Time-of-flight spectroscopy and energy-time uncertainty

Question

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?

Answer 1

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.

Answer 2

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.

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.