Showing that gauge fields become massless and massive

[creepypasta13]

Homework Statement

Consider a non-abelian gauge theory of SU(N) × SU(N) gauge fields coupled to N$^{2}$
complex scalars in the (N,N$^{_}$) multiplet of the gauge group. In N × N matrix notations,
the vector fields form two independent traceless hermitian matrices Bμ(x) =$\Sigma$$_{a}$ B$^{a}_{\mu}$(x) $\frac{\lambda}^{a}{2}$ and Cμ(x) = $\Sigma$$_{a}$ C$^{a}_{\mu}$(x) $\frac{\lambda}^{a}{2}$ , the scalar fields form a complex matrix $\Phi$, and the Lagrangian is
L = −(1/2) tr(B$_{\mu\nu}$B$^{\mu\nu}$ −(1/2) tr(C$_{\mu\nu}$C$^{\mu\nu}$) + tr(D$_{\mu}$$\Phi$$^{+}$D$^{\mu}$$\Phi$) - ($\alpha$/2)(tr($\Phi$$^{+}$$\Phi$))^2 - ($\beta$/2)(tr($\Phi$$^{+}$$\Phi$$\Phi$$^{+}$$\Phi$)) - (m^2)*(tr($\Phi$$^{+}$$\Phi$)

where

B$_{\mu\nu}$ = $\partial$$_{\mu}$B$_{\nu}$ - $\partial$$_{\nu}$B$_{\mu}$ - ig[B$_{\mu}$, B$_{\nu}$]
C$_{\mu\nu}$ = $\partial$$_{\mu}$C$_{\nu}$ - $\partial$$_{\nu}$C$_{\mu}$ - ig[C$_{\mu}$, C$_{\nu}$]
D$_{\mu}$$\Phi$ = $\partial$$_{\mu}$$\Phi$
+ igB$_{\mu}$$\Phi$ - ig$\Phi$C$_{\mu}$
D$_{\mu}$$\Phi$$^{+}$ = $\partial$$_{\mu}$$\Phi$$^{+}$
+ ig$\Phi$$^{+}$B$_{\mu}$ + igC$_{\mu}$$\Phi$$^{+}$

(2)
For simplicity, let both SU(N) factors of the gauge group have the same gauge coupling g.
Besides the local SU(N)×SU(N) symmetries, the Lagrangian (1) has a global U(1) phase
symmetry $\Phi$(x) → e$^{i\theta}$$\Phi$(x), but some of these symmetries become spontaneously broken
for m$^{2}$ = −μ$^{2}$ < 0. Specifically, for m$^{2}$ = −μ$^{2}$ < 0 but α, β > 0, the scalar potential has
a local maximum at $\Phi$ = 0 while the minima lie at
$\Phi$ = $\sqrt{\frac{\mu^{2}}{N\alpha + \beta}}$ × a unitary matrix. (3)

All such minima are related by SU(N) × SU(N) × U(1) symmetries to
$\Phi$ = $\sqrt{\frac{\mu^{2}}{N\alpha + \beta}}$ × 1$_{NxN}$


Now let’s be more specific: Show that the vector fields A$^{a}_{\mu}$ = $\frac{1}{\sqrt{2}}$(B$^{a}_{\mu}$ + C$^{a}_{\mu}$) remain massless while the orthogonal combinations X$^{a}_{\mu}$ = $\frac{1}{\sqrt{2}}$(B$^{a}_{\mu}$ - C$^{a}_{\mu}$) become massive.
Hint: Fix the unitary gauge in which the $\Phi$(x) matrix is hermitian up to an overall
phase, $\Phi$$^{+}$(x) = $\Phi$(x) × e$^{-2i\theta(x)}$. Explain why this gauge condition is non-singular for $\Phi$(x) near the minima (3).

The Attempt at a Solution

From what I've seen in textbooks, to see what new massive and massless field arise, you just substitute the new field (ie $\phi$ -> $\phi$e$^{-i\theta}$)
But I'm confused with this problem. Do I just substitute the A$^{a}_{\mu}$ in place of B and C in equation (1)? Or expand out (1) using the $\Phi$$^{+}$(x) = $\Phi$(x) × e$^{-2i\theta(x)}$, and then factor it out to get A$^{a}_{\mu}$ and X terms?

I don't see how by doing the latter, we will see a mass term in front of A$^{a}_{\mu}$? In equation (1), the only term the mass term is in front of is tr($\Phi$$^{+}$$\Phi$)

 

[creepypasta13]

The corrections in the typos in the OP are:
B$_{\mu}$(x) = $\Sigma$$_{a}$B$^{a}_{\mu}$(x)$\frac{\lambda^{a}}{2}$

C$_{\mu}$(x) = $\Sigma$$_{a}$C$^{a}_{\mu}$(x)$\frac{\lambda^{a}}{2}$

 

[micromass]

Thread pending moderation.