# Does a photon travel in a straight line?

1. Nov 24, 2009

### DavidSullivan

Noob question: if an atom in a vacuum releases a photon does it have a direction? Does it propagate in a straight line?

My answer was always "of course!" until I read Feynman's QED and now I'm confused. What I'm trying to get my head around is the idea that light "travels all possible paths probabilistically." (My words). If that's the case, shouldn't a photon propagate out from the atom in all directions as an expanding sphere?

My Google skills seem to be failing me, so if someone can point my to anything on the Web regarding this idea I'd appreciate it.

Oh, you might have guessed I am not a physicist, so anything beyond Calc III is going to be over my head.

Thanks, David

2. Nov 24, 2009

### RUTA

Feynman was probably talking about his path integral formalism. In that approach to quantum field theory you integrate from initial state A to final state B over all possible paths between them to obtain the transition amplitude from which you get the probability for A --> B. Do a search for "Feynman path integral."

3. Nov 24, 2009

### Neo_Anderson

You're somewhat correct: the photon will propagate outwards as an expanding sphere at speed c. Once you look at it, however, its wavefunction collapses, causing a definite position and detection of the photon particle.

4. Nov 25, 2009

### Fredrik

Staff Emeritus
RUTA is right. Note that Feynman doesn't actually say that the photon takes all possible paths. He's just saying that every path from the emission event A to the detection event B contributes to the amplitude for "detection at B given emission at A". QM doesn't tell you what "actually happens". It just tells you how to calculate probabilities of possibilities.

Yes, there are "interpretations" of QM that at least try to specify what they think QM is telling us about what "actually happens", and what Neo is saying is consistent with the one that for some unfathomable reason has become known as "the Copenhagen interpretation". It isn't very popular around here. There are lots of threads about interpretations, so you can check out some of those if you want to know more. There's been several just in the last few weeks.

Last edited: Nov 25, 2009
5. Nov 27, 2009

### frustr8photon

good answers. Frerik I thought Copenhagen interpretation was pretty standard and I suppose you would say "popular"...where do you mean by not popular "around here"? just curious.

6. Nov 27, 2009

### DrChinese

I think of Copenhagen as being the original QM formalism. The Copenhagen interpretation sometimes goes by other names, although there are those who claim that standard QM is not the same thing. There are a lot of vocal proponents of various interpretations around here - meaning folks who comment on this forum.

Those that tend to focus on the formalism tend to be quieter (since the interpretations all follow the formalism in terms of predictions, and science currently has no way to otherwise distinguish). So you may see a non-representative sample of opinions.

7. Nov 27, 2009

### Fredrik

Staff Emeritus
There's no universally accepted definition of the CI. Most people here seem to think that it says that QM doesn't apply to measuring devices, and that wavefunction collapse is a mysterious physical process that replaces a superposition with a state that's exactly an eigenstate when a measurement is performed. This version of the CI is just ridiculous.
See my posts #4, #6 and #11 in this thread for more.

The really short version is that I think the definition above is a huge misunderstanding of the original ideas, and that the currently popular "ensemble interpretation" is very close to what the Copenhagen guys had in mind back then. So maybe we should call that "the Copenhagen interpretation". Nah...it's too late for that. It would be like trying to change the word "gay" back to its original meaning.

Last edited: Nov 27, 2009
8. Nov 28, 2009

### DavidSullivan

This is what I thought, so I'm feeling a little better. It's curious to see graphics of a photon as a squiggly little line still used so frequently. Easier than an expending sphere of energy I guess.

Thanks all for the thoughtful answers, and I'm going to stay out of the way regarding the Copenhagen Interpretation. :-)

-David

9. Nov 28, 2009

### lightarrow

and what happens if I measure the atom's recoil's momentum?

10. Nov 29, 2009

### dmtr

It doesn't matter how do you perform the measurement. What matters is the resulting information that you have obtained. If you've obtained the exact same information the wave function would collapse for you in the exact same way.

Last edited: Nov 29, 2009
11. Nov 29, 2009

### lightarrow

And what happens to the still moving away spherical photon?

