Shape of the wave of a single photon

[Mike Smith Cosmos]

If you will read back, I have repeatedly stated that one cannot know what the photon's actual "shape" is while it's out in free space. All you know is that it satisfies the wave equation. Thus, I have no answers for these questions.

 

From a semantic aspect, I don't see how a peak amplitude can go to zero. That describes a flat line.

I do , and have in principle accepted what you say about the shape of a photon. However I find it odd that something so fundamental in the atomic/quantum arena , as a photon ,has as yet remained ill defined as to its ' physical ' nature. Surely that is part of the significance of the probability wave, pointing to where the photon has or does not have a 'presence'. I would have thought that mapping of the photon would shed some of these mysteries that seem to haunt quantum mechanics. ( things being here , not here , both places at the same time etc etc) .However be that as it may, if that is the case ,so be it.

 

My question 1) however, irrelevant as to the shape of the photon, is surely describing the probability of it existing at a certain position even if it's shape is unknown ! Is that not so ?

 

My question 2) is asking as regards the PEEK amplitude as opposed to its INSTANTANEOUS amplitude which varies throughout the cycle. I am asking does this (peek amplitude) remain constant as per the German Wikipedia quote OR vary PEEKs as per probability wave shape. This is not semantics but a genuine concern of these conflicting quotes. Unless my brain has totally rotted ( which is not out of the question) .

Amplitude was considered as an instantaneous value on the one hand ,that changed continuously with time, and described as a PEEK amplitude for the extremity of an individual swing either positive or negative , high or low , about a mean position which could ,but not necessarily be zero

 

So I hate to say it , these two questions still stand. Maybe they are currently unanswerable, or unknown. But I have heard others ask similar questions before, and they are reasonable things to ask.

 

Mike

Edited October 21, 2013 by Mike Smith Cosmos

[swansont]

 

Phase is fixed function of the geometric of the experiment (distance between slits and detector). The peak and minimum (cosine = +1 or -1) are visible in the cosine shaped interference pattern on the detector.

 

Which, if it's a probability distribution, is exactly what you'd expect.

 

My question 1) however, irrelevant as to the shape of the photon, is surely describing the probability of it existing at a certain position even if it's shape is unknown ! Is that not so ?

How is asking if the photon is zero-dimensional unrelated to its shape?

 

There is this, though: photons can be localized to areas corresponding to their wavelength, e.g. in a waveguide or cavity. If the photon wavelength is too large, the photon cannot exist in the structure. But that's not a free-space condition.

 

 

My question 2) is asking as regards the PEEK amplitude as opposed to its INSTANTANEOUS amplitude which varies throughout the cycle. I am asking does this (peek amplitude) remain constant as per the German Wikipedia quote OR vary PEEKs as per probability wave shape. This is not semantics but a genuine concern of these conflicting quotes. Unless my brain has totally rotted ( which is not out of the question) .

Amplitude was considered as an instantaneous value on the one hand ,that changed continuously with time, and described as a PEEK amplitude for the extremity of an individual swing either positive or negative , high or low , about a mean position which could ,but not necessarily be zero

 

So I hate to say it , these two questions still stand. Maybe they are currently unanswerable, or unknown. But I have heard others ask similar questions before, and they are reasonable things to ask.

 

Mike

You may be misusing amplitude of you are saying it varies throughout the cycle. The amplitude of a wave is the value of the greatest value; for a physical wave it is the height of the peak, wherever that peak is.

 

I didn't say it was unreasonable to ask. My answer, though, is unchanged. AFAIK, you do not have enough information to answer it. To have any hope of doing so, you have to do a measurement, which destroys the photon, and is not keeping with the question of what the photon's shape is when it's not interacting.

[DParlevliet]

 

Which, if it's a probability distribution, is exactly what you'd expect.

 

So at a minimum the cosine is -1, and then the difference between a1 and a2 determines if A is zero or not. The more a1 and a2 differs between the two phases, the higher A will be. So the cosine shape of the interference is always the same, But the hight of this cosine depends on the difference of amplitude between two phases.

Edited October 21, 2013 by DParlevliet

[swansont]

 

So at a minimum the cosine is -1, and then the difference between a1 and a2 determines if A is zero or not. The more a1 and a2 differs between the two phases, the higher A will be. So the cosine shape of the interference is always the same, But the hight of this cosine depends on the difference of amplitude between two phases.

 

So you have been claiming. How do you show this without assuming it first?

[Enthalpy]

Amplification is also absolutely necessary [to a laser], which you don't have with a single atom sitting in a cavity.

True and authentic lasers have been built with one single atom in a cavity, and operated.

Population inversion is then easier to achieve, as it reduces to "the atom is excited".

An emitter in a cavity is a laser. A gain medium is nothing more than an emitter, even a single excited atom. [...]

Concentrated on single-atom lasers, I forgot to add: "an emitter whose population is inverted", which is a condition to lase. My mistake.

[DParlevliet]

So you have been claiming. How do you show this without assuming it first?

