Power of a photon? Energy of a photon

[sirius0]

The electrical field I work in people often confuse power and energy. I.e. In Australia we get billed peak power kW or MW (kilo Watts, Mega Watts) and our energy is charged at so many cents per kWhr (kilo Watt hour). The kWhr is convertible to joules. Explaining the difference between power and energy to accounting departments and others has been a frequent pleasure in my career.

Now photons. I was taught that photons are wave packets and in this way they are also able to behave as particles in certain contexts. In addition to this I was taught that the higher the frequency of the electromagnetic photon the higher the ammount of energy imparted to say an electron as it selects a new energy level in an atom.

BUT isn't the wavelength a function of time given that the photon must travel at light speed? Is our thinking about photons totally right or is the 'energy' difference between say red and blue light actually a power difference. Could it be that all photons have the same amount of energy (somewhat redefined though)? Could it be that ALL photons are the same just red or blue shifted versions of each other? These are just some thoughts; they don't sit well with me though as the energy levels in atoms seem to disprove what I am saying. But the longer wave lengths must take more time to impart energy surely or am I being too Newtonian? What have I missed?

I can see in my language that I am borderline ATM here but I don't know where my thoughts will lead me. The thought process is similiar to one that lead me to develop an ATM I can't be sure that this line won't as well. I honestly don't have a concious agenda so if you see that I am wrong save me before I fall.......Ahhh!

[cjl]

No - the energy per photon is what determines the frequency. The power is dependent on two things: energy per photon, and photons per second. Obviously, with the same number of photons per second, the higher energy photons will impart more energy per second, I.E. more power.

[sirius0]

No - the energy per photon is what determines the frequency. The power is dependent on two things: energy per photon, and photons per second. Obviously, with the same number of photons per second, the higher energy photons will impart more energy per second, I.E. more power.

Yes this makes cold hard sense; but i am asking about the photons themselves.
Surely a photon with a wavelength of say 380nm (violet) will take only approximately half the time at light speed to pass a point than say a photon with a wavelength of say 750nm (red)? Not because of speed but length!

i mean if I am right then "Obviously, with the same number of photons per second, the higher power photons will impart more energy per second, I.E. more power." would still work as a statement. I mean i know the standard theory but I am wondering why the time for a wavelength hasn't been considered. Is there a power component to photons? If there is then this could help us have energy conservation on a cosmic scale as per the queries raised in this post http://www.bautforum.com/showthread.php?t=48114

An excellent question that still has me thinking (perhaps this thread should have been a post in that one?)

[swansont]

Yes this makes cold hard sense; but i am asking about the photons themselves.
Surely a photon with a wavelength of say 380nm (violet) will take only approximately half the time at light speed to pass a point than say a photon with a wavelength of say 750nm (red)? Not because of speed but length!

How would one measure this? You have to interact with the photon in order to measure it.

[Ken G]

i mean if I am right then "Obviously, with the same number of photons per second, the higher power photons will impart more energy per second, I.E. more power." would still work as a statement.

You were originally making an observation of another way to think about things, but there is no new physics here. (The power idea, I mean-- the above statement is incorrect, your whole point before is that all photons may to some degree be considered as having the same power while being absorbed.) Thus your approach makes no new predictions at all, but it may indeed have pedagogical value (value in how we teach and understand concepts). You are essentially (correctly) stating that the energy of a photon is proportional to the energy delivered over a single oscillation (for photons of similar "sharpness"), so what is the same is the energy delivered per unit time as the photon is being absorbed. However, note that this is a classical picture of the photon and ignores energy quantization (and indeed, the very concept of "photon" is an idealization which requires an infinite time and hence zero power in the first place). Thus the physics behind this view is unmeasurable, as pointed out by swansont, and probably obfuscates the quantum mechanics. Still, from a classical perspective, there may be some way to shape this observation into something pedagogically interesting, it's an interesting point to consider.

