LASER - diameter

[quiet]

Suppose it were possible to build a laser device capable of fulfilling the
following condition.

In each period  [math]T[/math] emits only one photon.

[math]T=\dfrac{1}{\nu}[/math]

[math]\nu \ \rightarrow[/math] frequency

Question:

Does the theory allow to calculate the diameter of the emitted beam?

Edited December 8, 2018 by quiet

[Strange]

Surely the diameter depends on all sorts of factors such as they type of laser, it’s size, the collimation of the beam, how far away you measure it. 

I’m not sure that the pulse rate would change this. 

[quiet]

1 hour ago, Strange said:

I’m not sure that the pulse rate would change this. 

Hello, Strange. Thanks for responding. I understand what you mentioned. The idea is to advance a little more towards the fundamentals of the theory of light, which includes everything known about light, regardless of the design of the emitting device. Then I will give another form to the question.

Do the theoretical foundations specify a lower limit for the diameter? That is to say that for a given frequency, the beam with a photon in each cycle must have a diameter that does not exceed a certain measure.

[John Cuthber]

1 hour ago, quiet said:

That is to say that for a given frequency, the beam with a photon in each cycle must have a diameter that does not exceed a certain measure.

 

No

Because you can always put the beam through a beam expander.

 

There may be a minimum diameter due to diffraction effects, below which the beam direction is too poorly defined to be a beam.

[quiet]

43 minutes ago, John Cuthber said:

No

Because you can always put the beam through a beam expander.

There may be a minimum diameter due to diffraction effects, below which the beam direction is too poorly defined to be a beam.

Hi John. Thanks for answering.

Is that answer referring to the spread in a vacuum?

[John Cuthber]

Diffraction means that a "beam" will spread out whatever the medium is.

Forcing the light through a small hole makes the beam spread more.

[quiet]

Help me to understand. There can be diffraction without something material that causes the diffraction?

[Strange]

The value of the diameter depends how you define it: https://en.wikipedia.org/wiki/Beam_diameter#Width_definitions

[quiet]

44 minutes ago, Strange said:

The value of the diameter ...

I can not understand what relationship that wikipedia page has with a laser that propagates in a vacuum with only one photon per cycle.

Edited December 8, 2018 by quiet

[Strange]

3 minutes ago, quiet said:

I can not understand what relationship that wikipedia page has with a laser that propagates in a vacuum with only one photon per cycle.

Why would that make any difference? You still have to define how you are measuring the width. You would obviously need many of your single photon pulses to make the measurement.

Are you asking if a photon has a size?

[quiet]

My initial question, which I have reiterated later, is not about measuring something in practice. It is referred to what the fundamental physical laws imply regarding a beam in a vacuum that has only one photon per cycle.

[Strange]

7 minutes ago, quiet said:

My initial question, which I have reiterated later, is not about measuring something in practice. It is referred to what the fundamental physical laws imply regarding a beam in a vacuum that has only one photon per cycle.

But whether you are measuring or calculating the diameter, you still need to define what it is that you are measuring/calculating. The beam does not have a sharp edge so there are multiple ways of saying what the diameter is. Your calculation (if it were possible) would need to match that definition.

What difference do you expect pulses of single photons to make, compared to a normal laser.

[quiet]

6 hours ago, Strange said:

What difference do you expect pulses of single photons to make, compared to a normal laser.

I hope the following.

1. That the beam propagate in a vacuum without suffering diffraction.

2. From what is expressed in item 1, I hope that the beam diameter remains constant when it propagates in a vacuum.

3. For the two previous expectations, I hope that each photon has a finite diameter determined by fundamental electrodynamic laws, in the version of the corresponding electrodynamics, classical or quantum.

4. For the three previous expectations, I hope that the photon in the vacuum, in the far zone with respect to the emitting device, appears as a cylindrical entity whose length measures a wavelength, with a finite radius equal to the wavelength divided by a geometric constant. What geometric constant? Nothing strange. Maybe the radius equals the wavelength divided by [math]4 \ \pi[/math].

