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I Boltzmann bosons factor two

  1. Jan 30, 2017 #1
    Introduction and warm up

    Suppose [itex]H_0[/itex] is some Hamilton's operator that has eigenvectors [itex]|\psi_n\rangle[/itex] for [itex]n\in\mathbb{N}[/itex] with some eigenvalues [itex]E_n[/itex] so that
    [tex]
    H_0|\psi_n\rangle = E_n|\psi_n\rangle,\quad \forall n\in\mathbb{N}.
    [/tex]

    Suppose we define a new Hamilton's operator by setting [itex]H=H_0\otimes \textrm{id}+\textrm{id}\otimes H_0[/itex]. Now one possible set of eigenvectors is given by a product [itex]|\psi_n\rangle\otimes |\psi_{n'}\rangle[/itex] for all [itex](n,n')\in\mathbb{N}^2[/itex], and these have eigenvalues [itex]E_n+E_{n'}[/itex].

    Suppose we insist that only those vectors [itex]|\psi\rangle[/itex], which satisfy the symmetry property
    [tex]
    (\langle x|\otimes\langle y|)|\psi\rangle = (\langle y|\otimes \langle x|)|\psi\rangle
    [/tex]
    are allowed. If the original [itex]H_0[/itex] was interpreted as acting on a state of some particle, now [itex]H[/itex] will be acting an a pair of bosonic particles.

    Now the acceptable eigenvectors will be (up to a factor) [itex]|\psi_n\rangle\otimes |\psi_{n'}\rangle + |\psi_{n'}\rangle\otimes |\psi_n\rangle[/itex].

    Suppose we would like to write down a formula for the set of pairs [itex](E,|\psi\rangle)[/itex] which would contain all the eigenvalues and eigenvectors. There is two obvious ways to write down the set. The simplest formula for the set of pairs is
    [tex]
    \Big\{\big(E_n+E_{n'}, |\psi_n\rangle\otimes |\psi_{n'}\rangle + |\psi_{n'}\rangle\otimes |\psi_n\rangle\big)\;\Big|\; n,n'\in\mathbb{N}\Big\}
    [/tex]
    However, this formula contains some pairs in redundant way, and we might also write down the set of pairs as
    [tex]
    \Big\{\big(E_n+E_{n'}, |\psi_n\rangle\otimes |\psi_{n'}\rangle + |\psi_{n'}\rangle\otimes |\psi_n\rangle\big)\;\Big|\; n,n'\in\mathbb{N},\; n\leq n'\Big\}
    [/tex]
    These two formulas define the same set, because the elements of these two sets are the same. As long as we are only interested in the set itself, there is no trouble in sight.

    The problems start here

    Suppose we would like to apply Boltzmann's distribution to this pair of bosons under the assumption that they interact with warm surroundings of some temperature [itex]T[/itex]. There are two obvious options available.

    The first option is that we begin by allowing the domain of the index pair [itex](n,n')[/itex] to be the full set [itex]\mathbb{N}^2[/itex]. There does not seem to be anything wrong with this choice alone. Once this decision has been made, it seems reasonable to define, perhaps axiomatically, the probability distribution to be proportional to the function
    [tex]
    (n,n')\mapsto e^{-\frac{E_n+E_{n'}}{T}},\quad\forall n,n'\in\mathbb{N}
    [/tex]
    Suppose we regret the choice of domain, and would like to shrink it with the additional costraint [itex]n\leq n'[/itex]. In the shrunk domain the new probability distribution, which would be equivalent to the just axiomatically defined one, would be proportional to
    [tex]
    (n,n')\mapsto \left\{\begin{array}{ll}
    e^{-\frac{2E_n}{T}},\quad & n=n' \\
    2e^{-\frac{E_n+E_{n'}}{T}},\quad & n<n' \\
    \end{array}\right.
    [/tex]