[Fazor]

Whenever you reply to such a question Ken, I can count on two things: A) that I'll only understand about every 5th word, and B) you'll use the word Pedagogically at least once. It's nice having something you can count on, makes one feel at home.

Have there been many, or any single-photon experiments yet? from what I can tell they aren't really able to single out one lone photon. hmm... i think i got some googling and wiki-surfing to do.

[sirius0]

You were originally making an observation of another way to think about things, but there is no new physics here. (The power idea, I mean-- the above statement is incorrect, your whole point before is that all photons may to some degree be considered as having the same power while being absorbed.) Thus your approach makes no new predictions at all, but it may indeed have pedagogical value (value in how we teach and understand concepts). You are essentially (correctly) stating that the energy of a photon is proportional to the energy delivered over a single oscillation (for photons of similar "sharpness"), so what is the same is the energy delivered per unit time as the photon is being absorbed. However, note that this is a classical picture of the photon and ignores energy quantization (and indeed, the very concept of "photon" is an idealization which requires an infinite time and hence zero power in the first place). Thus the physics behind this view is unmeasurable, as pointed out by swansont, and probably obfuscates the quantum mechanics. Still, from a classical perspective, there may be some way to shape this observation into something pedagogically interesting, it's an interesting point to consider.

Yes I am just thinking. I got a bit defensive with the bolding due to a need to asert the validity of the question rather than defending some point ( I was changing cjl's statement to show that power or energy in packets per second would still summ up to power). As I read and think I realise that my initial question is "What is the point where the concept of a photon ceases to be classical and starts having to be the new physics (quantum sr etc)"

I point out though that I am not saying -(anything just asking; but with poorly managed language)- that the power of photons is the same I am asking if the difference in power can explain what we call the difference in energy between say a red and a blue photon. Yes probably no new physics but perhaps shaking the definition of what a photon should be regarded as (is that what pedagogical means?)

I am starting to wonder if the difference between a blue and red photon is the tau? Or time dilation

[Grey]

Yes this makes cold hard sense; but i am asking about the photons themselves.
Surely a photon with a wavelength of say 380nm (violet) will take only approximately half the time at light speed to pass a point than say a photon with a wavelength of say 750nm (red)? Not because of speed but length!

No, actually. Although photons are sometimes drawn as wave packets with a length, they interact, like all fundamental particles, as though they have no size at all. Photons of differing wavelengths don't have different "lengths" in any way that we can measure. And if a photon is absorbed, it doesn't take any longer for it to happen if it has a longer wavelength.

[sirius0]

Ok enough pennies have dropped! I was effectively saying that a shortwave radio photon is longer than a football field!

KenG perhaps for teaching purposes this question serves to demonstrate the difference between frequency as "How many cycles per second" versus "Frequency is the rate of change of amplitude or perhaps phase"
Meaning that a photon represents a packet of energy with a rate of change of cycle that if extrapolated to display a whole cycle would have a wavelength of say 450nm if it was violet/blue
In other words a photon is a snapshot of a small segment of wavelength perhaps.

I thank all for your answers that have set me straight.

I sill think that a red photon being a time dilated blue might be interesting.

[Jeff Root]

Photons of differing wavelengths don't have different "lengths" in
any way that we can measure. And if a photon is absorbed, it doesn't
take any longer for it to happen if it has a longer wavelength.

I question whether this is true. I don't see how the time for
absorption could be measured, so I'd say that it is unknown as
yet whether photons have length. Am I wrong? Has absortion
time been measured? I'd like to hear how that can be done,
and the results.

