# Understanding photon direction

1. Oct 9, 2015

### JFS321

I have been trying to better understand the concept of a photon (I know...a thousand threads on this alone) and the direction that it propagates. I understand the time-varying field explanation for em waves, but here is where I break down. If an electric field extends in all directions simultaneously, why should we consider that photons have a given direction (unless we physically manipulate it at a given point in space), and also describe photons as occurring in a finite amount? (I'm thinking that a spherical wave could be divided infinitely).

Or, maybe I can word it this way -- why would a vibrating charge not also radiate an em wave in all directions? I can't understand how a spherical wave could be considered as discrete packets traveling in random directions. Thanks in advance!

Last edited by a moderator: Oct 9, 2015
2. Oct 9, 2015

### Matterwave

Photons are a totally different way of thinking of light than EM waves. This is the wave particle duality in QM, and is spread throughout. In the current field theory machinery, Photons are excitations of the background EM field.

The electric field extends in all directions, but it also points in a specific direction (it is a vector field). The direction of propagation of a transverse EM wave is going to be both perpendicular to the electric field and the magnetic field. So even in a purely field theory approach, there can certainly be a "direction of propagation" of an EM wave. It is the direction in which the disturbances in the background EM field propagates. This is exactly analogous to a wave on a string. The direction of the wave is down the string, even though at every point on the string, the particles on the string are only going up and down and not along the string.

Photons and EM waves are just 2 different paradigms with which we view electro magnetic phenomena. Some phenomena are much easier described with one versus the other.

A vibrating charge will, in fact radiate in (roughly) all directions. The specific nature of the vibration governing how the radiation propagates. The simplest is dipole radiation.

3. Oct 10, 2015

### vanhees71

There is no wave-particle duality in modern quantum theory. Photons are special states of the quantized electromagnetic fields, namely one-photon Fock states. They have not much in common with classical particles.

There is only one description of electromagnetic phenomena (and not different paradigms), and this is QED, which has a classical limit known for 150 years now, called classical electrodynamics. A vibrating charge will lead to a coherent state of the electromagnetic field.

4. Oct 10, 2015

### JFS321

Hmm, this gives me more to think about. Thanks.

So, if a single photon is emitted by an atom, the photon may or may not interact with neighboring atoms depending upon its direction. How, then, is that reconciled with the spherical wave nature of light? It seems that a single photon would simultaneously disturb all regions of space as it passed by in a spherical fashion.

I think lasers are a good example here. If all lasers diverge, what we see must be a product of the countless photons radiating in a common direction and the divergence is much too small for our eyes to detect. This is why I cannot understand how a single direction can be reconciled for just a single photon.

5. Oct 10, 2015

### Staff: Mentor

You are ascribing classical properties like direction and even number of photons to quantum stuff that doesnt necessarily have such properties.

Its described by QFT (Quantum Field Theory) and is not amenable to easy visualisation.

Thanks
Bill

6. Oct 10, 2015

### Jeff Rosenbury

I'm an EE, not a QED. My understanding is that a quantum wave is a probability wave, not an EM wave.

Thus a photon travels according to quantum probability, not according to its EM field. However, the photon has an EM field, so it affects the EM field as it travels. While it's a good shorthand for us EEs to think the two are identical, they are not.

Back in the day we thought quantum waves expanded until they were "observed", then they collapsed. This expansion could be viewed as adding a spherical component since it involves the expansion in three dimensions of the probability field. However, I understand current thinking is that the matrix equations govern things nicely and we humans (or at least I) simply don't understand them well.

This does lead to a problem of the direction of the photon not being determined until it reaches its destination. This is an aspect of quantum entanglement which I think is still an active field of study (i.e., we don't know). To illustrate: Suppose a supernova sends a photon a billion lightyears to an observer on earth. How does it know how much momentum (including direction) to send with (as part of) the photon? How did the supernova know the mirrors in the telescope in advance? (Which affect the quantum probability of where it lands.)

There is a tendency for EEs to want to believe in hidden variables to explain this. While hidden variables haven't been completely ruled out, the simple explanation of them has been. The effect happens even with photon which are too weak to carry the energy associated with the data they would need to carry. It makes such hypothesis unlikely enough that I don't think any are in the standard model.

