 Solution

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

Let

g

and

g^{'}

Φ = (0, 0, v + h)^{T}

(T^{a})_{b c} = - i ϵ_{ab c}

, the kinetic term becomes

\frac{1}{2} (D_{μ} Φ)^{T} (D^{μ} Φ) = \frac{1}{2} (\partial h)^{2} + \frac{1}{2} g^{2} (v + h)^{2} [W_{μ}^{1} W^{1 μ} + W_{μ}^{2} W^{2 μ}] .

Thus, in the basis

(W^{1}, W^{2}, W^{3}, B)

M_{V}^{2} = g^{2} v^{2} diag (1, 1, 0, 0), m_{W^{\pm}} = gv = \frac{g m}{2 λ}, m_{W^{3}} = m_{B} = 0.

Here

W^{\pm} = (W^{1} \mp i W^{2}) / 2

. The two massless neutral gauge bosons may be rotated to any orthonormal basis; no neutral mass term selects a unique mixing angle.

Expanding the same kinetic term to first order in

h

gives

L_{hVV} = g^{2} v h [(W^{1})^{2} + (W^{2})^{2}] = 2 g^{2} v h W_{μ}^{+} W^{- μ} .

h W_{μ}^{+} W_{ν}^{-} : 2 i g^{2} v g_{μν} = \frac{2 i m _{W}^{2}}{v} g_{μν},

and, in the real basis,

h W_{μ}^{a} W_{ν}^{b} : 2 i g^{2} v δ^{ab} g_{μν}

for

a, b \in {1, 2}

. All such three-point Feynman vertices involving

W^{3}

B

vanish. The factor two in the real basis comes from the two identical gauge boson fields; in the charged basis it is already explicit in

L_{hVV}

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