Another way to interpret this question is: Is there a 'heuristic' reason that all closed nondegenerate $2$-forms in $2n$ dimensions are locally equivalent? (It becomes more reasonable to ask this when you realize that nothing like this holds for, say, $3$-forms in general dimension, even for $3$-forms in dimension $5$.) There is a 'function count' heuristic reason for this local equivalence, which goes like this:

If $\omega$ is a closed $2$-form in dimension $n$, then, locally $\omega = \mathrm{d}\alpha$ for some $1$-form, by Poincaré's Lemma.

Now, in any local coordinates, $\alpha = a_i(x) \mathrm{d} x^i$, so $\alpha$ (and hence $\omega$) 'depends' on a choice of $n$ functions of $n$ variables, the $a_i$. However, local coordinate changes also depend on $n$ functions of $n$ variables, so you might expect that, by choosing coordinates cleverly, you could bring $\alpha$ into some standard form, just as you can put a (nonzero) vector field into a standard local form in so-called 'flow-box coordinates'.

It turns out that you can't quite do this for $1$-forms (another piece of evidence that you shouldn't mentally identify $1$-forms with vector fields as people often do when they first start thinking about differential forms); there is an algebraic invariant, the rank $\rho(p)$ of $\mathrm{d}\alpha$ as a tensor at each point $p$. However, if this rank is locally constant, then, yes, you can choose coordinates so as to reduce $\mathrm{d}\alpha$ to a standard normal form. That is the content of Darboux' Theorem.

To reduce $\alpha$ itself to standard form, you'd need a little more information. That is the content of the Pfaff Theorem. However, because $\alpha$ wasn't uniquely determined by $\omega$ (you could always add a term $\mathrm{d}f$ to $\alpha$), this extra information gets thrown away. (In fact, it is this 'gauge' ambiguity that shows that you should still have one arbitrary function left over in your choice of normalizing coordinates once you put $\omega$ in normal form. Hence the arbitrary function, i.e., the arbitrary choice of a Hammiltonian, in finding vector fields whose flows preserve $\omega$.)

If you try to do the same count for, say, closed $3$-forms in dimensions above $4$, you'll see immediately that there is no hope for such a simple normal form.

Another way to interpret this question is: Is there a 'heuristic' reason that all closed nondegenerate $2$-forms in $2n$ dimensions are locally equivalent? (It becomes more reasonable to ask this when you realize that nothing like this holds for, say, $3$-forms in general dimension, even for $3$-forms in dimension $5$.) There is a 'function count' heuristic reason for this local equivalence, which goes like this:

If $\omega$ is a closed $2$-form in dimension $n$, then, locally $\omega = \mathrm{d}\alpha$ for some $1$-form, by Poincaré's Lemma.

Now, in any local coordinates, $\alpha = a_i(x) \mathrm{d} x^i$, so $\alpha$ (and hence $\omega$) 'depends' on a choice of $n$ functions of $n$ variables, the $a_i$. However, local coordinate changes also depend on $n$ functions of $n$ variables, so you might expect that, by choosing coordinates cleverly, you could bring $\alpha$ into some standard form, just as you can put a (nonzero) vector field into a standard local form in so-called 'flow-box coordinates'.

It turns out that you can't quite do this for $1$-forms (another piece of evidence that you shouldn't mentally identify $1$-forms with vector fields as people often do when they first start thinking about differential forms); there is an algebraic invariant, the rank $\rho(p)$ of $\mathrm{d}\alpha$ as a tensor at each point $p$. However, if this rank is locally constant, then, yes, you can choose coordinates so as to reduce $\mathrm{d}\alpha$ to a standard normal form. That is the content of Darboux' Theorem.

To reduce $\alpha$ itself to standard form, you'd need a little more information. That is the content of the Pfaff Theorem. However, because $\alpha$ wasn't uniquely determined by $\omega$ (you could always add a term $\mathrm{d}f$ to $\alpha$), this extra information gets thrown away. (In fact, it is this 'gauge' ambiguity that shows that you should still have one arbitrary function left over in your choice of normalizing coordinates once you put $\omega$ in normal form. Hence the arbitrary function, i.e., the arbitrary choice of a Hamiltonian, in finding vector fields whose flows preserve $\omega$.)

If you try to do the same count for, say, closed $3$-forms in dimensions above $4$, you'll see immediately that there is no hope for such a simple normal form.

