Electroweak symmetry breaking

Electroweak symmetry breaking is the Higgs-field breaking $SU(2{)}_L\times U(1{)}_Y\to U(1{)}_Q$. It gives masses to the $W$ and $Z$ bosons while leaving the photon massless.

With the hypercharge convention $Q=T_3+Y$, the lower component of the Higgs doublet has $T_3=-1/2$ and $Y=1/2$. Thus $Q⟨\varphi ⟩=0$: the vacuum expectation value is electrically neutral. This is why one combination of the neutral gauge fields remains the massless photon, as the electroweak doublet gauge-boson mass matrix shows.

Continue the chosen Higgs field potential example with $g=3/5$ and $g^{\prime }=4/5$. The tree-level electroweak gauge-boson masses are $m_W=72\mathrm{GeV}$, $m_Z=120\mathrm{GeV}$ and $m_A=0$. In the $(W^3,B)$ basis,Its orthonormal eigenvectors $(4/5,3/5{)}^T$ and $(3/5,-4/5{)}^T$ have eigenvalues zero and $14400{\mathrm{GeV}}^2$. Thus $\cos {\theta }_W=3/5=m_W/m_Z$. This is an exact model check, not a fit to measured electroweak interaction parameters.

A diagonal charged-fermion Yukawa coupling expands asThe Yukawa coupling remains a free parameter: the mechanism relates a fermion mass to its scalar interaction but does not predict the numerical mass. This example concerns charged fermions; the minimal Standard Model has no right-handed neutrino and does not account for neutrino oscillations without an extension.

An electroweak doublet with $Y=-1/2$ has component charges $(0,-1)$ under $Q=T^3+Y$. Its upper-component vacuum therefore preserves electromagnetism. The canonical covariant-derivative term gives $m_W=g_{2}v/2$, $m_Z=v\sqrt{g_1^2+g_2^2}/2$ and a massless photon. Neutral eigenvectors are proportional to $(g_2,-g_1)$ for $Z$ and $(g_1,g_2)$ for the photon. This is the conjugate-doublet convention to the usual positive-hypercharge Higgs doublet and gives the same mass ratio and electric coupling.

A thermal change in the electroweak Higgs sector as the early Universe cools. Its order and strength depend on the particle content, couplings and effective description. A polynomial landau free energy is a simplified model and does not, by itself, establish a model-independent Higgs-mass bound.

If $\lambda /a=x=m_H^2/m_0^2$ and $c/a=A+x$, then $\gamma =[72{\pi }^{2}x(A+x){]}^{-1}$ and ${\varphi }^{\ast }/T^{\ast }=(6\pi x{)}^{-1}$. Requiring both $\gamma <1$ and ${\varphi }^{\ast }>T^{\ast }$ gives $x_{-}<x<1/(6\pi )$, where $x_{-}=[-A+\sqrt{A^2+1/(18{\pi }^2)}]/2$. This is an open interval for the specified model. Failing the strength condition does not alone eliminate its first-order jump.

Renormalization-group evolution can change a Higgs squared-mass parameter from positive at a high input scale to negative at lower scales, making a symmetry-breaking scalar expectation energetically favorable. In the Minimal supersymmetric Standard Model, the large top Yukawa coupling can drive the up-type Higgs soft squared mass downwards as the scale is lowered. Actual breaking also requires the other Higgs-potential parameters to give a stable suitable minimum; the running of one mass alone is not sufficient.

The electroweak scale is the characteristic energy associated with electroweak symmetry breaking. With the Standard Model Higgs doublet normalized by $⟨H⟩=(0,v/\sqrt{2}{)}^T$, its vacuum expectation value satisfies $v=(\sqrt{2}{G}_F{)}^{-1/2}\simeq 246\mathrm{GeV}$ and $m_W=gv/2$. Its smallness compared with possible ultraviolet scales motivates the hierarchy problem.

For a Higgs doublet with hypercharge $1/2$ and vacuum expectation value $(0,v/\sqrt{2}{)}^T$, its gauge-covariant kinetic term gives the neutral gauge-boson mass matrixin the $(W^3,B)$ basis. Its eigenvalues are zero and $(g^2+g^{\prime 2})v^2/4$, and the Weinberg angle rotation gives the photon and Z boson. The two charged electroweak gauge bosons have $m_W^2=g^2{v}^2/4$. The zero eigenvalue expresses the unbroken electric-charge gauge symmetry.

For the Higgs doublet vacuum $(0,v{)}^T/\sqrt{2}$, the neutral gauge-boson mass matrix in $(W^3,B)$ is $(v^2/4)\left(\begin{array}{cc}{g}^2& -g{g}^{\prime }\\ -g{g}^{\prime }& g^{\prime 2}\end{array}\right)$. Its zero eigenvector is proportional to $(g^{\prime },g)$, the photon; the perpendicular vector is the Z boson. The charged fields $W^{\pm }=(W^1\mp i{W}^2)/\sqrt{2}$ have the stated W boson mass. Thus $m_W=m_Z\cos {\theta }_W$, where ${\theta }_W$ is the Weinberg angle. The unbroken charge generator annihilates the vacuum and explains the zero eigenvalue.