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A distant star like the sun, thousands of light years away, could be so faint that only one photon might arrive per square meter every few hundred seconds. How can we think about such an arriving photon in wave packet terms?

Years ago, in a popularisation entitled “Quantum Reality”, Nick Herbert suggested that the photon probability density function in such a case would be a macroscopic entity, something like a pancake with a diameter of metres in the direction transverse to motion, but very thin. (I know the wave packet is a mathematical construct, not a physical entity).

I have never understood how such a calculation could have been derived. After such a lengthy trip, tight lateral localisation suggests a broad transverse momentum spectrum. And since we know the photon’s velocity is c, the reason for any particular pancake “thickness” in the direction of motion seems rather obscure.

(Herbert then linked the wave packet width to the possibilities of stellar optical interferometry).

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Just to add a little to the wave packet thickness issue. In the direction of motion, E = cp. Since E = hf, if the photon frequency is well-defined then so is its momentum and by the uncertainty principle its localisation would be rather undefined in the direction of travel. Wouldn't this make the pancake thick? – Nigel Seel Jan 21 '11 at 20:35
up vote 2 down vote accepted

first of all, the shape of the wave function of a photon that is emitted by an atom is independent of the number of photons because the photons are almost non-interacting and the atoms that emit them are pretty much independent of each other. So if an atom on the surface of a star spontaneously emits a photon, the photon is described by pretty much the same wave function as a single photon from a very dim, distant source. The wave function of many photons emitted by different atoms is pretty much the tensor product of many copies of the wave function for a single photon: they're almost independent, or unentangled, if you wish.

The direction of motion of the photon is pretty much completely undetermined. It is just a complete nonsense that the wave function of a photon coming from distant galaxies will have the transverse size of several meters. The Gentleman clearly doesn't know what he is talking about. If the photon arrives from the distance of billions of light years, the size of the wave function in the angular directions will be counted in billions of light years, too.

I think it's always the wrong "classical intuition" that prevents people from understanding that wave functions of particles that are not observed are almost completely delocalized. You would need a damn sharp LASER - one that we don't possess - to keep photons in a few-meter-wide region after a journey that took billions of years. Even when we shine our sharpest lasers to the Moon which is just a light second away, we get a one-meter-wide spot on the Moon. And yes, this size is what measures the size of the wave function. For many photons created in similar ways, the classical electromagnetic field pretty much copies the wave function of each photon when it comes to the spatial extent.

Second, the thickness of the wave packet. Well, you may just Fourier-transform the wave packet and determine the composition of individual frequencies. If the frequency i.e. the momentum of the photon were totally well-defined, the wave packet would have to be infinitely thick. In reality, the width in the frequency space is determined up to $\Gamma$ which is essentially equal to the inverse lifetime of the excited state. The Fourier transform back to the position space makes the width in the position space close to $c$ times the lifetime of the excited state or so.

It's not surprising: when the atom is decaying - emitting a photon - it is gradually transforming to a wave function in which the photon has already been emitted, aside from the original wave function in which it has not been emitted. (This gradually changing state is used in the Schrödinger cat thought experiment.) Tracing over the atom, we see that the photon that is being created has a wave function that is being produced over the lifetime of the excited state of the atom. So the packet created in this way travels $c$ times this lifetime - and this distance will be the approximate thickness of the packet.

An excited state that lives for 1 millisecond in average will create a photon wave packet whose thickness will be about 300 kilometers. So the idea that the thickness is tiny is just preposterous. Of course, we ultimately detect the photon at a sharp place and at a sharp time but the wave function is distributed over a big portion of the spacetime and the rules of quantum mechanics guarantee that the wave function knows about the probabilistic distribution where or when the photon will be detected.

The thickness essentially doesn't change with time because massless fields or massless particles' wave functions propagate simply by moving uniformly by the speed $c$.

