[TommyJ]

Many experiments convincingly lead to the result that electromagnetic waves, although they are waves, have a particle-like nature. These particle-like components are called photons. The photon model of electromagnetic waves consists of three basic postulates:
1. Electromagnetic waves consist of discrete, massless units called photons. A photon travels in vacuum at the speed of light.
2. Each photon has energy where is the frequency of the wave and h is a universal constant called Planck’s constant. The value of Planck’s constant is In other words, the electromagnetic waves come in discrete “chunks” of energy.
3. The superposition of a sufficiently large number of photons has the characteristics of a continuous electromagnetic wave.
Coming just to every-day human detection.
Infrared radiation, with its relatively long wavelength and low photon energy, produces effects in tissue similar to those of microwaves. Infrared is absorbed mostly by the top layer of your skin and simply warms you up, as you know from sitting in the sun or under a heat lamp.
In contrast, ultraviolet photons have enough energy to interact with molecules in entirely different ways, ionizing molecules and breaking molecular bonds. Interactions of ultraviolet radiation with matter are best understood from the photon perspective, with the absorption of each photon being associated with a particular molecular event.
Visible light is at a transition point in the electromagnetic spectrum. The energy of photons of visible light is large enough to cause molecular transitions—which is how the eye detects light. The bending of light by the lens of the eye requires us to think of light as a wave.
Color Vision. The cones, the color-sensitive cells in the retina of the eye, each contain one of three slightly different forms of a light-sensitive photopigment. A single photon of light can trigger a reaction in a photopigment molecule, which ultimately leads to a signal being produced by a cell in the retina. The color vision is a result of the differential response of the three types of cones containing these three different pigments.
Creating the impression of a color Computer monitors and color TVs can create millions of different colors by combining light from pixels of only three colors: red, green, and blue. These are called RGB displays. REASON We’ve seen that there are three different types of cones in the eye. By using differing amounts of three pure colors, we can independently stimulate each of the cone types and thus mimic the response of the eye to light of almost any color. ASSESS The fact that there are three primary colors of light—red, green, and blue— is a function of our physiology, not basic physics.
At the highest energies of the electromagnetic spectrum we find x rays and gamma rays.

[Eternal Student]

Hi.
   That sounds reasonable, TommyJ.

Fine details may be debatable

A single photon of light can trigger a reaction in a photopigment molecule, which ultimately leads to a signal being produced by a cell in the retina.

    Some sources of information suggest upto 3 photons are required, one photon on it's own may not be enough to cause a single pigment molecule to trigger.   Whatever the precise details, it's fair to say photopigments can be triggered by just a few photons even if it isn't exactly one.   Whether your brain can sensibly present that information to you in a way you understand is a different matter.  Your brain may process stimulus from the optic nerve in a way similar to "signal averaging" or completely ignore a single stimulation of one pigment molecule if it is not repeated soon or accompanied by signals from near-by photopigment molecules.
   Some more discussion was covered by TNS here:  https://www.thenakedscientists.com/articles/questions/how-many-photons-does-it-take-see-object.

    I think I can see how your comment links with the main topic in this thread.  Our eyes work a certain way and this is some evidence for the particle nature of light.
    There may be some interesting links with "Olber's paradox".  Light intensity will generally fall off as the square of the distance from the source.   However, there should eventually be a distance where the photons are spread so thinly that there are whole patches of sky where a detector of finite cross-sectional area can be placed and 0 photons would hit it in a second.   When we look up at the night sky we don't see light everywhere, just in a few paces and those are the stars we believe are close.  If the Universe was infinite and similar to waht we can see near-by then we should always find a star along any line through space.  The "thinning down of photons" as described above is one way to resolve that paradox.  The Hubble "deep field" photographs give some indication of what happens if we do collect photons over enough time (we see stars and whole galaxies in regions that superficially appeared black).

Best wishes.

[TommyJ]

When we look up at the night sky we don't see light everywhere, just in a few paces and those are the stars we believe are close.  If the Universe was infinite and similar to waht we can see near-by then we should always find a star along any line through space.  The "thinning down of photons" as described above is one way to resolve that paradox.  The Hubble "deep field" photographs give some indication of what happens if we do collect photons over enough time (we see stars and whole galaxies in regions that superficially appeared black).

