# Boson statistics

## Main Question or Discussion Point

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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Bill_K
I think it doesn't fail, so I'm sure someone will explain to me why I'm wrong! 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
= (100|100,0> + |99,1>)/√2.

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Ken G
Gold Member
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.

DrDu
I think it doesn't fail, so I'm sure someone will explain to me why I'm wrong! 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.

DrDu
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?

"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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I think it doesn't fail, so I'm sure someone will explain to me why I'm wrong! 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
= (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?

Cthugha