Must photons be absorbed or emitted one by one?

  • Thread starter xxfzero
  • Start date
  • Tags
    Photons
In summary, the conversation discussed the concept of photon echo experiments and the behavior of atoms and photons during light-matter interactions. It was mentioned that in these experiments, atoms are excited into a "superposition state" of ground state and 1st excited state, and the expectation of energy for this state is 1/2*hbar*ω. It was also mentioned that statements like "atoms can only absorb photons as a whole" refer to the fact that no addend in the superposition represents a state with a partially-absorbed photon. The conversation also touched on the use of rotation wave approximation in describing the behavior of photons and atoms during interaction. In summary, the RWA corresponds to keeping only the terms which conserve energy,
  • #1
xxfzero
16
0
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.
 
Physics news on Phys.org
  • #2
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.
 
  • #3
Timo said:
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.
 
Last edited:
  • #4
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.
 
  • #5
Timo said:
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?
 
  • #6
xxfzero said:
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.
 
  • Like
Likes xxfzero
  • #7
Cthugha said:
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?
 

1. What are photons?

Photons are tiny particles of light that have no mass and travel at the speed of light.

2. How are photons created?

Photons are created through electromagnetic interactions, such as the movement of electrons, or through nuclear reactions.

3. Do photons always have to be absorbed or emitted one by one?

No, photons can be absorbed or emitted in groups, depending on the energy of the source and the material it is interacting with.

4. Can photons be destroyed?

Photons cannot be destroyed, but they can be absorbed by matter and their energy can be transferred to other particles.

5. What happens when a photon is absorbed?

When a photon is absorbed, its energy is transferred to the absorbing material, causing changes in the material's properties, such as heating or exciting electrons.

Similar threads

  • Atomic and Condensed Matter
Replies
23
Views
2K
  • Atomic and Condensed Matter
Replies
3
Views
999
  • Atomic and Condensed Matter
Replies
6
Views
1K
Replies
4
Views
2K
  • Atomic and Condensed Matter
Replies
3
Views
2K
  • Atomic and Condensed Matter
Replies
10
Views
1K
  • Atomic and Condensed Matter
Replies
9
Views
2K
  • Atomic and Condensed Matter
Replies
6
Views
1K
  • Atomic and Condensed Matter
Replies
8
Views
2K
  • Atomic and Condensed Matter
Replies
9
Views
6K
Back
Top