Consider a particle detector whose principal component is a conteiner with monoatomic gas whose atoms do not interact, except by collisions. Assume that the conteiner walls are perfectly reflective to the atoms, though transparent to light. Let the gas be at the room temperature and the average free path of an atom exceede by much its de Broglie wavelength. Thus, the atoms would obey the Maxwell-Boltzmann statistics. Inside the conteiner are placed the charged plates of a capacitor, but as long as the atoms are neutral, non-ionized, the field had no influence.
Consider that we send upon this conteiner photons able to dislocate an electron from the atom, creating an electron-ion pair. The interaction can be described as
(1) |ph> |An> --> q |phscat> |An,scat> + p |ph'inel> |A'n,inel> + s |ionn> |electronn>.
An is the atom characterized by the velocity vn. The mplitudes q, p, s, are functions of vn. ph' is a photon exiting the conteiner with a lowered energy after inellastic scattering, and A' is the atom with modified center-of-mass linear momentum. All the states |ph'inel> |A'n,inel>, |ionn>, and |electronn>, depend on vn .
I have a couple of questions:
1) Does the transformation (1) describe sufficiently the interaction of the photon with the gas? Should we take in consideration the interaction of the photon with a single atom? It seems plausible that for describing the interaction of the photon with the detector, the microscopic state of the entire gas has to appear in the formula.
2) What is the initial state of the gas? Assume that the photons emitted in collisions and lost to the environment of the conteiner through the transparent walls, bring a negligible modification to the gas state. In this case it is plausible that the gas be described by a wave-function,
(2) α|A1>|A2> . . .|AN> + α'|A'1>|A'2> . . .|A'N> + α"|A"1>|A"2> . . .|A"N> + α"'|A"'1>|A"'2> . . .|A"'N> +. . .
where in each product, the atoms have another set of velocities compatible with the Maxwell-Boltzmann distribution at the given temperature.
However, for maintaining constant the temperature of the gas, the conteiner has to be kept in a thermostat. Thus the gas before the visit of the photon is in a mixture of states, and not in a pure state.
3) If the gas is in a mixture of states, each time a photon enters the conteiner it finds the gas in the same microstate? The most plausible answer seems to be "No". Then the equation (1) should be replaced by
(3) |ph> Σs ηs (Σn |ψs,n>)(Σn is the state s with the atom no. n having undergone the transformation (1).
Am a little confused, you talk about a particle detector and yet you are interested in the effect of a photon?
Seems that the photon energy would go to produce the electron ion pair plus any residual kinetic energy.
The electron and the ion would go to the plates, presumably indicating detection of the photon.
Depends on the incoming energy as to further avalanch effects or not.
What question are you really interested in?
Yes, Juan, I consider the photon also a particle despite not having rest-mass.
"Seems that the photon energy would go to produce the electron ion pair plus any residual kinetic energy."
Correct! The photon is supposed to dislocate the electron and if the photon is entirely absorbed, then, if there remains some energy besides the one needed to ionize the atom, it goes the kinetic energy of the electron and of the center-of-mass of the ion. The latter one is much smaller than the former, because the mass of the electron is more than 3 orders of magnitude smaller than the ion mass.
The electron and the ion will be accelerated in the electric field and collide with other atoms, producing additional electron-ion pairs. In all, an avalanche would result, the ions (electrons) would fall on the negative (positive) plate of the capacitor, reducing its charge and therefore the difference of potential. This effect would be recorded.
However, my point in the question is that, if we want to describe the state of the gas before and immediately after the interaction photon-atom, the equation (1) is not sufficient, the equation (3) is the correct one.
"What question are you really interested in?"
Very smart of you asking me this. But I'll answer you in a private message.
With kind regards!
What happens inside the detector would depend on a lot of things, the initial photon energy, the density of atoms,
mean free path of the ionizing charges, etc. You could even have auger effects.
Yes, Juan, but as a major feature, the detector is an open system. It can't have a pure quantum state as some people claim. Moreover, a Maxwell-Boltzman gas is a classic gas, with independent molecules, because the mean free path exceeds by far the Broglie wavelength. Each microstate of the gas is a product of states of the individual atoms.The issue is that the microstates differ from one another by the set of linear momenta of the atoms. Also, there are microstates of higher probability, and microstates of lower probability, depending on the set of velocities. E.g. microstates in which most of the atoms have very low velocities and a few atoms very high velocities, are improbable. The same in the vice-versa case.
Anyway, all these microstates don't form a coherent quantum superposition, because, as I said, we have an open system (connected with the thermostat).
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