Let me put some findings.

The process of detecting light destroys the photon concerned. However for gamma-rays and cosmic rays this is not necessarily true.
For example, when a gamma-ray Comptonscatters on an electron, it will lose energy, but can carry on and do it again. This can produce a series of detectable events which in principle will form a track pointing back in the arrival direction of the photon.
In practice, more indirect information can be obtained. A typical “Compton Telescope” has just two detection layers. In the upper layer a Compton scattering takes place and in the second layer the photon is absorbed. If the incoming photon has energy E1 and scatters through an angle θ then the energy of the scattered photon E2 is given by
(E1 − E2)/ (E1*E2) = (1/ (mc^2) · (1 − cos θ),
where m is the mass of the electron.
After measuring the initial and final energies, and so the scattering angle θ is known, but not the azimuth.
What is possible to get is a ring of possible photon directions. Over time, with the telescope in different orientations, many photons from the same astronomical source may be seen, in which case the intersection of the various rings finally pins down the source position.

This summing of different paths from the same source is used also in telecommunication systems, in order to minimize data burst losses.

[hamdani yusuf (OP)]

Why did you think a distant metal plate would affect the process?

Because the positrons are supposed to produce gamma ray only when they hit electrons, which are abundant on metal surface.

After a second thought, I realized that determining the direction of positron-electron collisions can be done better using a small ball of positron source and a small ball of electron source, separated by some distance inside the PET scanner. We can observe if the gamma rays coming out from the target still have random direction.

[hamdani yusuf (OP)]

The photon model of electromagnetic waves consists of three basic postulates:
1. Electromagnetic waves consist of discrete, massless units called photons. A photon travels in vacuum at the speed of light.
2. Each photon has energy where is the frequency of the wave and h is a universal constant called Planck’s constant. The value of Planck’s constant is In other words, the electromagnetic waves come in discrete “chunks” of energy.
3. The superposition of a sufficiently large number of photons has the characteristics of a continuous electromagnetic wave.

Let's compare these postulates to this article I quoted in another thread.

https://en.wikipedia.org/wiki/Planck_constant

The Planck constant, or Planck's constant, is a fundamental physical constant denoted h, and is of fundamental importance in quantum mechanics. A photon's energy is equal to its frequency multiplied by the Planck constant. Due to mass–energy equivalence, the Planck constant also relates mass to frequency.

In metrology it is used, together with other constants, to define the kilogram, an SI unit.[1] The SI units are defined in such a way that, when the Planck constant is expressed in SI units, it has the exact value h = 6.62607015×10⁻³⁴ J⋅Hz⁻¹.[2][3]

At the end of the 19th century, accurate measurements of the spectrum of black body radiation existed, but predictions of the frequency distribution of the radiation by then-existing theories diverged significantly at higher frequencies. In 1900, Max Planck empirically derived a formula for the observed spectrum. He assumed a hypothetical electrically charged oscillator in a cavity that contained black-body radiation could only change its energy in a minimal increment, E, that was proportional to the frequency of its associated electromagnetic wave.[4] He was able to calculate the proportionality constant from the experimental measurements, and that constant is named in his honor. In 1905, Albert Einstein determined a "quantum" or minimal element of the energy of the electromagnetic wave itself. The light quantum behaved in some respects as an electrically neutral particle, and was eventually called a photon. Max Planck received the 1918 Nobel Prize in Physics "in recognition of the services he rendered to the advancement of Physics by his discovery of energy quanta".

Confusion can arise when dealing with frequency or the Planck constant because the units of angular measure (cycle or radian) are omitted in SI.[5][6][7][8][9] In the language of quantity calculus,[10] the expression for the value of the Planck constant, or a frequency, is the product of a numerical value and a unit of measurement. The symbol f (or ν), when used for the value of a frequency, implies cycles per second or hertz as the unit. When the symbol ω is used for the frequency's value it implies radians per second as the unit. The numerical values of these two ways of expressing the frequency have a ratio of 2π. Omitting the units of angular measure "cycle" and "radian" can lead to an error of 2π. A similar state of affairs occurs for the Planck constant. The symbol h is used to express the value of the Planck constant in J⋅s/cycle, and the symbol ħ ("h-bar") is used to express its value in J⋅s/rad. Both represent the value of the Planck constant, but, as discussed below, their numerical values have a ratio of 2π. In this article the word "value" as used in the tables means "numerical value", and the equations involving the Planck constant and/or frequency actually involve their numerical values using the appropriate implied units.