7. Oct 10, 2015

### Staff: Mentor

Electromagnetic radiation propagates as a nice spherical wave just as described by classical electrodynamics. Photons only come into the picture when the radiation interacts with matter; we find that the radiation always delivers its energy and momentum in discrete chunks landing in a fairly tightly constrained area, and when this happens we say that "a photon landed there". The probability of a photon landing at a given location depends on the intensity of the radiation at that point.

The key thing to note here is that QM says nothing about where the photon is before that interaction; in particular you cannot think of the photon as a particle traveling on some definite trajectory from the source to target. It's not the case that the photon is moving on some path but we don't know what it is; it has no position and no path. It is tempting to think of light as a stream of moving photons the way that a river is a stream of water molecules moving by... But that model is hopelessly misleading.

A good non-technical introduction is Feynman's "QED: the strange science of light and matter" which shows both how we calculate the probability of the photon landing somewhere and the general futility of imagining photons as small particles of light flying through space.
What you're seeing with a laser is a coherent wave propagating from an aperture of finite width. It diverges because the aperture is not of infinite width; if it were, we'd have an idealized plane wave.

Last edited: Oct 10, 2015
8. Oct 10, 2015

### JFS321

Thanks for all the replies. Nugatory, what you describe is what my mental model came to after reading multiple sources. But, I am still unsure as to why em waves (of the proper energy) would not affect all electrons in the vicinity because of the spherical nature. In other words -- what leads us to say that a photon "missed" a target and thus was not absorbed? Based on replies it seems that a photon should be considered to be an infinitely small section of the spherical wave--this, too, would jive with the common notion that light does indeed travel in a straight line. Are there issues with this?

9. Oct 10, 2015

### cgk

JFS321, photons should not be considered as anything resembling classical particles. They are not small balls moving through space.

The classical EM picture of waves is perfectly fine. You should keep it. The QM picture only comes in if your energy scales get small enough to notice individual $\hbar\omega$ units of energy, because in QM only integer multiples of this can be absorbed or created (so if you insist on a classical analogy, it is better to think of a photon as "one bucket of wave" rather than "one particle moving about").

How does it work? You know that in a cavity, you can decompose the EM field into a set of eigenmodes with unique frequency. In classical EM, the coefficients of these modes evolve according to an equation equivalent to an Harmonic oscillator. If the EM-field is quantized, these become quantum mechanical oscillators, with energies which can only take values of $(n + 1/2)\hbar \nu\;(n\in\mathbb{N}_0)$. What consequences does this have? Apart from eigenmodes having the integer+1/2 multiples of the possible energies, not many.

10. Oct 10, 2015

### andrewkirk

I am so glad to hear you say that! [the bold was added by me]

I provisionally came to that view a while back, reading between the lines in what people had said about various experiments and working through the mathematical model in my mind. But until now I had not seen anybody say that in such unequivocal terms, and my QM texts are woefully silent on the topic - indeed they barely mention photons at all. So I assumed it was just a wacky personal interpretation I had. I now feel licensed to believe it with confidence.

I find the whole thing so much easier to understand when I look at it that way.

11. Oct 10, 2015

### Staff: Mentor

I do have to caution you that I've made some serious oversimplifications here (for example: there are photon-photon interactions; and Feynman's brilliance just means that "QED: The strange theory..." is a brilliant popularization) so none of this is a substitute for really working through a serious QED or QFT text. But oversimplifications and all, I find it a better starting point than the notion that photons are little particles of light.

12. Oct 10, 2015

### Mentz114

There are no photons in flight. As Nugatory says

We can only know where the photon was after some interaction.

13. Oct 10, 2015

### Jeff Rosenbury

It's my understanding that photons do travel. Specifically the probability of interaction is related to the sum of all possible travel paths. In addition they carry momentum between the source and the destination particles. Finally the amount of photons "traveling" through a region will affect the probability of a test charge interacting with their electric field, which is how we measure field strength.

So while they have no observable effects without interaction (like everything else in the universe), making the claim they don't travel seems a bit odd.

I can easily foresee this turning into a discussion about what the word "travel" means though.

14. Oct 10, 2015

### Staff: Mentor

You are misunderstanding Feynman's "sum over all possible paths" formalism. It says that if we calculate the phase shift over each possible path and then at the destination vectorially add the amplitudes from each path we will - somewhat amazingly - get the right probability. It does not say that the particle actually moves along each or any of these paths; indeed, if we try thinking in those terms (which is completely unnecessary in the mathematical formalism) we will end up with the same contradictions and problems that we encounter if we naively interpret the double-slit experiment as saying that the particle "went through both slits".

