The 1st part of the question doesn't really need QFT, it can be asked in mechanics too. Consider a harmonic oscillator of mass m with two coordinates x, y and a symmetric potential

V(x, y) = k(x^2 + y^2) / 2

We can define the variables

z = (x + iy) / sqrt(2)

z* = (x - iy) / sqrt(2)

The conjugate momenta of x and y are

p = m dx / dt

q = m dy / dt

Straightforward computation shows

{x + iy, p + iq} = 0

{x + iy, p - iq} = 2

{x - iy, p + iq} = 2

{x - iy, p - iq} = 0

So the conjugate momentum of z is

w = m dz* / dt

and the conjugate momentum of z* is

w* = m dz / dt

However, {z, w*} = {z*, w} = 0, so this commutation relation *is* trivial.

Regarding conserved quantities it depends on the potential which you haven't specified. For the free theory with equal masses for phi1 and phi2, the full internal symmetry group is O(4) since we can rewrite the theory in terms of 4 real fields. This yields dim O(4) = 6 conserved quantities.

The Belifante-Rosenfeld tensor actually yields only 3 conserved quantities in the usual sense: the angular momenta. These correspond to rotation symmetry. The other 3 quantities correspond to generators of the Lorentz group which don't commute with time translation: the boosts. So these quantities are not conserved unless you introduce explicit time dependence into their definition.

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