Self Same Photon

[Guest]

I'm a beginner. One of the reasons I bought a telescope was because I wanted to see galaxies with my own actual eyes. I won't explain why but one thing that occurred to me was this. Suppose I'm looking at the M33 galaxy through my telescope. Photons are landing on my retina and forming an image. Now M33 is about 3 million light years away so lets take one of those photons. Is it a sensible question to ask if that photon is the actual self same photon which left M33 galaxy 3 million years ago, travelled across billions of miles of empty space, through our atmosphere, reflecting off the mirror in my telescope, through the glass in the eyepiece and through my eyeball before reaching my retina? That would be cool if it were true. 

I've asked this question elsewhere and got different answers.

Cheers

Steve

 

[Starlight 1]

I see it the same way as you good post Steve.

[johnrt]

1 hour ago, woodblock said:

I'm a beginner. One of the reasons I bought a telescope was because I wanted to see galaxies with my own actual eyes. I won't explain why but one thing that occurred to me was this. Suppose I'm looking at the M33 galaxy through my telescope. Photons are landing on my retina and forming an image. Now M33 is about 3 million light years away so lets take one of those photons. Is it a sensible question to ask if that photon is the actual self same photon which left M33 galaxy 3 million years ago, travelled across billions of miles of empty space, through our atmosphere, reflecting off the mirror in my telescope, through the glass in the eyepiece and through my eyeball before reaching my retina? That would be cool if it were true. 

I've asked this question elsewhere and got different answers.

Cheers

Steve

 

I think the general consensus is no, the mirror in your reflector absorbs the photon and re-emits a new one. Sorry! But you are still observing M33 as it was 3 million years ago, who cares if it's exaclly the same photon?

[MARS1960]

42 minutes ago, johnrt said:

I think the general consensus is no, the mirror in your reflector absorbs the photon and re-emits a new one. Sorry! But you are still observing M33 as it was 3 million years ago, who cares if it's exaclly the same photon?

Credit for this explanation goes to Chad Orzel.

 

This is a tricky question to answer, because in many ways it doesn't make sense to talk about a definite path followed by a single photon. Quantum mechanics is inherently probabilistic, so all we can really talk about are the probabilities of various outcomes over many repeated experiments with identically prepared initial states. All we can measure is something like the average travel time for a large number of single photons passing through a block of glass one after the other.

The transmission of light through a medium is easier to explain in a classical sense, where you think of the light beam as a wave that drives oscillations in the atomic dipoles making up the material. Each atom then re-radiates its own waves at the same frequency, but slightly out of phase. The sum of the initial wave and the re-radiated wave is a wave that lags behind the incoming wave a little bit, which explains the reduced speed. A beam of light entering a block of material tends to continue in the same direction because forward scattered light from any individual atom tends to interfere constructively with forward scattered light from other atoms in the material, while light scattered off to the sides mostly interferes destructively, and cancels out.

Carrying this picture over to the quantum regime, you would say that a single photon entering the material will potentially be absorbed and re-emitted by each of the atoms making up the first layer of the material. Since we cannot directly measure which atom did the absorbing, though, we treat the situation mathematically as a superposition of all the possible outcomes, namely, each of the atoms absorbing then re-emitting the photon. Then, when we come to the next layer of the material, we first need to add up all the wavefunctions corresponding to all the possible absorptions and re-emissions, and when we do that, we find that just as in the classical wave case, the most likely result is for the photon to continue on in the same direction it was originally headed. Then we repeat the process for all the atoms in the second layer, and the third, and so on.

At any given layer, the probability of being absorbed then re-emitted by any individual atom is pretty tiny, but there are vast numbers of atoms in a typical solid, so the odds are that the photon will be absorbed and re-emitted during the passage through the glass are very good. Thus, on average, the photon will be delayed relative to one that passes through an equal length of vacuum, giving rise to the lower observed transmission velocity.

Of course, it's not possible to observe the exact path taken by any photon-- that is, which specific atoms it scattered from-- and if we attempted to make such a measurement, it would change the path of the photon to such a degree as to be completely useless. Thus, when we talk about the transmission of a single photon through a refractive material, we assign the photon a velocity that is the average velocity determined from many realizations of the single photon experiment, and go from there.

