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Wikipedia claims here that the 2 out of 3 property is the following relationship between unitary, orthogonal, symplectic, and general linear complex groups:

$U(n)=Sp(2n,R)∩O(2n)∩GL(n,C)$

Intuitively I would think the symplectic group would contain the Hamiltonian dynamics of the physical system considering Hamiltonian mechanics is defined on a symplectic manifold. In searching for this connection I stumbled upon this answer to an unrelated question here. In the top rated answer, Cristoph seems to confirm my intuition by claiming this is indeed the role of the symplectic component. His answer goes further to claim that the orthogonal component governs the probability and $GL(C)$ is the connecting piece between these two aspects of quantum mechanics.

This seems very satisfying in light of the connection between the Schrodinger equation and its polar decomposition in the de Broglie-Bohm formulation of quantum mechanics where the real part corresponds to a continuity equation for the probability current and the phase corresponds to the Hamilton-Jacobi equation with a quantum correction.

$\psi =Re^{iS/ \hbar}$

$\frac{\partial R}{\partial t} = - \frac{1}{2m}[R \nabla^2 S + 2 \nabla R \cdot \nabla S]$

Continuity equation representing "orthogonal" group contribution as probability

$\frac{\partial S}{\partial t} = - [\frac{|\nabla S|^2}{2m} + V - \frac{\hbar^2}{2m} \frac{\nabla^2 R}{R}]$

Hamilton-Jacobi equation representing "symplectic" group contribution as dynamics.

$i \hbar \frac{\partial}{\partial t} \psi = (- \frac{\hbar^2}{2m} \nabla^2 + V) \psi = H \psi $

Schrodinger equation representing general linear complex group with the unitary time evolution operator, $\psi(t) = e^{iHt/\hbar} \psi(0)$, representing the unitary group.

However, my mathematical intuition leads me to a different conclusion. The unitary group is the set of transformations that preserve the standard inner product on $V(n,C)$. Along these lines I would interpret the intersection to include only matrices that preserve the inner product of $GL(2n,R)$ and $Sp(2n,R)$ simultaneously. Considering that the complex numbers can be represented as the following matrices, I feel this intersection can be interpreted by how it preserves the inner product:

$ a+ ib$ is equivalent to $ \left( \begin{array}{ccc} a & -b \\ b & a \end{array} \right) $

A complex vector in $V(2,C)$ could be expressed as the following matrix for example:

$ \left( \begin{array}{ccc} a + ib \\ c + id \end{array} \right) \simeq \left( \begin{array}{ccc} a & 0 & -b & 0 \\ 0 & c & 0 & -d \\ b & 0 & a & 0 \\ 0 & d & 0 & c\end{array} \right)$

Then the inner product would be calculated as:

$\left( \begin{array}{ccc} a & 0 & -b & 0 \\ 0 & c & 0 & -d \\ b & 0 & a & 0 \\ 0 & d & 0 & c\end{array} \right) \cdot \left( \begin{array}{ccc} a & 0 & b & 0 \\ 0 & c & 0 & d \\ -b & 0 & a & 0 \\ 0 & -d & 0 & c\end{array} \right) = \left( \begin{array}{ccc} a^2 + b^2 & 0 & 0 & 0 \\ 0 & c^2 + d^2 & 0 & 0 \\ 0 & 0 & a^2+b^2 & 0 \\ 0 & 0 & 0 & c^2 +d^2 \end{array} \right)$

Along the diagonal blocks you have the standard inner product on the vector space, $V(2n,R)$, which the orthogonal group, $O(2n)$, preserves and in this case $n = 2$ so we have a 4 dimensional matrix. On the off diagonal blocks you have the inner product on a symplectic vector space of dimension $2n$ which the group $Sp(2n,R)$ preserves. The inner product on this 4 dimensional space represents the standard inner product on $V(n,C)$ and thus the 4 dimensional matrices that preserve this inner product must preserve the inner products for these 2 components of the 4 dimensional matrices. I know this is far from a proof but I hope I've illustrated how my mathematical intuition interprets this intersection in terms of conditions on the inner product of these matrices. Is this a mathematically correct interpretation of the group intersection?

Which of these is the correct way to interpret the connection between the symplectic group and the unitary group in quantum mechanics? If the mathematical connection is correct, does the symplectic structure of Hamiltonian mechanics have any connection with the form of the operators besides the operators unitarily representing the Galilei group?

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  • $\begingroup$ The relation between the symplectic group Sp(2m,R) and unitary operators acting on a space of wavefunctions is that there is an infinite-dimensional unitary representation of the symplectic group carried on the space of wavefunctions $\psi(q)$ or $\psi(p)$. Mathematicians call it the Metaplectic rep. Ray optics uses the finite-dimensional matrix defining rep of Sp(2,R) or Sp(4,R) whilst diffractive optics in the Fresnel approximation is the infinite dimensional unitary rep. $\endgroup$ – Stephen Blake Sep 24 '15 at 9:55
  • $\begingroup$ This answer confuses me since the metaplectic group has no finite dimensional representation (en.wikipedia.org/wiki/Metaplectic_group) while the representation of U(n) in terms of this group intersection suggests it has a representation over a finite vector space in terms of a symplectic group in general. Perhaps I'm not knowledgeable in representation theory to appreciate how this is the relationship for unitary groups in general. I do appreciate that this is a representation of a symplectic group on a unitary group, but I feel this relationship is more general than representation. $\endgroup$ – Daniel Kerr Sep 24 '15 at 17:16
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This answer outlines how the defining matrix representation of the symplectic group Sp(2m,R) is ray optics, whilst the infinite-dimensional unitary rep of Sp(2m,R) carried on the space of wavefunctions is diffractive optics in the Fresnel approximation. The outline is for Sp(2,R) (cylindrical lenses) but the generalization to Sp(2m,R) is reasonably straightforward.

