Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Bose-Einstein Probabilities

  1. Apr 18, 2017 #1
    This paper https://arxiv.org/abs/quant-ph/9911101 says this:
    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?
     
  2. jcsd
  3. Apr 18, 2017 #2

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    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.
     
  4. Apr 18, 2017 #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?
     
    Last edited: Apr 18, 2017
  5. Apr 19, 2017 #4

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    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.

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

    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.
     
  6. Apr 19, 2017 #5

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    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##.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Bose-Einstein Probabilities
Loading...