Another way to interpret this question is: Is there a 'heuristic' reason that all closed nondegenerate $2$-forms in $2n$ dimensions locally equivalent? (It becomes more reasonable to ask this when you realize that nothing like this holds for, say, $3$-forms in general dimension, even for $3$-forms in dimension $5$.) There is a 'function count' heuristic reason for this local equivalence, which goes like this:

If $\omega$ is a closed $2$-form in dimension $n$, then, locally $\omega = \mathrm{d}\alpha$ for some $1$-form, by Poincaré's Lemma.

Now, in any local coordinates, $\alpha = a_i(x) \mathrm{d} x^i$, so $\alpha$ (and hence $\omega$) 'depends' on a choice of $n$ functions of $n$ variables, the $a_i$. However, local coordinate changes also depend on $n$ functions of $n$ variables, so you might expect that, by choosing coordinates cleverly, you could bring $\alpha$ into some standard form the same way that you can bring a (nonzero) vector field into standard form in so-called 'flow-box coordinates'.

It turns out that you can't quite do this for $1$-forms (another piece of evidence that you shouldn't mentally identify $1$-forms with vector fields as people often do when they first start thinking about differential forms); there is an algebraic invariant, the rank $\rho(p)$ of $\mathrm{d}\alpha$ as a tensor at each point $p$. However, if this rank is locally constant, then, yes, you can choose coordinates so as to reduce $\mathrm{d}\alpha$ to a standard normal form. That is the content of Darboux' Theorem.

To reduce $\alpha$ itself to standard form, you'd need a little more information. That is the content of the Pfaff Theorem. However, because $\alpha$ wasn't uniquely determined by $\omega$ (you could always add a term $\mathrm{d}f$ to $\alpha$), this extra information gets thrown away. (In fact, it is this 'gauge' ambiguity that shows that you should still have one arbitrary function left over in your choice of normalizing coordinates once you put $\omega$ in normal form. Hence the arbitrary function, i.e., the arbitrary choice of a Hammiltonian, in finding vector fields whose flows preserve $\omega$.)

If you try to do the same count for, say, closed $3$-forms in dimensions above $4$, you'll see immediately that there is no hope for such a simple normal form.

Another way to interpret this question is: Is there a 'heuristic' reason that all closed nondegenerate $2$-forms in $2n$ dimensions are locally equivalent? (It becomes more reasonable to ask this when you realize that nothing like this holds for, say, $3$-forms in general dimension, even for $3$-forms in dimension $5$.) There is a 'function count' heuristic reason for this local equivalence, which goes like this:

If $\omega$ is a closed $2$-form in dimension $n$, then, locally $\omega = \mathrm{d}\alpha$ for some $1$-form, by Poincaré's Lemma.

Now, in any local coordinates, $\alpha = a_i(x) \mathrm{d} x^i$, so $\alpha$ (and hence $\omega$) 'depends' on a choice of $n$ functions of $n$ variables, the $a_i$. However, local coordinate changes also depend on $n$ functions of $n$ variables, so you might expect that, by choosing coordinates cleverly, you could bring $\alpha$ into some standard form, just as you can put a (nonzero) vector field into a standard local form in so-called 'flow-box coordinates'.

It turns out that you can't quite do this for $1$-forms (another piece of evidence that you shouldn't mentally identify $1$-forms with vector fields as people often do when they first start thinking about differential forms); there is an algebraic invariant, the rank $\rho(p)$ of $\mathrm{d}\alpha$ as a tensor at each point $p$. However, if this rank is locally constant, then, yes, you can choose coordinates so as to reduce $\mathrm{d}\alpha$ to a standard normal form. That is the content of Darboux' Theorem.

To reduce $\alpha$ itself to standard form, you'd need a little more information. That is the content of the Pfaff Theorem. However, because $\alpha$ wasn't uniquely determined by $\omega$ (you could always add a term $\mathrm{d}f$ to $\alpha$), this extra information gets thrown away. (In fact, it is this 'gauge' ambiguity that shows that you should still have one arbitrary function left over in your choice of normalizing coordinates once you put $\omega$ in normal form. Hence the arbitrary function, i.e., the arbitrary choice of a Hammiltonian, in finding vector fields whose flows preserve $\omega$.)

If you try to do the same count for, say, closed $3$-forms in dimensions above $4$, you'll see immediately that there is no hope for such a simple normal form.