Cheers LM

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Thanks for this very helpful answer. If a series of stellar processes eventually result in a photon exiting the star on its way to earth, then perhaps at some initial point the photon wave function is spherically symmetric about its point of origin. However, it will soon encounter a stray proton in space in one direction or another. Won't such occasional interactions localise the photon's wave function pretty quickly (cf. a cloud chamber)? – Nigel Seel Jan 22 '11 at 10:29
Dear @Nigel Seel, thanks for your interest. The wave function of a photon can't be entirely spherically symmetric - because a photon carries a spin (transverse polarization) and one can't comb the sphere. However, it's morally true that the wave function occupies almost the whole solid angle $4\pi$. And no, the number of interactions with charged particles (e.g. proton) that a photon coming from star has participated in is exactly zero. If it were not zero, it would change the direction. So no, the photons are not measured at all before they arrive to the telescope. No interference is lost. – Luboš Motl Jan 22 '11 at 10:37


I would like to make another suggestion concerning this question. I have not read the 1985 book by Nick Herbert, but from online links I see that he was discussing the "Hanbury Brown and Twiss" mechanism for optical interferometry. Although this sounds like quite a mouthful and not very fundamental - in fact it is.

The phenomenon has to do with photon ie boson quantum entanglement. The entanglement is between two photons from opposite sides of a star, allowing the star width to be measured. This technique was counter-intuitive apparently even to quantum physicists at the time. Here is a quote from Penrose 2004 (p598) on the matter:

"When their method was first proposed, it met with great opposition from many (even distinguished) physicists, who argued that 'photons can only interfere with themselves, not with other photons'; but they had overlooked the fact that the 'other photons' were part of a boson entangled whole."

It is possible that NH is making some geometric claim about the shape of this photon entangled state.

You can follow the Wikipedia links on the HTB effect, or raise another Stack question.


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Excellent link at… which I think must be what Herbert was getting at. Thanks for the reference. +1. – Nigel Seel Jan 22 '11 at 15:03

The answers posted by Lubos and Nigel are not bad but I think there is a point that needs to be added. The "photon" we detect on our photographic plate cannot be identified with a single photon that was emitted by a particular atom in the distant star. It must be thought of as the superposition of trillions of photons emitted at the same time from all over the surface of the star. These photons do not maintain separate identities as they spread through space.

In fact, these photons cannot do anything that cannot also be accounted for by ordinary light waves. I used to believe the argument that the power density of the light from a distant star was far too weak to provide the necessary energy for the reduction of a single atom silver bromide to metallic silver, the minimum energy needed to make a dot on the photographic plate. But last year I realized that this argument is invalid: to calculate the energy required for the transition, we must consider not a single atom but the thermodynamics of an entire silver bromide crystal. If we treat the crystal as a solid solution of silver bromide/metallic silver, then at the extremely low concentrations of metallic silver present in the unexposed crystal, the equilibrium point is shifted so far to the left that the conversion is virtually spontaneous. So you don't really need a full "photon's" worth of energy to drive the transition.

I explain this in more detail in these articles on my physics blog, "Why I Hate Physics": The Collapse of the Wave Function, and also Quantum Siphoning

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I don't know what the business about dyes has to do with my "Theory". I only claimed to analyze the thermodynamics of the silver bromide=> metallic silver conversion. You call my analysis "amateurish": are you saying it's wrong? – Marty Green Nov 8 '11 at 8:24
Okay. So what exactly is the role of light in the photographic process if not the reduction of silver bromide? – Marty Green Nov 8 '11 at 10:18
Okay. So what exactly is the role of light in the photographic process if not the reduction of silver bromide? – Marty Green Nov 8 '11 at 13:08
For the record, I see that I have two comments in a row which are both identical. Just to explain how this happened: Georg said it was "silly" to think that silver bromide was converted to silver by light. So I asked: what then is the role of light in the process? Georg gave some kind of answer which didn't address my question, so I repeated the question. Then Georg came back and erased his previous answer, leaving my two questions just sitting there. – Marty Green Nov 8 '11 at 20:42
Additional for the record: Georg had a series of comments where he ridiculed my explanation, but he has erased them all now. – Marty Green Nov 9 '11 at 9:53

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