I'm currently reading the book Quantum Fields on a Lattice by I. Montvay and G.Münster, and in section 6.1 they describe lattice actions for various higgs models. And I got confused at the moment where they describe the unitary gauge action for $SU(2)$-higgs theory. In 6.16 they say that higgs field can be represented as $φ_x = ρ_xα_x$ , $ρ_x \ge 0$ , $α_x\in SU(2)$ and $φ_x^+φ_x=ρ_x^21$. Where $ρ_x$ is length of higgs field at given point, and $α_x$ is angular components of the higgs field. And in 6.24 they define unitary gauge $α_x' = 1$ , $φ_x' = \begin{pmatrix} ρ_x & 0 \\ 0 & ρ_x \\ \end{pmatrix}$. Higgs field becomes a diagonal matrix with entries being higgs length(modulus), thus getting a preferred direction and breaking gauge symmetry. But what is the meaning of this $ρ_x$ ? It is almost always non-zero, it can be equal to zero only if the Higgs field $φ_x$ itself at this point is equal to zero. And you can measure it without going to unitary gauge, because it's gauge invariant quantity. So what you should use to measure higgs VEV or higgs propagator for example?
1 Answers
Your answer is here, up to language and normalizations. That is, your authors adopt the 2×2 hermitian matrix representation of the complex Higgs doublet $\phi$, by arraying it and its conjugate in entries of a transposed two vector, (Longhitano's thesis): $$ \varphi \equiv (\tilde{\phi}, \phi)=(i\tau_2 \phi^*, \phi). \tag{6.1,6.2} $$ I have suppressed throughout the entirely superfluous subscripts x.
This matrix may be written as $$ \varphi = \rho \alpha , \leadsto \\ \varphi^\dagger \varphi = \rho^2 \alpha^\dagger \alpha = \rho^2 \mathbb{I}, \tag{6.15, 6.16} $$ since the as are unitary and unimodular. The latter matrix is L- and hypercharge gauge invariant, since it is only variant under the global R, custodial, symmetry!
The ρ is the unshifted radial variable of the Higgs field, the only one uninvolved in the Goldstone phenomenon, basically the v.e.v. plus the "debris" Higgs particle excitation. Tastefully, the authors call it the σ, mapping it to the legendary O(4)/O(3) σ-model of Gell-Mann and Levy... $\langle \rho^2\rangle= 1$ so $\langle \varphi^\dagger \varphi\rangle= \mathbb{I}$.
In unitary gauge, $ \alpha =\mathbb{I}$, so $\langle \varphi\rangle= \mathbb{I}$.
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