[ollypenrice]

Yes, it's great, isn't it! The cure for thinking of photons as billiard balls is to read QED, a slim but alarming volume by the mighty Richard Feynman. 

http://www.amazon.com/QED-Strange-Theory-Light-Matter/dp/0691024170

Looking at a school ray diagram of a stream of billiard ball photons striking a mirror at 45 degrees and bouncing off again in pious good order at 45 degrees would leave you thinking that the only part of the mirror that was doing anything was the bit they bounced off.  How wrong you'd be... 

Olly

[MARS1960]

34 minutes ago, ollypenrice said:

Yes, it's great, isn't it! The cure for thinking of photons as billiard balls is to read QED, a slim but alarming volume by the mighty Richard Feynman. 

http://www.amazon.com/QED-Strange-Theory-Light-Matter/dp/0691024170

Looking at a school ray diagram of a stream of billiard ball photons striking a mirror at 45 degrees and bouncing off again in pious good order at 45 degrees would leave you thinking that the only part of the mirror that was doing anything was the bit they bounced off.  How wrong you'd be... 

Olly

It sure is extremely interesting.

At the price of the paperback i think i just might have a read, if i can make any sense of it will be a bonus .

[Guest]

My physics is rather shaky. Some things other people have said are -

A photon will only be absorbed if there is a valid transition in the atom other wise it will pass straight through.

If a photon is absorbed and another one emitted then it would have to be in the same direction and at the same wavelength otherwise telescopes and spectrographs wouldn't work.

Under normal circumstances when a photon is absorbed it is because there is a valid transition available and that is what gives rise to absorption lines in the spectra so we know what kind of gas it has passed through.

 

[cloudsweeper]

9 minutes ago, woodblock said:

My physics is rather shaky. Some things other people have said are -

A photon will only be absorbed if there is a valid transition in the atom other wise it will pass straight through.

If a photon is absorbed and another one emitted then it would have to be in the same direction and at the same wavelength otherwise telescopes and spectrographs wouldn't work.

Under normal circumstances when a photon is absorbed it is because there is a valid transition available and that is what gives rise to absorption lines in the spectra so we know what kind of gas it has passed through.

 

So many things going on here!  I'm not knowledgeable enough to give a definitive answer, but yes, for absorption and re-emission an electron needs to be excited to a higher level.  When it relaxes back, it emits a photon.  I doubt it would be in the same direction as the original incident photon.  When light passes through a gaseous medium, you pick up absorption lines in the original path through that medium.  But in other directions, you would see emission lines.  And I don't think these processes take place when light strikes glass.

But is there not a "simpler" kind of absorption going on as well?  The sort when a photon ceases to be, and its energy is transferred to a minuscule amount of thermal energy.  Is this not what happens when light falls in intensity on passing through glass for example?  And is it not carried out through absorption by atoms/molecules (increasing their thermal agitation), as opposed to absorption by orbiting electrons?

Do photons constantly change as a wavefront advances?  I confess not to have heard of that mechanism.

Within my limited depth of learning, I would therefore say that excluding photons that actually have been emitted further down the line on the way to your eye, there is a fair chance that you would intercept a fair few of the original ones.

And just to complicate matters, remember that a photon is anyway only a model we use to try to understand a phenomenon.  E/M radiation is a wave/particle dual according to QM, and how it behaves is determined by how we observe or detect it.  So heaven only knows what it is you are seeing!

(Please feel free to correct any of this!  As I said, I am far from being an expert on these matters!)

Doug.

 

 

[cgarry]

This is a good view on the subject of refraction, though not the original subject of reflection:

 

 

[ollypenrice]

To the OP I'd just say that the infomrmation comes directly from the source. That should keep everyone happy!

Olly

[Guest]

Ask a simple question eh?

[andrew s]

48 minutes ago, woodblock said:

Ask a simple question eh?

I am sure that if the OP had asked the question on PhysicsForums.com he would have been told that it was meaningless! The response would have been along the lines " Within Quantum Field Theory at "M33" an excitation would have been created/added to the EM field and at his "retina" one one would have need destroyed/removed. QFT has nothing to say about what goes on between these two events. Classical EM theory provides a framework describing the propagation of light between M33 and the eye but photons are not part of the classical theory"

I much prefer the answers here but unfortunately QFT or even old style quantum theory is beyond simple visualisation/description:- the answer is in the mathematical formalism. In modern QFT particle/wave duality is out and photons are just a name for the field excitation's. Sad but true.

