SO(3) Representation Consider an SO(3) invariant Lagrangian density $$\mathcal{L}=\frac{1}{2}\sum_{a=1}^{3}(\partial_{\mu}\phi_{a}\partial^{\mu}\phi_{a}-m^{2}\phi_{a}^{2})$$ Denote the three SO(3) generators by $T_{a}$, and the corresponding Noether charges by $Q_{a}$. In the quantised theory, there are three types of excitation, one for each field, and a single particle excitation may be written $|k,a\rangle$ where $k$ denotes its four momentum and $a$ its type. It is possible to show that $$Q_{a}|k,b\rangle=\sum_{c=1}^{3}(T_{a})_{cb}|k,c\rangle$$ Apparently this proves that the single particle states form a three-dimensional representation of SO(3). However I don't understand where the final conclusion comes from. Could anyone help? Please try to not to stray into too much mathematical language if possible :)
 A: The logic is as follows. Fix a momentum $k$, and take the space of single particles with this momentum. Any state in this space can be written as a linear combination
$$| \psi \rangle = \alpha | k, 1 \rangle + \beta | k, 2 \rangle + \gamma | k, 3 \rangle$$
so you can see already that if these states are a representation, whatever that means, it's sure going to be a three dimensional one, because the space of one particle states with fixed momentum is three dimensional.
Now, a warning. When physicists say that some vectors are a representation, what they really mean is that you have some operators on that space of vectors which form a representation of some matrices. So we really need to look at the $Q_a$, not so much at the states.
Another warning: the $Q_a$ really are a representation of the Lie Algebra of $SO(3)$. This is a fancy way of saying that they act just like the generators $T_a$ of $SO(3)$: they're representing them, not the actual group elements. What does it mean to say they're a representation? We know that the generators obey the commutation relations
$$[T_a, T_b] = \sum_c i \epsilon_{abc} T_c.$$
(Remember that the $T_a$ are $3\times3$ matrices, not operators on Hilbert space.) The $Q_a$ being a representation means that they too should fulfill these conditions: we need to check that 
$$[Q_a, Q_b] = \sum_c i \epsilon_{abc} Q_c.$$
To do that, we take the left hand side, act on some state $|k, d \rangle$, use the action of the $Q_a$ (that you wrote in your question) and the commutation relations of the $T_a$ (which, again, are just matrices), and see if we get the right hand side. I'll let you do the math, but this is the general idea.
