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\PhiC_{\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 -> \phie^{-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]

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