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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 $\phi_a$.

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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).

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  • $\begingroup$ 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? $\endgroup$ Nov 12, 2019 at 17:39
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    $\begingroup$ 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. $\endgroup$ Nov 12, 2019 at 17:40
  • $\begingroup$ 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. $\endgroup$ Nov 12, 2019 at 17:55

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