Model with SU(2) gauge symmetry and SO(3) global symmetry

[jack476]

Homework Statement

Full context:
We consider a model for an $SU(2)$ doublet with the following Lagrangian:
$L = -\frac{1}{4}F^a_{\mu\nu}F^a_{\mu\nu}+(D_{\mu}\varphi)^{\dagger}(D_{\mu}\varphi)+\mu^2 \varphi^\dagger \varphi -\lambda(\varphi^\dagger \varphi)^2$ where $D_\mu \varphi = \partial_\mu \varphi -\frac{1}{2}A^a?\mu \tau^a \varphi$ and $\tau^a$ are the generators of the Lie algebra of $SU(2)$ (the Pauli matrices) and the $A^a_\mu$ are real-valued functions.

We choose the following as the ground state: $A^{a,(v)}_\mu = 0, \varphi^{(v)} = \frac{1}{\sqrt{2}}\begin{bmatrix}0 \\ \varphi_0 \end{bmatrix}$ where $\varphi_0 = \frac{\mu}{\sqrt{2\lambda}}$, and we consider small excitations about the ground state with the form $\varphi = \frac{1}{\sqrt{2}}\begin{bmatrix}\theta^1 +i\theta^2 \\ \varphi_0 + \chi +i\theta^3\end{bmatrix}$. Notice that under the $SU(2)$ transformation $\varphi \to \omega(x)\varphi$ where $\omega = I+i\tau^a u^a(x)$ where the $u^a$ are small real functions, to linear order in the fields, the $\chi$ function does not change, but the $\theta$ functions each pick up a new term which will depend on the $u$ functions, so only the $\chi$ function represents a physical scalar boson field and the $\theta$ fields are arbitrary "fictitious" bosons. To quadratic order in the fields, we can eliminate these from the Lagrangian by substituting $\varphi$ into the Lagrangian, expanding, and making a change of the gauge field variables that gets rid of the $\theta$ terms, or alternatively by fixing a unitary gauge in which $\varphi = \frac{1}{2}\begin{bmatrix}0 \\ \varphi_0 + \chi \end{bmatrix}$ and then expanding the Lagrangian. Either way, the quadratic-order Lagrangian turns out to be $L^{(2)} = -\frac{1}{4}(\partial_{\mu}A^a_{\nu}-\partial_{\nu}A^a_{\mu})^2 + \frac{1}{2}(\partial_\mu \chi)^2+\frac{1}{8}g^2 \varphi_0^2 A^a_{\mu}A^a_{\mu} -\mu^2\chi^2$. In this case, all three of the vector boson fields $A^a_\mu$ have the same mass, $m_V = \frac{g\varphi_0}{2}.$

We can write the $SU(2)$ doublet in the form $\begin{bmatrix}\eta^1+i\eta^2 \\ u + i\eta^3\end{bmatrix}$. Then the scalar potential for the model can be written $-\mu^2(u^2+\eta^a\eta^a) +\lambda(u^2+\eta^a\eta^a)^2$, and this potential has a global $SO(3)$ invariance where $u$ is a scalar that transforms according to the trivial representation of $SO(3)$ (singlet) and the $\eta^a$ are the components of a vector in $\mathbb{R}^3$ that transform according to the fundamental representation of $SO(3)$ (triplet). This global $SO(3)$ invariance is not spontaneously broken by the chosen ground state $\varphi^{(v)} = \frac{1}{\sqrt{2}}\begin{bmatrix}0 \\ \varphi_0\end{bmatrix}$.

Problem statement:
1.) Choose a transformation rule for the gauge vector fields so that the Lagrangian for the model is invariant under global $SO(3)$ transformations.
2.) Show that the equality of the masses of the vector bosons in the quadratic Lagrangian is a consequence of the fully unbroken global $SO(3)$ symmetry.

Source: Valery Rubakov, Classical Theory of Gauge Fields, problem 6.6.

Relevant Equations

Everything is given in the problem statement.

1.) The rule for the global $SO(3)$ transformation of the gauge vector field is $A^i_{\mu} \to \omega_{ij}A^j_{\mu}$ for $\omega \in SO(3)$.

