Are clicks proof of single photons?

[lightarrow]

Are you implying something like - "click" in detector=photon ?

You'd think so, if its a photon dectector.

Because it's written in the kit label? Or because it's written in some books? The fact it's a "photon detector" say nothing. Firstly you should explain what exactly is a photon for you, then by which mechanism your photon interacts with a photon detector expelling an electron.
Then you will notice that with a classical EM field is much simpler...

 

[zonde]

Very obscure.

But does it matter if based on this we can make testable predictions?
I would say that you don't have to look far for an example.

What are k and $$\phi$$ ?

k is wavenumber and $$\phi$$ is phase shift along particular path.

 

[lightarrow]

k is wavenumber and $$\phi$$ is phase shift along particular path.

So you are describing nothing else than the classical EM field. No need for photons at all. We have come to an agreement

 

[A. Neumaier]

I don't quite understand - aren't photons discrete quantum matter? So photons are setting it off?
Classical light = incident classical external radiation field?

Classical light = light in a mixture of coherent quantum states.

 

[A. Neumaier]

You mean - photons must be wave packets to speak about phase difference?
Well, but if we speak about soliton waves rather than wave packets? They are rather very similar.

Photons must satisfy the Maxwell equations. These don't have soliton solutions.

 

[Q-reeus]

From #110:
Originally Posted by Q-reeus:
'Well if there is no energy storage between 'hits', the implication seems to be either an actual collapse of all the energy into the screen of what could be a very extended single field quanta, or a purely probabilistic event where energy conservation is just a time-averaged thing. Please clarify.'

"QM claims energy conservation only in the ensemble average, which is usually equated (with practical success but no fully convincing demonstration) with the time average..."

Context for this exchange was in #105.
Disappointed no-one took issue with the claim conservation of energy is merely statistical at the quantum level. Motivated by an entry in another thread (#14 in https://www.physicsforums.com/showthread.php?t=250361) claiming it always exactly holds (which is my understanding), let's look at this again. Suppose as per #105 the screen is highly efficient but presents a very small cross-section relative to an approximately hemispherical surface that has as radius r the distance from the slit or slits to the screen. Let the mean rate of photons passing through the slits be one per minute. Let the ratio of screen area A_s to 2*pi*r² be say 0.001. Then on a photon-as-particle picture an event rate of roughly once every 1000 minutes on average applies. What about the situation with photon-as-continuous-spherical-wave? Realistically if for each 'hit' some tiny fraction .001 of a single photon energy can actually be absorbed by the screen, we might expect exponential decay-rate dissipation as heat with a time constant in the nano-second range. For all practical purposes there is simply no memory between hits, and given the vast insufficiency in available energy for an electron ejection event ('click') by a single photon-as-spherical-wave, how could we ever expect one - ever?
Further, even taking the position that energy can in fact be 'borrowed' on time scales of ~ 1000 minutes in order to statistically match the photon-as-particle count rate, how would the screen 'know' how to get the stats right - when to eject an electron, given the total obliteration of evidence between hits? Is there some quantum ghost that hovers about, doing a count? In solving one 'quantum mystery' (interference), two 'mysteries' at least as serious seem to have taken it's place. Does M & W cover this situation?

 

[A. Neumaier]

From #110:
Originally Posted by Q-reeus:
'Well if there is no energy storage between 'hits', the implication seems to be either an actual collapse of all the energy into the screen of what could be a very extended single field quanta, or a purely probabilistic event where energy conservation is just a time-averaged thing. Please clarify.'

"QM claims energy conservation only in the ensemble average, which is usually equated (with practical success but no fully convincing demonstration) with the time average..."

Context for this exchange was in #105.
Disappointed no-one took issue with the claim conservation of energy is merely statistical at the quantum level. Motivated by an entry in another thread (#14 in https://www.physicsforums.com/showthread.php?t=250361) claiming it always exactly holds (which is my understanding),

Conservation of energy holds on the Feynman diagram level. But a Feynman diagram doesn't describe a real process in time - only a contribution to a scattering amplitude between t=-inf and t=+inf. Thus this is irrelevant for the interpretation of processes that happen in time.

Given the Schroedinger equation i hbar d/dt psi(t) = H psi(t) , one can easily prove that d/dt psi(t)^*A psi(t) = 0 whenever A commutes with H. This is the case for any function of H. Therefore, <f(H)> is time-independent for any function of H. This is the strongest conservation statement concerning energy that can be proved. It is only a statistical conservation law.

