= Solution
For an infinitesimal spacetime-dependent parameter, the action variation has the <Noether current> form
$$
\delta S=-\int d^4x\,(\partial^\mu\alpha^a)J_\mu^a.
$$
In canonical quantization, apply the corresponding transformation inside an equal-time correlation function. Integrating the resulting <Ward-Takahashi identity> over a thin time interval around an insertion gives
$$
\delta\varphi_m(x)=i\alpha^a[\varphi_m(x),Q^a],
\qquad
Q^a=\int d^3y\,J_0^a(x^0,\mathbf y).
$$
Comparison with $\delta\varphi_m=i\alpha^aT^a_{mn}\varphi_n$ yields
$$
\boxed{[\varphi_m,Q^a]=T^a_{mn}\varphi_n}
$$
in the conventions of the question. Current conservation makes $Q^a$ time independent when the surface flux vanishes, so the same charge generates the symmetry at every time.
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