tom.stoer said:
I don't think you have 10 conserved quantities.
The invariance under the Poincare’ group implies that the energy-momentum 4-vector
<br />
P_{ a } = \int d^{ 3 } x \ T_{ 0 a } = \int d^{ 3 } x \left( \frac{ \partial \mathcal{ L }}{ \partial ( \partial_{ 0 } \phi ) } \ \partial_{ a } \phi - \eta_{ 0 a } \mathcal{ L } \right) , \ \ (1)<br />
and the angular momentum tensor
<br />
M_{ ab } = \int d^{ 3 } x \ \left( T_{ 0 b } x_{ a } - T_{ 0 a } x_{ b } + \frac{ \partial \mathcal{ L } }{ \partial ( \partial_{ 0 } \phi ) } \ \Sigma_{ a b } \phi \right) , \ \ (2)<br />
are CONSTANTS OF MOTION. These are the (4+6=10) conserved Noether CHARGES. It is very easy to show that
\frac{d}{dx^{ 0 }} P_{ a } = \frac{d}{dx^{ 0 }} M_{a b} = 0 . \ \ (3)
The commutation relations for rotations L, boosts K, 3-momentum P with the Hamiltonian H are
[Li,H] = [Pi,H] = 0
[Ki,H] = -i Pi
That means that the boosts K do not commute with H and can therefore not be 'conserved charges'.
This is very common misunderstanding. The components M_{ i 0 } has an EXPLICIT time dependence which has to be accounted for when writing Heisenberg (Poisson) equation of motion. So, you need to write
<br />
\frac{ d }{ dx^{ 0 } } M_{ i 0 } = \partial_{ 0 } M_{ i 0 } + [ i P_{ 0 } , M_{ i 0 } ] .<br />
The conservation of M_{ i 0 }, [Eq(3)], therefore implies
<br />
[ i P_{ 0 } , M_{ i 0 } ] = - \partial_{ 0 } M_{ i 0 } = - \partial_{ 0 } \int d^{ 3 } x \ \left( - \pi \partial_{ i } \phi \right) \ x^{ 0 } = \int d^{ 3 } x \ \pi \partial_{ i } \phi = P_{ i } .<br />
So, the non-vanishing commutator [ i H , M_{ i 0 } ] DOES NOT mean that M_{ i 0 } is NOT CONSERVED.
Sam
