The Geometry of Ensembles on Configuration Space

Abstract

A description of ensembles on configuration space incorporates at least two geometrical structures which arise in a natural way: a metric structure, which derives from the natural geometry associated with a space of probabilities, and a symplectic structure, which derives from the symplectic geometry associated with a Hamiltonian description of motion. We show that these two geometrical structures give rise to a Kähler geometry. We first consider probabilities P and introduce the information metric. This leads to information geometry, a Riemannian geometry defined on the space of probabilities. We then bring in dynamics via a Hamiltonian formalism defined on a phase space with canonically conjugate coordinates P and S. This leads to more geometrical structure, a symplectic geometry defined on this phase space. The next step is to extend the information metric, which is defined over the space of probabilities only, to a metric over the full phase space. This requires satisfying certain conditions which ensure the compatibility of the metric and symplectic structures. These conditions are equivalent to requiring that the space have a Kähler structure. In this way, we are led to a Kähler geometry. This rich geometrical structure allows for a reconstruction of the geometric formulation of quantum theory. One may associate a Hilbert space with the Kähler space and this leads to the standard version of quantum theory. Thus the theory of ensembles on configuration space permits a geometric derivation of quantum theory.

Appendices

Appendix 1: Symplectic Geometry, Compatibility Conditions, and Kähler Structure

We consider a finite space, but similar relations hold for infinite dimensional spaces. A symplectic vector space is a vector space V that is equipped with a bilinear form ${\varOmega } : V \times V \rightarrow R$ that is [22]:

a. Skew-symmetric: $\varOmega (u,v) = -\varOmega (v,u)$ for all $u,v~\epsilon ~V$,

b. Non-degenerate: if $\varOmega (u,v) = 0$ for all $v~\epsilon ~V,$ then $u=0$.

The standard space is $\mathfrak {R}^{2n}$, and typically ${\varOmega }$ is chosen to be the matrix

$$\begin{aligned} \varOmega _{ab} =\left( \begin{array}{cc} 0 &{} {\mathsf{1}} \\ -{\mathsf{1}} &{} 0 \end{array} \right) , \end{aligned}$$

(6.58)

where ${\mathsf{1}}$ is the unit matrix in n dimensions.

Consider the dual space $V^*$. The symplectic structure can be identified with an element of $V^* \times V^*$, so that $\varOmega (u,v) = \varOmega _{ab}u^a v^b$. Since the spaces V and $V^*$ are isomorphic, there is a $\varOmega ^{ab}$ that is the dual of $\varOmega _{ab}$. This $\varOmega ^{ab}$ can be identified with an element of $V \times V$. The convention is to set [22]

$$\begin{aligned} \varOmega ^{ab} = -(\varOmega ^{-1})^{ab} =\left( \begin{array}{cc} 0 &{} {\mathsf{1}} \\ -{\mathsf{1}} &{} 0 \end{array} \right) , \end{aligned}$$

(6.59)

so that $\varOmega ^{ac}\varOmega _{cb}=-\delta ^{a}_{~b}.$

We assume there is a metric in the space, $g_{ab}=g_{ba}$, and a corresponding inverse metric $g^{ab}$ with $g_{ab}g^{bc}=\delta _{ac}$ (indices are raised and lowered with $g_{ab}$ and $g^{ab}$). The metric also defines a map $V \rightarrow V^*$ to the dual space in an obvious way. Therefore, the space has two linear operators that induce maps $V \rightarrow V^*$, $\varOmega _{ab}$ and $g_{ab}$. They will be related by an equation of the form $\varOmega _{ab} = g_{ac} J^c_{~b}$ for some choice of linear operator $J^c_{~b}$. This is Eq. (6.22), the first of the Kähler conditions.

The relations $\varOmega ^{ac}\varOmega _{cb}=-\delta ^{a}_{~b}$ and $\varOmega _{ab} = g_{ac} J^c_{~b}$ lead to the condition $J^a_{~s} J^s_{~c} = -\delta ^{a}_{~c}$, that is, that $J^a_{~b}$ is a complex structure. This is Eq. (6.24), the third of the Kähler conditions.

Finally, $\varOmega _{ab} = g_{ac} J^c_{~b}$ and $J^a_{~s} J^s_{~c} = -\delta ^{a}_{~c}$, together with the symmetries $-\varOmega _{cb} = \varOmega _{bc}$ and $g_{ba}=g_{ab}$, lead to $g_{cd} = J^a_{~c} g_{ab} J^b_{~d}$ which is Eq. (6.23), the second of the Kähler conditions.