Edited December 9, 2018 by quiet

[Strange]

6 hours ago, quiet said:

2. From what is expressed in item 1, I hope that the beam diameter remains constant when it propagates in a vacuum.

Even a laser will diverge. For example, the laser used for measuring the distance to the moon has a diameter of about 6km when it reaches the moon.

The relationship between diameter, wavelength and distance is given by: [math]w^2(z)=w_0^2 \left(1+\left(\frac{z \lambda}{\pi w_0^2}  \right )^2\right)[/math]

Quote

3. For the two previous expectations, I hope that each photon has a finite diameter determined by fundamental electrodynamic laws, in the version of the corresponding electrodynamics, classical or quantum.

The "size" of a photon depends what you doing. They are often treated as point particles. Or you can use the interaction cross section, but that depends what they are interacting with.

6 hours ago, quiet said:

4. For the three previous expectations, I hope that the photon in the vacuum, in the far zone with respect to the emitting device, appears as a cylindrical entity whose length measures a wavelength, with a finite radius equal to the wavelength divided by a geometric constant. What geometric constant? Nothing strange. Maybe the radius equals the wavelength divided by 4 π .

You seem to be thinking of photons as classical objects, rather than having a probability distribution.

[quiet]

5 hours ago, Strange said:

Even a laser will diverge. For example, the laser used for measuring the distance to the moon has a diameter of about 6km when it reaches the moon.

Between the laser aimed at the Moon and the theoretical case that I have raised there is an essential and critical difference. The cross section of the laser aimed at the Moon encompasses [math]n[/math] photons. In the theoretical case that interests me, the cross section covers only one photon. To facilitate verbal expression let's symbolize R a part of the cross section. Which part ? The corresponding part, say, at 99% probability of finding the photon. Are you claiming that R widens while the photon travels, even though there is only one photon in the cross section?

 

5 hours ago, Strange said:

The relationship between diameter, wavelength and distance is given by: [math]w^2(z)=w_0^2 \left(1+\left(\frac{z \lambda}{\pi w_0^2}  \right )^2\right)[/math]

The equation seems typically Pythagorean, as the vector sum of two mutually perpendicular components, one in the direction of propagation and another transversal. If the photons belonging to the same section are in phase, I begin to think that there is a slight repulsion between them, responsible for the transversal component in the vectorial sum. Magnetic repulsion? Electric repulsion? Both? The widening of the beam seems an interesting subject, if there is a transverse repulsion. If the propagation of the photon does not have the least relation with the polarization of the vacuum, why would the photons repel?
 

5 hours ago, Strange said:

You seem to be thinking of photons as classical objects, rather than having a probability distribution.

To formulate the spatiotemporal probability distribution of finding the photon, you first need the photon to exist. No matter how hard I try to believe that the photon is a propagating probability, that is, to believe that the photon is made of something called probability, I find it inconceivable.

So I think it is made of electromagnetic field, that the constitutive field of the photon occupies a finite volume related to the wavelength and that, without the polarization of the vacuum in the region where the photon is present, the photon could not exist and could not spread.

I prefer to believe that the fields [math]\vec{E}[/math] and [math]\vec{H}[/math] they are both consequences of the same cause. What cause? A polarization wave that propagates in a vacuum. The polarization wave does not imply movement of electric charges. Simply, at each vacuum point the charge density varies locally as a function of time. If the wave is sinusoidal, the charge density at the point varies sinusoidally.

In the computer you can see the animation of a wave that spreads. Your view informs that something is moving in the direction of propagation, but no luminance cell on the monitor moves. Simply the brightness of each cell varies sinusoidal with time, alternating between two colors, one for the negative half cycle and another for the positive half cycle. If instead of thinking of two colors you think of two signs of linked charge, no charge moves but the effect is an electromagnetic wave that propagates in a vacuum.

To formulate probabilities you need first that the photon exists. And nothing forbids its existence to be based on a pair of linked, equal and opposite charges, which appear to be in motion without there actually being movement of charges. The movement of energy, the linear moment and the spin are real. The movements of charges and fields are apparent. Thus I conceive the propagation in the vacuum and the existence of the photon. Then yes, an existing photon can meet probabilistic conditions.