 

What do I assume? I fill in the standard formula of interference.

[swansont]

True and authentic lasers have been built with one single atom in a cavity, and operated.

Population inversion is then easier to achieve, as it reduces to "the atom is excited".

 

Concentrated on single-atom lasers, I forgot to add: "an emitter whose population is inverted", which is a condition to lase. My mistake.

Which works by putting excited atoms into the cavity. Not the same atom as the one which emitted the photon. Apples and oranges. In the scenario in question, there is no gain. There is, however, a photon in the cavity, so it's not a half-meter long.

What do I assume? I fill in the standard formula of interference.

And you assume that it is not due to a phase difference, and that the wave is a pure frequency. (Which we know is not true)

[DParlevliet]

-

Edited October 21, 2013 by DParlevliet

[imatfaal]

 

Phase is fixed function of the geometric of the experiment (distance between slits and detector). The peak and minimum (cosine = +1 or -1) are visible in the cosine shaped interference pattern on the detector.

 

But the point of the double slit experiment is that the pattern on the detector is an interference pattern and not the simple expected pattern from a beam of light. Secondly how would the phase manifest in a measurement like this

[Mike Smith Cosmos]

 

There is this, though: photons can be localized to areas corresponding to their wavelength, e.g. in a waveguide or cavity. If the photon wavelength is too large, the photon cannot exist in the structure. But that's not a free-space condition.

 

I didn't say it was unreasonable to ask.

I have been thinking about this continuous wave. I may have been mistaken in thinking this!

I suppose each photon of light is coming from a different atom or at least from a different energy band jump. Thus there is unlikely to be any synchronous jump just as the previous photon has fled. It is more likely to be a flurry of separate photons unconnected with each other. The only hope of a continuous wave would be by any detecting " thing" whatever that is receiving photons sequentially and turning the stream of photons into a resonant form of sine wave. Whether this is what happens I do not know, or whether detection becomes the reverse of transmission ,with the arrival of a non synchronised stream of photons gives rise to an unsynchronised absorbsion of electron changing energy bands.

 

So the idea of a continuous light beam as a pure sine wave is unlikely UNLESS synchronisation does in fact take place at the light source ?

 

Mike

[Strange]

Thus there is unlikely to be any synchronous jump just as the previous photon has fled. It is more likely to be a flurry of separate photons unconnected with each other.

My understanding is that this is normally true. But do the photons produced by a laser have coherent phase?

[swansont]

My understanding is that this is normally true. But do the photons produced by a laser have coherent phase?

 

The coherence is limited; photons will eventually not be coherent with ones emitted earlier.

 

http://en.wikipedia.org/wiki/Coherence_time

http://en.wikipedia.org/wiki/Coherence_length

[Mike Smith Cosmos]

My understanding is that this is normally true. But do the photons produced by a laser have coherent phase?

 

 

 

The coherence is limited;

 

 

 

That is unless the photons carry, or participate in there own medium or work within the Higgs mechanism ( Higgs Field ) Virtual particles clouds ,quantum fluctuations etc etc There are so many Fields existing, including electro-magnetic fields, The idea of a vacuum of nothingness seems to be fading with all the recent discoveries . This would make a HUGH difference as in cases like Water , Air etc. Here, the medium, regulates the waves, so from various different winds/breezes a coherent continuous synchronous wave is generated.

Edited October 22, 2013 by Mike Smith Cosmos

[swansont]

That is unless the photons carry there own medium along with them or work within the Higgs mechanism ( Higgs Field )

 

Which is conjecture and not part of mainstream physics. Accepted physics says no medium and no interaction with the Higgs.

 

The idea of a vacuum of nothingness seems to be fading.

The idea of a vacuum as nothing died years ago.

[Mike Smith Cosmos]

 

 

Which is conjecture and not part of mainstream physics. Accepted physics says no medium and no interaction with the Higgs.

 

 

The idea of a vacuum as nothing died years ago.

 

But surely if vacuum is NOT empty it must be a medium of some sort or another, unless you are saying that E-M waves ( photons) have absolutely NO interaction with any of these fields, particles , bits and pieces. ?

 

Frank Wilczec ( Nobel Prize Winner ) in his book " The Lightness of Being" , seemed to say there was a " Grid " making up this mixture of fields and virtual particles which he indicated was very near to being a medium . surely he is considered fairly mainstream .

Edited October 22, 2013 by Mike Smith Cosmos

[DParlevliet]

But the point of the double slit experiment is that the pattern on the detector is an interference pattern and not the simple expected pattern from a beam of light. Secondly how would the phase manifest in a measurement like this

 

The interference is caused by difference in path length between both waves, which cause difference in phase. The result you see in the interference formula above.

[Enthalpy]

[single-atom laser] Which works by putting excited atoms into the cavity. Not the same atom as the one which emitted the photon. Apples and oranges. In the scenario in question, there is no gain. There is, however, a photon in the cavity, so it's not a half-meter long.

