 Solution

ID: past-exam-of-the-advanced-quantum-field-theory-for-particle-physics-course-of-the-university-of-oxford/2021/3/a/i/solution

Put

r^{2} = Φ^{T} Φ

. The potential has its minimum at

\frac{d V}{d r} = - m^{2} r + 4 λ r^{3} = 0, v^{2} = \frac{m ^{2}}{4 λ} .

Φ_{0} = (0, 0, v)^{T}

. Since the weak hypercharge is zero,

Q = T^{3}

T^{3} Φ_{0} = 0

. Neither

T^{1}

nor

T^{2}

annihilates

Φ_{0}

. The stabilizer subgroup is consequently

S U (2) \times U (1)_{Y} ⟶ U (1)_{T^{3}} \times U (1)_{Y} .

In particular, spontaneous symmetry breaking has not reduced the two neutral gauge generators to a single one:

Y

annihilates every component and remains unbroken independently of

T^{3}

\frac{\partial ^{2} V}{\partial ϕ _{i} \partial ϕ _{j}}_{Φ_{0}} = (- m^{2} + 4 λ v^{2}) δ_{ij} + 8 λ (Φ_{0})_{i} (Φ_{0})_{j} = 8 λ v^{2} δ_{i 3} δ_{j 3} .

With canonical kinetic normalization

\frac{1}{2} (\partial h)^{2}

, the radial Higgs boson therefore has

m_{h}^{2} = 8 λ v^{2} = 2 m^{2}, m_{h} = 2 m .

The two tangential zero eigenvalues are Goldstone modes, absorbed by the two massive gauge bosons through the Higgs mechanism.

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