# Single photon energy and multiple observers

1. Jul 3, 2014

### afcsimoes

I am not a student. I am a retired Chemical Engineer and a retired IT Auditor (ex-ISACA member). My questions are a product of my thoughts about Physics and the Cosmos and my talks with a closer friend. Please answer me.
Imagine an atom in a position P in the Universe and n observers, all them at a distance of d from it. An excited electron of the atom decays to a lower energy level and a photon is produced, whose energy is E (and is equal to the value of the difference between the electron's energetic levels before and after, isn't it?). (Remainder: a photon is an electro-magnetic wave that expands spherically from the point where the electronic transition has occurred).
Question 1: How many observers will detect the photon?
Question 2: What will be the measured value of the photon's energy that each observer will detect?
Question 3: What will be the value of the sum of all the measured energies by the n observers?
Question 4: When n goes to infinity what will be the value of the sum of the question 3?
Question 5: How this last value compares with the energy E of the electronic transition?

2. Jul 3, 2014

### Staff: Mentor

Hi, afcsimoes, and welcome to PF!

Yes.

Not really. An electromagnetic wave is a classical approximation, and is composed of many photons, not one single one. (Even that is not quite correct, but I don't know how much detail you want to go into about the actual math behind all this--there isn't a good way to accurately describe it just using ordinary language.)

More precisely, the photon's wave function expands spherically from the source, until it is detected. See below.

One.

The one observer that detects the photon will detect energy E. Other observers will detect nothing.

E. See above.

E. See above.

See above.

3. Jul 3, 2014

### fluidistic

I don't want to hijack the thread but the answer surprises me a lot! Let's say n=1, you mean that the observer will necessarily detect the photon?

As a side-note, I would like to mention to the OP that I don't think we can really imagine an atom at a "position P". Unless we make approximations of course.

4. Jul 3, 2014

### Staff: Mentor

Good point. I was assuming that n was large enough that some observer does detect the photon. A more correct answer, covering all values of n including n=1 and recognizing that for any finite n, however large, there is some chance the photon won't be detected, would be "Zero or one". The point I was trying to make was that it's never going to be *more* than one.

I was assuming that we were making an appropriate approximation for purposes of this discussion. But it's good to be clear about which approximations we are making, yes.

5. Jul 4, 2014

### afcsimoes

Approximation level to answer the questions

The lower limit of the approximations we must have is, I think, the Plank uncertainty.
Is this enough to justify that only one observer will detect the photon? Can one say, preview or calculate, in any way, who will be that one?

Last edited: Jul 4, 2014
6. Jul 4, 2014

### afcsimoes

In quantic language: the wave function describes, I think, the density of the probability for the photon's existence.

I has forgot the Plank uncertainty. Because of it the value of d can be different for each observer.
Can this be enough to justify that only one will detect the photon?
Are there any way to previously know who will detect the photon?

7. Jul 4, 2014

### Staff: Mentor

I assume you mean the Heisenberg uncertainty principle? This has nothing to do directly with the Planck length.

If you mean Heisenberg uncertainty, yes, I think that's what fluidistic was getting at: if we imagine the atom being precisely at position P, its momentum is completely uncertain and it could be anywhere in the universe in the next instant. The best classical approximation we can make is a coherent state whose position is centered on P and whose momentum is centered on zero, with the product of the spreads in position and momentum being $\hbar$. But for something the size of an atom, such a state remains localized at P, within measurement error, for quite long enough to run this experiment, so it works fine as an approximation.

At most one observer, yes. It's possible, as fluidistic pointed out, that no observer detects a photon (unless we have a continuous infinity of observers so that every single point in a sphere surrounding the atom is covered by an observer). But it's not possible that more than one observer detects a photon, because you specified that only one transition between atomic energy levels takes place, so only one photon is emitted.

No. All quantum mechanics can tell you is the probability that some observer will detect a photon (vs. no observer detecting one). It can't tell you which observer that will be, because, again, you specified a single atomic transition that emits a photon with a spherically symmetric wave function, so there is an equal probability for any observer to detect the photon.

Sort of. All this is a matter of interpretation. Another interpretation is that the wave function (more precisely, its square) describes the density of probability for the photon to be *detected*, without making any claim about its "existence". And there are other interpretations besides those two. The key thing is the prediction about what will be observed: that's all the math of quantum mechanics actually tells you.

Again, I assume you mean Heisenberg uncertainty here; see above. This won't affect the results of the experiment (except that, if we got really precise, it might slightly change the probability that no observer detects a photon, instead of one).

The fact that at most one observer will detect a photon has nothing to do with Heisenberg uncertainty: it has to do with how you specified the problem. See above.

No. See above.

8. Jul 5, 2014