous energy transition and the whole beam of light is regarded as being in a pure state. As for the polarisation, the antennae are all lined up to be parallel and produce a single polarisation plane. (Or circular / elliptical , I guess)

3. Sep 11, 2015

### pellman

Thanks

4. Sep 12, 2015

### vanhees71

Quantum mechanically laser light is (nearly) a coherent state and as such a pure state, while "thermal" light is in a mixed state, which is in the case of ideal black-body radiation is described by the thermal-equilibrium Statistical Operator
$$\hat{R}=\frac{1}{Z} \exp(-\beta \hat{H}),$$
where $\beta=1/k_{\text{B}} T$ with $T$ the temperature, and the partition sum
$$Z=\mathrm{Tr} \exp(-\beta \hat{H}).$$
The invention of the laser was so ground breaking for science and applications alike, because

(a) after all one had a source of (nearly perfectly) coherent light. Indeed the coherent state with a high intensity can be very accurately described also as a coherent classical electromagnetic wave. The most recent breakthrough in this direction is the possibility to have such light sources also in the X-ray regime, where the coherent light can be used for high-precision measurements with applications reaching from materials science (like condensed-matter physics) to biology.

but also because

(b) it provided a possibility to create sources of single photons and entangled two- (and even few-) photon Fock states, which brought many "gedanken experiments" a la Einstein, Podolsky, and Rosen to the realm of real experiment with the possibility to check the astonishing quantum effects related with entanglement, among other things the disprove of the possibility of a local deterministic hidden-variable theory (Bell's inequality) to an overwhelming degree of significance. Also this branch, which is the real "quantum optics", i.e., optics that cannot be understood within classical Maxwell theory (and it is less trivial to find observables realizable in the real world that prove the quantum nature of electromagnetic waves than even some modern quantum textbooks suggest, but that's another story).

5. Sep 12, 2015

### atyy

But some pure states have thermal properties, so couldn't thermal light also be described by a pure state?

Here are some examples of pure thermal states
http://arxiv.org/abs/1302.3138
http://arxiv.org/abs/1309.0851
http://www2.yukawa.kyoto-u.ac.jp/~new-noneq2015.ws/Presentation/Sugiura.pdf

Last edited: Sep 12, 2015
6. Sep 12, 2015

### vanhees71

That sounds interesting, but it seems also to be a bit misleading to call these states "pure states", because there's an random process underlying, defining stochastic pure states rather than what you call a "pure state" in the usual sense of standard quantum theory. I've not yet understood, what's the point of this construction (despite that it seems to be a pretty interesting mathematical idea to do Monte-Carlo Simulations of equilibrium many-body systems).

7. Sep 12, 2015

### atyy

The point of the construction is similar to Boltzmann kinetic theory: how do we reconcile irreversibility in thermodynamics with the reversible laws of mechanics?

So can a pure quantum system thermalize without averaging? The existence of pure thermal states suggests that it is possible in many cases.

To add to the above references (which did not mention the approach to equilibrium), here is another one which talks about thermalization without ensemble or time averaging (strong thermalization): http://arxiv.org/abs/1007.3957 (http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.106.050405).

Last edited: Sep 12, 2015
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