The simplest solution to this is Φ Φ 1 = Φ Φ 2 ≡ ≡ Φ Φ ( x ⊥ ⊥ ) which leads to (note the equality in this case is not a weak one in that no constraint need be imposed after the Poisson bracket is worked out) { H 2 , H 1 } = 0 (see Todorov, and Wong and Crater ) with the same x ⊥ ⊥ defined above.