The meaning of $i$ in the Yang-Mills

Question

In Yang mills theory (including maxwell's theory), the generators of $SU(n)$ are complex matrices acting on the spinor $\psi$.

Now the phase of $\psi$ is also given by $e^{i\theta}$.

Why do we use this same $i$ in both cases? In other words, for the $SU(n)$ it would be possible to define a different $i$ to the phase, by choosing a matrix $$\begin{bmatrix} 0 & 1 \\ -1 & 0\end{bmatrix}$$ And adding some extra parameter to the spinor $\psi^a$.

Is there some close link which explains why Yang-Mills has to affect the phase like this?

Long story short, in QM we make the change $k\rightarrow i \partial$, and I'm wondering what the explanation (mathematical or physical) why this is the same $i$ that appears in the complex generators of $SU(n)$. Because from first sight these seem like they don't have anything to do with each other.

Answer 1

The symbol for the Lie group SU(n) means "special unitary" group. So the elements in this group are unitary. Often we represent such an element in exponential form $\exp(i\alpha_n\tau_n)$ where $\tau_n$ represents the generators. For the two-dimensional representation of SU(2) they are given by the Pauli matrices. These generators are Hermitian matrices. For the exponential expression of a unitary transformation the exponent must be anti-Hermitian. That is why we need to $i$ in the exponent.

Answer 2

In physically important example, $SU(2)$, the use of the imaginary here simply depends upon the fact that physics convention of coordinatising $Spin(3)$ with the isomorphic complex group, $SU(2)$. So the use of the imaginary here for the Yang-Mills theory over $SU(2)$ is spurious.

This argument doesn't generalise to higher $SU(n)$.