They have been built with YAG, where one single emitting atom stays indefinitely in the cavity, gets excited and emits. The resonance changes the transition lifetime with this single atom. Exactly as in any laser; call it gain if you wish. Slightly more exotic: antiresonating cavities are used to prevent the de-excitation of atoms.

 

By the way, the image of a bunch of photons travelling through the cavity can hold in superradiant mode, but in most continuous-mode lasers, the coherence is hugely longer than the cavity, which means that the photons just stand in the cavity as a standing wave, and the lasing atoms de-excite at random places without any propagation.

 

The photon is as long as the cavity, because of the mirrors. No worry with that!

 

What I strongly disagree is if someone desires - sorry if I misinterpret - a photon size that differs from the size of the wave packet. Until a photon is absorbed, I see no reason nor means to define a photon size other than by the wave packet - there is no hidden variable. The photon can get shorter when it's absorbed, but not before. And the length of the wave packet, as well as the linewidth, agree fully with the lifetime of the atomic transition.

 

We touch something fundamental in QM here: I dare to claim that the photon is not emitted briefly at a random moment whose distribution is defined by the lifetime of the atomic transition, but that the photon is emitted - or in equivalent words, the wave packet lasts - over the whole duration of the transition. Only this is consistent with with the observed linewidth. Then, this photon can be absorbed in a short time or place.

[swansont]

But surely if vacuum is NOT empty it must be a medium of some sort or another, unless you are saying that E-M waves ( photons) have absolutely NO interaction with any of these fields, particles , bits and pieces. ?

 

If you can find a way to measure if we're stationary or moving with respect to the medium (and at what speed) and still have relativity work, be my guest.

 

Frank Wilczec ( Nobel Prize Winner ) in his book " The Lightness of Being" , seemed to say there was a " Grid " making up this mixture of fields and virtual particles which he indicated was very near to being a medium . surely he is considered fairly mainstream .

I would caution anyone reading pop-sci books about the illusion that they are learning science, vs. learning about science. "Very near to being a medium" can mean a few different things, depending on the context.

They have been built with YAG, where one single emitting atom stays indefinitely in the cavity, gets excited and emits.

 

Link? All the ones I know about feed excited atoms into the cavity.

[Strange]

 

But surely if vacuum is NOT empty it must be a medium of some sort or another, unless you are saying that E-M waves ( photons) have absolutely NO interaction with any of these fields, particles , bits and pieces. ?

There is a difference between a "medium" that photons may or may not interact with and a medium that is required for the transmission of EM waves (i.e. the classical aether). There is absolutely no evidence, and no theoretical need, for the latter.

[DParlevliet]

Let me show graphical. The formula for interference amplitude A is (a is the amplitude of the original wave):

 

 

In QM the propability is the quadrature of A.

 

When you look to the interference pattern on the detector (a is the ampltiude of the original wave):

 

So the shape and min/max position are independent of the wave amplitude. But the minimum and maximum value depends on the variation of the amplitude with phase. If a =1 without variation, then the maximum is 4, minimum is zero. If a differs with phase, then the maximum is less then 4 and the minimum is larger then zero.

[swansont]

(a is the ampltiude of the original wave)

We don't have a wave, we have a photon.

So the shape and min/max position are independent of the wave amplitude. But the minimum and maximum value depends on the variation of the amplitude with phase. If a =1 without variation, then the maximum is 4, minimum is zero. If a differs with phase, then the maximum is less then 4 and the minimum is larger then zero.

Which of these scenarios is seen experimentally? If you have a laser diffraction pattern, the signal does, in fact go to zero. Thus, a does not vary.

[DParlevliet]

We don't have a wave, we have a photon.

 

Which of these scenarios is seen experimentally? If you have a laser diffraction pattern, the signal does, in fact go to zero. Thus, a does not vary.

 

And a photon is a wave(packet), otherwise there is interference

 

If it is always goes to zero (in the double slit experiment), then always a1=a2, so a single photon is/has a wave which ampltitude does not change with phase/time.

Edited October 23, 2013 by DParlevliet

[swansont]

 

If it is always goes to zero (in the double slit experiment), then always a1=a2, so a single photon is/has a wave which ampltitude does not change with phase/time.

 

I'm glad you finally agree.

[DParlevliet]

So I suppose now we agree that a photon is/has a wave with constant amplitude, before and after. As Mike told, that is already right after it is emitted.

And when looking to another result:

Now another result of the double-slit. Suppose the distance between the slits is one kilometre, Then there will be interference too. Full interference is only possible when the waves through both slits have the same amplitude. So the amplitude of (one) photon wave is equal at both slits, at a kilometre distance....

 

then the conclusion is that a photon is/has a wave which on large distances (in extreme: in the whole universe) is a wave with constant amplitude.

Edited October 23, 2013 by DParlevliet

[swansont]

We already know that E = hv, so a constant amplitude should not come as a surprise. Classically, the amplitude (field strength) is related to the energy.