You have a photomultiplier tube pointed at a distant star, exactly 100 light years away. It's perfectly set up so that nothing can get into the tube unless it came from that star. Every hour or so, you get a click in the detector.

Each time you hear a click, are you justified in saying that exactly 100 years ago, a photon was emitted from that star?

Or to put it more precisely: let's say your photomultiplier tube is pre-filtered so for all intents and purposes it only responds to a certain arbitrary atomic transition, say something in the potassium spectrum. The next time you hear a click, are you justified in saying that exactly 100 years ago, a particular potassium atom on that star made that particular transition from higher to lower energy?

Obviously, if I'm asking the question, I don't think you are allowed to make that connection. But I wonder what people think?

EDIT: We can sharpen up the question a little more if we look, for example, at the sun's corona. As Wikipedia explains,

The Fraunhofer lines are typical spectral absorption lines. These dark lines are produced whenever a cold gas is between a broad spectrum photon source and the detector. In this case, a decrease in the intensity of light in the frequency of the incident photon is seen as the photons are absorbed, then re-emitted in random directions, which are mostly in directions different from the original one. This results in an absorption line, since the narrow frequency band of light initially traveling toward the detector, has been turned into heat or re-emitted in other directions. By contrast, if the detector sees photons emitted directly from a glowing gas, then the detector often sees photons emitted in a narrow frequency range by quantum emission processes in atoms in the hot gas, resulting in an emission line. In the Sun, Fraunhofer lines are seen from gas in the outer regions of the Sun, which are too cold to directly produce emission lines of the elements they represent.

If we look at the sun directly, we see absorption lines: but I want to look for emission lines. So let's look at the sun's corona; and let's set up our photomultiplier with appropriate filters and whatever so that we can say, with 90% confidence, that a click in the detector came from a particular potassium-atom transition in the corona. Then the question is perfectly clear: when we hear a click, can we say (with 90% confidence) that exactly 8 minutes ago, a particular potassium atom made a transition from a higher to a lower energy level,thereby emitting a photon...the exact same photon that was just captured in our photomultiplier?

I think the question is pretty clear, and my answer is of course: No. And I think the people who say yes are the same people who would say that in the two-slit experiment, the photon must have passed through either one slit or the other one. Or the people who say that in the Stern-Gerlach experiment, the silver atom jumps into either the spin-up or the spin-down states at the moment when it passed through the magnetic field.

I wonder what people think about this.

  • $\begingroup$ A) Why every hour? Every $10^{-\text{something very big}}$ seconds is more like it. B) It will be impossible to eliminate outside noise. C) Particles don't necessarily travel in straight lines - a straight line is simply the most likely path. See this. I'd put this into some sort of an answer, but I think someone more qualified than I can organize it much better than I can, and can explain it better. $\endgroup$
    – HDE 226868
    Dec 6, 2014 at 22:51
  • 1
    $\begingroup$ Can you explain why you are unsure that this is possible? The way you have set up the question it sounds as if the only logical conclusion would be that 100 years ago an atomic transition of potatssium took place at that star. Depending on where abouts on the start it happened there would be a little bit of uncertainty about the timing. $\endgroup$
    – tom
    Dec 6, 2014 at 23:02
  • $\begingroup$ How do you propose ensuring that your tube can discriminate between photons emitted from that particular star, and not any point in between (say, by reflection off of a speck of dust)? Purely from an optics perspective, I think you would need an imaging system the size of our solar system in order to determine the distance from which the photon originated (via defocusing or a rangefinder type apparatus) $\endgroup$ Dec 7, 2014 at 0:49

2 Answers 2


This is a perfect example of the collapse of the wave function. The "potassium" atom makes a transition from higher energy to lower energy, sending out a speherical wavefront into the universe. One hundred years later, that spherical wavefront crosses the photomultiplier tube. Since the energy of that wave is spread over hundreds of square light years, there is only one chance in a gazillion that it is captured and caused a click to be heard. But there are jillions of potassium atoms in that star, emitting trillions of photons every hour or so. Since (1 jillion) x (1 trillion) = 1 gazillion, the result is that every hour or so you get a click in the detector. THAT is the collapse of the wave function, and it happens at the moment of detection.

Or does it? It seems that the doubters out there want the wave function to collapse twice...once at the point of emission and again at the point of detection. In the wave picture...and this is the quantum-mechanical wave picture, mind you...the emission of photon by a potassium transition has a characteistic linewidth associated with a finite emission time - in other words, a wave train. That wave train overlaps with the wave trains of the jillions of other potassium atoms in the star, creating a continuous output.The wave at the photodetector contains contributions from all those jillions of potassium emissions. THAT is the wave that "collapses" into a photon, causing a click in the photodetector.

But that wave cannot be traced back and associated with any particular emission event on the star. Or can it? If you think it can, then you are really saying that the wave function collapses twice...once at the point of emission, and once at the point of detection.

