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Peskin & Schroeder about spontaneous symmetry breaking

Physics Asked on January 11, 2021

I am confused about how Peskin & Schroeder derive the matrix (21.39):
begin{equation}
gF^a_{ i}=frac{v}{2} left(begin{array}{ccc}
g & 0 & 0
0 & g & 0
0 & 0 & g
0 & 0 & -g’
end{array}right),
end{equation}

where $F^a_{ i} equiv T^a_{ij}phi_{0j}$ ($T^a_{ij}$ are the real anti-symmetric generators of the group $SU(2) times U(1)$, and $phi_{0j}$ is the $j$-th vacuum expectation value). Just above for an example, they say that "$T^1 phi_0$ equals $v/2$ times a unit vector in the $phi_1$ direction", but according to their definition of $T^a$, one has $T^1=-ifrac{sigma^1}{2}$ (where $sigma^1$ is the first Pauli matrix), and so
begin{equation}
T^1 phi_0 = frac{1}{sqrt{2}}
left(begin{array}{cc}
0 & -i/2
-i/2 & 0
end{array}right)
left(begin{array}{c}
0
v
end{array}right)
=
-frac{1}{sqrt{2}}left(begin{array}{c}
-iv/2
0
end{array}right),
end{equation}

which is not normalized to $1$. Furthermore, this matrix is a column of two lines, and I don’t understand how to recover their matrix (21.39) from two lines-vectors.

One Answer

This is for the expansion of the covariant derivative term $D_muphi_i$ expanded near the vacuum state. The perturbation field $phi-phi_0=frac{1}{sqrt{2}}left(begin{array}{c} phi^{2}-iphi^{1} h+iphi^{3} end{array}right)$ is to be reorganized as four scalar fields $chi=left(begin{array}{c} phi^1 phi^2 phi^3 h end{array}right)$.

The terms in $frac12(partial_mu phi_i+gA^a_mu T^a_{ij}phi_j)^2$ include a first order perturbation term $partial^mu chi_i (g A^a_mu T^a_{i0}phi_0)$.

The massive Higgs boson field $h$ is always orthogonal to $F^a_i=gT^a_{i0}phi_0$ so they turn the matrix to 4x3 instead of 4x4.

Your calculation of $T^1phi_0$ is perfectly fine up to a sign. It's just that the resulting complex-2D field should be turned into the 3D real scalar field $chi$. So $partial chi$ being paired to $left(begin{array}{c} vg/2 0 0 end{array}right)$ is to be treated as $partial(phi-phi_0)$ being paired to $frac{1}{sqrt{2}}left(begin{array}{c} -ivg/2 0end{array}right)$, which is the first row of the 4x3 matrix.

Correct answer by C Tong on January 11, 2021

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