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Quasi-integrals of motion

By construction, for a Hamiltonian in BGNF $H_2$ 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 $H_2$ as

\quad H_2({\mbox{\protect\boldmath$z$}}) = \frac{1}{2} {\mb...
...t\boldmath$\cdot$}J^{-1}L {\mbox{\protect\boldmath$z$}} \quad.
\end{displaymath} (16)

and decompose $L$ by means of the Jordan-Chevalley decomposition [25] into its diagonalizable and nilpotent parts $D$ and $N$:
\quad L = D + N \quad.
\end{displaymath} (17)

Existence and uniqueness of this decomposition are assured by the Jordan normal form theorem for matrices. Define the diagonalizable component $I({\mbox{\protect\boldmath$z$}})$ and the nilpotent component $K({\mbox{\protect\boldmath$z$}})$ of $H_2({\mbox{\protect\boldmath$z$}})$ by
...cdot$}J^{-1}N {\mbox{\protect\boldmath$z$}} \quad,
\end{array}\end{displaymath} (18)

such that $H_2({\mbox{\protect\boldmath$z$}})=I({\mbox{\protect\boldmath$z$}})+K({\mbox{\protect\boldmath$z$}})$. We are now in the position to prove the main

Theorem: For a Hamiltonian $H({\mbox{\protect\boldmath$z$}})$ in generalized normal form the diagonalizable part $I({\mbox{\protect\boldmath$z$}})$ of $H_2({\mbox{\protect\boldmath$z$}})$ is a formal integral of motion.

For the proof we must show that the Poisson bracket of $I$ with $H_m$ vanishes for all $m\geq 2$. We start with $m=2$ and then proceed to the case $m>2$.

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

\mbox{\rm ad}_{\left\{H_2,I\right\}} = \mbox{\rm ad}_I\mbox{\rm ad}_{H_2} - \mbox{\rm ad}_{H_2}\mbox{\rm ad}_I \quad.
\end{displaymath} (19)

This expression is zero if $\mbox{\rm ad}_{H_2}$ and $\mbox{\rm ad}_{I}$ commute. In order to show that the latter is the case we first remark that the matrices $L$, $D$ and $N$ are infinitesimally symplectic [26], i.e. they satisfy $M^tJ+JM=0$ (for $M=L,D,N$). Direct computation shows that for an infinitesimally symplectic matrix $M$ the Lie operator adjoint to the quadratic polynomial $
P({\mbox{\protect\boldmath$z$}}) = \frac{1}{2}{\mbox{\protect\boldmath$z$}} \mbox{\protect\boldmath$\cdot$}J^{-1}M {\mbox{\protect\boldmath$z$}}
$ can be written as $
\mbox{\rm ad}_P(\cdot) = D_{\mbox{\protect\footnotesize\protect\boldmath$z$}}(\cdot) \mbox{\protect\boldmath$\cdot$}M {\mbox{\protect\boldmath$z$}}
$. Thus we have for the Lie operators ${\cal D}_m$ and ${\cal N}_m$ adjoint to $I$ and $K$:
{\cal D}_m...
...math$\cdot$}N {\mbox{\protect\boldmath$z$}} \quad.
\end{array}\end{displaymath} (20)

It is one of the key advantages of this formulation of the theory that we can characterize all the important operators ${\cal A}_m$, ${\cal D}_m$ and ${\cal N}_m$ which operate in a space of the high dimension ${2n+m-1 \choose 2n-1}$ by matrices of the considerably smaller dimension $2n$: $L$, $D$ and $N$.

We now show that for two commuting matrices $M_1$, $M_2$ the corresponding Lie operators $\mbox{\rm ad}_{P_1}$, $\mbox{\rm ad}_{P_2}$ (defined as above) commute as well:

$\displaystyle \mbox{\rm ad}_{P_1}\mbox{\rm ad}_{P_2}(\cdot)$ $\textstyle =$ $\displaystyle D_{\mbox{\protect\footnotesize\protect\boldmath$z$}} \left( D_{\m...$}M_2{\mbox{\protect\boldmath$z$}} \right)
M_1 {\mbox{\protect\boldmath$z$}}$  
  $\textstyle =$ $\displaystyle \sum_{\mu,\nu=1}^{2n}
\frac{\partial^2}{\partial z_\mu \partial z...}{\partial z_\nu}(\cdot)
  $\textstyle =$ $\displaystyle \mbox{\rm ad}_{P_2}\mbox{\rm ad}_{P_1}(\cdot) \quad.$ (21)

Because $LD-DL=0$ this implies that the right hand side of (27) is zero, and thus $\left\{H_2,I\right\}=0$.

For $m>2$ we proceed in the following way: We show that diagonalizability and nilpotence of the matrices $D$ and $N$ carry over to the corresponding Lie operators ${\cal D}_m$ and ${\cal N}_m$; these properties then imply that the null spaces of ${\cal D}_m$ and ${\cal D}_m^*$ coincide and that ${\cal D}_m^* H_m=0$, which in turn means $\left\{H_m,I\right\} = {\cal D}_m H_m = 0$.