The unit of Planck constant is J⋅s/cycle. If there are 2 cycles in a chirp of electromagnetic wave, the energy would be twice as much. It's implicitly assumed that the value of cycle must be an integer. Otherwise there would be no observed quantization of energy transfer.

The word quantum is the neuter singular of the Latin interrogative adjective quantus, meaning "how much". "Quanta", the neuter plural, short for "quanta of electricity" (electrons), was used in a 1902 article on the photoelectric effect by Philipp Lenard, who credited Hermann von Helmholtz for using the word in the area of electricity. However, the word quantum in general was well known before 1900,[2] e.g. quantum was used in E.A. Poe's Loss of Breath. It was often used by physicians, such as in the term quantum satis. Both Helmholtz and Julius von Mayer were physicians as well as physicists. Helmholtz used quantum with reference to heat in his article[3] on Mayer's work, and the word quantum can be found in the formulation of the first law of thermodynamics by Mayer in his letter[4] dated July 24, 1841.

In 1901, Max Planck used quanta to mean "quanta of matter and electricity",[5] gas, and heat.[6] In 1905, in response to Planck's work and the experimental work of Lenard (who explained his results by using the term quanta of electricity), Albert Einstein suggested that radiation existed in spatially localized packets which he called "quanta of light" ("Lichtquanta").[7]

The concept of quantization of radiation was discovered in 1900 by Max Planck, who had been trying to understand the emission of radiation from heated objects, known as black-body radiation. By assuming that energy can be absorbed or released only in tiny, differential, discrete packets (which he called "bundles", or "energy elements"),[8] Planck accounted for certain objects changing color when heated.[9] On December 14, 1900, Planck reported his findings to the German Physical Society, and introduced the idea of quantization for the first time as a part of his research on black-body radiation.[10] As a result of his experiments, Planck deduced the numerical value of h, known as the Planck constant, and reported more precise values for the unit of electrical charge and the Avogadro–Loschmidt number, the number of real molecules in a mole, to the German Physical Society. After his theory was validated, Planck was awarded the Nobel Prize in Physics for his discovery in 1918.

https://en.wikipedia.org/wiki/Quantum#Etymology_and_discovery

Do you agree with my underlined statement above?

[Bored chemist]

Do you agree with my underlined statement above?

No.
There are "many" cycles in a photon. There's no reason for it to be an integer.

[hamdani yusuf (OP)]

There may be some interesting links with "Olber's paradox".  Light intensity will generally fall off as the square of the distance from the source.   However, there should eventually be a distance where the photons are spread so thinly that there are whole patches of sky where a detector of finite cross-sectional area can be placed and 0 photons would hit it in a second.   When we look up at the night sky we don't see light everywhere, just in a few paces and those are the stars we believe are close.  If the Universe was infinite and similar to waht we can see near-by then we should always find a star along any line through space.  The "thinning down of photons" as described above is one way to resolve that paradox.  The Hubble "deep field" photographs give some indication of what happens if we do collect photons over enough time (we see stars and whole galaxies in regions that superficially appeared black).

Alternatively, the light from extremely distant galaxies might be absorbed by intergalactic media, which convert it into heat which is detected as microwave background.

[hamdani yusuf (OP)]

Do you agree with my underlined statement above?

No.
There are "many" cycles in a photon. There's no reason for it to be an integer.

Can a photon have a fraction number of cycles, such as 1.2345?
For example, a photon has frequency of 1 THz, and 1.2345 cycle. What's its energy?

[alancalverd]

After a second thought, I realized that determining the direction of positron-electron collisions can be done better using a small ball of positron source and a small ball of electron source, separated by some distance inside the PET scanner. We can observe if the gamma rays coming out from the target still have random direction.

