Bose-Einstein Probabilities

  • #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
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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
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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.

even allowing that we must replace ##HT+TH## with ##HT##?

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

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
vanhees71
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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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