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Take the Lagrangian with one fermion: $$ \mathcal{L} = -\frac{1}{4}F^{\mu\nu}_aF^a_{\mu\nu} + \bar{\psi}(i\gamma^\mu D_\mu - m)\psi$$ where the gauge covariant derivative $D_\mu = \partial_\mu+i\frac{g}{2}t^aW^a_\mu$. The Lagrangian is invariant under a local $SU(2)$ transformation: $$ \psi(x) \rightarrow \exp \left[-i\theta^a(x)t^a \right]\psi(x) $$ $$W^a_\mu(x) \rightarrow W^a_\mu(x) +\frac{1}{g}\partial_\mu\theta^a(x) + \epsilon^{abc}\theta^b(x)W^c_\mu(x)$$

Often, we say that $W_\mu^a$ transforms according to the adjoint representation of $SU(2)$ but how can we say that based on the previous equation?

Hunter
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KoObO
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1 Answers1

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Note that the finite transformation of: $$ W^a_\mu \to W^a_\mu + \frac{1}{g} \partial_\mu \theta^a + \epsilon^{abc} \theta^b W^c_\mu $$ is: $$ W^a_\mu t^a \to g W_\mu^a t^a g^{-1} + \frac{i}{g} \partial_\mu g \tag{1} $$ where: $$ g = \exp(-i \theta^a t^a) \;\;\; \text{and} \;\;\; [t^a,t^b] = i \epsilon^{abc} t^c $$ Thus, the first term on the right-hand side of equation $(1)$ transforms under the adjoint representation of the Lie group. The second term does not transform under the adjoint representation, but it should be easy to verify that the transformed gauge field still takes values in the Lie algebra (hint: looking at infinitesimal transformations is the easiest method to verify this).

In case you want more information on the adjoint representation of the Lie group, it might be worth looking at this question.

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Hunter
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  • Thank you for your answer. I think my problem is that I misunderstand what "transforms under" means really. Tell me if I'm wrong : Let's take $SU(3)$, for the $\mathbf{3}$ with Dynkin indices (1,0), a state transforms like : $\psi(x) \rightarrow g\psi(x)$. For the $\bar{\mathbf{3}}$ (0,1), a state tranforms like : $\phi(x) \rightarrow \phi(x) g^{-1}$. And for the adjoint representation (1,1) : $\mathcal{O} \rightarrow g \mathcal{O} g^{-1}$.

    But then, if I come back to $SU(2)$, because the $\mathbf{2}$ and $\mathbf{\bar 2}$ are equivalent, they should transform in the same way ?

     – KoObO Apr 18 '14 at 08:41
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    @KoObO I am not really familiar with Dynkin indices to be honest. However, a field that takes values in the Lie algebra, i.e. $\phi \in \mathfrak{g}$, will transform under the adjoint representation as $\phi \to g \phi g^{-1}$ and will take again values in the adjoint representation, i.e. $g \phi g^{-1} \in \mathfrak{g}$. The easiest way to understand this is by look at my answer given here. The whole point of the transformation property of the gauge field is that – Hunter Apr 18 '14 at 09:01
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    they live in the Lie algebra before and after a transformation. This is probably why people say that the gauge fields transform under the adjoint representation. – Hunter Apr 18 '14 at 09:03
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    @KoObO Finally, it worth mentioning that you should be careful not to confuse the adjoint representation of a Lie group, and the adjoint representation of a Lie algebra. – Hunter Apr 18 '14 at 09:05
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    Ok I think I've understand. Thank you for yours answer(s) ! – KoObO Apr 18 '14 at 09:22
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    @KoObO no problem! Also, if you want to know the answer to your first comment (related to Dynkin indices), then I think it is a very valid question to ask here. Personally, I would be interested to see an answer! – Hunter Apr 18 '14 at 09:28
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    Here is the question : http://physics.stackexchange.com/questions/108919/question-about-representations-and-transformations-under-an-sun-lie-group – KoObO Apr 18 '14 at 09:41