Regards Andrew

[Stu]

1 hour ago, andrew s said:

I am sure that if the OP had asked the question on PhysicsForums.com he would have been told that it was meaningless! The response would have been along the lines " Within Quantum Field Theory at "M33" an excitation would have been created/added to the EM field and at his "retina" one one would have need destroyed/removed. QFT has nothing to say about what goes on between these two events. Classical EM theory provides a framework describing the propagation of light between M33 and the eye but photons are not part of the classical theory"

I much prefer the answers here but unfortunately QFT or even old style quantum theory is beyond simple visualisation/description:- the answer is in the mathematical formalism. In modern QFT particle/wave duality is out and photons are just a name for the field excitation's. Sad but true.

Regards Andrew

Did anyone hear the 'whooshing' sound as Andrew's post shot straight over my head? ??

Great thread; excellent posts by the OP and everyone else. I shall retire to let my poor brain try to absorb some of this ??

[Paul M]

I'm deflated and exhausted.

I thought "Oh, what a charming question" by the OP. I used to ponder the exact same question. As a boy I shone a torch into my eye in a dark room while looking in a mirror (my stomach is truly churning considering that in conjunction with THIS) to see if I could see the photons going in and hitting my retina. Daft I know but we were all young once..

So I really thought I knew the answer to this one....

Strangely, I've read Feynman's QED too. Must have gone further over my head than I realised.

Who among us would have bothered ever buying a telescope if they knew all they'd be seeing was some weird virtual reality stuff!

 

 

[andrew s]

1 minute ago, Paul M said:

 

Who among us would have bothered ever buying a telescope if they knew all they'd be seeing was some weird virtual reality stuff!

 

 

We all would. It is not "some weird virtual reality" just a very weird reality!!! Just because the physics is weird it is still wonderful as is the pleasure of observing the heavens.

Regards Andrew

[PhotoGav]

Very interesting stuff. I have just returned from a lecture entitled 'Particle Physics In A Nutshell', given by a local physicist who worked at CERN. After the event I asked her this question having seen it here this morning and having not had a clue about the answer. She quickly and simply replied 'yes', and went on to add 'that's why you can measure redshift'. Who was I to argue! So, from the closest thing I have to an expert, that photon that left M33 around 3 million years ago, along with all its pals, really has travelled that huge distance just to land on your retina and be interpreted by your brain, probably as a faint fuzzy smudge. I think in this instance I can be allowed to use the word AWESOME! My question is this - what happens to that photon once it has been detected on your retina?

[Starlight 1]

13 minutes ago, PhotoGav said:

Very interesting stuff. I have just returned from a lecture entitled 'Particle Physics In A Nutshell', given by a local physicist who worked at CERN. After the event I asked her this question having seen it here this morning and having not had a clue about the answer. She quickly and simply replied 'yes', and went on to add 'that's why you can measure redshift'. Who was I to argue! So, from the closest thing I have to an expert, that photon that left M33 around 3 million years ago, along with all its pals, really has travelled that huge distance just to land on your retina and be interpreted by your brain, probably as a faint fuzzy smudge. I think in this instance I can be allowed to use the word AWESOME! My question is this - what happens to that photon once it has been detected on your retina?

Sound good I was reading about redshift today trying to fine a answer.

[Guest]

Thanks everyone. Thrilling isn't it?

I just ordered my copy of QED.

Cheers

Steve

 

[cloudsweeper]

@PhotoGav - re redshift:

In my earlier post I mentioned absorption lines.  These occur at fixed wavelengths in the visible spectrum, depending on the gaseous medium the incident light has passed through.

They show up as dark lines, i.e. thin strips where there is an absence of light.

Now think of a vehicle with a siren blaring, going past you then speeding away - the frequency of the siren drops.  To put it another way, wavelength increases.  This is the Doppler Effect.  When light from a galaxy is analysed, the absorption lines observed are seen not to be where they are expected, but shifted to the red end of the spectrum (ROYGBIV, remember), hence redshift, since the red end is the lower frequency end.