The emitted positrons in a solid or liquid have a range of about 0.5 mm which limits the spatial resolution of PET. Havoing an energy of between 0.5 and 2 MeV (depending on the nuclide) you would need a very strong electric field to determine which way they move before annihilating, and since the p-e pair is electrically neutral, there's no much you can do to orient it in space. What we do know is that the coincident photon pairs are spherically distributed and randomly timed, so the hypothesis that they are randomly distributed seems to be supported.

[alancalverd]

For example, a photon has frequency of 1 THz, and 1.2345 cycle. What's its energy?

E = hf

You are wasting your time trying to apply continuum analytics to a particle model. They are two different models.

[hamdani yusuf (OP)]

After a second thought, I realized that determining the direction of positron-electron collisions can be done better using a small ball of positron source and a small ball of electron source, separated by some distance inside the PET scanner. We can observe if the gamma rays coming out from the target still have random direction.

The emitted positrons in a solid or liquid have a range of about 0.5 mm which limits the spatial resolution of PET. Havoing an energy of between 0.5 and 2 MeV (depending on the nuclide) you would need a very strong electric field to determine which way they move before annihilating, and since the p-e pair is electrically neutral, there's no much you can do to orient it in space. What we do know is that the coincident photon pairs are spherically distributed and randomly timed, so the hypothesis that they are randomly distributed seems to be supported.

Can a positron move further in air, or better yet, vacuum?
If the source of positron can be concentrated in a small volume, so does the source of electron next to it, then the gamma rays coming from electron source as the product of p-e collision must come from positron moving from positron source to electron source, i.e. the direction is determined. Positron moving to other directions would produce gamma ray somewhere else.

[hamdani yusuf (OP)]

For example, a photon has frequency of 1 THz, and 1.2345 cycle. What's its energy?

E = hf

You are wasting your time trying to apply continuum analytics to a particle model. They are two different models.

Why Schrodinger developed wave mechanics?
What's the unit of h?

[Bored chemist]

For example, a photon has frequency of 1 THz, and 1.2345 cycle. What's its energy?

E=hf

[hamdani yusuf (OP)]

The symbol h is used to express the value of the Planck constant in J⋅s/cycle

Let's do some unit analysis.
The time unit as numerator there, indicates that the value is an accumulated quantity over time, which is second. It's similar to mAh in battery capacity, or kWh in energy consumption.
The cycle as denominator means that the value is the rate of a quantity, which is per cycle.
When a quantity of radiation has value of 6.62607004 x 10^-34 J. s/cycle, it means that
6.62607004 x 10^-34 Joule of energy is transfered in 1 second interval for each cycle.
If the frequency is 1 Hz, then there is 1 cycle in a second. Thus, in 1 second interval  the energy transfered is 6.62607004 x 10^-34 Joule.

If the frequency is 1 THz, there are 1 trillion cycles in a second. So the energy transfer accumulated in 1 second is 6.62607004 x 10^-22 Joule.

[alancalverd]

Why Schrodinger developed wave mechanics?

to model the probability distribution of a particle.

What's the unit of h?

joule.second So if you multiply by frequency (second ^-1) you get the energy of the photon, though it's more usually quoted in eV rather than joules. Alternatively if you know the photon energy and use E = hc/λ you can predict its diffraction properties.

[TommyJ]

Do you agree with my underlined statement above?

Agree, in these terms (in case we are on the same page

To explain the radiation equation Planck made an assumption that energy in discrete quantities (electron oscillations in atom). Each discrete 'bit' of energy exists as a whole number multiple of minimum energy.

- Energy exists as a discrete quantity
- Exists any minimum quantity of energy for any frequency of oscillation E=hf. E can be greater by a factor of a some positive integer.
If oscillation is greater than the minimum, there exists an integer multiple of minimum quantity. E=nhf, where n is an integer number.

[evan_au]

If there are 2 cycles in a chirp of electromagnetic wave
...1.2345 cycle

If you try to generate an electromagnetic wave with a small number of cycles (eg 1 or 2), you generate a signal with an extremely wide bandwidth, so the frequency of the photon(s) is poorly defined, which means that the energy of the photons is also poorly defined. E=hf is still true, but the problem is that you don't know f accurately.

If you want a precisely measured frequency, you have to measure over many cycles (eg thousands or billions).

