# Show Lagrangian is invariant under infinitesimal $SO(3)$ transformation

Suppose we have the Lagrangian density for a triplet of real scalar fields, $$L = \sum_{a=1}^3 \left[ \frac{1}{2}\partial_\mu\phi_a\partial^\mu\phi_a - \frac{1}{2}\phi_a\phi_a \right].$$

How do you show that it's invariant under the infinitesimal $$SO(3)$$ transformation $$\phi_a \to \phi_a + \theta \epsilon_{abc}n_b\phi_c,$$

where $$\theta$$ is a constant and $$n_b$$ a unit vector. I'm not really sure actually what to do after simply substituting $$\phi_a$$.

This is easiest done not using index notation. Define the column vector

$$\vec{\phi} = \begin{pmatrix} \phi_1 \\ \phi_2 \\ \phi_3 \end{pmatrix}$$

The derivative acts on the column vector as $$\mathbf{1}_3\partial_\mu$$ where $$\mathbf{1}_3$$ is the identity matrix on the same vector space in which $$\vec{\phi}$$ lives. That is,

$$\partial_\mu \vec{\phi} = \begin{pmatrix} \partial_\mu \phi_1 \\ \partial_\mu \phi_2 \\ \partial_\mu \phi_3 \end{pmatrix}$$

Then we can write the lagrangian as

$$\mathcal{L}=\frac{1}{2}(\partial_\mu \vec{\phi})^T (\partial_\mu \vec{\phi}) - \frac{1}{2}\vec{\phi}^T \vec{\phi}$$

Now it can be seen basically by inspection that for $$R\in SO(3)$$ we have that

$$\vec{\phi} \stackrel{SO(3)}{\longmapsto} R\vec{\phi} \tag{(a)}\\ \vec{\phi}^T \stackrel{SO(3)}{\longmapsto} \vec{\phi}^T R^T$$

Under this transfermation, both terms in the lagrangian both terms pick up an $$R^TR = \mathbf{1}_3$$ sandwiched in the middle. Hence the lagrangian is invariant.

Relating to the transformation law using index notation:

$$\phi_a \to \phi_a + \theta \epsilon_{abc}n_b\phi_c$$

A general $$SO(3)$$ transformation can be written as

$$R = e^{-iL_an_a \theta}$$

so that an infinitesimal transformation can be written as

$$R = 1 -iL_an_a \theta + \mathcal{O}(\theta^2)$$

Now, in the fundamental representation the generators are given by

$$(L^a)_{bc} = -\epsilon_{abc}$$

We therefore see that an infinitesimal SO(3) transformation on $$\phi_a$$ is given as

$$\phi_a \to R_{ab}\phi_b = \phi_a + i\theta\epsilon_{abc}n_c \phi_b + \mathcal{O}(\theta^2) \tag{b}$$

So we see that proving the lagrangian is invariant under (b) is a special case of proving it is invariant under (a).

• Thank you, but how would you relate this to the $+\theta \epsilon_{abc}n_b\phi_c$ part. It seems as if you ignored that, no? Nov 12, 2019 at 17:39
• I'll add an edit. In short they're identical, but you can show it for any SO(3) matrix, not just the infinitesimal transformations. Nov 12, 2019 at 17:40
• While this is correct, I think it misses the point of the exercise, which is presumably to build up lie groups from their generators. Edit: your edit makes this comment irrelevant. Nov 12, 2019 at 17:55