This shows that consistency requirements imply that a space with both symplectic and metric structures must have a Kähler structure.

Appendix 2: Generalized Markov Mappings

1.1 Constants of the Motion and Dynamics

To derive the canonical transformation that generalizes the Markov mapping of Eq. (6.13), the dimensionality of the original phase space was increased by two in a trivial way. This led to two constants of the motion, $P^{m+1}$ and $S^{m+1}$. $P^{m+1}$ was set to $P^{m+1} \approx 0$, with corresponding constraints for the $\tilde{P}^k$ of the form

$$\begin{aligned} \tilde{P}^m\approx & {} kP^m,\nonumber \\ \tilde{P}^{m+1}\approx & {} (1-k)P^m. \end{aligned}$$

(6.60)

These are precisely the conditions that are needed to get a generalization of the Markov mapping of Eq. (6.13). When $k=1/2$, $\tilde{P}^m \approx \tilde{P}^{m+1}$, which is expected because in this case there should be invariance under the re-labeling $m \leftrightarrow m+1$. To fix the value of $S^{m+1}$, notice that $S^{m+1} \approx c$ leads to constraints for the $\tilde{S}^k$ of the form

$$\begin{aligned} \tilde{S}^m\approx & {} S^m+(1-k) c,\nonumber \\ \tilde{S}^{m+1}\approx & {} S^m-k c. \end{aligned}$$

(6.61)

We can argue once more that there should be invariance under the re-labeling $m \leftrightarrow m+1$ in the case when $k=1/2$. But this can only be satisfied if $c=0$. We can conclude therefore that the constants of the motion must satisfy

$$\begin{aligned} P^{m+1}= & {} (1-k) \tilde{P}^m - k \tilde{P}^{m+1} \approx 0,\nonumber \\ S^{m+1}= & {} \tilde{S}^m - \tilde{S}^{m+1} \approx 0. \end{aligned}$$

(6.62)

The corresponding constraints for the unprimed coordinates are of the form

$$\begin{aligned} \tilde{P}^m\approx & {} kP^m,~~~~~\tilde{P}^{m+1} \approx (1-k)P^m,\nonumber \\ \tilde{S}^m\approx & {} S^m,~~~~~~~\tilde{S}^{m+1} \approx S^m. \end{aligned}$$

(6.63)

We now need to check that the dynamics on the $2(n+1)$-dimensional phase space (with coordinates with tildes) reproduces precisely the dynamics on the original 2n-dimensional phase space (with coordinates without tildes). Using Eqs. (6.33–6.34), one can show that

$$\begin{aligned} \frac{\partial \tilde{P}^i}{\partial t}= & {} \frac{\partial H}{\partial S^i},~~~~~~~~ \frac{\partial \tilde{P}^m}{\partial t} = \frac{\partial H}{\partial S^m}\,k,~~~~~ \frac{\partial \tilde{P}^{m+1}}{\partial t} = \frac{\partial H}{\partial S^m}\,(1-k),\nonumber \\ \frac{\partial \tilde{S}^i}{\partial t}= & {} -\frac{\partial H}{\partial P^i},~~~~~ \frac{\partial \tilde{S}^m}{\partial t} = -\frac{\partial H}{\partial P^m},~~~~~ \frac{\partial \tilde{S}^{m+1}}{\partial t} = - \frac{\partial H}{\partial P^m}. \end{aligned}$$

(6.64)

These equations lead to

$$\begin{aligned} \frac{\partial P^i}{\partial t}= & {} \frac{\partial \tilde{P}^i}{\partial t} = \frac{\partial H}{\partial S^i},~~~~~~~~ \frac{\partial P^m}{\partial t} = \frac{\partial \tilde{P}^m}{\partial t} + \frac{\partial \tilde{P}^{m+1}}{\partial t} = \frac{\partial H}{\partial S^m}\nonumber \\ \frac{\partial S^i}{\partial t}= & {} \frac{\partial \tilde{S}^i}{\partial t} = -\frac{\partial H}{\partial P^i},~~~~~ \frac{\partial S^m}{\partial t} = k\,\frac{\partial \tilde{S}^m}{\partial t} + (1-k)\,\frac{\partial \tilde{S}^{m+1}}{\partial t} = -\frac{\partial H}{\partial P^m},\qquad \qquad \end{aligned}$$

(6.65)

which are the correct equations of motion for the original space.