This relates back to the Heisenberg uncertainty principle - the more you attempt to constrain one parameter of a quantum variable, the more it becomes uncertain in another parameter.

The 2018 Nobel prize in physics was awarded for research into ultra-short laser pulses. Unlike the more common continuous-wave laser pointers, ultrashort laser pulses have an ultrawide bandwidth.

See: https://en.wikipedia.org/wiki/Ultrashort_pulse

[Bored chemist]

Let's do some unit analysis.
The time unit as numerator there, indicates that the value is an accumulated quantity over time, which is second. It's similar to mAh in battery capacity, or kWh in energy consumption.

That is not a valid analysis (for a number of reasons).
The Joule is a unit of work- i.e. force times distance and a force is a mass times an acceleration.
So Joule is  Kg m² /sec²

So the unit of the Planck constant is
Kg m² / sec

The "time" part is in reciprocal seconds, not seconds.
And what the hell does the product of an area with a mass mean?
Because, whatever that is, h has units of "the rate of change of whatever".

If the frequency is 1 THz, there are 1 trillion cycles in a second. So the energy transfer accumulated in 1 second is 6.62607004 x 10-22 Joule.

No
I can transmit (in principle) any rate of power I like at 1THz.
I am not restricted to sending 6.62607004 x 10^-22 Joule per second i.e. 6.62607004 x 10^-22 W

It is more correct to say
If the frequency is 1 THz, there are 1 trillion cycles in a second. So the energy transfer accumulated in 1 second
 Photon is 6.62607004 x 10-22 Joule.

It does not take a second to send a photon.
It would be very weird if it did.
What sort of coincidence would it be if the duration of a photon was a sixtieth of a sixtieth of a twenty-fourth of the time it takes  the Earth to rotate on its axis?

[hamdani yusuf (OP)]

If there are 2 cycles in a chirp of electromagnetic wave
...1.2345 cycle

If you try to generate an electromagnetic wave with a small number of cycles (eg 1 or 2), you generate a signal with an extremely wide bandwidth, so the frequency of the photon(s) is poorly defined, which means that the energy of the photons is also poorly defined. E=hf is still true, but the problem is that you don't know f accurately.

If you want a precisely measured frequency, you have to measure over many cycles (eg thousands or billions).

This relates back to the Heisenberg uncertainty principle - the more you attempt to constrain one parameter of a quantum variable, the more it becomes uncertain in another parameter.

The 2018 Nobel prize in physics was awarded for research into ultra-short laser pulses. Unlike the more common continuous-wave laser pointers, ultrashort laser pulses have an ultrawide bandwidth.

See: https://en.wikipedia.org/wiki/Ultrashort_pulse

At a glance, the pulse doesn't have a uniform wavelength. It looks longer at beginning and shorter at the end.
But as long as it starts and finish at 0, the number of cycles must be an integer multiple of a half.

[hamdani yusuf (OP)]

So the unit of the Planck constant is
Kg m2 / sec

The "time" part is in reciprocal seconds, not seconds.
And what the hell does the product of an area with a mass mean?
Because, whatever that is, h has units of "the rate of change of whatever".

You're missing the cycle part. Although it's dimensionless, it doesn't mean that it's not there. Planck constant has the same unit as angular momentum, when the angle is measured in cycles.

The joule-second (symbol J⋅s or J s) is the product of an SI derived unit, the joule (J), and an SI base unit, the second (s).[1] The joule-second is a unit of action or of angular momentum. The joule-second also appears in quantum mechanics within the definition of Planck's constant.[2] Angular momentum is the product of an object’s moment of inertia, in units of kg⋅m2 and its angular velocity in units of rad⋅s−1. This product of moment of inertia and angular velocity yields kg⋅m2⋅s−1 or the joule-second. Planck's constant represents the energy of a wave, in units of joule, divided by the frequency of that wave, in units of s−1. This quotient of energy and frequency also yields the joule-second (J⋅s).

https://en.wikipedia.org/wiki/Joule-second

If we try to analyze a derived unit using only its base unit equivalent, we can easily get confused of the meaning.
https://en.wikipedia.org/wiki/SI_derived_unit
How do we get the meaning of this unit?
kg−1⋅m−2⋅s4⋅A2
or this?
kg⋅s−3