Boson Statistics: Why Does the Rule Fail?

In summary: So, if you want to understand what you describe as paradox, you first need to define the process which leads to the 99 photons with horizontal polarization. If it is absorption and reemission, what is the state of the other particle after emission? How do you make sure that the polarization of reemitted photon matches the polarization of the other 99 photons? If it is scattering, how do you make sure the polarization of the scattered photon matches the polarization of the other 99 photons?The thing is, there is no paradox, it is just a very special and extreme case, where you applied a rather simple and (usually) correct rule in a situation where it does not hold and does not apply. So you need to
  • #1
phonon44145
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It is said that is there already n bosons in a particular quantum state, the probability of another boson joining them is (n+1) times larger than it would have been otherwise. But if we apply this rule to calculate probability for one horizontally (H) polarized photon to join a bunch of n=99 diagonally (D) polarized photons, we'll get (99+1) * |< D|H>|^2 = 100 * (1/2) = 50. Why does the rule fail?
 
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  • #2
I think it doesn't fail, so I'm sure someone will explain to me why I'm wrong! :smile:

To make it easier, let me switch Horizontal and Diagonal. The rule is a*|n> = √(n+1)|n+1>.

Suppose we have a state with 99 Horizontally polarized and zero Vertically polarized photons, written |99,0>. And we add one Diagonally polarized photon to it. The creation operator for a diagonally polarized photon is aD* = (aH* + aV*)/√2. We get
aD*|99,0> = (aH*|99,0> + aV*|99,0>)/√2
= (100|100,0> + |99,1>)/√2.
 
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  • #3
phonon44145 said:
It is said that is there already n bosons in a particular quantum state, the probability of another boson joining them is (n+1) times larger than it would have been otherwise.
How does that argument work? Sounds like some kind of stimulated emission process, but I would have thought of at as a physical process that is stimulated by the n photons, not a statistical probability effect.
 
  • #4
Bill_K said:
I think it doesn't fail, so I'm sure someone will explain to me why I'm wrong! :smile:

To make it easier, let me switch Horizontal and Diagonal. The rule is a*|n> = (n+1)|n+1>.
This should be root of n+1.
 
  • #5
phonon44145 said:
It is said that is there already n bosons in a particular quantum state, the probability of another boson joining them is (n+1) times larger than it would have been otherwise.
What does "otherwise" mean exactly?
 
  • #6
"What does "otherwise" mean exactly?"
This is a good question, and the one that I think lies at the root of the matter. If you read almost any textbook, they will simply say "the probability increases". My understanding is, what increases by (n+1) is the conditional probability of transition from |H> to |D> given that the same transition was made by n other bosons. Or formally,

Prob(D | n photons in state H) = (n+1) Prob (D | zero photons in state H)

(where the vertical bar means "conditioned on"). Is that the right way to put it? And if yes, would that mean the ONLY way to resolve the paradox is to say that conditional probability on the right is NOT equal 1/2?
 
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  • #7
Bill_K said:
I think it doesn't fail, so I'm sure someone will explain to me why I'm wrong! :smile:

To make it easier, let me switch Horizontal and Diagonal. The rule is a*|n> = √(n+1)|n+1>.

Suppose we have a state with 99 Horizontally polarized and zero Vertically polarized photons, written |99,0>. And we add one Diagonally polarized photon to it. The creation operator for a diagonally polarized photon is aD* = (aH* + aV*)/√2. We get
aD*|99,0> = (aH*|99,0> + aV*|99,0>)/√2
= (10|100,0> + |99,1>)/√2.

I replaced 100 with 10 as amplitude is given by sqrt 100 (not that it changes much). If the probabilities were to relate as 100:1, that would solve the problem. However, what happened to normalization? Looks like the amplitudes 10/√2 and 1/√2 must be attenuated by a normalizing constant?
 
  • #8
phonon44145 said:
It is said that is there already n bosons in a particular quantum state, the probability of another boson joining them is (n+1) times larger than it would have been otherwise.

It is not that easy. Bosonic final state stimulation results in the probability amplitude for processes which lead to bosons ending up in the same state and being indistinguishable to be proportional to n+1. The n is the stimulated emission/scattering/whatever effect, while the 1 is the spontaneous emission/scatteing/whatever.

Photons do not just join and you cannot just add some photon to 99 photon state by will. You need to define some process which does. So you first need to know how you want to make your horizontally polarized photon interact with something that can turn it into being in the same state the photons with horizontal polarization are in. This might be absorption and reemission or something similar. Just superposing the photons or arranging them in some random fashion on a beamsplitter does not work.
 

1. What is Boson Statistics?

Boson Statistics is a theory in quantum mechanics that describes the behavior of particles that have integer spin (such as photons and gluons). It is based on the principles of symmetry and probability, and is used to determine the properties and interactions of these particles.

2. What is the "Rule" in Boson Statistics?

The "Rule" in Boson Statistics refers to the principle that identical particles with integer spin are allowed to occupy the same quantum state. This means that they can exist in the same location and have the same energy level at the same time.

3. How does the Rule fail?

The Rule in Boson Statistics fails when the particles are in a state of degeneracy, meaning that there are more particles than available energy levels. In this case, the particles cannot all occupy the same energy level and must be distributed among the available levels, leading to a deviation from the expected behavior.

4. What are some real-life examples where the Rule fails?

One example of the Rule failing is in the behavior of Bose-Einstein condensates, which are collections of bosons cooled to extremely low temperatures. In this state, the particles become indistinguishable and the Rule fails, causing unusual behaviors such as superfluidity and superconductivity.

Another example is in the study of black holes, where the particles that make up the Hawking radiation are bosons. The Rule fails in this case, causing the radiation to behave in unexpected ways.

5. How is the failure of the Rule important in scientific research?

The failure of the Rule in Boson Statistics is important because it allows scientists to better understand the behavior of particles in extreme conditions, such as at very low temperatures or in the presence of strong gravitational fields. It also provides insights into the fundamental principles of quantum mechanics and the nature of matter and energy.

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