In that Pythagorean type equation, transverse repulsion would be expected if the photons are in phase because, in this case, the constituent charges of a photon could be slightly repelled with the constituent charges of the neighboring photons. Why would they repel each other? In absolute vacuum the propagation is perfectly straight. Where there is a gravitational field, the trajectory is curved. If the gravitational field is not very intense, the curvature is slight. Then, the imbalance caused by the curvature between electrical repulsion and magnetic attraction is slight. This can give as a result a slight repulsion between neighboring photons, which causes the thickening of the beam as it propagates. 

[Strange]

7 minutes ago, quiet said:

Between the laser aimed at the Moon and the theoretical case that I have raised there is an essential and critical difference. The cross section of the laser aimed at the Moon encompasses n photons. In the theoretical case that interests me, the cross section covers only one photon. To facilitate verbal expression let's symbolize R a part of the cross section. Which part ? The corresponding part, say, at 99% probability of finding the photon. Are you claiming that R widens while the photon travels, even though there is only one photon in the cross section?

A photon will have a probability of being somewhere within that radius. The probability will be highest near the centre and almost zero outside the beam. 

7 minutes ago, quiet said:

No matter how hard I try to believe that the photon is a propagating probability, that is, to believe that the photon is made of something called probability, I find it inconceivable.

If it is "made" of anything, I guess that would be electromagnetism. (But that is a pretty meaningless concept.) It is a quantum of the electromagnetic field.

It is not made of probability, but a probability function defines where you will detect it.

7 minutes ago, quiet said:

So I think it is made of electromagnetic field, that the constitutive field of the photon occupies a finite volume related to the wavelength and that, without the polarization of the vacuum in the region where the photon is present, the photon could not exist and could not spread.

It is normally treated as having zero size (a point particle). 

7 minutes ago, quiet said:

To formulate probabilities you need first that the photon exists. And nothing forbids its existence to be based on a pair of linked, equal and opposite charges, which appear to be in motion without there actually being movement of charges.

...

Then, the imbalance caused by the curvature between electrical repulsion and magnetic attraction is slight. This can give as a result a slight repulsion between neighboring photons, which causes the thickening of the beam as it propagates. 

Do we need to ask the moderators to move this to Speculations if you are coming up with your own model of the photon?

7 minutes ago, quiet said:

Why would they repel each other?

They don't.

[quiet]

Any location that moderators deem appropriate will be fine. I prefer that the moderators decide, because they will do better than me.

[swansont]

13 hours ago, quiet said:

I hope the following.

1. That the beam propagate in a vacuum without suffering diffraction.

2. From what is expressed in item 1, I hope that the beam diameter remains constant when it propagates in a vacuum.

Beam divergence can only be zero from a plane wave. And finite source will have divergence. Lasers often have gaussian beams, Strange has given the beam diameter info for that.

 

I can't recall any connection where the beam diameter has any relation to the intensity. Further, there are reasons to think the "one photon per cycle" would be impossible to achieve. Photon counting statistics does not lend itself to that result.

[quiet]

6 minutes ago, swansont said:

Beam divergence can only be zero from a plane wave ...

... are reasons to think the "one photon per cycle" would be impossible to achieve.

Thanks for point out that.

[studiot]

1 hour ago, quiet said:

In the theoretical case that interests me, the cross section covers only one photon.

For a single photon any discussion of beam divergence is meaningless.

A single photon neither diverges nor converges. It goes where it goes.

There are, however, some  multi slit / diffraction experiments where the photon appears to pass through more than one point (slit), or not, depending upon how the experiment is conducted.

 

The only laser beam divergences I have looked at closely are alignment lasers.

Their beams neither diverge nor converge, but vary in width (diameter) in about a 20: 1 ratio from max to min and back again ,almost in a standing wave type configuration.

[quiet]

6 minutes ago, studiot said:

For a single photon any discussion of beam divergence is meaningless.

A single photon neither diverges nor converges. It goes where it goes.

Thanks for point out this.

9 minutes ago, studiot said:

Their beams neither diverge nor converge, but vary in width (diameter) in about a 20: 1 ratio from max to min and back again ,almost in a standing wave type configuration.

Very interesting phenomenon.