    The second option is that we begin by allowing the domain of the index pair [itex](n,n')[/itex] to be the set [itex]\{(n,n')\;|\;n,n'\in\mathbb{N},\; n\leq n'\}[/itex]. There does not yet seem to be anything wrong with this choice either. Once this decision has been made, it seems reasonable to define, perhaps axiomatically, the probability distribution to be proportional to the function
    [tex]
    (n,n')\mapsto e^{-\frac{E_n+E_{n'}}{T}},\quad\forall n,n'\in\mathbb{N},\; n\leq n'
    [/tex]
    Suppose we regret the choice of domain, and would like to extend it to the full set [itex]\mathbb{N}^2[/itex]. In the extended domain the new probability distribution, which would be equivalent to the just axiomatically defined one, would be proportional to
    [tex]
    (n,n')\mapsto \left\{\begin{array}{ll}
    e^{-\frac{2E_n}{T}},\quad & n=n' \\
    \frac{1}{2}e^{-\frac{E_n+E_{n'}}{T}},\quad & n\neq n' \\
    \end{array}\right.
    [/tex]

    The question: Now we have two different Boltzmann's distributions, which both look reasonable, but since they are different, they cannot be both right. So is one of these right, and the other one wrong? Which way around would be the correct answer?
     
    Last edited: Jan 30, 2017
  2. jcsd
  3. Jan 31, 2017 #2

    DrClaude

    User Avatar

    Staff: Mentor

    You have to use the proper partition function. In the first case, the partition function will have factor 1/2 to account for the over-counting of indistinguishable cases.
     
  4. Jan 31, 2017 #3
    While using the domain [itex]\mathbb{N}^2[/itex], in the first case the partition function will be
    [tex]
    Z(T) = \sum_{n,n'=0}^{\infty} e^{-\frac{E_n+E_{n'}}{T}}
    [/tex]
    and the probability distribution will be
    [tex]
    p(n,n') = \frac{1}{Z(T)}e^{-\frac{E_n+E_{n'}}{T}},\quad n,n'\in\mathbb{N}
    [/tex]
    Did you mean to claim that the partition function should be something different, with some factor [itex]\frac{1}{2}[/itex] somewhere? I don't believe that that would be right, because bringing such factors in would imply that
    [tex]
    \sum_{n,n'=0}^{\infty}p(n,n')=1
    [/tex]
    would no longer hold.

    Equivalently, while using the domain [itex]\{(n,n')\;|\;n,n'\in\mathbb{N},n\leq n'\}[/itex], in the first case the partition function will be
    [tex]
    Z(T) = \sum_{n=0}^{\infty} e^{-\frac{2E_n}{T}} + 2\underset{n<n'}{\sum_{n,n'=0}^{\infty}} e^{-\frac{E_n+E_{n'}}{T}}
    [/tex]

    In the second case, which will not be equivalent to the first one, while using the domain [itex]\{(n,n')\;|\;n,n'\in\mathbb{N},n\leq n'\}[/itex], the partition function will be
    [tex]
    Z(T) = \underset{n\leq n'}{\sum_{n,n'=0}^{\infty}} e^{-\frac{E_n+E_{n'}}{T}}
    [/tex]

    Did you mean that this second option looks more right than the first one? The second option has some kind of over-counting removed, and it looks nice, but how do you know that the over-counting was wrong in the first place?
     
    Last edited: Jan 31, 2017
  5. Jan 31, 2017 #4

    DrClaude

    User Avatar

    Staff: Mentor

    The sum in the partition should run over all states. If you can't distinguish between ##(n,n')## and ##(n',n)##, then these two cases correspond to the same state, hence the sum you have there is not correct: you are over-counting.

    See, e.g., http://theory.physics.manchester.ac.uk/~judith/stat_therm/node84.html
     
  6. Jan 31, 2017 #5
    I explained two different ways to define the Boltzmann's distribution, and put forward a question that which one of them is right. Your answer is the same as that the first option was wrong, and the second option was right.
     
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