Shape of the wave of a single photon

[Strange]

How do you get constructive interference.when there is only one wave from one slit?

 

That's the point I'm trying to make. If photons are fundamentally quantised then a single photon can only go through one slit and arrive at one point on the screen.

[Mike Smith Cosmos]

That's the point I'm trying to make. If photons are fundamentally quantised then a single photon can only go through one slit and arrive at one point on the screen.

No , I think if the photon has a wave aspect, which we believe it behaves like a wave AND a particle. Then the wave aspect will mean it has a wave front ( like a wave on water passing through a gap, and In fact a second gap a little distance away). Beyond the two gaps the water waves start anew from the gaps , (both waves).

 

These waves interfere into peaks and troughs.

 

I just walked the dog by the ducks in the canal. I threw in two identical mushrooms. The two waves started off as rings . As they crossed ,hey,ho, interference. Peaks and troughs.

 

So the wave aspect of the photon must at least have a wavefront bigger than the distance between the two slits!

 

Mike

Edited October 19, 2013 by Mike Smith Cosmos

[DParlevliet]

 

That's the point I'm trying to make. If photons are fundamentally quantised then a single photon can only go through one slit and arrive at one point on the screen.

Photons are not fundamentally quantised. They have both waves or particle proporties. This shows in the double slit, so therefore this experiment is always used in QM explanations. But why, and how they interact, nobody knows. How it acts in the double slit with the particle, is a mystery.

That is the Copenhagen interpretation, who also forbids to think about it. It is what the experiment shows and thats all. If you can make a formula that's enough.

 

There is another minority view of Boglie-Bohm, who states that particle and wave are connected and are physicly real. In this explanation the particle goes throught one slit but is pushed by the two interfering waves to where it will be detected.

[swansont]

 

That's the point I'm trying to make. If photons are fundamentally quantised then a single photon can only go through one slit and arrive at one point on the screen.

It is experimentally confirmed that single photons will interfere. It goes through both slits

[swansont]

 

But there are parts of the detector where the photon is never localized. These are the places where two waves from both slits arrive which cancel each other fully. That is only possible if they have opposite sign (180 degree phase shit) and equal amplitude. If not then the cancelling is not complete.

 

The photon is never there, and never going to be there. When you detect it, you detect the whole photon. The wave nature tells you where it can end up, which is the result of the interference, but the detection itself shows no cancellation effects. And the location depends only on the wavelength, as far as the light's properties are concerned.

[DParlevliet]

The wave nature tells you where it can end up, which is the result of the interference, but the detection itself shows no cancellation effects. And the location depends only on the wavelength, as far as the light's properties are concerned.

 

So the place where it can end up (or not) tells you something about the wave. The location depends on the wavelength because of the phase shift between tho waves, and the amplitude of both waves. See the German Wiki formula of the sum of two waves:

If the amplitude of both waves differ (a1, a2) the interference will be different. If phase difference is 180 degree, cos = -1, A = 0 if a1=a2. Then there never a photon will end up. If they differ from each other (like when the amplitude is not constant with time/phase) A will have some value, so some propability of photon absorption.

[swansont]

So the place where it can end up (or not) tells you something about the wave. The location depends on the wavelength because of the phase shift between tho waves, and the amplitude of both waves. See the German Wiki formula of the sum of two waves:

If the amplitude of both waves differ (a1, a2) the interference will be different. If phase difference is 180 degree, cos = -1, A = 0 if a1=a2. Then there never a photon will end up. If they differ from each other (like when the amplitude is not constant with time/phase) A will have some value, so some propability of photon absorption.

Since the photon interferes with itself, how can the amplitude be different? For single photons in general, amplitude has no meaning.

[Mike Smith Cosmos]

Since the photon interferes with itself, how can the amplitude be different? For single photons in general, amplitude has no meaning.

The question I still have dogging me is.

 

Is the frequency of the photon light wave the same as the frequency of the probability location wave or different .?

 

Or are the two waves totally different as one ( light ) is per time reference, whereas probability wave ( de broglie ) is per position/distance reference ?

 

Mike

[DParlevliet]

Since the photon interferes with itself, how can the amplitude be different? For single photons in general, amplitude has no meaning.

 

The photon-wave through one slit interferes with its copy through the other slit, but with a phase = time shift. So if the original photon-wave amplitude changes with time, you will see that in the interference.