-- Jeff, in Minneapolis

[Ken G]

I think there might actually be considerable truth to saying that a red photon requires a longer time to absorb than a blue. The way it works is that the longer the time you devote to absorbing the photon, the more precisely you can know what "photon" you were absorbing in the first place. A very short measurement may trap a lot of X-ray photons, but it will be unable to detect with the same efficiency a radio photon. It's tricky business, how to even ask the question properly is not so trivial. A "photon" is an eigenstate of a frequency measurement, but such a measurement requires a time much greater than the period of the wave to be precise. I think a "short" measurement will experience a very low efficiency at detecting a wide range of frequencies, but a "long" measurement will detect with higher precision if the photon frequency is unknown, and higher efficiency if the frequency is known.

[Gsquare]

In other words a photon is a snapshot of a small segment of wavelength perhaps.....
...
,...
. "What is the point where the concept of a photon ceases to be classical and starts having to be the new physics (quantum sr etc)"

A few years ago I remember the superluminal light topic coming up which, when viewed from a classical wave, can be approached. There can be a 'distortion' in the wave (not just phase vs, group velocity) which can alter the signal velocity.

There were claims of achieveing superluminal velocity using microwaves which met with some stiff resistance from others. ( Italian microwaves ) .
See for ex.here:

http://focus.aps.org/story/v5/st23

http://prola.aps.org/abstract/PRL/v84/i21/p4830_1

Apparently, the amount of resistance is directly proportional to the incredulousity you claim to have achieved.

G^2

- Consciousness: that uncomfortable state of awareness between naps. -

[Jeff Root]

I think there might actually be considerable truth to saying that
a red photon requires a longer time to absorb than a blue.

That is what I would expect.

The way it works is that the longer the time you devote to absorbing
the photon, the more precisely you can know what "photon" you were
absorbing in the first place. A very short measurement may
trap a lot of X-ray photons, but it will be unable to detect with
the same efficiency a radio photon. It's tricky business, how to
even ask the question properly is not so trivial. A "photon" is
an eigenstate of a frequency measurement, but such a measurement
requires a time much greater than the period of the wave to be
precise. I think a "short" measurement will experience a very low
efficiency at detecting a wide range of frequencies, but a "long"
measurement will detect with higher precision if the photon
frequency is unknown, and higher efficiency if the frequency is
known.

I think there may be a problem with that analysis. But it depends
on my description of what a photon is, which is entirely different
from yours. I hope that although my description is different, it
is not incompatible.

I think a photon is that "thing" which is, for example, detected
by a photomultiplier tube. A photon hits the photocathode and a
cascade of electrons results, giving a visible result. Limiting
the light level hitting the cathode reveals individual photons.
This works for a wide range of photon energies, but low-energy
photons cannot be detected in this way. By extrapolation, it is
obvious that all electromagnetic radiation consists of photons
of whatever energy. But photons with low energies can only be
detected in large numbers, so talking about "measurement" of the
properties of individual photons with low energies and low
frequencies can only be a thought experiment or mathematical
derivation, not anything real or physical. When such thought
experiments or mathematical derivations accurately predict the
behavior of large numbers of photons which can be measured,
then that is fine, but it isn't really the same as detecting or
describing the behavior of individual photons.

It is like determining the average interatomic distances in a
block of material whose atoms connot be studied individually,
and inferring the mass of the individual atoms from the mass
of the block and the number of atoms.

We can have a monochromatic source generate photons with a
precisely-measured frequency by measuring large numbers of
photons over a long time period, then reduce the photon rate
for observing the behavior of individual photons.

For the range of frequencies over which individual photons can
be detected (or even outside that range), is it possible to
measure the absorption time? If so, how can it be done?

-- Jeff, in Minneapolis

[Bob]

Photons have real lengths. Knowing the wavelength is essential in designing antennas, waveguides, diffraction gratings, and in xray crystallography.

[sirius0]

Photons have real lengths. Knowing the wavelength is essential in designing antennas, waveguides, diffraction gratings, and in xray crystallography.

Yes but this was the other side of my question. Does a photon have the whole wavelength enclosed? All those examples prove the wave nature of light. i am begining to think that the photon is a snapshot of a portion of wavelength. The frequency of the photon is the rate of change of amplitude rather than 1/T because we may not have a whole wavelength in that photon.

