Bose-Einstein Probabilities

In summary, With classical statistics, there are four possible outcomes with equal probabilities, while with Bose-Einstein statistics, there are only three possible outcomes with equal probabilities. When considering all possible states, the probabilities for each outcome are normalized to 1/3.
  • #1
This paper https://arxiv.org/abs/quant-ph/9911101 says this:
With classical statistics, i.e., where the particles are distinguishable, there are four possible outcomes:
##HH, HT, TH, TT##
Since all four outcomes are a priori equally likely, the probability for HH is 1/4. This is applicable to tossing
macroscopic coins, where quantum effects are negligible.

With Bose–Einstein statistics, where the allowable states must be symmetric under exchange, there are only three possible outcomes:
##HH, \quad (HT+TH)/ \sqrt{2}, \quad TT##
Consequently, the probability for HH increases to 1/3.
If we normalize those terms, don't we get 1/4 , 1/2 , 1/4 as the probabilities, since ##|HT\rangle## and ##|TH \rangle## are indistinguishable?
 
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  • #2
Swamp Thing said:
If we normalize those terms, don't we get 1/4 , 1/2 , 1/4 as the probabilities, since ##|HT\rangle## and ##|TH \rangle## are indistinguishable?

No. What ##|HT\rangle## and ##|TH\rangle## being "indistinguishable" means is that they are one outcome, not two. So they only count once when we are counting outcomes.
 
  • #3
Thanks. But how are we to deal with the ##1/ \sqrt{2}##, even allowing that we must replace ##HT+TH## with ##HT##?

The amplitudes are 1/4, ##1/4*1/ \sqrt{2}##, 1/4 which give
1/16 , (1/16)*(1/2), 1/16 when squared.
Normalizing, I seem to see
0.4, 0.2, 0.4
and no 1/3 anywhere. I'm sure I'm going wrong somewhere, but where?
 
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  • #4
Swamp Thing said:
how are we to deal with the ##1/ \sqrt{2}##,

We don't have to do anything to "deal" with it. It's there so that each of the 3 distinguishable outcomes have properly normalized state vectors associated with them. And as normalized, they all have equal amplitudes. That's what tells you that all three outcomes have equal probabilities.

Swamp Thing said:
even allowing that we must replace ##HT+TH## with ##HT##?

Where do we do that? I don't understand.

Swamp Thing said:
The amplitudes are 1/4, ##1/4*1/ \sqrt{2}##, 1/4

No, they aren't. Each of the three state vectors ##|HH\rangle##, ##\left( |HT\rangle + |TH\rangle \right) / \sqrt{2}##, and ##|TT\rangle## are normalized, i.e., they are all unit vectors. So the amplitude of each, if we are looking at an equal combination of each of the state vectors, is 1/3, because there are three vectors and each one is a unit vector taken by itself.
 
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  • #5
What's given is a basis consisting of three orthonormalized vectors,
$$|b_1\rangle=|HH \rangle, \quad |b_2 \rangle=\frac{1}{2}(|HT \rangle+|TH \rangle), \quad |b_3 \rangle = |TT \rangle.$$
It's wrong to say that these are all allowable states, but these are given by all statistical operators in this 3D Hilbert space. What we can read from the above quoted text snippet is only that all states are equally probable. That's not complete information, and we need some objective criterion of how to associate a state with this information. The answer is provided by information theory: First one defines the Shannon-Jaynes-von-Neumann entropy as a measure for the missing information, given the state,
$$S[\rho]=-\mathrm{Tr}(\hat{\rho} \ln \hat{\rho}).$$
Then we want a state of minimal prejudice, i.e., with maximal entropy, compatible with the information. Here, we have given the probabilities for the outcomes of measurements to be
$$P_j=\langle b_j|\hat{\rho}|b_j \rangle.$$
This we have to minimize with these constraints and the constraint that ##\mathrm{Tr} \hat{\rho}=1##, i.e., we introduce Lagrange multipliers for these constraints
$$\tilde{S}[\rho]=\sum_j \lambda_j \langle b_j|\hat{\rho}|b_j \rangle+\lambda \mathrm{Tr} \hat{\rho}-\mathrm{Tr}(\hat{\rho} \ln \hat{\rho}).$$
Variation of the statistical operator gives
$$\delta \tilde{S}[\rho]=\sum_j [\lambda_j + \lambda-1] \delta \rho_{jj}-\mathrm{Tr} [\delta \hat{\rho} \ln \hat{\rho}] =0.$$
Since we can vary the 9 matrix elements of ##\hat{\rho}## independently now (thanks to the Lagrange parameters), the expression can only be 0 if ##[\ln \hat{\rho}]_{ij}=0## for ##i \neq j##, i.e., ##\hat{\rho}## is diagonal in the above given basis. After some algebra thus we get
$$\hat{\rho}=\sum_{j=1}^{3} P_j |b_j \rangle \langle b_j|, \quad \sum_{j=1}^3 P_j=1.$$
if all the ##P_j## are equal, as said in the above text snippet, we have necessarily ##P_j=1/3##.
 
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1. What are Bose-Einstein probabilities?

Bose-Einstein probabilities refer to the statistical distribution of particles that obey Bose-Einstein statistics, which describes the behavior of a group of identical particles that can occupy the same quantum state.

2. How are Bose-Einstein probabilities different from other statistical distributions?

Bose-Einstein probabilities are different from other statistical distributions, such as Fermi-Dirac and Maxwell-Boltzmann, as they take into account the quantum nature of particles and allow for multiple particles to occupy the same quantum state.

3. What is the significance of Bose-Einstein probabilities in physics?

Bose-Einstein probabilities have been used to explain various phenomena in physics, such as the behavior of superfluids and the Bose-Einstein condensate, and have also been applied in fields such as cosmology and condensed matter physics.

4. How are Bose-Einstein probabilities calculated?

Bose-Einstein probabilities can be calculated by using the Bose-Einstein distribution function, which takes into account the number of particles, the energy levels, and the temperature of the system.

5. Can Bose-Einstein probabilities be observed in everyday life?

While Bose-Einstein probabilities are more commonly observed in the behavior of particles at very low temperatures, they can also be seen in everyday life in the form of Bose-Einstein condensates in certain materials and the behavior of lasers.

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