# Ryder's SU(2) example

1. Nov 23, 2015

### ShayanJ

In section 3.5 of his textbook Quantum Field Theory, Ryder discusses an example of a non-Abelian gauge theory. He considers a 3D internal space and rotations in this space.

At first he shows that the fields in this internal space transform like $\delta \vec \phi=-\vec \Lambda \times \vec \phi$ under a rotation $\vec \Lambda$ in the internal space. Then he shows that $\delta(\partial_\mu \vec \phi)=-\vec \Lambda \times (\partial_\mu \vec \phi)-(\partial_\mu \vec \Lambda)\times \vec \phi$.

He then introduces the covariant derivative $D_\mu \vec \phi=\partial_\mu \vec \phi+g\vec W_\mu \times \vec \phi$ and demands that $\delta(D_\mu \vec \phi)=-\vec \Lambda \times (D_\mu \vec \phi)$ which gives him the transformation rule for $\vec W_\mu$ which is $\delta \vec W_\mu=-\vec \Lambda \times \vec W_\mu+\frac 1 g \partial_\mu \vec \Lambda$.

Now here is where my confusion begins. He writes $\delta(D_\mu \vec \phi)=\delta(\partial_\mu \vec \phi)+g(\delta \vec W_\mu)\times \vec \phi+g\vec W_\mu \times (\delta \vec \phi)$ and then using the above expressions, he verifies that $\delta(D_\mu \vec \phi)=-\vec \Lambda \times (D_\mu \vec \phi)$.
But it seems to me that without using $\delta(D_\mu \vec \phi)=\delta(\partial_\mu \vec \phi)+g(\delta \vec W_\mu)\times \vec \phi+g\vec W_\mu \times (\delta \vec \phi)$, I still should be able to get the same conclusion with just replacing the transformed quantities in the expression for the covariant derivative, so I write:
$(D_\mu \vec \phi)_{transformed}=\partial_\mu \vec \phi-\vec \Lambda \times (\partial_\mu \vec \phi)-(\partial_\mu \vec \Lambda)\times \vec \phi+g\left( \vec W_\mu-\vec\Lambda\times \vec W_\mu+\frac 1 g \partial_\mu \vec \Lambda \right) \times \left( \vec\phi-\vec\Lambda\times \vec \phi \right)$.

But when I do the calculations, the result is far from what Ryder gets(which should already be obvious from my expression).

What's wrong with this method?
Thanks

2. Nov 23, 2015

### nrqed

Nothing is wrong with your method if you keep in mind that we are supposed to work in first order of the transformation parameter $\Lambda$. If you drop all your terms with two powers of $\Lambda$ and then look at the difference between your expression and $D_\mu \phi$(because you want the variation $\delta D_\mu \phi$) then you should get his expression.