I was just thinking that maybe we only get photons when EM enters a quantised region such as an atom. Electrons etc respond to light in a quantised way and so photons are a quantised portion of the incident light which in free space just is one (perhaps continous) field inducing another as per theory.

That would mean that the photon size etc is defined by the energy levels available to an electron. This could be tested perhaps by seeing if an electron in free space can have its electron volts increased (speed increased) in a quantised way (photon ) or a continuos way (wave theory).

Perhaps that other thread that led me down this path (http://www.bautforum.com/showthread.php?t=48114)
could be satisfied because red shifted light may not lose it's energy but actually have more quantised photon levels available on arriving at atoms (due to red shifting as a result of expansion) but these photons are less in energy each conserving energy in this way.

[sirius0]

For the range of frequencies over which individual photons can
be detected (or even outside that range), is it possible to
measure the absorption time? If so, how can it be done?

-- Jeff, in Minneapolis

Is the absorbtion time equivalent to the time for a quantum leap? I mean is absorbtion of a photon the same event as an electron moving from one energy level to another?

I recall that this is mean't to be nearly or actually instant?

[Ken G]

I think there may be a problem with that analysis. But it depends
on my description of what a photon is, which is entirely different
from yours.

I don't see anything wrong in your description, nor do I see any problem or even difference in the definition of a photon from mine. A photon is a quantum, but what it's a quantum of depends on what quantum you are measuring. Typically, that's energy, but if you are saying that really it just has to be a quantum of some kind, that's a correct distinction.

For the range of frequencies over which individual photons can
be detected (or even outside that range), is it possible to
measure the absorption time? If so, how can it be done?

By in effect having a "shutter" that opens and closes. If you can get the shutter to close fast enough, you lose efficiency at detecting lower frequency photons, moreso than higher frequency photons. Also, you introduce uncertainty into the frequencies you are observing, if you open and
close the shutter in a time less than the wave period.

[sirius0]

I think, thanks to the replies, that I can almost completely conclude that the direction of the my initial question is wrong. Although it had the same intent as where I have landed.

I thought I had something new that stemmed from the thread http://www.bautforum.com/showthread.php?t=48114
But it would appear that i have simply continued to try and solve it.
That is that red shifted light may only appear to lose energy because we are only looking at photons but are unable to count them.

I say this because I recall the FRED Ultra-Violet laser. That was being used for research where I studied. This used a FrEquency Doubling crystal. I can't recall the exact nanometers of the initial laser light but I believe it was red.

In short the crystal has some kind of dual refrative indices that can be configured to sum the red light wavelengths so precisely that they merge to become Ultra Violet. This then can be used to 'burn' a grating longitudally in optical fibre making a narrow pass filter that can be used to detect strain and or temperature. My point here is that there is no increase of energy from red to UV but there must then be less photons of more energy each. This is an example of a testable and used phenomina in a lab. I think it is only reasonable to posit that a single photon at the blue end of the spectrum can be split into more than one photon at the red end when red shifted. Surely FRED is reversible?

[swansont]

Yes, you can do the opposite. It's often called down-conversion, and all of the possibilities of combining or splitting two frequencies fall under the general description of four-wave mixing, of which three-wave mixing is a subset. Frequency doubling (second harmonic generation) is just a special case of mixing two different frequencies to get a higher-frequency photon. All part of nonlinear optics. Your mixing material needs a nonlinear susceptibility for this to work, and this requires an absence of inversion symmetry, so it can happen at a surface. (if memory serves)


Googling points to this wiki article
http://en.wikipedia.org/wiki/Nonlinear_optics

[sirius0]

Yes, you can do the opposite. It's often called down-conversion,

Thank you swansont. My degree had a large component of non-linear optics but things get rusty as I haven't worked in that area just yet.