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Must photons be absorbed or emitted one by one?

by xxfzero
Tags: absorbed, emitted, photons
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xxfzero
#1
Jul31-14, 08:43 AM
P: 4
Good evening.

Usually we say that during light-matter interaction, energy of light should be transferred piece by piece, rather than continuously, with each piece equals hbar*ω. But in photon echo experiment, we say atoms are excited into a "superposition state" of ground state and 1st excited. For simplicity, we suggest a (1/sqrt(2)) (|0>+|1>) superposition state, thus expectation of energy of this state is 1/2*hbar*ω (incident light is on resonance). So it seems that each atom absorbs half photon, is that correct?

Thanks a lot.
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Timo
#2
Jul31-14, 09:09 AM
P: 324
To keep the discussion simplified: The statement that the photons get fully absorbed refers to the state after measurement, which is either absorbed or non-absorbed. It does not refer to the probability distribution before the measurement. A non-QM example: A dice does role only integer numbers, despite 3.5 being the expectation value of a single dice roll.
xxfzero
#3
Jul31-14, 09:52 AM
P: 4
Quote Quote by Timo View Post
To keep the discussion simplified: The statement that the photons get fully absorbed refers to the state after measurement, which is either absorbed or non-absorbed. It does not refer to the probability distribution before the measurement. A non-QM example: A dice does role only integer numbers, despite 3.5 being the expectation value of a single dice roll.
Thank you for your answer first.

I have understood what you say about atom state after measurement, but if I do not care these atoms, and measure the unabsorbed photons, what will happen? For simplicity, I suggest 1 photon impinge on 1 atom, and cross section is large enough so that this atom is excited by photon, and exactly excited into the superposition state above. Then sensor after sample cell measures these unabsorbed photons, how many photons will be detected? I guess half chance to get 0, and half chance to get 1. Does it means that during flight from atom to sensor, the light field is a superpostion state of 0 photon and 1 photon?

Thank you very much.

Timo
#4
Aug1-14, 03:21 AM
P: 324
Must photons be absorbed or emitted one by one?

Assume a picture where there are three distinct time regimes: An incoming photon before interaction with an atom, a time after the interaction but before measurement, and a time after measurement. After the interaction, the system indeed is in a state that can be called a superposition of the photon being absorbed and the photon not being absorbed. The crucial point that is referred to by statements like "atoms can only absorb photons as a whole" is that no addend in the superposition represents a state with a partially-absorbed photon. And, more importantly but also a bit more technical, that the overlap between your superposition state and a state with a half-absorbed photon is zero.

Pseudo-mathematically: Assume fully-absorbed state, half-absorbed state and non-absorbed state were three orthogonal states |0>, |0.5>, |1>, respectively. Then, <0.5| (A |0> + B |1> ) = 0. Note that |0.5> would probably be a photon with half the original energy, in which case <0|0.5>=0 and <1|0.5> can be justified.
xxfzero
#5
Aug13-14, 03:22 PM
P: 4
Quote Quote by Timo View Post
Assume a picture where there are three distinct time regimes: An incoming photon before interaction with an atom, a time after the interaction but before measurement, and a time after measurement. After the interaction, the system indeed is in a state that can be called a superposition of the photon being absorbed and the photon not being absorbed. The crucial point that is referred to by statements like "atoms can only absorb photons as a whole" is that no addend in the superposition represents a state with a partially-absorbed photon. And, more importantly but also a bit more technical, that the overlap between your superposition state and a state with a half-absorbed photon is zero.

Pseudo-mathematically: Assume fully-absorbed state, half-absorbed state and non-absorbed state were three orthogonal states |0>, |0.5>, |1>, respectively. Then, <0.5| (A |0> + B |1> ) = 0. Note that |0.5> would probably be a photon with half the original energy, in which case <0|0.5>=0 and <1|0.5> can be justified.
Thank you, Timo.

Actually, I am haunted by another problem for long time. This one is also in such a "an atom probably absorbs photon" situation. When incident light is resonant with transition, usually we use rotation wave approximation (RWA), which says that positive frequency part mainly interact with atom and excite population to 1st excited state (and then form Rabi oscillation) while negative part just add a small rapid oscillation. What does this mean in photon's fashion, the atom at ground state only absorbs positive frequency photon? Or this cannot be stated in photon's fashion?
Cthugha
#6
Aug13-14, 06:42 PM
Sci Advisor
P: 1,659
Quote Quote by xxfzero View Post
What does this mean in photon's fashion, the atom at ground state only absorbs positive frequency photon? Or this cannot be stated in photon's fashion?
When going from the classical description to the quantum description you go from the positive and negative frequency fields to photon annihilation and destruction operators. If you do so, you get 4 main terms:

1) transition from the ground state to the excited state and a photon gets destroyed
2) transition from the ground state to the excited state and a photon gets created
3) transition from the excited state to the ground state and a photon gets destroyed
4) transition from the excited state to the ground state and a photon gets created

Obviously, only processes 1) and 4) conserve energy and the other two terms do not. The RWA now corresponds to keeping only the terms which conserve energy. The other two terms are strongly suppressed and are discarded. Such non-conserving processes are only important on really short timescales which are equivalent to the timescale over which the rapid oscillation in the RWA averages out to 0.
xxfzero
#7
Aug14-14, 03:51 AM
P: 4
Quote Quote by Cthugha View Post
When going from the classical description to the quantum description you go from the positive and negative frequency fields to photon annihilation and destruction operators. If you do so, you get 4 main terms:

1) transition from the ground state to the excited state and a photon gets destroyed
2) transition from the ground state to the excited state and a photon gets created
3) transition from the excited state to the ground state and a photon gets destroyed
4) transition from the excited state to the ground state and a photon gets created

Obviously, only processes 1) and 4) conserve energy and the other two terms do not. The RWA now corresponds to keeping only the terms which conserve energy. The other two terms are strongly suppressed and are discarded. Such non-conserving processes are only important on really short timescales which are equivalent to the timescale over which the rapid oscillation in the RWA averages out to 0.
Thank you, Cthugha.

I can understand your meaning. My previous picture is atoms at ground state absorb some amplitude of "positive frequency part" and transit to excited state, and then the amplitude of positive frequency part of light field is weaker than negative frequency part, and as a result, light field is no longer REAL field, but a COMPLEX field.

I'm sure what you mean is correct, but I cannot find what's wrong with my previous picture. It must be something inconsistent. Could you please point it out?


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