But all of these effects involve an interaction with something else. Except at the endpoints, the classical electromagnetic field carries energy and momentum just fine - no photons needed. The photons only come into the picture when we ask questions like "why are individual dots appearing at the detector at the destination?" or "why does the black-body spectrum indicate that the interaction at the source is transferring energy in discrete amounts?"

15. Oct 10, 2015

### Staff: Mentor

As Nugatory correctly says you are misunderstanding the sum over histories approach. They take all paths which is not travelling - when objects travel they take one path by the definition of travelling. Here we face the problem of using words for a theory where many of our usual words are simply not applicable.

Here is the bottom line about QM and QFT. They do not have any properties until observed to have those properties. They do not travel, carry momentum etc etc until observed to have it. All classical pictures along those lines are incorrect.

At the beginner level there are a couple of books that give you the gist of what's happening in QFT. One is Fields of Colour:
https://www.amazon.com/Fields-Color-theory-escaped-Einstein/dp/0473179768

The other is the already mentioned Feynmans book on QED:
https://www.amazon.com/QED-Strange-Princeton-Science-Library-ebook/dp/B00BR40XJ6

The first book is the easiest but the least accurate. The second book is a bit more difficult and more accurate. But neither book gives the full detail of what happening in QFT. Basically everything you have read about QFT outside a QFT textbook is almost certainly wrong - and that includes some well respected beginner textbooks. Feynman understood this and lamented that learning physics was like that. However admitted it can't be taught any other way.

We all must start somewhere and those two books will give you a bit of the flavour you can use to start your journey to a full understanding - if you decide to go that far.

Thanks
Bill

16. Oct 10, 2015

### andrewkirk

Is the Feynman a technical book or a popularisation? I couldn't tell from the Amazon page because the 'Look Inside' feature didn't get past the end of the introduction, so there was no chance to gauge the level. I wouldn't want to waste my time on a non-technical book.

I see you didn't mention Ballentine. Does his book not venture into QED or QFT?

Thank you
Andrew

17. Oct 10, 2015

### Staff: Mentor

Its a popularisation. But it gets the detail about as correct as you can get at that level. Its a very unique book in that regard. The only criticism I have of it is he says when you learn the full detail you will not have to unlearn anything he says which isn't exactly true. For example, the true account of what's going on during refraction etc is a lot more complex than what he says:

But this is the well known problem in physics I alluded to in my post. What's really going on cant be explained at the beginner level in a number of areas of physics. Its lamentable its like that - but unfortunately true. QFT is another example. Beginner books speak about Feynman diagrams etc as if virtual particles etc actually exist. They don't - but detailing the true situation to start with will likely confuse rather than illuminate so they take a couple of liberties with the truth.

Ballentine is a book on advanced QM, not QFT. Its meant for after an intermediate book on QM like Griffiths and pulls no punches eg states are positive operators of unit trace - not elements of a vector space as less advanced books tell you. In doing that you get the full theory - from the entrancing beauty of the true origin of Schroedingers equation to the difficult problems of interpretation.

For QFT you don't really need advanced QM. A number of books have started to appear that can be tackled after a book at the level of Griffiths. My favourite is the following:

Thanks
Bill

Last edited by a moderator: May 7, 2017
18. Oct 10, 2015

### Jeff Rosenbury

I don't think I'm misunderstanding it. I agree with your statement, which agrees with mine.

Are there any non-quantum objects or objects not made of quantum parts?

Lacking non-quantum objects, and accepting there is no quantum traveling, there is no traveling at all -- anywhere. So the word "travel" has no meaning. It is fiction. I guess this proves Warren Buffet was right when he said of the airline travel industry, "[H]e would have done his successors a huge favor by shooting Orville down."

Or possibly the word "travel" is a more complex idea than we think?

Is traveling only a biological approximation of a more complex system? Or does it have fundamental meaning such as the evolving shape of probability fields? Such a question is perhaps better answered elsewhere. I think it has more to do with the philosophy of science than with the standard model.

