Effect of wavelength on photon detection

Question

When some photon detector detects a photon, is it an instantaneous process (because a photon can be thought of as a point particle), or does the detection require a finite amount of time depending on the wavelength of the photon?

EDIT: I guess what I am wondering is if a photon has a wavelength and travels at a finite speed, then if a photon had a wavelength of 300,000,000m, would its interaction with the detector last 1s? Or does the uncertainty principle say that a photon with wavelength 300,000,000m (and therefore energy E), it cannot be known exactly when it hit the detector with an accuracy better than 1s. Or is it more like this: suppose there is a stream of photons moving towards the detector with wavelengths of 300,000,000m and they reach the detector at a rate of 10 photons/second and the detector has a shutter speed such that the shutter is open for 1s at a time, then it would record 10 photon hits (records all the photons). But if the shutter speed is only 0.5s, then it would record 2.5 hits on average?

EDIT2: I'm not interested in the practical functioning of the detector and amplification delays. I'm looking at and ideal case (suppose the photon is 'detected' the instant an electron is released from the first photomultiplier plate). It is a question regarding the theory of the measurement, not the practical implementation.

Answer 1

It is hard to answer this because the question seems to be based on confused premises.

A photon hitting something happens pretty much instantaneously. What happens after that in the process of creating some kind of signal as a result of the photon hitting the detector can take various amounts of time, depending on the type of detector.

In a photomultiplier, there is a cascade of electrons hitting a sequence of plates. The system is arranged such that each plate will emit more electrons than it received, evenutally amplifying the direct effect of the photon hitting the first plate many times.

In a semiconductor photodiode, some of the charges have to diffuse out of the depletion layer before current can be observed externally.

I don't remember the exact mechanism of a CdS cell, but it eventually allows more current to pass for the same applied voltage.

All these secondary effects take time, and of course also have some upper frequency content. Therefore the result will never be a infinitely thin pulse, but a signal that rises and falls over time, with the peak coming some time after the photo actually crashed into the detector. This has nothing to do with photons, and is all about frequency limitations and the response times of other physical phenomena.

Answer 2

To address your revised question, here's a press release and paper describing the production of a laser pulse with duration 67×10^-18 s. That corresponds to having localized an ensemble of photons within about 20 nm of free space. The photons in the pulse had energies in the range 55–130 eV, or wavelengths of 10–20 nm. So the pulse is compressed into roughly a single wavelength. This is a minimum-uncertainty wavepacket, and is probably the best that you'll be able to do without involving shorter-wavelength photons.

That paper (or the literature that it cites) should give you an idea of the techniques involved in measuring these very short time intervals.

You seem more interested in the long-wavelength limit. For instance, suppose you have an AM radio receiver tuned to a station at 1 MHz (λ = 300 m). Does it take one microsecond for the receiver to respond to a single radio photon?

Here you have again the problem of practical details intruding on your theory of measurement. Measuring the 4 neV of energy deposited by a single radio photon would be an impossible feat of calorimetry. A real radio interacts with a coherent stream of radio photons, to produce large-scale coherent motion of the conduction electrons in the receiver circuit. To measure this coherent motion you have to wait for it to finish. If you could construct a minimum-uncertainty wavepacket of AM radio photons, your "detection" would consist of watching the electrons in the antenna move through the receiver as the radio pulse passed by, which would take about a microsecond.

The coherent motion of the free charges in the receiver probably has a quantum-mechanical representation as an ensemble of phonons in the degenerate Fermi gas of conduction electrons. You might be able to make an argument that the phonons spring into existence along the entire length of the antenna and leak into the receiver. But in practice there are so many phonons involved that the problem is essentially classical, so nobody treats it that way. Without some mechanism to distinguish between phonons that appear instantaneously and phonons that evolve into existence as the photons that produce gradually vanish, your question just doesn't have an answer. If we can't detect a single radio photon, so we don't get to learn how long it takes to detect.

I think that your desire to consider a "theory of measurement" that is divorced from any "practical implementation" is wrongheaded and unphysical. Consider the history of the Bohr-Einstein debates, in which every intellectual advance arose as the result of either a thought experiment or a real experiment.

Answer 3

My experience with photon detectors doesn't really extend to your example wavelength. A photon with $\lambda = 3\times10^8$ m would have an energy of $E = hc/\lambda \approx 4\times10^{-15}$ eV, which is a wavelength you might find in a thermal distribution with temperature $T = E/k = 50$ pK. (To my surprise, there's a claim of a laboratory experiment reaching 100 pK.) I can't think of any detector that would interact with a femtoelectronvolt photon, so I can't speak to your specific example.