The proof is by direct calculation. First, if $A^i_{\mu} \to \omega_{ij}A^j_{\mu}$ then $F^i_{\mu \nu} \to \omega_{ij}F^j_{\mu\nu}$, so $F^i_{\mu\nu}F^i_{\mu\nu} \to \omega_{ij}\omega_{ik}F^j_{\mu\nu}F^k_{\mu\nu} = \delta_{jk}F^j_{\mu\nu}F^k_{\mu\nu} = F^j_{\mu\nu}F^j_{\mu\nu}$, and in general "dot product" terms $x^iy^i$ are invariant under $SO(3)$ transformations. so the part with the strength tensor is invariant. Now we expand the kinetic energy: $(D_{\mu}\varphi)^{\dagger}(D_{\mu}\varphi) = (\partial_{\mu}u)^2 + (\partial_{\mu}\eta^i)^2 +g\big(\epsilon^{ijk}(\partial_{\mu}\eta^i)A_{\mu}^j\eta^k+(\partial_{\mu}u)A^i_\mu\eta^i-uA^i_{\mu}\partial_{\mu}\eta^i\big)+\frac{g^2}{4}A^a_{\mu}A^a_{\mu}(u^2+\eta^i\eta^i)$> The dot product terms are all invariant. The "cross product" $\epsilon^{ijk}A^j_{\mu}\eta^k$ transforms to $\epsilon^{ijk}\omega_{jl}\omega_{km}A^l_{\mu}\eta^m$ and by the rotation invariance of the cross product this is equal to $\omega_{il}\epsilon^{ljk}A^j_{\mu}\eta^k$, so $\epsilon^{ijk}(\partial_{\mu}\eta^i)A_{\mu}^j\eta^k \to \omega_{ij}\omega_{il}\epsilon^{ljk}(\partial_{\mu}\eta^j)A_{\mu}^j\eta^k = \epsilon^{ijk}(\partial_{\mu}\eta^i)A_{\mu}^j\eta^k$. The invariance of dot products under $SO(3)$ means that the potential is also invariant. Therefore the Lagrangian is invariant under global $SO(3)$ transformations of the form $u \to u, \eta^i \to \omega_{ij}\eta^j, A^i_\mu \to \omega_{ij}A^j_\mu$.

Simple enough. The second part is really where I'm having trouble.

For simplicity, and without losing generality, we can spontaneously break the $SO(3)$ symmetry down to $SO(2)$ by choosing the ground state $$\varphi^{(v)} = \frac{1}{\sqrt{2}}\begin{bmatrix}0 \\ \phi_0+iv\end{bmatrix}$$
Let's consider small excitations about this ground state with the form $$\varphi = \frac{1}{\sqrt{2}}\begin{bmatrix}\theta^1+i\theta2 \\ \phi_0+\chi+i(v+\theta^3)\end{bmatrix}$$ Let $\omega(x) = I+i\tau^a u^a$ be an SU(2) gauge transformation where the $u^a$ are small. Then to linear order:
$$\omega\varphi = \frac{1}{\sqrt{2}}\begin{bmatrix}\theta^1+\varphi_0u^2-vu^1+i(\theta^2+\varphi_0u^1+vu^2) \\ \varphi_0+ \chi +vu^3+i(v+\theta^3 - \varphi_0u^3)\end{bmatrix}$$
The $\chi$ function has picked up a non-constant term $vu^3$. This means that as a result of spontaneously breaking the $SO(3)$ global symmetry, the $\chi$ function has been demoted from a Higgs field to a fictitious field which we will need to eliminate from the Lagrangian.