Is the frequency of the photon light wave the same as the frequency of the probability location wave or different .?

 

Yes, the same frequency.

[swansont]

The photon-wave through one slit interferes with its copy through the other slit, but with a phase = time shift. So if the original photon-wave amplitude changes with time, you will see that in the interference.

 

Your evidence for this?

Yes, the same frequency.

No. The description of a wave packet cannot be a pure tone, i.e a single frequency, but when it interacts it is.

[DParlevliet]

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....

Your evidence for this?

No. The description of a wave packet cannot be a pure tone, i.e a single frequency, but when it interacts it is.

My first or second sentence?

 

Let's say the base frequency of the fourier

[swansont]

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....

What is "full interference?

 

My first or second sentence?

Both.

[DParlevliet]

What is "full interference?

 

 

Both.

 

That at the black lines no photons arrive, because both waves cancel each other.

 

1: that is the principle of the double slit, it introduces different lentgh path so a phase/time shift. For instance with 2.5 period shift between both waves, they cancel each other. 2.5 period = 2,5E-15 s (~green).

2: if the amplitude is differs in 2,5E-15 s change, there is no full cancel anymore.

[Mike Smith Cosmos]

No. The description of a wave packet cannot be a pure tone, i.e a single frequency, but when it interacts it is.

 

O.K. Are you saying the Wave Packet is or has a frequency ,identical to the fundamental frequency of light sine wave coming from the original light source ( say an electron changing energy level ) BUT the wave packet is shaped in one way or another by a probability shape [ for location purposes] much as i previously illustrated as (B) in my sketches at # 71 .

 

But in addition the wave packets are of pure sine wave [ for time purposes] of the self same frequency but of constant amplitude., BUT nonetheless existing as a succession of wave packets ,as previously illustrated in (A) in my sketches at # 71 .

Edited October 20, 2013 by Mike Smith Cosmos

[swansont]

That at the black lines no photons arrive, because both waves cancel each other.

 

1: that is the principle of the double slit, it introduces different lentgh path so a phase/time shift. For instance with 2.5 period shift between both waves, they cancel each other. 2.5 period = 2,5E-15 s (~green).

2: if the amplitude is differs in 2,5E-15 s change, there is no full cancel anymore.

The photon, where it is detected, is 100% detected. How can there be a partial cancellation?

O.K. Are you saying the Wave Packet is or has a frequency ,identical to the fundamental frequency of light sine wave coming from the original light source ( say an electron changing energy level ) BUT the wave packet is shaped in one way or another by a probability shape [ for location purposes] much as i previously illustrated as (B) in my sketches at # 71 .

 

But in addition the wave packets are of pure sine wave [ for time purposes] of the self same frequency but of constant amplitude., BUT nonetheless existing as a succession of wave packets ,as previously illustrated in (A) in my sketches at # 71 .

No. The wave packet is not and cannot be a pure sine wave, since that must be infinite in extent.

[DParlevliet]

The photon, where it is detected, is 100% detected. How can there be a partial cancellation?

 

It is partial cancellation of the wave, not the photon. The quadrature of the wave is the propability that the photon is absorbed at that prosition. If both waves cancel each other fully this propability is zero (no photons detected). If they cancel partial the propabity becomes higher (some photons detected).

[swansont]

 

It is partial cancellation of the wave, not the photon. The quadrature of the wave is the propability that the photon is absorbed at that prosition. If both waves cancel each other fully this propability is zero (no photons detected). If they cancel partial the propabity becomes higher (some photons detected).

 

How do you definitively show that the pattern is due to an amplitude difference?

[DParlevliet]

Because in the formula above only a1 and a2 has influence on A

[swansont]

Because in the formula above only a1 and a2 has influence on A

 

Only if you ignore the phase difference. In doing that, the formula makes the assumption you are trying to prove. It's a circular argument.

[DParlevliet]

Only if you ignore the phase difference. In doing that, the formula makes the assumption you are trying to prove. It's a circular argument.

 

Cosine gives only the shape, depending on the position on the detector. Suppose the detector position where the phase difference is 450 degree (2.5 period) then Cosine = -1 and if a1=a2 then A=0. So no propability to absorb photons. If a1 and a2 differ at that position, A gets a value above zero, so some propability.