Further, even taking the position that energy can in fact be 'borrowed' on time scales of ~ 1000 minutes in order to statistically match the photon-as-particle count rate, how would the screen 'know' how to get the stats right

Firing at random with the correct rate gives the correct statistics.

 

[Q-reeus]

It is only a statistical conservation law...

Alright let's take that as so. There still seems to be a monumental gulf in terms of what Heisenberg's Uncertainty Principle can allow here - ie delta E * delta t >= h. Taking the delta E to be hf (photo-electron ejection energy), and delta t as the time between 'clicks' ~ 1000 minutes, this comes to hf*1000*60 ~ h*6*10¹⁴*6*10⁴ = h*3.6*10¹⁹ , (using f~ 6*10¹⁴ Hz for typical visible light). Many orders of magnitude greater than h! And is this a totally inapplicable use of the HUP? I find it unreasonable to imagine quantum energy-time fluctuations at this level. It also seems hard to see how one does not in fact have a net gain in energy here, as there is no apparent mechanism to cancel the energy of the incident photons (unlike say for monochromatic radiation interacting with classical oscillators).

..Firing at random with the correct rate gives the correct statistics.

But how does that statement physically explain a situation where as I have claimed there is a screen total state reset between photon hits? Each hit might as well be the first. If 0.001 of the energy needed to generate a click is available that time, that's all that is available every other time, unless what physical process intervenes exactly?

 

[A. Neumaier]

Alright let's take that as so. There still seems to be a monumental gulf in terms of what Heisenberg's Uncertainty Principle can allow here - ie delta E * delta t >= h.

This is not Heisenberg's uncertainty relation, since t is not an observable in the usual QM sense.

Taking the delta E to be hf (photo-electron ejection energy), and delta t as the time between 'clicks' ~ 1000 minutes,

You cannot choose arbitrary energies and times in such a relation. delta E is the _uncertainty_ in measuring the energy E = h*nu, and delta t is the uncertainty in the timing of a measurement (i.e., of the order of the duration of a click).

Each hit might as well be the first.

Just like each throw of a die may be the first. This doesn't invalidate correctly getting the statistics of dice or clicks.

 

[Q-reeus]

..You cannot choose arbitrary energies and times in such a relation. delta E is the _uncertainty_ in measuring the energy E = h*nu, and delta t is the uncertainty in the timing of a measurement (i.e., of the order of the duration of a click)...

But the relationship is more than just a measurement phenomenon - does it not lie at the heart of say the ZPF energy spectrum?
I'd still like an explanation of how photo-ejection does not simply add to net energy with photons-as-spherical-waves picture. Where does cancellation of these incident photon energies ever occur?

Just like each throw of a die may be the first. This doesn't invalidate correctly getting the statistics of dice or clicks.

Sure, but in this case the energy deficit is enormous - where does the 0.999 fraction needed to make up the deficit come from? When the incident energy is relatively high, 'normal' statistics are OK, but here? The implied linearity of count rate vs photon incident rate seems way suss here for reasons already given (exponential seems more likely), but is perfectly natural in the particle picture.

 

[A. Neumaier]

I'd still like an explanation of how photo-ejection does not simply add to net energy with photons-as-spherical-waves picture. Where does cancellation of these incident photon energies ever occur?

? Energy is added continuously, while it is released erratically in bunches. What's the problem?

Sure, but in this case the energy deficit is enormous

The energy deficit is that of a single particle energy - a nothing compared to the energy floating in a detector.

 

[Q-reeus]

? Energy is added continuously, while it is released erratically in bunches. What's the problem?

Firstly, have we not previously agreed energy is not added continually (what you meant surely when saying 'No memory is needed', as in #39, or 'No storage is needed' as in #107)? You have not before challenged my view that any relatively minuscule energy absorbed at a particular hit is lost as heat in an extremely brief time scale - many orders of magnitude smaller than mean time between photon hits. Maybe you meant something else here.
Proceeding on that basis, the problem as I see it is owing to the highly statistical (ie random) relation between incident photons and photo-ejection, there no evident phase cancellation mechanism. Each incident photon in this picture simply expands forever as a spherical wavefront, with integral of ExB = constant = hf, and photo-ejection seems quite incapable of altering that. Compare that to wave interference when a classical monochromatic field interacts with classical oscillators as was related in #148 - we there have a detailed balance according to Poynting theorem.

The energy deficit is that of a single particle energy - a nothing compared to the energy floating in a detector.