We'll fix the gauge and set $\theta^1=\theta^2=\theta^3=0$. Then to linear order in the fields:
$$D_{\mu}\varphi = \frac{1}{\sqrt{2}}\begin{bmatrix}0 \\ \partial_{\mu}\chi \end{bmatrix} -\frac{ig}{2\sqrt{2}}A^a_{\mu}\tau^a\begin{bmatrix}0 \\ \varphi_0+iv\end{bmatrix}$$
Then to quadratic order, $(D_{\mu}\varphi)^\dagger (D_{\mu}\varphi) = \frac{1}{2}(\partial_{\mu}\chi)^2-\frac{1}{2}gvA^3_{\mu}\partial_{\mu}\chi+\frac{1}{8}g^2(v^2+\varphi^2_0)A^a_{\mu}A^a_{\mu}$. For the $a=3$ component, we have: $$\begin{align*}\frac{1}{2}(\partial_{\mu}\chi)^2 -\frac{1}{2}gvA^3_\mu\partial_{\mu}\chi+\frac{1}{8}g^2(v^2+\varphi_0^2)A^3_{\mu}A^3_{\mu} &= \frac{1}{8}g^2v^2\left(A^3_{\mu}-\frac{2}{gv}\partial_{\mu}\chi\right)^2+\frac{1}{8}g^2\varphi_0^2A^3_{\mu}A^3_{\mu}\\
&\equiv \frac{g^2v^2}{8}B^3_{\mu}B^3_{\mu}+\frac{g^2\varphi_0^2}{8}A^3_{\mu}A^3_{\mu}
\end{align*}$$
And from here I don't really know how to proceed. I could factor this into a product of complex fields, but there seems to be some ambiguity in how I can do that. For instance, I could set $Z^{\pm}_{\mu} = B^3_{\mu}\pm i\frac{\varphi_0}{v}A^3_{\mu}$ so that $\frac{g^2v^2}{8}B^3_{\mu}B^3_{\mu}+\frac{g^2\varphi_0^2}{8}A^3_{\mu}A^3_{\mu} = \frac{g^2v^2}{8}Z^{+}_{\mu}Z^{-}_{\mu}$ or I could do $Z^{\pm}_{\mu} = \frac{vB^3_{\mu} \pm i\varphi_0A^3_{\mu}}{\sqrt{\varphi_0^2+v^2}}$ so that $\frac{g^2v^2}{8}B^3_{\mu}B^3_{\mu}+\frac{g^2\varphi_0^2}{8}A^3_{\mu}A^3_{\mu} = \frac{g^2(v^2+\varphi_0^2)}{8}Z^{+}_{\mu}Z^{-}_{\mu}$.

So how do I finish this off and show that the three vector fields do not all have the same mass? Am I making a mistake somewhere?

 

[jack476]

I figured it out.

By Noether's theorem, there is a conserved current $j^a_{\mu}$ associated with this global $SO(3)$ symmetry, and there is an energy associated with this current that is proportional to $j^a_{\mu}A^a_{\mu}$. Let $t^{ij}_a$ be component $i, j$ of the $a$th generator of the Lie algebra of $SO(3)$. Then the Noether current is:
$$\begin{align*} j^a_{\mu} &= \frac{\partial L}{\partial(\partial_{\mu}A^i_{\nu})}t^{ij}_aA^j_{\nu} + \frac{\partial L}{\partial(\partial_{\mu}\eta^i)}t^{ij}_a\eta^j\\
&= \epsilon^{aij}\big(-F^i_{\mu\nu}A^j_{\nu}+2(\partial_{\mu}\eta^i)\eta^j+g\eta^j\epsilon^{ikl}A^k_{\mu}\eta^l-uA^i_{\mu}\eta^j\big)\end{align*}$$
Break the symmetry with an arbitrary ground state $\varphi^{(vz)} = \begin{bmatrix}v^1+iv^2 \\ \varphi_0+iv^3\end{bmatrix}$ and consider small excitations where $u = \varphi_0+\chi$ and $\eta^i = v^i+\theta^i$. To quadratic order in the fields, $$j^a_{\mu}A^a_{\mu} = 2\epsilon^{aij}v^j A^a_{\mu}\partial_{\mu}\theta^i+g\epsilon^{aij}\epsilon^{ikl}v^l v^j A^a_{\mu} A^k_{\mu}-\varphi_0 \epsilon^{aij}v^jA^a_{\mu}A^i_{\mu}$$
This quantity is zero only if all three $v^i$ are zero, and if any of the $v^i$ are nonzero then in general the interaction energy between the gauge field and the Noether current contributes different amounts of mass to each of the three vector bosons, after making a change of variables that gets rid of the $\partial_{\mu}\theta^i$ terms.