19. Oct 11, 2015

### andrewkirk

It's veering into philosophy, but I think that 'travel' can have a well-defined meaning when applied to an observer. An observer 'travels' when they make a series of observations at different locations. If I am sitting looking out the window of a moving train, every time I open my eyes after blinking I am making a new observation of a slightly different location. I could argue that, if there are no observations, there is no travel. One merely goes to sleep in one place and wakes up in another.

It seems to me that it's not accident that travel in literature generally refers to a traveller making observations along the way, by visiting palaces, cathedrals and temples, climbing mountains etc. In contrast, if one travels by Star Trek Teleport, Doctor Who Transmat or Harry Potter Portkey or Apparating, one tends not to use the word 'travel'. One just disappears in one place and appears in another.

A consequence of this definition is that one cannot necessarily talk about 'travel' by an object that is not an observer, such as an inanimate object. However, one could compromise and say that an object has 'travelled' if one has observed it at different stages of its journey. I guess we use that when we describe a ball curving through the air in cricket, baseball or soccer.

20. Oct 11, 2015

### Staff: Mentor

That does not follow.

In the sum over histories approach at the macro level virtually all paths cancel leaving just the one path. This is known as the principle of least action and is the foundation of classical physics. At the atomic level path cancellation does not occur and you cant speak of it having an actual path - or many other classical properties for that matter.

In the case of photons, for technical reasons you will really only find in a QFT textbook (and an advanced one at that - its actually quite deep - its due to the fact position is not an observable for photons) this path cancellation doesn't really work - even though beginner texts may suggest it does.

Thanks
Bill

Last edited: Oct 11, 2015
21. Oct 11, 2015

### Staff: Mentor

My view is simpler. When you say you travel from A tp B you mean by some well defined path where you have a position and a velocity at all points along the path. In QM such is not the case. Particles travel along all paths and the property it has along that path is not position but rather a complex number which introduces a magical added complication - phase. Most of the time if you vary the path a little bit the phase will change so given any path you can find a nearby path that is 180% out of phase so cancels. That will be the case except for one special path - the path where a small change does not change the phase and instead of cancelling you get reinforcement. That leaves just one path - the path from the principle of least action. That however is classically. If the paths are not long enough so phase cancellation can occur you have the quantum domain and cant speak of an actual path.

Here is the mathematical detail.

You start out with <x'|x> (the square is the probability of it initially being at x' and later being observed at x) then you insert a ton of ∫|xi><xi|dxi = 1 in the middle to get ∫...∫<x|x1><x1|......|xn><xn|x> dx1.....dxn. Now <xi|xi+1> = ci e^iSi (the square gives the probability of it being at xi and a very short time later observed to be at xi+1) so rearranging you get
∫.....∫c1....cn e^ i∑Si.

Focus in on ∑Si. Define Li = Si/Δti, Δti is the time between the xi along the jagged path they trace out. ∑ Si = ∑Li Δti. As Δti goes to zero the reasonable physical assumption is made that Li is well behaved and goes over to a continuum so you get S = ∫L dt.

What that weird integral says is in going from point A to B it follows all crazy paths and what you get at B is the sum of all those paths. Now if the path is long compared to how fast the exponential 'turns', since the integral in those paths is complex most of the time a very close path will be 180% out of phase so cancels out. The only paths we are left with is those whose close paths are the same and not out of phase so reinforce rather than cancel.

Now it does not say that all paths will cancel and you will only get one. No paths may cancel for example if the path is short compared to the frequency e^iSi 'turns' so no close paths will cancel. That is the case, for example, inside a Hydrogen atom and in that domain QM rules.

Also note the assumption made here - position is an observable. That is not the case for photons so the method does not work. Beginner texts will not tell you that - which is another example of you don't get the full truth at the start. This really is a maddening part of physics - but you get used to it after a while.

Technical aside. S is of course the action and L the Lagrangian. What this says is the only path that exists is the path that when you do a small change in path S does not change. This is the principle of least action - the basis of classical physics in higher dynamics. In fact from symmetry and this principle you can basically derive all of classical physics - even the existence of mass - strange but true. You can find the detail in Landau - Mechanics. A very beautiful and rewarding book to study by a master physicist - highly recommended. It will likely be life changing - it was for me.

Thanks
Bill

Last edited: Oct 11, 2015
22. Oct 11, 2015

### JFS321

Wonderful stuff. What we don't know is as fascinating as what we do know. I am going to order a copy of Feynman's text today.