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I am a bit confused about spectral and temporal filtering of light.

I consider a single-photon source of a bandwidth of $1\,$nm at $1550$ nm.

That means, that the wavelength of the emitted photons should be gaussian distributed around $1550$nm. The bandwidth of $1$nm gives rise to a coherence time of about $8$ps.

That means, if I place a single-photon detector with a perfect timing resolution behind the triggered source, I expect to see an uncertainty of the photon arrival times of $8$ps.

If I now send the photons through a dispersive medium, for example $100$km of a dispersive fiber with $18$ps/(nm km), i expect the peak of arrival times to become widened to about $1800$ps.

Now, the spectral bandwidth of the photons has still not changed, their coherence time due to their energy-uncertainty is still $8$ps. Now i can divide the time-axis into bins of $8$ps and get much more spectral information about the photon. in fact, I can for example divide the $1800$ps - peak into 225 $8$ps bins. Would this type of "spectrometer" give me a spectral resolution of $~\sim 4.5$pm?

I am a bit confused about this, because the arrival of a photon at a very specific time lets me deduce its wavelength due to the deterministic chromatic dispersion. However, isn't this also the same as spectral filtering? If I had before filtered the spectrum down to $4.5$pm, the coherence time would be huge (in fact, about $1.8$ ns). Therefore my question: Where is the mistake in these ideas? Would this type of spectrometer work?

If I had a source of energy-anticorrelated photon pairs that I send through the dispersive medium, would I be able to find the pairs in the anti-correlated bins?

Mechanix
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  • photons are elementary particles, and will have an energy h*nu. The wavelength describes the light pulse (i.e. the space variation). I think there is a confusion between the dxdp and dtdE heisenberg uncertainty, but I cannot locate where in the narrative. – anna v Mar 09 '19 at 17:48
  • Maybe i should have stayed in the timing and frequency picture to be less confusing. However, the question is still the same: Is the time binning in the end equivalent to spectral filtering? Because in this case, both timing and energy would be much better known than before the dispersive medium. That, i don't understand. – Mechanix Mar 10 '19 at 00:38

2 Answers2

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Have a look at my answer here.

It gives the reasoning on why individual photons cannot be constrained to pass sequential slits and thus defeat the momentum uncertainty.

The pulse with the accuracy you describe is a classical electromagnetic pulse. Quantum mechanically it is made up by the superposition of zillions of photons of the given frequency with the given width in space and time.

For individual photons a dispersive medium is similar to running the gauntlet of the series of slits envisaged by the OP in the linked question. The probability for an individual photon to remain in the main pulse in space ,given by the $dxdp$ Heisenberg uncertainty is independent of the probability of remaining in the pulse as given by the $dtdE$ , these are different quantum mechanical operators.

I think this is where the confusion comes.

anna v
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  • So would you say that the coherence time of the single photon becomes longer due to the dispersion, such that the possible time bins become larger in the same fashion? – Mechanix Mar 10 '19 at 13:01
  • I think so, that something similar happens with trying to constrain the time sequence for a single photon. – anna v Mar 10 '19 at 13:03
  • time energy experiments are very few , here is one https://www.researchgate.net/publication/13380482_High-visibility_interference_in_a_Bell-inequality_experiment_for_energy_and_time , where they have interference in time in the small length and dispersion of their experiment. You are thinking kilometers, and consecutive interferences as the photon interacts with the dispersive medium would have the same effect of removing photons from the sample under consideration, as far as can see. – anna v Mar 10 '19 at 14:45
  • I think the temporal filtering for single photons is kind of equivalent to the following case in the dxdp uncertainty. If i have a tiny detector sitting at the screen, where the interference pattern is recorded and this detector fires, then i know the position very precisely and also the momentum, that the particle had, due to the position of the detector. However, i will not be able to predict this behavior or prepare a state always reaches this detector. So does uncertainty only hold for ensembles? – Mechanix Mar 10 '19 at 20:52
  • Uncertainty is measured in ensembles because it is a probability distribution for the ensemble, but it is imposed in nature to the individual measurement being not predictable due to the intrinsic probabilistic nature of quantum mechanics at dimensions commensurate to h_bar, for pairs of variabes. – anna v Mar 11 '19 at 04:41
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If I understand the question correctly, then it can be answered purely on the basis of Fourier theory. The notion of a single photon and the quantum mechanics associated with it does not seem to enter here.

So if I may rephrase it slightly, the question is what happens when considering a laser pulse with a wavelength bandwidth of 1 nm at 1550 nm (which corresponds to a pulse width of 8 ps).

A nice way to think about this is to plot time and frequency on two perpendicular axes and consider the pulse as a blob on this two-dimensional space. (See green blob in figure below.) However, due to the Fourier relationship between time and frequency (which comes down to the Heisenberg uncertainty for quantum states), the spread in frequency multiplied by the duration in time is always larger than a certain area on this two-dimensional plane.

Dispersion of pulse

So, if the pulse passes through a dispersive fibre, then the shape of the blob will become twisted (blue blob in the figure). However, if I would now try to divide the pulse along time into multiple short bins, the frequency for each of these bins will spread out to satisfy the requirement for the minimum area as shown by the yellow blob in the figure (this is due to Heisenberg uncertainty).

In other words, if you would try to make multiple time-bin measurements of a pulse that was spread out by dispersion, you would find that the uncertainty in the frequency would blur the information that you are trying to gain about the spectrum.

flippiefanus
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  • Thank you for your reply, flippiefanus! I don't fully understand this graphic picture. If the blob gets elongated in direction of time (that's what dispersion seems to do), that would mean that the uncertainty in frequency becomes smaller, right? – Mechanix Mar 12 '19 at 12:52
  • Or let me ask another question: If i apply dispersion to the pulse and then send it into a michelson interferometer, where i can measure the coherence length of a pulse rather directly by moving a mirror, will i observe the same as without applying the dispersion before? – Mechanix Mar 12 '19 at 15:20
  • I added a figure to explain what I meant. Hopefully it is clearer now. The coherence length also obeys the restrictions of the widths between the time and frequency. – flippiefanus Mar 13 '19 at 04:43