W and Z Boson Masses from the Electroweak Lagrangian

1. The Higgs Kinetic Term: The Source of Mass

The masses of the gauge bosons arise from the kinetic term in the Higgs Lagrangian, which we introduced previously:

The key is the covariant derivative, $D_{\mu }$, which enforces gauge invariance by coupling the Higgs field $H$ to the gauge fields. For a field transforming as an $SU(2{)}_L$ doublet with hypercharge $Y$, the covariant derivative is:

Here, $W_{\mu }^a$ (with $a=1,2,3$) are the three $SU(2{)}_L$ gauge fields with coupling $g$, and $B_{\mu }$ is the $U(1{)}_Y$ gauge field with coupling $g^{\prime }$. For the Higgs doublet, the hypercharge is $Y=1/2$.

After spontaneous symmetry breaking, the Higgs field acquires its VEV, which we can write in the "unitary gauge" where the Goldstone boson components have been gauged away:

Here, $v\approx 246$ GeV is the constant VEV, and $h(x)$ is the physical Higgs boson field, representing fluctuations around that minimum.

To find the masses of the gauge bosons, we can ignore the fluctuation $h(x)$ for a moment and just substitute the constant VEV into the kinetic term. When we do this, the partial derivative term ${\partial }_{\mu }H$ vanishes, as the VEV is constant in spacetime. The kinetic term then becomes a source of pure interaction terms between the gauge fields and the Higgs VEV:

This expression, which contains only gauge fields and the constant $v$, will reveal the mass terms. The following reading provides the detailed derivation we will now walk through.

2. Deriving the Boson Masses

Let's evaluate the expression for the mass terms. The operator acting on the VEV is:

Applying this to the VEV gives:

Now, we take the Hermitian conjugate of this vector and multiply it by the vector itself. The factor of $(-i)$ gives $(-i{)}^{†}(i)=i(-i)=1$.

This expression contains two distinct parts, one for the charged bosons and one for the neutral bosons.

2.1 The Charged Bosons: $W^{+}$ and $W^{-}$

Let's analyze the first part. We define the charged W boson fields as the linear combinations:

Using these definitions, the term becomes:

A mass term for a complex vector boson has the form $M_W^2{W}^{-\mu }{W}_{\mu }^{+}$. Comparing our result to this standard form, we can directly read off the squared mass of the W boson:

We have successfully generated a mass for the charged weak bosons!

2.2 The Neutral Bosons: $Z$ and $\gamma$

Now for the second part of our expression:

This expression describes the masses of the neutral gauge bosons. However, the presence of the cross-term $W_{\mu }^3{B}^{\mu }$ tells us that $W_{\mu }^3$ and $B_{\mu }$ are not the physical particles we observe; they are not mass eigenstates. The physical particles are the linear combinations that diagonalize this expression.

We can write this in matrix form. A mass term for two real fields ${\varphi }_1,{\varphi }_2$ is $\frac{1}{2}({\varphi }_1{\varphi }_2)M^2\left(\begin{array}{c}{\varphi }_1\\ {\varphi }_2\end{array}\right)$. Our Lagrangian term is therefore:

The squared masses of the physical fields are the eigenvalues of the matrix $M_{\text{neutral}}^2=\frac{v^2}{4}\left(\begin{array}{cc}{g}^2& -g{g}^{\prime }\\ -g{g}^{\prime }& g^{\prime 2}\end{array}\right)$.

Let's find the eigenvalues:

• Determinant: $\det (M_{\text{neutral}}^2)={\left(\frac{v^2}{4}\right)}^2(g^2{g}^{\prime 2}-(-g{g}^{\prime }{)}^2)=0$. Since the determinant is zero, one of the eigenvalues must be zero. This corresponds to a massless boson.
• Trace: $\text{Tr}(M_{\text{neutral}}^2)=\frac{v^2}{4}(g^2+g^{\prime 2})$. Since the trace is the sum of the eigenvalues, this must be the value of the non-zero eigenvalue.

So, the squared masses of our two physical neutral bosons are:

We have found one massive neutral boson, the $Z$ boson, and one massless neutral boson, which we identify as the photon ($\gamma$).

3. The Weinberg Angle

The physical fields $Z_{\mu }$ and $A_{\mu }$ (the photon field) are the eigenvectors of the neutral mass matrix. This relationship is typically parameterized by a rotation, defining the weak mixing angle, or Weinberg angle, ${\theta }_W$:

where the angle is defined such that it diagonalizes the mass matrix. This gives the relations:

Using this definition, we can express the $Z$ boson mass in a very elegant way in relation to the $W$ boson mass:

This relationship, $M_W=M_Z\cos {\theta }_W$, is a fundamental prediction of the Standard Model and has been verified with incredible precision at particle colliders.

Conclusion

Key Takeaways:

• The masses of the W and Z bosons are generated from the kinetic term of the Higgs Lagrangian, $(D_{\mu }H{)}^{†}(D^{\mu }H)$, after substituting the Higgs vacuum expectation value $v$.
• The masses are directly proportional to the VEV and the respective gauge couplings: $$M_W=\frac{gv}{2}\text{and}{M}_Z=\frac{v}{2}\sqrt{g^2+g^{\prime 2}}$$
• The physical neutral bosons (photon $\gamma$ and $Z$) are mixtures of the original $W^3$ and $B$ fields. This mixing is described by the Weinberg angle, ${\theta }_W$.
• The theory correctly predicts one massless gauge boson (the photon) and a precise relationship between the W and Z masses: $M_W=M_Z\cos {\theta }_W$.

We have now seen how the electroweak theory, through the Higgs mechanism, can account for the massive nature of the weak force carriers while preserving the gauge symmetry of the Lagrangian and the masslessness of the photon.

Preview of the Next Lesson:

The Higgs mechanism also gives mass to the fundamental matter particles—the quarks and leptons. In our next lesson, we will explore this by introducing Yukawa couplings to the Higgs field and showing how they generate fermion masses after symmetry breaking. This will complete our picture of mass generation in the Standard Model.