I have two complex scalar massive fields realizing the foundamental representation of SU(2)

$$ \varphi = \begin{pmatrix} \varphi_1 \\ \varphi_2 \\ \end{pmatrix} \qquad \varphi^{\dagger} = (\varphi^*_1,\varphi^*_2)$$ I know that the lagrangian invariant under a SU(2) gauge transformation is

$$\mathscr L = \left(D_\mu\varphi\right)^{\dagger}D^\mu\varphi-m^2\varphi^\dagger\varphi+\frac{1}{2g^2}\text{Tr}\left(F_{\mu\nu}F^{\mu\nu}\right)$$

Where I used the gauge vector fields $A_\mu = A_{\mu}^a\tau_a \ a=1,2,3$ ($\tau_a$ are the generators of SU(2) satisfying $[\tau_a,\tau_b] = i\varepsilon_{ab}^{\ \ \ \ c}\tau_c$) in order to introduce

\begin{align} D_\mu & = \partial_\mu +A_\mu \\ F_{\mu\nu} & =[D_\mu,D_\nu] = \partial_{[\mu}A_{\nu]} +[A_\mu,A_\nu] \end{align} Now I want to find the EOM of $A^a_\mu$.

Since $F_{\mu\nu} = F^a_{\mu\nu}\tau_a$, where $F^a_{\mu\nu} = \partial^{\ }_{[\mu}A^a_{\nu]} +i\varepsilon_{bc}^{\ \ \ \ a}A_\mu^bA_\nu^c$, I can rewrite the previous lagrangian in terms of the three fields $A^a_\mu$. From now on I will call $\tilde{F}_{\mu\nu}\equiv \partial_{[\mu}A_{\nu]}$ and of course $\tilde{F}^a_{\mu\nu}\equiv \partial^{\ }_{[\mu}A^a_{\nu]}$

First I computed

\begin{align} \frac{1}{2g^2}\text{Tr}\left(F_{\mu\nu}F^{\mu\nu}\right)& = {1 \over2g^2}F^a_{\mu\nu}F^{b\mu\nu}\text{Tr}(\tau_a\tau_b) = {1 \over 4g^2}F^a_{\mu\nu}F^{b\mu\nu}\delta_{ab} \\ & = {1 \over4g^2}\left(\tilde{F}^a_{\mu\nu} +i\varepsilon_{bc}^{\ \ \ \ a}A_\mu^bA_\nu^c\right)\left(\tilde{F}_a^{\mu\nu} +i\varepsilon^{df}_{\ \ \ \ a}A^\mu_dA_f^\nu\right) \\ & ={1\over4g^2}\left\{\tilde{F}^a_{\mu\nu}\tilde{F}_a^{\mu\nu}+2i\varepsilon_{bc}^{\ \ \ a}\tilde{F}_a^{\mu\nu}A^b_\mu A^c_\nu-A^b_\mu A^c_\nu\left(A_b^\mu A_c^\nu-A_c^\mu A_b^\nu\right)\right\} \end{align} Therefore I can write \begin{align} \mathscr L & = \partial_\mu\varphi^\dagger\partial^\mu\varphi - m^2\varphi^\dagger\varphi-A^a_\mu\varphi^\dagger\tau_a\partial^\mu\varphi-A^a_\mu A^{b\mu} \varphi^\dagger\tau_a\tau_b\varphi+A^a_\mu\partial^\mu\varphi^\dagger\tau_a\varphi \\ &+ {1\over4g^2}\left\{\tilde{F}^a_{\mu\nu}\tilde{F}_a^{\mu\nu}+2i\varepsilon_{bc}^{\ \ \ a}\tilde{F}_a^{\mu\nu}A^b_\mu A^c_\nu-A^b_\mu A^c_\nu\left(A_b^\mu A_c^\nu-A_c^\mu A_b^\nu\right)\right\} \end{align}

I get the EOM of $A^a_\mu$

\begin{align} \partial_\mu\tilde{F}_a^{\mu\nu}&+i\varepsilon_{abc}\left((\partial_\mu A^{b\mu})A^{c\nu}+A^{b\mu}\partial_\mu A^{c\nu}\right) +i\varepsilon_{abc}A^b_\mu\tilde{F}^{c\mu\nu} + A^b_\mu\left(A^\nu_a A^\mu_b-A^\nu_b A^\mu_a\right) = \\ & = g^2\left(\partial^\nu\varphi^\dagger\tau_a\varphi -\varphi^\dagger\tau_a\partial^\nu\varphi-A^{a\nu}\varphi^\dagger\varphi\right) \end{align}

However our professor gave us a compact solution of the form

$$\left(D_\mu F^{\mu\nu}\right)_a = g^2\left(\left(D^\nu\varphi\right)^\dagger\tau_a\varphi-\varphi^\dagger\tau_a\left(D^\nu\varphi\right)\right)$$

The right hand side perfectly matches. I can't understand the left hand side because explicitly

$$D_\mu F^{\mu\nu} = \partial_\mu F^{\mu\nu}+A_\mu F^{\mu\nu}$$

Which is the $a$ component of this term? I mean, the first term is simply

$$\partial_\mu F^{\mu\nu} = \partial_\mu F_a^{\mu\nu}\tau^a = \left(\partial_\mu\tilde{F}_a^{\mu\nu}+i\varepsilon_{abc}\left((\partial_\mu A^{b\mu})A^{c\nu}+A^{b\mu}\partial_\mu A^{c\nu}\right) \right)\tau^a$$

which is fine; the second one is

$$A_\mu F^{\mu\nu} = A_{a\mu}F_b^{\mu\nu}\tau^a\tau^b = {1\over4}A_{a\mu}F^{a\mu\nu} + {i\over4}\varepsilon^{ab}_{\ \ \ \ c}A_{a\mu}F_b^{\mu\nu}\tau^c$$

and explicitely is a mess! I know this is really tedious, but I can't understand what I'm doing wrong.


1 Answer 1


You have written

$$D_\mu F^{\mu\nu} = \partial_\mu F^{\mu\nu} + A_\mu F^{\mu\nu}$$

which is incorrect. It should be

$$D_\mu F^{\mu\nu} = \partial_\mu F^{\mu\nu} + [A_\mu,F^{\mu\nu}]$$

If you are curious why, it is because $D_\mu$ is not quite the thing you have defined it as. Actually, it is defined as

$$D_\mu\Omega = \partial_\mu\Omega + \rho_\Omega(A_\mu)\cdot\Omega$$

where $\rho_\Omega$ is the homomorphism corresponding to the representation (of the gauge group) to which $\Omega$ belongs, and $\cdot$ is the Lie algebra action. The field strength $F_{\mu\nu}$ belongs to the adjoint representation, while your scalar field $\varphi$ belongs to the fundamental.


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