1.2 Invariance of the Kähler Metric Under Generalized Markov Mappings

The metric of the Kähler space is given by

$$\begin{aligned} g_{ab}=\left( \begin{array}{cc} {\mathsf{G}} &{} {\mathsf{A}}^{T} \\ {\mathsf{A}} &{} ({\mathsf{1}}+{\mathsf{A}}^2){\mathsf{G}}^{-1} \end{array} \right) , \end{aligned}$$

(6.66)

where ${\mathsf{G}}=\mathrm {diag}(\alpha /2P^i)$ and the $n \times n$ matrix A satisfies ${\mathsf{G}}{\mathsf{A}}{\mathsf{G}}^{-1} ={\mathsf{A}}^{T}$. For the calculations in this Appendix it is convenient to introduce the matrix ${\mathsf{B}}$ with matrix elements given by

$$\begin{aligned} B_{jk}=\sqrt{{P_j}/{P_k}}\;A_{jk}. \end{aligned}$$

(6.67)

It is straightforward to show that ${\mathsf{B}}$ is a symmetric matrix, $B_{jk}=B_{kj}$.

To restrict the form of ${\mathsf{B}}$, it will be sufficient to consider the invariance of the metric under the particular generalized Markov mapping which corresponds to the inverse canonical transformation given by Eq. (6.34). After taking into consideration the constraints, Eq. (6.35), the generalized Markov mapping can be written in the form

$$\begin{aligned} P= & {} (P^i,P^m) \rightarrow ~~ \tilde{P} = (\tilde{P}^i, \tilde{P}^m, \tilde{P}^{m+1}) := (P^i, kP^m, (1-k)P^m),\nonumber \\ S= & {} (S^i,S^m)\, \rightarrow ~~ \tilde{S} = (\tilde{S}^i, \tilde{S}^m, \tilde{S}^{m+1})\,:= (S^i, S^m, S^m), \end{aligned}$$

(6.68)

where $i=1,\ldots ,m-1$.

As a first step, look at the contribution to the line element $d\sigma ^2$ from the mixed terms $dP^k dS^k$. In terms of the coordinates without tildes,

$$\begin{aligned} d\sigma ^2&= \sum _{i=1}^{m-1}\left\{ B_{ii}dP^i dS^i + \sqrt{\frac{P^i}{P^m}}B_{im}dP^i dS^m + \sqrt{\frac{P^m}{P^i}}B_{mi}dP^m dS^i \right\} \nonumber \\&\quad + B_{mm}dP^m dS^m. \end{aligned}$$

(6.69)

There is a corresponding expression for the coordinates with tildes, and with the help of Eq. (6.68) it can be rewritten in terms of coordinates without tildes. This leads to

$$\begin{aligned} d\sigma ^2&=\sum _{i=1}^{m-1}\left\{ \tilde{B}_{ii}dP^i dS^i + \left[ \sqrt{\frac{P^i}{kP^m}}\tilde{B}_{im} + \sqrt{\frac{P^i}{(1-k)P^m}}\tilde{B}_{i(m+1)}\right] dP^i dS^m \right\} \nonumber \\&\quad + \sum _{i=1}^{m-1}\left\{ \left[ \sqrt{\frac{k^3P^m}{P^i}}\tilde{B}_{mi} + \sqrt{\frac{(1-k)^3P^m}{P^i}}\tilde{B}_{(m+1)i}\right] dP^m dS^i \right\} \nonumber \\&\quad + \left[ k\tilde{B}_{mm} + \sqrt{\frac{k^3}{1-k}}\tilde{B}_{m(m+1)} \right. \nonumber \\&\quad \left. + \sqrt{\frac{(1-k)^3}{k}}\tilde{B}_{(m+1)m} + (1-k)\tilde{B}_{(m+1)(m+1)}\right] dP^m dS^m. \end{aligned}$$

(6.70)

Equate terms in Eqs. (6.69) and (6.70) proportional to the same $dP^a dS^b$, where $a,b=1,\ldots ,m$. This leads to the four relations

$$\begin{aligned} B_{ii}&= \tilde{B}_{ii}, \nonumber \\ B_{im}&= \sqrt{\frac{1}{k}}\tilde{B}_{im} + \sqrt{\frac{1}{(1-k)}}\tilde{B}_{i(m+1)}, \nonumber \\ B_{mi}&= \sqrt{k^3}\tilde{B}_{mi} + \sqrt{(1-k)^3}\tilde{B}_{(m+1)i}, \nonumber \\ B_{mm}&= k\tilde{B}_{mm} + \sqrt{\frac{k^3}{1-k}}\tilde{B}_{m(m+1)} + \sqrt{\frac{(1-k)^3}{k}}\tilde{B}_{(m+1)m}\nonumber \\&\quad + (1-k)\tilde{B}_{(m+1)(m+1)}. \end{aligned}$$