Firstly, ray optics must be put into Hamiltonian form. Fermat's principle of least time gives the action, \begin{equation} S=\int dt = \int \frac{nds}{c}=\int dz\frac{n}{c}\sqrt{1+\dot{x}^{2}}=\int Ldz \end{equation} where $n$ is refractive index, c is the vacuum velocity of light, $ds$ is arc length, $\dot{x}=dx/dz$. The z-coord is now in the role of time in the action and $L$ is the Lagrangian. The canonically conjugate momentum is, \begin{equation} p=\frac{n}{c}\frac{dx}{ds} \end{equation} and using natural units $c=1$, the canonical momentum is the direction cosine of the ray in air when $n=1$.

In optics, one is interested in the input/output relations. These are canonical transformations $p=p(P,Q), q=q(P,Q)$ where the $q$'s are the coordinates $x$. The canonical transformations are generated by functions of the "astride" variables such as $G(q,Q)$ with $p=\partial G/\partial q$, $P=-\partial G/\partial Q$.

For paraxial optics, the generators are quadratic functions of the astride variables. For example, \begin{equation} G(q,Q)=-\frac{1}{2D}(Bq^{2}-2qQ+CQ^{2})\ . \end{equation} The quadratic generators give rise to linear canonical transformations. \begin{equation} \left[ \begin{array}{c} p\\ q \end{array} \right]= \left[ \begin{array}{cc} B & -A\\ -D & C \end{array} \right] \left[ \begin{array}{c} P\\ Q \end{array} \right]= \end{equation} The $2\times 2$ transfer matrices are the defining representation of the symplectic group Sp(2,R) carried on the two dimensional vector space (p,q) because they satisfy the defining property of the symplectic group that the antisymmetric polarity matrix transforms trivially under Sp(2,R). \begin{equation} \left[ \begin{array}{cc} B & -A\\ -D & C \end{array} \right]^{T} \left[ \begin{array}{cc} 0 & 1\\ -1 & 0 \end{array} \right]\left[ \begin{array}{cc} B & -A\\ -D & C \end{array} \right]=\left[ \begin{array}{cc} 0 & 1\\ -1 & 0 \end{array} \right] \end{equation} These $2\times 2$ transfer matrices are used in the practical computations of paraxial ray optics. Hence, ray optics in the paraxial approximation is the defining rep of Sp(2,R) carried on the vector phase space $(p,q)$. A lens may be thought of as a concrete realization of a $g\in$ Sp(2,R).

Now we seek the infinite-dimensional unitary rep of Sp(2,R) carried on the space of wavefunctions $\psi(q)$. In this rep, the phase space variables $(p,q)$ become operators acting on the space of wavefunctions. The momentum operator is, \begin{equation} \hat{p}=-\frac{i}{\omega}\frac{d}{dq} \ . \end{equation}
The canonical momentum $p$ has dimensions of second/metre and $q$ has dimensions of metre, so the constant $1/\omega$ is needed with dimensions of second. (It turns out that $\omega$ is the angular frequency of the light being diffracted.) The constant $\omega$ is in the role of Planck's constant. From now on, use natural units $c=1$ and $\omega=1$.

Here is a heuristic way that suggests the form of the infinite-dimensional transformation matrix rep of Sp(2,R). \begin{equation} \langle q|\hat{p}|Q\rangle=-i\frac{d}{dq}\langle q|Q\rangle=\langle q|\hat{\frac{\partial G}{\partial q}}|Q\rangle=\frac{\partial G}{\partial q}\langle q|Q\rangle \end{equation} Integrating, \begin{equation} \langle q|Q\rangle \sim e^{iG(q,Q)} \ . \end{equation} More careful analysis shows that the infinite-dimensional unitary rep of sp(2,R) is given by the infinite-dimensional matrices (aka kernel), \begin{equation} [U(g)]^{q}_{\ Q}=\frac{1}{\sqrt{-iD}}e^{-\frac{i}{2D}(Bq^{2}-2qQ+CQ^{2})}\ . \end{equation} The transformation of the input wavefunction $\psi(Q)$ to the output wavefunction $U\psi(q)$ is, \begin{equation} U\psi(q)=\int [U(g]^{q}_{\ Q}\psi(Q)\frac{dQ}{\sqrt{2\pi}} \end{equation} where the measure is $dQ/\sqrt{2\pi}$. This formula is how the lens $g\in$ Sp(2,R) diffracts the input wave field $\psi(Q)$ in the Fresnel approximation. Note that we are not doing quantum mechanics here, this is just the classical Fresnel diffraction of light by a lens. $\psi(q)$ is one component of (say) the classical electric field of the light.

To summarize, a lens is a $g\in$ Sp(2,R). The way the lens bends the rays of light in the paraxial approximation is handled by the $2\times 2$ matrix rep of Sp(2,R) carried on the 2-d vector space $(p,q)$. The way the lens diffracts light in the Fresnel approximation is handled by the infinite-dimensional unitary rep $[U(g)]^{q}_{\ Q}$ carried on the space of wavefunctions $\psi(q)$.

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  • $\begingroup$ Like I said for your previous comment, this symplectic symmetry group is explicitly being represented in a unitary group. The 2 out of 3 property implies that the unitary group, which we use for the representation of the Galilei group in quantum mechanics and Sp(2m,R) in optics as you've just explain, inherently already has a symplectic component to it. In this question I'm trying to get at if the Hamiltonian mechanics of the Galilei group is projected on to this symplectic component in the unitary representation. This is still a good answer as I think it highlights an important structure. $\endgroup$ – Daniel Kerr Sep 27 '15 at 0:16

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