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In the Standard Model, electroweak unification is based on the symmetry breaking of $ \text{SU(2)}\times \text{U(1)}_Y \rightarrow \text{U(1)}_{EM}$ by the VEV of a complex doublet $H$ with hypercharge $\frac{1}{2}$.

The Lagrangian is

$$\mathcal{L}=-\frac{1}{4}(W^a_{\mu\nu})^2-\frac{1}{4}B^2_{\mu\nu}+(D_\mu H)^\dagger(D_\mu H)+m^2H^\dagger H-\lambda(H^\dagger H)^2,$$

where $W^a_\mu$ are the $\text{SU(2)}$ gauge bosons and $W^a_{\mu\nu}$ are given by $W^a_{\mu\nu}=\partial_\mu W^a_\nu-\partial_\nu W^a_\mu+gf^{abc}W^b_\mu W^c_\nu$, and $B_\mu$ is the $\text{U(1)}$ gauge bosons and $B_{\mu \nu}=\partial_\mu B_\nu-\partial_\nu B_\mu$.

The covariant derivative is $$D_\mu H=\partial_{\mu}H-igW^a_\mu \tau^aH-\frac{1}{2}ig'B_\mu H.$$

After $H$ gets its VEV $$H_0=\begin{pmatrix} 0 \\ \frac{v}{\sqrt{2}} \end{pmatrix},$$ the mass term of the gauge bosons comes from $$(D_\mu H_0)^\dagger(D_\mu H_0)=\frac{v^2}{8}\left[ g^2(W^1_\mu)^2+g^2(W^2_\mu)^2+g'^2(g'B_\mu-gW^3_\mu)^2 \right].$$

My questions are why we define W bosons as $W^\pm_\mu=\frac{1}{\sqrt{2}}(W^1_\mu\mp iW^2_\mu)$ rather than just $W^1_\mu$ and $W^2_\mu$ respectively, and how to see the charges of $W^\pm$ and $Z$ bosons?

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    $\begingroup$ Look not at the Higgs sector but on the gauge field self-interaction. It will give you interaction between photon and $W^\pm$ $\endgroup$ – OON Dec 1 '16 at 14:22
  • $\begingroup$ @ OON The gauge field self-interaction comes from the term $-\frac{1}{4}(W^a_{\mu\nu})^2$, but I expand this term and have the interaction of three W bosons looking like $gWW\partial W$ and that of four W bosons looking like $g^2 WWWW$. I don't know how to get the interaction between photon and $W^\pm$ $\endgroup$ – William Huang Dec 2 '16 at 1:42
  • $\begingroup$ Electromagnetic field is not an original $B$-field but superposition of $B_\mu$ and $W_\mu^3$. Here it will appear from $W_\mu^3=\cos\theta_W Z_\mu + \sin\theta_W A_\mu$ where $\tan\theta_W=g_2/g_1$. $\endgroup$ – OON Dec 2 '16 at 6:04
  • $\begingroup$ @OON So should I transform $W^a_\mu , B_\mu$ basis to $W^\pm_\mu , A_\mu , Z_\mu$ basis and then expand the term $−\frac{1}{4}(W^a_{\mu \nu})^2−\frac{1}{4}(B_{\mu \nu})^2$ and find the interaction term looking like $AW^\pm$ ? But there is still no two interaction. If not, which term I should looking for? $\endgroup$ – William Huang Dec 2 '16 at 9:51
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    $\begingroup$ Ok, write thoroughly the $\frac{1}{4}(W_{\mu\mu}^a)^2$ through components $W_\mu^\pm$, $Z_\mu$ and $A_\mu$. Then omit for simplicity all terms involving $Z$ and $W^\pm$ self-interaction. Then compare it with the following - if we want $W^\pm$ to be electrically charged its kinetic term should take the form $\vert D_\mu W_\nu^\pm-D_\nu W_\mu^\pm\vert^2$ where $D_\mu=i\partial_\mu\mp eA_\mu$. $\endgroup$ – OON Dec 2 '16 at 10:45
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The only thing You need is the explicit form of the electric charge generator $Q$, the only one "unbroken" by the Higgs VEV: $$ Q = t_{3} + \frac{Y}{2}, $$ where $t_{3}$ is the weak isospin and $Y$ is the weak hypercharge.

For $SU_{\text{L}}(2)$ bosons, namely, $W_{1},W_{2},W_{3}$, the weak hypercharge $Y$ is zero by the definition (note also that the hypercharge of $U_{Y}(1)$ gauge field, which is $B \sim g_{1}Z - g_{2}A$).

In order to define the isospin $t_{3}$, we have to calculate the commutator $$ \tag 1 [T_{3},T_{i}] = t_{3}T_{i}, $$ where $T_{3} \equiv \sigma_{3}$, with $\sigma_{3}$ being the Pauli matrix, is the isospin generator, while $T_{i}$ is the $SU_{\text{weak}}(2)$ generator associated with given boson. $(1)$ is the standard charge algebra.

From $W_{1},W_{2}, W_{3}$ bosons you can construct three states with given eigenvalues of isospin: $t_{3} = +1, t_{3} = -1, 0$. These states corresponds to generators $$ \sigma_{+} = \frac{1}{2}\left(\sigma_{x} + i\sigma_{y}\right), \quad \sigma_{-} = \frac{1}{2}\left(\sigma_{x} - i\sigma_{y}\right), \quad \sigma_{0} = \sigma_{3} $$ We have $$ [T_{3},\sigma_{\pm}] = \pm \sigma_{\pm}, \quad [T_{3},\sigma_{3}] = 0 $$ Therefore, the combination $W_{1} - iW_{2}$ has the charge $Q = +1$, $W_{1}+iW_{2}$ has the charge $Q = -1$, and $W_{3} \sim g_{1}Z + g_{2}A$ has the charge $Q = 0$ (as well as $Z, A$ separately).

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  • $\begingroup$ Why we need the commutator to define the isospin $t_3$ ? $\endgroup$ – William Huang Dec 2 '16 at 1:44
  • $\begingroup$ @WilliamHuang Because vector bosons transform as adjoint representation of $SU(2)$ - not as 2-component complex vectors but as traceless hermitian 2x2 matrices. $\endgroup$ – OON Dec 2 '16 at 6:12
  • $\begingroup$ @OON Do you mean $W^a_{\mu \nu} \rightarrow W^a_{\mu \nu}-f^{abc}\alpha^b W^c_{\mu \nu}$ under $SU(2)$ transformation? But I still cannot realize the connection between the adjoint representation and the commutator and $T_3$. $\endgroup$ – William Huang Dec 2 '16 at 9:40
  • $\begingroup$ @WilliamHuang $[T_a,T_b]=f^{abc}T_c$. $\endgroup$ – OON Dec 2 '16 at 10:38
  • $\begingroup$ @OON But now why do we need $[T_3, T_i]=t_3T_i$ rather than $[T_3, T_i]=t_3T_j$ and why does $f^{abc}$ relate to the isospin? I have a lot of confusion about isospin. Very thanks for more explanation. $\endgroup$ – William Huang Dec 3 '16 at 16:14

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