- #1
jostpuur
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Introduction and warm up
Suppose H_0 is some Hamilton's operator that has eigenvectors |\psi_n\rangle for n\in\mathbb{N} with some eigenvalues E_n so that
<br /> H_0|\psi_n\rangle = E_n|\psi_n\rangle,\quad \forall n\in\mathbb{N}.<br />
Suppose we define a new Hamilton's operator by setting H=H_0\otimes \textrm{id}+\textrm{id}\otimes H_0. Now one possible set of eigenvectors is given by a product |\psi_n\rangle\otimes |\psi_{n'}\rangle for all (n,n')\in\mathbb{N}^2, and these have eigenvalues E_n+E_{n'}.
Suppose we insist that only those vectors |\psi\rangle, which satisfy the symmetry property
<br /> (\langle x|\otimes\langle y|)|\psi\rangle = (\langle y|\otimes \langle x|)|\psi\rangle<br />
are allowed. If the original H_0 was interpreted as acting on a state of some particle, now H will be acting an a pair of bosonic particles.
Now the acceptable eigenvectors will be (up to a factor) |\psi_n\rangle\otimes |\psi_{n'}\rangle + |\psi_{n'}\rangle\otimes |\psi_n\rangle.
Suppose we would like to write down a formula for the set of pairs (E,|\psi\rangle) 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
<br /> \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\}<br />
However, this formula contains some pairs in redundant way, and we might also write down the set of pairs as
<br /> \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\}<br />
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 T. There are two obvious options available.
The first option is that we begin by allowing the domain of the index pair (n,n') to be the full set \mathbb{N}^2. 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
<br /> (n,n')\mapsto e^{-\frac{E_n+E_{n'}}{T}},\quad\forall n,n'\in\mathbb{N}<br />
Suppose we regret the choice of domain, and would like to shrink it with the additional costraint n\leq n'. In the shrunk domain the new probability distribution, which would be equivalent to the just axiomatically defined one, would be proportional to
<br /> (n,n')\mapsto \left\{\begin{array}{ll}<br /> e^{-\frac{2E_n}{T}},\quad & n=n' \\<br /> 2e^{-\frac{E_n+E_{n'}}{T}},\quad & n<n' \\<br /> \end{array}\right.<br />
The second option is that we begin by allowing the domain of the index pair (n,n') to be the set \{(n,n')\;|\;n,n'\in\mathbb{N},\; n\leq n'\}. 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
<br /> (n,n')\mapsto e^{-\frac{E_n+E_{n'}}{T}},\quad\forall n,n'\in\mathbb{N},\; n\leq n'<br />
Suppose we regret the choice of domain, and would like to extend it to the full set \mathbb{N}^2. In the extended domain the new probability distribution, which would be equivalent to the just axiomatically defined one, would be proportional to
<br /> (n,n')\mapsto \left\{\begin{array}{ll}<br /> e^{-\frac{2E_n}{T}},\quad & n=n' \\<br /> \frac{1}{2}e^{-\frac{E_n+E_{n'}}{T}},\quad & n\neq n' \\<br /> \end{array}\right.<br />
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?
Suppose H_0 is some Hamilton's operator that has eigenvectors |\psi_n\rangle for n\in\mathbb{N} with some eigenvalues E_n so that
<br /> H_0|\psi_n\rangle = E_n|\psi_n\rangle,\quad \forall n\in\mathbb{N}.<br />
Suppose we define a new Hamilton's operator by setting H=H_0\otimes \textrm{id}+\textrm{id}\otimes H_0. Now one possible set of eigenvectors is given by a product |\psi_n\rangle\otimes |\psi_{n'}\rangle for all (n,n')\in\mathbb{N}^2, and these have eigenvalues E_n+E_{n'}.
Suppose we insist that only those vectors |\psi\rangle, which satisfy the symmetry property
<br /> (\langle x|\otimes\langle y|)|\psi\rangle = (\langle y|\otimes \langle x|)|\psi\rangle<br />
are allowed. If the original H_0 was interpreted as acting on a state of some particle, now H will be acting an a pair of bosonic particles.
Now the acceptable eigenvectors will be (up to a factor) |\psi_n\rangle\otimes |\psi_{n'}\rangle + |\psi_{n'}\rangle\otimes |\psi_n\rangle.
Suppose we would like to write down a formula for the set of pairs (E,|\psi\rangle) 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
<br /> \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\}<br />
However, this formula contains some pairs in redundant way, and we might also write down the set of pairs as
<br /> \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\}<br />
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 T. There are two obvious options available.
The first option is that we begin by allowing the domain of the index pair (n,n') to be the full set \mathbb{N}^2. 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
<br /> (n,n')\mapsto e^{-\frac{E_n+E_{n'}}{T}},\quad\forall n,n'\in\mathbb{N}<br />
Suppose we regret the choice of domain, and would like to shrink it with the additional costraint n\leq n'. In the shrunk domain the new probability distribution, which would be equivalent to the just axiomatically defined one, would be proportional to
<br /> (n,n')\mapsto \left\{\begin{array}{ll}<br /> e^{-\frac{2E_n}{T}},\quad & n=n' \\<br /> 2e^{-\frac{E_n+E_{n'}}{T}},\quad & n<n' \\<br /> \end{array}\right.<br />
The second option is that we begin by allowing the domain of the index pair (n,n') to be the set \{(n,n')\;|\;n,n'\in\mathbb{N},\; n\leq n'\}. 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
<br /> (n,n')\mapsto e^{-\frac{E_n+E_{n'}}{T}},\quad\forall n,n'\in\mathbb{N},\; n\leq n'<br />
Suppose we regret the choice of domain, and would like to extend it to the full set \mathbb{N}^2. In the extended domain the new probability distribution, which would be equivalent to the just axiomatically defined one, would be proportional to
<br /> (n,n')\mapsto \left\{\begin{array}{ll}<br /> e^{-\frac{2E_n}{T}},\quad & n=n' \\<br /> \frac{1}{2}e^{-\frac{E_n+E_{n'}}{T}},\quad & n\neq n' \\<br /> \end{array}\right.<br />
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?
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