The mind-blowing thing about this is not that the galaxies are zooming away at speeds proportional to their distances (Hubble's Law), but it is the space itself that is expanding.

Quote:  My question is this - what happens to that photon once it has been detected on your retina?

I believe this is just a case of energy conversion where the photon ceases to be and its energy is converted to a tiny amount of heat/electrical impulse in nerves or whatever (I don't know any biology!).  It doesn't help to hang onto the notion of a particle (like a tiny billiard ball) - it would be hard to imagine a billiard ball disappearing when it struck a pocket.  The photon is better thought of as a quantum (or bundle) of energy.  

(Again, I am prepared to be corrected on any of this.  Fire away - it's all part of learning .)

BTW, for another fascinating topic, how about gravity not being a force, but rather a distortion of spacetime......

 

[Physicist13]

This also touches on the "measurement problem"..to say "the some photon" it would need to be "tagged in some way" then one meas made early on and anothor made much later..but act of detecting it will change its state...so no longer the same one   ...quantum non-demolition meas wld be needed... its the info that matters. 

[Guest]

I did an Open University module on Quantum Mechanics, much of which I have forgotten.  I hope I've got this right. The photon exists in what's called a superposition of quantum states. All we can calculate is the probabilities of it's being in each of these states. When we make a measurement the photon quantum wave function collapses into one of those quantum states according to the various probabilities. You don't know which one it will collapse into but if you take lots of photons in the same superposition of quantum states and measure them the numbers you get will correspond to the probabilities.

This harks back to the thing called entanglement. If you have an experiment where two photons are created and they are described by a single wave function. The wave function will say that if one is polarised one way then the other must be polarised the other way. According to entanglement, at the instant you measure the polarisation of one photon then the other photon will automatically have the opposite polarity. And the really weird thing is that it doesn't matter how far apart they are and this implies that it happens faster than the speed of light. In fact it happens instantly even if the two photons are at opposite sides of the galaxy.

There are two interpretations. The first one is that the opposite polarisation occurs at the moment of creation. So at the moment of creation photon A is polarised one way and photon B is polarised the other. Then all you are doing when you measure the polarisation of photon A is that you are just measuring it.

The quantum explanation is what I said earlier. The decision about polarisation of photon A occurs at the moment of measurement and it's wave function collapses at that moment. Because the two photons are tied together by the same wave function it means that photon B's polarisation is determined at that moment as well - instantaneously.

The physicist Steve Bell worked out a way of proving which was the true interpretation and it is a method based on probabilities so you have to repeat the experiment millions of time. In fact the experiment has been carried out and the quantum explanation is the winner.

Entanglement is practically unbelievable.  I hope I got all that right.

Cheers

Steve

 

[andrew s]

17 hours ago, PhotoGav said:

Very interesting stuff. I have just returned from a lecture entitled 'Particle Physics In A Nutshell', given by a local physicist who worked at CERN. After the event I asked her this question having seen it here this morning and having not had a clue about the answer. She quickly and simply replied 'yes', and went on to add 'that's why you can measure redshift'. Who was I to argue! So, from the closest thing I have to an expert, that photon that left M33 around 3 million years ago, along with all its pals, really has travelled that huge distance just to land on your retina and be interpreted by your brain, probably as a faint fuzzy smudge. I think in this instance I can be allowed to use the word AWESOME! My question is this - what happens to that photon once it has been detected on your retina?

Well I will argue as a simple "yes" goes against all modern understanding of QFT although understandable in a popular science context. There are other issues as GR used to explain the cosmological red shift is a classical theory and has no accepted quantum theory and the redshift Doppler effect is treated by classical electromagnatism not QFT. The problems and difficulties come when you mix elements from classical and quantum theory.

Regards Andrew

[Tiki]

21 hours ago, andrew s said:

I much prefer the answers here but unfortunately QFT or even old style quantum theory is beyond simple visualisation/description:- the answer is in the mathematical formalism. In modern QFT particle/wave duality is out and photons are just a name for the field excitation's. Sad but true.

Therefore it doesn't make sense to talk of an individual photon since no-one knows what (or where) they are.

 

On ‎2‎/‎9‎/‎2016 at 10:42, woodblock said:

 Is it a sensible question to ask if that photon is.....

Yes it is a sensible question , but the problems start to arise as soon as you try and specify 'which photon'.