 Solution

ID: past-exam-of-the-particle-physics-course-of-the-university-of-oxford/2022/5/b/solution

G = g^{2} + g^{'2}

and

s_{W} = sin θ_{W}

. First expand the charged term using the ladder operators:

τ^{+} W^{+} + τ^{-} W^{-} = \frac{( τ _{1} + i τ _{2} ) ( W ^{1} - i W ^{2} ) + ( τ _{1} - i τ _{2} ) ( W ^{1} + i W ^{2} )}{2} = 2 (τ_{1} W^{1} + τ_{2} W^{2}) .

Substituting the neutral gauge fields as well, the covariant derivative becomes

D_{μ} = \partial_{μ} - i g (τ_{1} W_{μ}^{1} + τ_{2} W_{μ}^{2}) - i [g (τ_{3} - Q s_{W}^{2}) + \frac{e g ^{'}}{G} Q] W_{μ}^{3} - i [- g^{'} (τ_{3} - Q s_{W}^{2}) + \frac{e g}{G} Q] B_{μ} .

Matching

W^{3}

g τ_{3}

requires

g s_{W}^{2} = \frac{e g ^{'}}{G} .

Matching

B

g^{'} (Q - τ_{3}) = g^{'} Y /2

requires

g^{'} s_{W}^{2} + \frac{e g}{G} = g^{'} .

For

g, g^{'} > 0

, elimination yields

sin^{2} θ_{W} = \frac{g ^{'2}}{g ^{2} + g ^{'2}}, tan θ_{W} = \frac{g ^{'}}{g}, θ_{W} = arctan \frac{g ^{'}}{g}, e = \frac{g g ^{'}}{g ^{2} + g ^{'2}} .

Equivalently,

e = g sin θ_{W} = g^{'} cos θ_{W}

and

1/ e^{2} = 1/ g^{2} + 1/ g^{'2}

. The conventional Weinberg angle lies in the first quadrant.

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