So the energy reservoir is the screen/detector 'electron sea'? OK, but then why is photoelectric effect so highly sensitive to frequency, with a very sharp lower frequency cutoff? Such pickiness seems very hard to explain in the very infrequent incident photon-as-wave view.

 

[A. Neumaier]

Firstly, have we not previously agreed energy is not added continually (what you meant surely when saying 'No memory is needed', as in #39, or 'No storage is needed' as in #107)? You have not before challenged my view that any relatively minuscule energy absorbed at a particular hit is lost as heat in an extremely brief time scale - many orders of magnitude smaller than mean time between photon hits.

Even tiny amounts of energy are taken in (randomly but nonlocally) by the detector as a whole. Otherwise we couldn't have energy conservation in the mean. It is only the ionization response that happens randomly, discretely, and locally.

Each incident photon in this picture simply expands forever as a spherical wavefront.

This is a problem only if you think in a photon particle picture. In the quantum field picture, the field ends wherever there is matter (or at least, it is modeled very differently there), and the waves are only where there is room for them. Just as with classical radiation.

So the energy reservoir is the screen/detector 'electron sea'? OK, but then why is photoelectric effect so highly sensitive to frequency, with a very sharp lower frequency cutoff?

The energy reservoir is the macroscopic object in which the discrete qubits (given by weakly bound electrons or by silver bromide molecules) are located. The macroscopic object takes up the energy and distributes it randomly; the electrons jump according to the quantum probabilities, which is exactly zero when the frequency is below the ionization threshold.

 

[Q-reeus]

Even tiny amounts of energy are taken in (randomly but nonlocally) by the detector as a whole. Otherwise we couldn't have energy conservation in the mean. It is only the ionization response that happens randomly, discretely, and locally.

Well it seems we have been having a long running communication problem, despite your excellent technical grasp of English. From #107:

Originally Posted by Q-reeus:
'But that's the real sticking point as I see it. This 'best case scenario' assumes the screen is not only capable of fully absorbing all incident radiation, but losslessly accumulating each hit for perhaps hours until sufficient energy is present to eject one electron. What's more in order to reproduce the interference pattern, a memory of the incident intensity is also dissipationlessly stored. This seems utterly incredible.' Your response:

"No storage is needed. Just a rate of response. That's the nature of probability - even tiny rates accumulate to an almost sure success over a long enough time."

Perhaps you can appreciate my confusion! Can I ask you then - does the screen accumulate essentially all incident energy (perhaps from a succession of many individual photons), until releasing on a statistical basis in one go - 'single click'. Or is incident energy rapidly dissipated after each photon hit (as I believe the case)?

 

[A. Neumaier]

Well it seems we have been having a long running communication problem, despite your excellent technical grasp of English. From #107:

Originally Posted by Q-reeus:
'But that's the real sticking point as I see it. This 'best case scenario' assumes the screen is not only capable of fully absorbing all incident radiation, but losslessly accumulating each hit for perhaps hours until sufficient energy is present to eject one electron. What's more in order to reproduce the interference pattern, a memory of the incident intensity is also dissipationlessly stored. This seems utterly incredible.' Your response:

"No storage is needed. Just a rate of response. That's the nature of probability - even tiny rates accumulate to an almost sure success over a long enough time."

Perhaps you can appreciate my confusion! Can I ask you then - does the screen accumulate essentially all incident energy (perhaps from a succession of many individual photons), until releasing on a statistical basis in one go - 'single click'. Or is incident energy rapidly dissipated after each photon hit (as I believe the case)?

The confusion is between the screen as a whole and the many embedded qubits that respond. In the older mails, I had concentrated on the qubits and their behavior. These have no memory but respond locally and independently, according to a simple stochastic law. But the total energy (a nonlocal quantity) is accumulated by the screen as a whole, essentially continuously. This ensures that, on the average, the energy needed for firing the qubits equals the energy provided by the photon field. For an ideal screen, there is no dissipation of energy, only its redistribution within the screen.

 

[Q-reeus]

The confusion is between the screen as a whole and the many embedded qubits that respond. In the older mails, I had concentrated on the qubits and their behavior. These have no memory but respond locally and independently, according to a simple stochastic law. But the total energy (a nonlocal quantity) is accumulated by the screen as a whole, essentially continuously. This ensures that, on the average, the energy needed for firing the qubits equals the energy provided by the photon field. For an ideal screen, there is no dissipation of energy, only its redistribution within the screen.

Well glad you have taken the trouble to clear that issue up - thanks. This has removed to a certain level any mystery re energy balance. Still have overall serious misgivings but I'll give this topic a rest for now!