The generalized normal form

We now turn to the generic case where $\L _m$ cannot be decomposed into the direct sum of the kernel and range spaces of ${\cal A}_m$. As a trivial example for the way in which this problem arises, consider a particle with a single degree of freedom ($n=1$) which in lowest order approximation is ``free'':

$$\quad H_2(q,p) = \frac{1}{2} p^2 \quad.$$ (9)

${\cal A}_m$ takes on the form ${\cal A}_m = p \frac{\partial}{\partial q}$, such that we get $\mbox{Ker}\left({\cal A}_3\right) = \mbox{span} \left\{ p^3 \right\}$ and $\mbox{Im}\left({\cal A}_3\right) = \mbox{span} \left\{ p^3,p^2q,pq^2 \right\}$, and obviously

$$\quad \L _3 \neq \mbox{Ker}\left({\cal A}_3\right) \oplus \mbox{Im}\left({\cal A}_3\right) \quad.$$

This shows that the normal form considered by Gustavson is not suitable for all types of $H_2$.

We circumvent this problem by using Fredholm's alternative for $\L _m$:

$$\quad \L _m = \mbox{Ker}\left({\cal A}_m^*\right) \oplus \mbox{Im}\left({\cal A}_m\right) \quad.$$ (10)

Here, as usual, the adjoint operator ${\cal A}_m^*$ is defined via $(R\vert{\cal A}_mS)=({\cal A}_m^*R\vert S) \; \forall \; R,S\in \L _m$, where $(\cdot\vert\cdot)$ is any suitable scalar product. Below we will specify a particular scalar product that will simplify the following expressions as much as possible.

In accordance with (18) it is natural to define a new normal form. Let $H$ be a Hamiltonian of type (1a) with an arbitrary quadratic contribution $H_2$. We say that $H$ is in generalized normal form up to order $m$ if

$${\cal A}_l^*(H_l) = 0 \quad \mbox{for} \quad l=3,4,\ldots,m.$$ (11)

$H$ is in generalized normal form if (19) holds for all $l\geq 3$.

Notice that (19) is not required to hold for $l=2$ -- in contrast to the corresponding definition (6) of the BGNF. The reason being that in general it is impossible to normalize $H_2$, since transforming $H_2$ implies changing ${\cal A}_m$ as well. For generic $H_2$ one has to expect ${\cal A}_2^*(H_2) \equiv \mbox{\rm ad}_{H_2}^*(H_2) \neq 0$. In Gustavson's case, however, (6) is always true for $n=2$, because the Poisson bracket of $H_2$ with itself vanishes.

In order to complete our definition of a normal form we have to specify the explicit form of the scalar product. For $R({\mbox{\protect\boldmath$z$}}) = \sum_{\vert{\mbox{\protect\footnotesize\prot... ...rotect\boldmath$z$}}^{\mbox{\protect\footnotesize\protect\boldmath$j$}}\in\L _m$ and $S({\mbox{\protect\boldmath$z$}})\in\L _m$ we set [24,22]

$$\quad (R\vert S) := \left( \sum_{\vert{\mbox{\protect\foo... ...2n}}z_{2n} } \right) S({\mbox{\protect\boldmath$z$}}) \quad,$$ (12)

where the bar denotes complex conjugation. It is easy to see that this product operation $(\cdot\vert\cdot)$ indeed has the properties of a scalar product. In [17] a somewhat different scalar product was introduced by choosing a special basis of $\L _m$ and defining it to be orthonormal. These two scalar products are identical up to a normalization factor. However, the definition given here paves way for a more general approach and is much easier to use. This becomes apparent when trying to derive an explicit expression for ${\cal A}_m^*$. In [17] this was only achieved for a very restricted case, namely the so-called ``mirror machine'' or ``magnetic bottle Hamiltonians''. See section 3 for a discussion of this class of systems.

We first write ${\cal A}_m$ in yet another form. Linearizing Hamilton's equations we obtain the Hamiltonian matrix $L=J\mbox{Hess}(H_2)$, with the $2n$-dimensional symplectic matrix $J = \left(\protect\begin{array}{@{}c@{\hspace*{0.1cm}} c@{}}0&\mbox{\rm id}_n\\ [-0.1cm]-\mbox{\rm id}_n&0\protect\end{array}\right)$ and the Hessian $\mbox{Hess}(H_2)$. Thus we have

$$\quad {\cal A}_m(\cdot) = D_{\mbox{\protect\footnotesize\pr... ...\protect\boldmath$\cdot$}L{\mbox{\protect\boldmath$z$}} \quad,$$ (13)

with the abbreviation ${\mbox{\protect\boldmath$a$}}\mbox{\protect\boldmath$\cdot$}{\mbox{\protect\boldmath$b$}}=\sum_{i=1}^{2n} a_ib_i$ for ${\mbox{\protect\boldmath$a$}},{\mbox{\protect\boldmath$b$}}\in{\bf C}^{2n}$. In order to find ${\cal A}_m^*$ we rewrite its definition $\left( {\cal A}_m^*R\vert S \right) = \left( R\vert{\cal A}_mS \right)$ as

$$\quad \left( {\cal A}_m^*R\vert S \right) = \left. \frac{d... ...ath$z$}}\right) \Big\vert S \right) \right\vert _{t=0} \quad,$$

where we have used the relation $\left(R\circ M^*\vert S\right) = \left(R\vert S\circ M\right)$ which holds for any linear mapping $M$ on ${\bf C}^{2n}$. Evaluating the time derivative yields $\left( {\cal A}_m^*R\vert S \right) = \left( D_{\mbox{\protect\footnotesize\pr... ...box{\protect\boldmath$\cdot$}L^*{\mbox{\protect\boldmath$z$}} \vert S \right)$, and we obtain ${\cal A}_m^*$ as

$$\quad {\cal A}_m^*(\cdot) = D_{\mbox{\protect\footnotesize\... ...rotect\boldmath$\cdot$}L^*{\mbox{\protect\boldmath$z$}} \quad.$$ (14)

This expression is identical with (21) after conjugating and transposing $L$.

In the form (22) ${\cal A}_m^*$ can easily be used for determining the splitting (18) of $\L _m$. For the example (17) considered in the beginning of this section we obtain ${\cal A}_m^* = q\frac{\partial}{\partial p}$ and therefore $\mbox{Ker}\left({\cal A}_3^*\right) = \mbox{span} \left\{ q^3 \right\}$.

The method for transforming a given Hamiltonian into its generalized normal form is exactly the same as the one described in the previous section; one only has to replace (12b) by

$$\quad H_m' \in \mbox{Ker}\left({\cal A}_m^*\right) \quad.$$ (15)

Since the splitting (18) holds for any $H_2$, we have proven that any Hamiltonian can be normalized according to the generalized definition (19).

Note that for a Hamiltonian of Gustavson's type (1) the two definitions of normal form coincide, because in this case ${\cal A}_m^* = -{\cal A}_m$. So if $H$ is in BGNF (up to order $m$) it is in generalized normal form (up to order $m$), too. The utility of the normal form will become evident in the next section.