Quasi-integrals of motion

By construction, for a Hamiltonian in BGNF is a formal integral of motion (see section 2.1). We now show how to find an analogous formal integral for a Hamiltonian in generalized normal form. Our results are similar to the findings of Meyer and Hall [19], but the proof differs in some details. We have tried to make the exposition as transparent as possible by focusing on just those aspects that are essential for the reasoning.

We write
as

(16) |

(17) |

(18) |

**Theorem:**
* For a Hamiltonian
in generalized normal form the
diagonalizable part
of
is a formal integral
of motion.
*

For the proof we must show that the Poisson bracket of with vanishes for all . We start with and then proceed to the case .

By virtue of Jacobi's identity for the Poisson bracket we have

(20) |

We now show that for two commuting matrices , the corresponding
Lie operators
,
(defined as above)
commute as well:

Because this implies that the right hand side of (27) is zero, and thus .

For we proceed in the following way: We show that diagonalizability and nilpotence of the matrices and carry over to the corresponding Lie operators and ; these properties then imply that the null spaces of and coincide and that , which in turn means .

Consider a unitary matrix that transforms into the diagonal matrix
.
Inserting twice the identity into the expression for
we get

With , and denoting in the new coordinates by , we obtain

Application of this transformed operator to any of the basis monomials of yields, because is diagonal, an eigenvalue equation with the eigenfunction and a certain eigenvalue -- thus diagonalizability of is shown.

Now consider any .
is a
polynomial in of degree less than or equal to ,
since by nilpotence there is some such that .
This polynomial is related to in the following way:

Iterating this expression and evaluating for we get

which implies nilpotence of , because (31) holds for all .

Identity of the null spaces of and is a direct
implication of diagonalizability: Application of the diagonalized operator
(cf. (30)) yields

So the eigenspaces corresponding to the eigenvalue 0 of and are identical.

Finally, we determine how acts on polynomials in . Notice that is nilpotent, because its adjoint is. With we obtain for any :

because and commute (since the corresponding matrices and commute; cf. (29)). is zero for and because is in generalized normal form. From this it follows for that

here, for the sake of notational convenience, we have again turned to the coordinates as defined above. Linear independence of the basis monomials then gives the result and thus

and we have proven the theorem.