12. Nov 29, 2009

### RUTA

Once you add a means by which the information CAN be obtained (doesn't necessarily mean someone actually HAS obtained it), the wave function is not spherical anymore. But, if you subsequently add a component to the experiment to "erase" that information, the wave function is restored -- it's called quantum eraser.

http://www.scientificamerican.com/media/inline/DD39218F-E7F2-99DF-39D45DA3DD2602A1_p95.gif

See the thread "Interpretations of quantum eraser experiment"

13. Nov 29, 2009

### apeiron

Can we talk about the propagating sphere having both a postive and a negative aspect. So at some point, with collapse, there is the location of the event. And at the same time, the collapse is creating the definite absence of an event across the rest of that lightcone, the rest of that sphere surface. And the sphere's interior I guess.

Obviously in modelling, we like to keep things simple by representing only the positive event and leaving out all the non-events.

But when it comes to interpretation of the formalism, is the fact that collapse leads to one definite photon path, and an accompanying multitude of all the paths that now definitely never happened, an important issue?

14. Nov 29, 2009

### edpell

You ask does the photon go in a straight line? I would restrict the question to if an atom at point A emits a photon and it is measured to arrive at point B did it take a straight line path between A and B? I would say that Feynman is correct to calculate the probability you much integrate over all possible paths including even the none straight paths. Now the probability of the odd ball paths (not straight, the further from straight the more odd ball) is small and get smaller as the path deviates further from the straight line path. So no not equal in all directions.

15. Nov 29, 2009

### ajw1

When I place a detector west from the photon source and I don't get a hit, should I conclude the wavefunction is now only propagating to N, E, S, Up and Down (so is partly collapsed)?

16. Nov 29, 2009

### edpell

I have never heard anyone use the idea of a null measurement to partially collapse a wavefunction. I like the idea. Sound reasonable to me.

17. Nov 29, 2009

### DrChinese

I believe that is akin to a weak measurement. In weak measurements, the state is constrained in varying degrees.

http://arxiv.org/abs/0911.0071

18. Nov 29, 2009

### enotstrebor

Thats because focus on the formalism avoids the problem of the multiplicity of interpretations especially since all equally valid.

Especially since there are so many interpretations.

Hm, let me see, if a model is a model of the particle object (one thing) shouldn't the model have one interpretation ?

Maybe the particle not only morphs between a particle and a wave (the wave particle duality problem) but morphs between a Copenhagen particle, a Many Worlds particle, etc.

So much confusion and so little time.

QM is a mathematical model of the interaction behavior (Q) of particles (P).

Copenhagen is an interpretation of a mathematical model that assumes the mathematical model of the interaction of the particle represents the particle itself rather than simply a model of the particle's behavior.

Although P (the particle) must imply Q (the behavior) the reverse is not required.

One does not actually start with a model of the photon and a model of matter and result in the photon particle interaction behavior.

All present models are inter-particle interaction models, i.e Q models not P models.

This is at the heart of all of the problems, like renormalization. Though it is a mathematical accomplishment, no one seems to ask, does the particle perform the mathematics of renormalization? If the particle can not, can the model be a model of the particle?

19. Nov 29, 2009

### eaglelake

I will first answer your questions and then try to explain "what quantum mechanics is all about".

Photons do not travel in straight lines. Quantum particles do not have trajectories as classical particles do.

Photons do not propagate in all directions as a spherical wave does. Photons are not waves.

Photons are always detected as particles. A photon propagates through space-time and interacts with detectors. It is not a wave. The quantum wave function associated with the experiment is a mathematical construct that allows us to calculate probabilities. It is defined in what mathematicians call a Hilbert space. It does not exist in space-time, as the photon does, and it does not interact with particle detectors. Quantum mechanics works just fine without a collapsing wave function.

Repeating the same experiment many times reveals the probability distribution of all possible results. Oftentimes, this probability distribution is reminiscent of interference patterns seen in wave optics.

The experiment doesn't tell us anything about how the photon gets from the atom to the detector. Quantum theory doesn't tell us anything either. In fact, we don't know anything about the photon until it is detected. If we assume the photon has a trajectory, then our theoretical calculations do not agree with experimental results. (Actually, if a particle has a trajectory, then it is correctly described by Classical Mechanics and there is no need for Quantum Mechanics.)

Any attempt to describe the behavior of the photon before it is detected is pure speculation and cannot be verified in the experiment.

If you are unhappy with this answer, then you are in good company. Einstein didn't like it either!

20. Nov 29, 2009

### dmtr

Yes. You are right. I stand corrected. It doesn't matter wherever you have the information or not. What matters is that the information is not in the wave function any more. For example dumping the 'which path' information to the environment would destroy the interference pattern as effectively as the measurement and observation of the 'which path' data.