How do I see if the equations of motion are satisfied?

[sayebms]

Homework Statement

(a) Calculate the Conserved currents $$K_{\mu \nu \alpha} $$ associated with the global lorentz transformation and express them in terms of energy momentum tensor.
(b) Evaluate the currents for $$L=\frac{1}{2}\phi (\Box +m^2)\phi$$. Check that these currents satisfy $$\partial_{\alpha} K_{\mu \nu \alpha}=0$$ on the equations of motion.
(c) what is the physical interpretation of the conserved quantity $$Q_i=\int d^3xK_{0i0}$$ associated with boosts?
(d) show that $$\frac{dQ_i}{dt}$$ can still be consistent with $$i\frac{\partial Q_i}{\partial t}=[Q_i,H]$$[/B]

Homework Equations

The Attempt at a Solution

I have calculated the part (a) as following (note epsilon is small)
$$j^{\mu}=- \frac{\partial L}{\partial (\partial_\mu \phi)}\epsilon ^{\rho}_{\nu}x^{nu} \partial_{\rho}\phi+\epsilon^{\mu}_{\nu}x^{nu}L=-\epsilon^{\rho}_{\nu} T^{\mu}_{\rho}x^{\nu}$$
but considering the anti symmetry of epsilon we have to write it as following:
$$K_{mu\nu\alpha}=x_{\nu}T_{\mu\alpha}-x_{\alpha}T_{\mu\nu}$$
is this correct?

and for part (b) do I just check the equations of motion and see if they become zero or do I actually find the currents one by one?

from equations of motion I know $$\frac{\partial L}{\partial (\partial_{\mu}\phi_n)}=0$$ which gives me $$\frac{\partial L}{\partial \phi}=0$$ and hence I just get $$(\Box + m^2)\phi=0$$ which is only held if phi is exponential. Am i doing it right?[/B]

 

[strangerep]

(b) [...] Check that these currents satisfy $$\partial_{\alpha} K_{\mu \nu \alpha}=0$$ on the equations of motion.

[...] and for part (b) do I just check the equations of motion and see if they become zero or do I actually find the currents one by one?

The question said "on" the equations of motion. That probably means $K_{\mu \nu\alpha}$ has zero divergence only if the equations of motion are also satisfied. I.e., when you calculate the divergence it probably won't be zero, but will have terms that vanish if the equations of motion are satisfied.