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We are asked to show that the following Lagrangian is invariant under the three $SU(2)$ transformations $\Phi \rightarrow \exp{({\frac{i}{2}{\alpha_j\sigma^j}}) \Phi}$, where $\Phi$ is a doublet complex scalar field$$\Phi =( \phi_1 \phi_2)^{T}$$ The given Lagrangian is

$$ \mathcal{L} = \partial_\mu\Phi^{\dagger}\partial_\mu\Phi-m^2\Phi^{\dagger}\Phi $$

I have re-written the Lagrangian as

$$ \mathcal{L} = g^{\mu\nu}\partial_\mu{\phi_i}^\dagger\partial_\nu{\phi_i}-m^2\phi^{\dagger}_i\phi_i $$

Where $i = 1,2$.

I derived $\delta\mathcal{L}=0$ using the fact that $\vec{\sigma}_j = \vec{\sigma}_j^\dagger$. My problem is finding $\delta\phi_i$ and $\delta\phi_i^\dagger$. How can I expand the exponential? And also, will I get 3 conserved currents for each complex field?

How would I calculate $$\alpha_j(j^\mu)_i = \frac{\partial\mathcal{L}}{\partial({\partial_\mu\phi_i})}\delta\phi_i+\frac{\partial\mathcal{L}}{\partial({\partial_\mu\phi^\dagger_i})}\delta\phi^\dagger_i$$

Edit 1:

Attempting to calculate $\delta\phi_i$ I expand the exponential to linear order since the parameter $\alpha_j \ll 1$. Hence $\delta\phi_i= I + i\alpha_j\sigma^j$. Assuming this is true, how does that summation look like? Would each j have a different charge?

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  • $\begingroup$ That was very helpful! I have edited the question. $\endgroup$ – fielder Oct 14 '19 at 20:21
  • $\begingroup$ I'm sorry, I shouldn't have used $\alpha$ for both the transformation parameter and the index of the fields. Does this edit make my question valid now ? $\endgroup$ – fielder Oct 14 '19 at 21:57
  • $\begingroup$ Yes. ..."..."........... $\endgroup$ – Cosmas Zachos Oct 26 '19 at 10:34
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You might be misunderstanding the notation. $$ \delta \Phi = \exp{({\frac{i}{2}{\alpha_j\sigma^j}}) \Phi} -\Phi +O(\alpha^2)= \frac{i}{2}\alpha_j\sigma^j \Phi, $$ so that $$ \delta \begin{pmatrix} \phi_1\\ \phi_2 \end{pmatrix} = \frac{i}{2} \begin{pmatrix} \alpha_3&\alpha_1-i\alpha_2\\ \alpha_1+i\alpha_2&-\alpha_3 \end{pmatrix} \begin{pmatrix} \phi_1\\ \phi_2 \end{pmatrix} , $$ which you may perform to get $\delta\phi_j$.

Now $$ \frac{i}{2} \alpha_j ~ (\partial^\mu \Phi^\dagger \sigma^j \Phi - \Phi^\dagger \sigma^j \partial^\mu \Phi) $$ represents three currents, an SU(2) adjoint triplet, the coefficients of the three parameters $\alpha_j$, normally omitted. Did you compute them? Indeed, integrating their zero components produces three isocharges, closing into an su(2) Lie algebra.

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