(6.71)

Since the matrix ${\mathsf{B}}$ is symmetric, $B_{im}=B_{mi}$, which leads to

$$\begin{aligned} \sqrt{\frac{1}{k}}\tilde{B}_{im} + \sqrt{\frac{1}{(1-k)}}\tilde{B}_{i(m+1)} = \sqrt{k^3}\tilde{B}_{mi} + \sqrt{(1-k)^3}\tilde{B}_{(m+1)i}. \end{aligned}$$

(6.72)

By symmetry, $\tilde{B}_{im}=\tilde{B}_{i(m+1)}$ at $k=1/2$, but this relation can only be satisfied if $\tilde{B}_{im}=\tilde{B}_{i(m+1)}=0$, which in turn implies $B_{im}=B_{mi}=0$. Since $B_{im}$ and $B_{mi}$ are independent of k, it follows that they must always be zero. This shows that the off-diagonal elements of the matrix B are zero.

Now look at the contribution to the line element $d\sigma ^2$ from terms proportional to $dP^a dP^a$ and $dS^a dS^a$. The terms proportional to $dP^a dP^a$ give the two relations

$$\begin{aligned} B_{ii}= & {} \tilde{B}_{ii}, \nonumber \\ B_{mm}= & {} k\tilde{B}_{mm} + (1-k)\tilde{B}_{(m+1)(m+1)}, \end{aligned}$$

(6.73)

while the terms proportional to $dS^a dS^a$ give the two relations

$$\begin{aligned} 1+B_{ii}^2= & {} 1+\tilde{B}_{ii}^{~2}, \nonumber \\ 1+B_{mm}^2= & {} k(1+\tilde{B}_{mm}^{~2})+(1-k)\left( 1+\tilde{B}_{(m+1)(m+1)}^{~2}\right) . \end{aligned}$$

(6.74)

Combining Eqs. (6.73) and (6.74) leads to

$$\begin{aligned} B_{ii}= & {} \tilde{B}_{ii}, \nonumber \\ B_{mm}= & {} \tilde{B}_{mm}= \tilde{B}_{(m+1)(m+1)}. \end{aligned}$$

(6.75)

Notice that this result is valid for arbitrary values of k. Since there is nothing special about the particular labels m and $(m+1)$, all the diagonal elements of the matrices ${\mathsf{B}}$ and $\tilde{{\mathsf{B}}}$ must be equal. Then

$$\begin{aligned} {\mathsf{B}}= & {} B {\mathsf{1}}_{m \times m}, \nonumber \\ \tilde{{\mathsf{B}}}= & {} B {\mathsf{1}}_{(m+1) \times (m+1)}, \end{aligned}$$

(6.76)

where ${\mathsf{1}}_{n \times n}$ is the $n \times n$ unit matrix and B still has to be determined.

To carry out this last step, use the relations

$$\begin{aligned} B(P,S)= & {} B(P^i,P^m,S^i,S^m), \nonumber \\ \tilde{B}(\tilde{P},\tilde{S})= & {} B(P(\tilde{P}),S(\tilde{S})) = B(\tilde{P}^i,\tilde{P}^m+\tilde{P}^{m+1},\tilde{S}^i,k\tilde{S}^m+(1-k)\tilde{S}^{m+1}).\nonumber \\ \end{aligned}$$

(6.77)

The functional form of B(P, S) must be the same as the functional form of $\tilde{B}(\tilde{P},\tilde{S})$, and these expressions must be both invariant under permutations and independent of k. The only functional form that seems to satisfy all these conditions appears to be $B(P,S)=B(\sum _i P^i)$. But $\sum _i P^i=1$, therefore one can conclude that B is a constant and the matrix

$$\begin{aligned} {\mathsf{B}}=B \mathbf 1 \end{aligned}$$

(6.78)

is a constant matrix proportional to the unit matrix. This in turn implies that

$$\begin{aligned} {\mathsf{A}}=A \mathbf 1 \end{aligned}$$

(6.79)

where A is a constant.