[swansont]

 

Cosine gives only the shape, depending on the position on the detector. Suppose the detector position where the phase difference is 450 degree (2.5 period) then Cosine = -1 and if a1=a2 then A=0. So no propability to absorb photons. If a1 and a2 differ at that position, A gets a value above zero, so some propability.

 

How do you go about proving the signal was due to an amplitude difference and not a phase difference?

[Mike Smith Cosmos]

No. The wave packet is not and cannot be a pure sine wave, since that must be infinite in extent.

O.K. Are you saying the Wave Packet is or has a frequency ,identical to the fundamental frequency of light sine wave coming from the original light source ( say an electron changing energy level ) BUT the wave packet is shaped in one way or another by a probability shape [ for location purposes] much as i previously illustrated as (B) in my sketches at # 71 .

 

But in addition the wave packets are of pure sine wave [ for time purposes] of the self same frequency but of constant amplitude., BUT nonetheless existing as a succession of wave packets ,as previously illustrated in (A) in my sketches at # 71 .

yes , well I sort of understand that , from the point of view of it being chopped, Quantised, so fourier analysis being not single tone. But what I am asking is the bit of sine wave in a packet , is it constant peek amplitude throughout the duration of time the sine wave is existing in an individual wave packet as opposed to some of these diagrams showing a rising peek amplitude going to a peak and falling back to zero amplitude.

 

Those particular type of waves are surely the probability wave that govern the position of the photon in some way or other. This statement has a somewhat tenuous aspect about it as nobody seems to be able to give a shape or size to the photon.

 

1) Is it a "point" particle of zero dimension, and is it the location of this point that is defined by the probability wave away from its average position. ?

 

2) And is the instantaneous amplitude which rises and falls with the intensity of the electric and magnetic fields (E-M field) a DIFFERENT WAVE that is defined by its frequency with respect to time. While the Peek Amplitude of this wave remains constant for the length of time of an individual photons existance ?

 

Yes /No to each question or a different explanation, please?

 

Edited October 21, 2013 by Mike Smith Cosmos

[Strange]

It goes through both slits

I know this is one interpretation. I feel uncomfortable with it because that also implies that a C60 molecule also "goes through both slits" (http://vcq.quantum.at/publications/all-publications/details/247.html - I believe they have also done this with larger molecules now).

 

Isn't an alternative interpretation that it is just due to the non-locality of quantum effects (c.f. entanglement) so that the probability of a photon taking a particular path is determined by the entire experimental setup (however large). This seems to fit better with the sum over histories approach in QED (and the fact that effrects are not localised in time as well; as in the delayed choice quantum eraser experiments).

Edited October 21, 2013 by Strange

[swansont]

I know this is one interpretation. I feel uncomfortable with it because that also implies that a C60 molecule also "goes through both slits" (http://vcq.quantum.at/publications/all-publications/details/247.html - I believe they have also done this with larger molecules now).

 

Isn't an alternative interpretation that it is just due to the non-locality of quantum effects (c.f. entanglement) so that the probability of a photon taking a particular path is determined by the entire experimental setup (however large). This seems to fit better with the sum over histories approach in QED (and the fact that effrects are not localised in time as well; as in the delayed choice quantum eraser experiments).

 

If you want to explain it in terms of probabilities you end up with the same result. When you send one photon at a time through the apparatus, you build up an interference pattern. That pattern goes away if you have "which path" information.

 

 

yes , well I sort of understand that , from the point of view of it being chopped, Quantised, so fourier analysis being not single tone. But what I am asking is the bit of sine wave in a packet , is it constant peek amplitude throughout the duration of time the sine wave is existing in an individual wave packet as opposed to some of these diagrams showing a rising peek amplitude going to a peak and falling back to zero amplitude.

 

Those particular type of waves are surely the probability wave that govern the position of the photon in some way or other. This statement has a somewhat tenuous aspect about it as nobody seems to be able to give a shape or size to the photon.

 

1) Is it a "point" particle of zero dimension, and is it the location of this point that is defined by the probability wave away from its average position. ?

 

2) And is the instantaneous amplitude which rises and falls with the intensity of the electric and magnetic fields (E-M field) a DIFFERENT WAVE that is defined by its frequency with respect to time. While the Peek Amplitude of this wave remains constant for the length of time of an individual photons existance ?

 

Yes /No to each question or a different explanation, please?

 

 

 

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.

[DParlevliet]

How do you go about proving the signal was due to an amplitude difference and not a phase difference?

 

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.