# What's the relationship between uncertainty principle and symplectic groups?

What's the relationship between uncertainty principle and symplectic groups? Does the symplectic groups mathematically capture anything fundamental about uncertainty principle?

• Why do you believe there is a relationship? – ACuriousMind Aug 11 '14 at 23:11
• You're not thinking of the Heisenberg group and its formulation in symplectic spaces, are you? See here and here. Symplectic groups are very different from the HG as the latter is nilpotent. – WetSavannaAnimal Aug 11 '14 at 23:29

The reason for the relationship is that the symplectic groups are defined by preserving an antisymmetric bilinear invariant, $$M A M^T = A$$ where $M$ is a matrix included into the symplectic group is the equation holds and $A$ is a non-singular antisymmetric matrix.
Where does the uncertainty principle enter? It enters because $A$ may be understood to be the commutator (or Poisson bracket) of the basic coordinates $x_i,p_i$ on the phase space. If we summarize $N$ coordinates $x_i$ and $N$ coordinates $p_i$ into a $2N$-dimensional space with coordinates $q_m$, their commutators are $$[q_m,q_n] = A_{mn}$$ with an antisymmetric matrix $A$. Consequently, the symplectic transformations may be defined as the group of all linear transformations mixing $x_i,p_i$, the coordinates of the phase space, that preserve the commutator i.e. all the uncertainty relations between the coordinates $q_m$.
Symplectic groups: Not directly. But symplectic manifolds (or even better: Poisson manifolds): Yes. The Poisson bracket $\{f,g\}_{PB}$ and the functions $f$, $g$ are the classical counterparts of the quantum commutator $[\hat{f},\hat{g}]$ and the operators $\hat{f}$, $\hat{g}$. The quantum commutator $[\hat{f},\hat{g}]$, in turn, gives rise to the Heisenberg uncertainty principle for the observables $\hat{f}$, $\hat{g}$.