# B Light as Wave

1. Jan 19, 2017

### _PJ_

I would really like some confirmation / clarification on the following points that I am trying to understand in terms of modern physics.

I would really appreciate if anyone might correct and complete my misunderstandings:

1 One can consider light as a wave.
2 As a particle, a photon may be considered as a single, "massless" particle.* Otherwise a single photon is a 'wave packet' - that is the waveform would be equivalent to that of combined, discrete frequencies.
3 A wave cannot have a single position, it's spread out - but fourier analysis can resolve a frequency and therefore momentum. Measurements of a point particle in terms of momentum or position will render inaccurate any determination of the other. THESE points preserve and support the uncertainty principle
4 A photon of light emitted from a source assuming flat spacetime - any mass or energy densities of the source ignored, will radiate spherically (Assuming 3 dimensions of space)

Okay if all is good and well with the above, then:

Alice is radially x metres from the source with a convenient light detector.
Bob is radially x metres but antipodean from Alice with his own convenient light detector.

Alice, Bob and the source are all at rest relative to each other and neither are distorting spacetime (they're on really good diets and very cold)

Surely both will detect light 'simultaneously'?

IN which case, they will detect light as a point, and since they know their location, then the energy must be uncertain?

Each person's detection would resolve the nature of the light for each person respectively, so they may later compare notes and decide the light was emitted as two separate photons, each with its own particular discrete energies.
In which case, the two photons should also be understandably entangled?

I am really comfortable with the Double-Split experiment and the thought of infinite possibilities for a 'particle's paths with the equivalent but opposing paths cancelling, and with the smaller scale contributions to the Hamiltonian for Feynman diagrams - but this which shoudl be a far simpler system to consider has my brain in knots!

A <-----------------O-----------------> B

If the consideration of the radial displacement as vector being obviously opposite is a factor, then this would entirely cancel equally? What if Charlie was also detecting but orthogonal to AB?

I have heard that there is a principle of "monogamous entanglement", so if AB and C all decide later their results indicated photons, all 3 cannot be entangled???
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I'm convinced there's some really trivial aspect I've overlooked that will clarify this whole picture, so I really welcome anyone who can help!
Thanks so much!
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*The nature of dimensionless non-extent point particle is a separat argument. Right now I'm just concerned with being a particle or wave-

2. Jan 19, 2017

### BvU

Rapid fire of questions, eh ? Good !
litterally: yes. As intended: one can describe many phenomona observed in the behaviour of light with the formalism developed for waves.
Sort of. frequency spectrum continuous, not discrete.
frequency is energy. Accuracy of simultaneous measurements is limited according to the uncertainty principle.
No. A disturbance in the electromagnetic field propagates with the speed of light. Can be in all directions but such a disturbance is a very, very long way from being a photon.

[caveat] This is an experimentalists answer, which I think may be good in your case. A theoretician might immediately and rightly refer you to quantumelectrodynamics.

I like the diet detail , don't understand the need for cold, but never mind.

Let me use the decay of a $\pi^0$ into $\gamma\gamma$ at rest in Alice and Bob's frame as example, so we have something slightly concrete to hold on to...
Two photons emerge in opposite directions

yes
one has nothing to do with the other. Any simultaneous measurement of $E$ and $t$ is subject to the uncertainty principle.
I don't know what you mean with the 'resolve' part, but yes: two photons,
Here we are leaving my personal comfort zone, but there are considerations here. You know what entanglement is ?
Duly noted. And somewhat noticeable.

what would cancel ?
In my example Charlie wouldn't observe a thing.

My best sincere guess is that you mix up different views: wave, photon in particular.
My pleasure. I hope a theoretician steps in and enlightens us all a little better, without becoming as hard to follow as you are in some of these quotes (no insult intended: aiming high and reaching for the stars is commendable).
Huh ?

3. Jan 19, 2017

### _PJ_

Thansk for this, can you offer any elaboration on this, since in my limitied understanding a photon IS a disturbance in the EM field... (I'm highlighting my misunderstanding, and seeking to learn a bit more here, not at all doubting or criticising your knowledge)

It was more of a joke really. But conveying the idea that A and B massenergy contributions (to ensure the light takes exactly the same time and they are precisely equidistant - no issues with geometry by potential wells around them etc) are equal was easier to make that just zero for them both.

I had skipped a step to the energy by the notion that the specific location lead to uncertainty in the velocity = uncertainty in momentum = uncertainty in energy (At least that was my thought process)

Either way, you're quite right and the E and t uncertainty relation is is something I'd completely neglected. Fortunately, though the end result is the same. The two can agree on the timing of their measurements (by virtue of ensuring they are equidistant from the source) - but the energy will be uncertain!

The entanglement I refer to is the quantum entanglement by virtue of the two photons individually detected by A and B having been created together by a single process.
My misgivings of the nature of a photon as wave I had thought might pose some relevance with this regard.

4. Jan 19, 2017

### BvU

Check out a few of the related links at the bottom. I think PF contains a lot of good info on this, but I don't have the most interesting ones at hand. Look around. We do a lot of physics with 1D wave packets and pretend they 'are' photons without hesitation. But 'how wide is a photon' triggers a whole discussion.

For photons $E=pc$ so the equals signs hold.

5. Jan 19, 2017

### _PJ_

Thanks so much! Really very helpful :)