I don't think that's right.

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    $\begingroup$ This is wrong: " The "potassium" atom makes a transition from higher energy to lower energy, sending out a speherical wavefront into the universe. One hundred years later, that spherical wavefront crosses the photomultiplier tube. Since the energy of that wave is spread over hundreds of square light years,". The quantum mechanical solution of the transition amplitude gives a PROBABILITY of finding the photon at the (x,y,z) at time t of the detector. The energy is not spread out. It comes in a quantum of energy that will excite a potassium atom to the analogous level it left. $\endgroup$
    – anna v
    Dec 7, 2014 at 15:19
  • $\begingroup$ It seems like I'm saying it didn't become a photon until it entered the detector, and you're saying it was a photon from the moment it left the potassium atom. $\endgroup$ Dec 7, 2014 at 15:28
  • $\begingroup$ I am using the standard quantum mechanical interpretation, which assigns probabilities of emission and detection and certainly no energy is spread out all over the universe in any decay. $\endgroup$
    – anna v
    Dec 7, 2014 at 15:36
  • $\begingroup$ If there is no energy spread out, then the energy must be concentrated. And in your "standard quantum mechanics", we have local conservation of energy, which means the energy must be somewhere. So you're saying the photon existed from the moment it left the potassium atom. I suppose you're saying if it encountered a double slit along the way, it must have passed through either one or the other of the slits. $\endgroup$ Dec 7, 2014 at 15:45
  • $\begingroup$ Yes for the double slit. Look at these cosmic gamma rays mpa-garching.mpg.de/lectures/ADSEM/WS0203_Obergaulinger.pdf . gamma rays are photons too. $\endgroup$
    – anna v
    Dec 7, 2014 at 16:25

There exists the KISS rule ( keep it simple stupid), and I will.

Let us take an excited potassium atom and solve the Schrodinger equation for the transitions available. This will give for this atom a wavefunction, in principle from -infinity to +infinity in space and time. The square of this wavefunction will give the probability for a photon to fall on a specific area centered at a specific (x,y,z) at time t. A probability.

The transition energy is quantized and it comes as an elementary particle in the standard model of particle physics, with the appropriate frequency to excite another potassium atom to that energy level.

If two potassium atoms existed 100 light years apart with nothing else in between there still will be a probability of identifying the photon coming from the direction of the sender potassium atom, within the aperture of the detector.

If a star existed in the location to have the coordinates and the star were not moving with respect to earth and there were no expansion of space and no gravitational wells on the direction the photon would have to take, the probability would be very very small, but the star has zillions of potassium atoms and we would know that that potassium photon came from that star by detecting the excitation, taking 100 years to arrive and knowing the location of the star from the general light of the star. That is the way x-rays and gamma-rays are identified to specific locations in the sky.

So my answer is that you would know that "a" potassium atom deexcited and sent a photon this way. Photons are elementary particles , as real as the cosmic ray protons or any other particles arriving from space.

Now spectral lines from 100 light years away are redshifted or blue shifted depending on the relative motion of the emitting body, that is how we measure star velocities, so in real life the spectrum will be shifted and the potassium atoms in the detector will not be excited, a diffraction grating would be used to see the frequencies of the arriving light and if there exist potassium lines ( although I have the impression that with higher atomic numbers than hydrogen it is absorption lines one sees).

  • $\begingroup$ You start by saying you are going to "keep it simple" but then you throw in so many extraneous complications that your answer is all but incomprehensible. Let me list the things you bring up that have nothing to do with the question: the motion of the star relative to the earth, the expansion of space, existence of gravitational "wells" (?), the redshift, the blueshift, and the standard model. And why do you have a second potassium atom...and why do you think there are potassium atoms inside the detector? And how do you "see" an absorption line with a photomultiplier? $\endgroup$ Dec 7, 2014 at 16:22
  • $\begingroup$ I am answering the question. The questioner has set potassium acceptors in the detector. For optical light you do not need a photomultiplier. Accumulation of photons will show absorption and emission lines. emission values (as in question) change due to redshift or blue shift and cannot match the potassium the questioner wants in the detector coolcosmos.ipac.caltech.edu/cosmic_classroom/cosmic_reference/… . The standard model comes in to emphasize that the photon is a particle. $\endgroup$
    – anna v
    Dec 7, 2014 at 19:29
  • $\begingroup$ Oh come on now. Does your potassium atom make an audible click when it absorbs a photon from the distant star? $\endgroup$ Dec 7, 2014 at 20:01
  • $\begingroup$ It could make a visible point spot on a film, as with single photon patterns , if I were clever enough and knew enough physical chemistry to design such a film with a potassium content. This was not the point of the question. I assumed I could. $\endgroup$
    – anna v
    Dec 8, 2014 at 4:08

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