For practical detection of a single photon, you have to have some sort of a machine that can take the microscopic energy involved and amplify it until it's visible with a macroscopic system. As far as I know, the time constants in these sorts of apparatus are entirely due to the macroscopic effects.

The most common tool for detecting single photons in the infrared, visible, and near-UV is the photomultiplier tube. A single photon liberates a single electron from the photocathode, with a few eV of leftover energy. This "photoelectron" is steered onto a metal sheet held a few hundred volts above the level of the cathode. Its collision with this "dynode" spits out three or four more electrons, which get steered onto another dynode a few hundred volts higher again. By chaining together several dynodes you can get 10⁷ electrons — a couple of picocoulombs of charge — from a single photon. A good fast PMT with this sort of gain responds to a single photon at the cathode by emitting a negative pulse from its anode of about 0.2 milliamperes that lasts for about 10 nanoseconds.

However, if you are able to rapidly digitize the pulse from a photomultiplier (with a fast oscilloscope or equivalent hardware), you may find that while charge is received over about ten nanoseconds, the shape of the pulse is reliable. I have heard of folks achieving timing stability of about 100 picoseconds, roughly the time it takes for light to travel an inch. This is still quite large compared to the wavelength of an optical photon, so any variation in the actual time required for the photon to interact with the cathode is negligible.

So, the answer to your question as originally asked is that the duration of a photon detector's measurement has very little correlation with any hypothetical duration of a single-particle photon-electron interaction.

Answer 4

The wavelength of a photon is closely related to the minimum possible uncertainty in its position. So, for a photon has a wavelength of $3x10^8m$, and if we assume that before the detection we already know as much about the photon as is possible, we still won't know exactly when it will be detected, and the uncertainty will be on the order of a second.

The interaction between the photon and the detector is itself in principle instantaneous, and in principle there is no limit on how accurately we can measure when the interaction occurred. It's all about what we can or can't predict in advance.

Answer 5

The simple answer is "no" it is not a point event. That violates the Heisenberg Uncertainty Principle. Re: Quantum Mechanics Claude Cohen-Tannoudji or Re: An Introduction to Theory and Applications of Quantum Mechanics Amnon Yariv.

Energy and time are conjugate variables, so the energy of a photon and the time it exists cannot both be determined to infinite precision / accuracy. Momentum and position are likewise conjugate variables and thus, the wavelength (momentum) and position cannot both be determined to infinite precision / accuracy. So, from this standpoint, you cannot accurately determine the "time" it takes to "absorb" the photon. You can't really think about this as "how many oscillations does it take to liberate an electron".

I think you are best thinking of this in two ways:

1. Antenna the detection of a photon is in some ways like an antenna. The photon's oscillations influence the motion of electrons in a material like inducing current in an antenna. Using this semi-classical approach, it is easy to compute / measure the lifetime of a transition and how it is related to the energy bandwidth of the transition. This implies that it is not a "point" event.
2. Slit diffraction with a photon counting device When you look at experiments with photon counting devices (e.g. PMT or APD) and you count one event at a time (see http://www.personal.psu.edu/agr126/doubleSlit.pdf) then over time, you see the wave-like properties of the photon.

So, it "point"-like behavior (e.g. you can detect a single electron liberated by a single photon). It has Hiesenberg-like behavior (e.g. you can't know the wavelength perfectly, so you can't know the energy perfectly, so you can't know the time required to "absorb" the photon perfectly). It also has wave-like behavior (e.g. the oscillating nature of the photon creates the quantum mechanical transition to a higher state to occur and liberate the photon).

Answer 6

Well I would argue a much simpler and shorter explanation:

That the measurement of a photon will collapse the wave function of the photon and therefore the system will no longer be a quantum mechanical one but classical. Before the measurement there will be a error due to the uncertainty principle, see it like the photon wave is hitting the detector but the photon is not yet detected. Once detected, the wave function is localized.

So to answer your question. It is an instantaneous process. But due to QM there will be a uncertainty in your measurement

Answer 7

Its hard to imagine a photon with a frequency as low as one cycle per second considering green light with a 500nm wavelength has a frequency of about 600 trillion cycles per second. Considering your question from the point of view of a photon traveling at c with a low frequency of 1 cps helps to imagine the process in slow motion. If a photon has an electromagnetic frequency then I imagine it frequently oscillating through cycles of positive and negative electromagnetic amplitudes. Depending on the distance and actual frequency of the source a photon could be impacting the electron at the detection screen in a positive or negative amplitude or maybe even in between. Couldn't this play a part in the overall accumulation of the photons? Just like crests and troughs of a wave individual photons could impact and convoluted.