Consider a unitary matrix $T$ that transforms $D$ into the diagonal matrix $\tilde{D}=TDT^{-1}$. Inserting twice the identity $T^{-1}T$ into the expression for ${\cal D}_m$ we get

{\cal D}_m(\cdot) = D_{\mbox{\protect\footnotesize\p...$\cdot$}T^{-1}TDT^{-1}T {\mbox{\protect\boldmath$z$}} \quad.

With $\tilde{{\mbox{\protect\boldmath$z$}}}=T{\mbox{\protect\boldmath$z$}}$, and denoting ${\cal D}_m$ in the new coordinates $\tilde{{\mbox{\protect\boldmath$z$}}}$ by $\tilde{{\cal D}}_m$, we obtain
\tilde{{\cal D}}_m(\cdot) = D_{\tilde{{\mbox{\protec...
\tilde{{\mbox{\protect\boldmath$z$}}} \quad.
\end{displaymath} (22)

Application of this transformed operator to any of the basis monomials $\tilde{{\mbox{\protect\boldmath$z$}}}^{\mbox{\protect\footnotesize\protect\boldmath$j$}}$ of $\L _m$ yields, because $\tilde{D}$ is diagonal, an eigenvalue equation with the eigenfunction $\tilde{{\mbox{\protect\boldmath$z$}}}^{\mbox{\protect\footnotesize\protect\boldmath$j$}}$ and a certain eigenvalue $\mu_{\mbox{\protect\footnotesize\protect\boldmath$j$}}$ -- thus diagonalizability of ${\cal D}_m$ is shown.

Now consider any $R\in\L _m$. $R\left(e^{Nt}{\mbox{\protect\boldmath$z$}}\right)$ is a polynomial in $t$ of degree less than or equal to $m(n_0-1)$, since by nilpotence there is some $n_0\in{\bf N}$ such that $N^{n_0}=0$. This polynomial is related to ${\cal N}_m$ in the following way:

\frac{d}{dt} R\left(e^{Nt}{\mbox{\protect\boldmath$z...
...{Nt}{\mbox{\protect\footnotesize\protect\boldmath$z$}}} \quad.

Iterating this expression and evaluating for $t=0$ we get
{\cal N}_m^{mn_0}(R) = \left. \frac{d^{mn_0}}{dt^{mn_0}}
\right\vert _{t=0}
= 0
\end{displaymath} (23)

which implies nilpotence of ${\cal N}_m$, because (31) holds for all $R$.

Identity of the null spaces of ${\cal D}_m$ and ${\cal D}_m^*$ is a direct implication of diagonalizability: Application of the diagonalized operator $\tilde{{\cal D}}_m$ (cf. (30)) yields

\left( \tilde{{\mbox{\protect\boldmath$z$}}}^{\mbox{...
...j$}}{\mbox{\protect\footnotesize\protect\boldmath$k$}}} \quad.

So the eigenspaces corresponding to the eigenvalue 0 of ${\cal D}_m^*$ and ${\cal D}_m$ are identical.

Finally, we determine how ${\cal D}_m^*$ acts on polynomials in $\L _m$. Notice that ${\cal N}_m^*$ is nilpotent, because its adjoint is. With ${\cal A}_m^* = {\cal D}_m^* + {\cal N}_m^*$ we obtain for any $H_m\in\L _m$:

0 & = & ({\cal N}_m^*)^{mn_0} H_m \\ [0.2cm]
& = & ({...
...}_m^* \right)^{mn_0-l}
\left( {\cal A}_m^* \right)^l H_m \quad,

because ${\cal A}_m^*$ and ${\cal D}_m^*$ commute (since the corresponding matrices $L_0^*$ and $D_0^*$ commute; cf. (29)). $\left( {\cal A}_m^* \right)^l H_m$ is zero for $m\geq 3$ and $l\geq 1$ because $H$ is in generalized normal form. From this it follows for $H_m(\tilde{{\mbox{\protect\boldmath$z$}}})=\sum_{\vert{\mbox{\protect\footnotes...
...mbox{\protect\boldmath$z$}}}^{\mbox{\protect\footnotesize\protect\boldmath$j$}}$ that

0 = \left( \tilde{{\cal D}}_m^* \right)^{mn_0} H_m
...$}}}^{\mbox{\protect\footnotesize\protect\boldmath$j$}} \quad;

here, for the sake of notational convenience, we have again turned to the coordinates $\tilde{{\mbox{\protect\boldmath$z$}}}$ as defined above. Linear independence of the basis monomials $\tilde{{\mbox{\protect\boldmath$z$}}}^{\mbox{\protect\footnotesize\protect\boldmath$j$}}$ then gives the result $h_{\mbox{\protect\footnotesize\protect\boldmath$j$}} \overline{\mu}_{\mbox{\protect\footnotesize\protect\boldmath$j$}} = 0$ and thus

\quad {\cal D}_m^* H_m = \sum_{\vert{\mbox{\protect\footnot...
= 0 \quad,

and we have proven the theorem.

next up previous contents
Next: Normalizing a magnetic bottle Up: Normal forms Previous: The generalized normal form   Contents
Martin_Engel 2000-05-25