In the context of chiral effective theories we usually deal with the pion field
\begin{equation} U= ie^{\frac{\pi^a \sigma^a}{2f}} \end{equation}
where $\pi^a=\big(\pi^1, \pi^2, \pi^3\big)$ are the Goldstone bosons after breaking the chiral symmetry into
\begin{equation} SU(2)_L\times SU(2)_R\rightarrow SU(2)_V, \end{equation}
$\sigma^a$ are the Pauli matrices and $f$ is just a constant. Here we are only dealing with the up and down quarks, hence the $SU(2)$ symmetry groups instead of a more general symmetry.
It is common in the literature [1] to gauge the chiral symmetry in order to simplify the calculations. One defines a covariant derivative using external fields and then uses those fields as a trick to obtain the conserved current.
My question is related to the introduction of the electroweak bosons into this scheme. In this paper [2] and this one [3] they embed the fields $W^a_\mu$ and $B_\mu$ into the covariant derivative like this
\begin{equation} D_\mu U=\partial_\mu U +ig\frac{\tau^a}{2}W^a_\mu U-ig'U\frac{t^3}{2}B_\mu \end{equation}
Which means that the embedding is done by identifying the weak force $SU(2)_L$ group with the chiral symmetry group $SU(2)_L$ and the hypercharge group $U(1)_Y$ with the third generator of $SU(2)_R$. This is confirmed by this book [4] which says that $U$ transforms under the electroweak symmetry $SU(2)_L \times U(1)_Y$ like
\begin{equation} U\rightarrow U'=e^{\frac{i}{2}\theta_L^a\sigma^a}Ue^{\frac{i}{2}\sigma^3\alpha} \end{equation}
which means that $U(1)_Y$ as indeed embedded into the third generator of $SU(2)_R$. My question is: Why is this the way to embed hypercharge into the chiral symmetry? In the standard model, the group $U(1)_Y$ can act on left and right handed fields and, furthermore, it can act independently on different right handed fields. On the other hand, the third generator of the chiral group $SU(2)_R$ can't act on left handed fields and acts on upper and lower components of the right handed doublet in a related way (i.e. not independent as the hypercharge transformation). All this evidence implies that the embedding they are using is nonsense, since it doesn't reproduce the standard model hypercharge group as we know it. What am I missing here?
[1] Starting at page 8 on https://arxiv.org/abs/hep-ph/9502366
[2] Equation 2 on https://arxiv.org/abs/hep-ph/9809237
[3] Equations 2 and 3 on https://arxiv.org/abs/hep-ph/9308276
[4] Equation 3.10 on Electroweak Effective Lagrangians, by José Wudka.
Answer
No, no evidence of nonsense, but they are a bit pedagogically smug and do not want to make it easy for readers who don't speak the language. You looked at the wrong equations in ref [3]. They explicitly tell you, in eqns (4-5), how the nonlinear would-be Goldstone boson operator $U$ transforms linearly, whereas its logarithm fields $\vec{\pi}$ transform linearly under the vector symmetries like charge, but non-linearly under the broken axial ones and hypercharge, instead: \begin{equation} g_{\rm L} = e^{\displaystyle{ i \vec{\alpha}\cdot\vec{\tau} / 2}} \;\;\in \; SU(2)_{\rm L} , \;\;\;\; g_{\rm R} = e^{\displaystyle{ i \beta \tau^3 / 2}} \;\;\in \; U(1)_Y , \\ U' = g_{\rm L} U g_{\rm R}^\dagger ~~. \\ \vec{\pi'} \cdot \vec{\tau} - \vec{\pi} \cdot \vec{\tau} \quad = v \vec{\alpha} \cdot \frac{\vec{\tau}}{2} - v \beta \frac{\tau^3}{2} - (\vec{\alpha}\times\vec{\pi})\cdot \frac{\vec{\tau}}{2} + \frac{\beta}{2}(\pi_2 \tau_1-\pi_1 \tau_2)+ \\ +\frac{1}{6v}[(\vec{\alpha}\cdot\vec{\pi}) (\vec{\pi}\cdot\vec{\tau})-(\vec{\alpha}\cdot\vec{\tau}) (\vec{\pi}\cdot\vec{\pi})]-\frac{\beta}{6v} [\pi_3(\vec{\pi}\cdot\vec{\tau})-\tau_3(\vec{\pi}\cdot\vec{\pi})] + O(\pi^3) . \end{equation} So all you need do is confirm how the absorbable goldstons transform.
(The fermion doublets displayed in your ref [2] for the covariant derivatives of right fermions... !? should have spooked you—they most probably did. The representation of the Higgs doublet and its conjugate as a 2×2 unitary matrix is standard, cf. Longhitano, (1980) PhysRev D22 1166–75; NucPhys B188 (1981) 118–54, but reparameterized in the alternate, Gürsey language. Longhitano demonstrates in (2.7-2.8) how to trade an identity on the doublet for a custodial right isorotation on the matrix, the heart of your question, but the translation is too technical and I am sticking to a demonstration of correctness below. This prestidigitative magic, often hidden, merits its own question.)
So inspect the two leading orders in the infinitesimal transformation. Under the SSBroken $T^3_L$ transformation ($\alpha_3\neq 0$, the rest vanishing), they of course transform nonlinearly, $$ \delta \vec \pi \cdot \vec \tau= v \alpha_3 \cdot \frac{ \tau _3}{2} + \frac{ \alpha_3}{2} (\pi_2 \tau_1-\pi_1 \tau_2) +\frac{\alpha^3}{6v}[ \pi_3 (\vec{\pi}\cdot\vec{\tau})- \tau_3 (\vec{\pi}\cdot\vec{\pi})] +... $$ The neutral goldston is shifted and the charged ones rotate.
Under a pure hypercharge transformation, ($\beta\neq 0$, the rest vanishing), $$ \delta \vec \pi \cdot \vec \tau = - v \beta \cdot \frac{ \tau _3}{2} + \frac{ \beta}{2} (\pi_2 \tau_1-\pi_1 \tau_2) -\frac{ \beta}{6v}[ \pi_3 (\vec{\pi}\cdot\vec{\tau})- \tau_3 (\vec{\pi}\cdot\vec{\pi})] +... ,$$ quite similar, but with telltale mismatched signs of the linear and nonlinear pieces.
Consequently, for a vector transformation, ($\beta=\alpha_3\equiv \theta$, the rest vanishing), $$ \delta \vec \pi \cdot \vec \tau= \theta (\pi_2 \tau_1-\pi_1 \tau_2) +... ,$$ so a linear rotation of only the charged goldstons.
Isn't this exactly what you know for the Higgs complex doublet with hypercharge 1 and $T_3$ of 1/2 for the positive goldston and -1/2 for the neutral one? $Y=2(Q-T_3)$, alright. Gauging it cannot give you anything different than the hypercharge identity operator you grew up with.
But let's do it anyway. From the action (7) of the same ref ([3]), setting the charged Ws equal to 0 as well as the gradient terms, you see the remaining term to be $$ \frac{v^2}{16} \operatorname{tr} (g\tau^3 W^3_\mu U -g' U \tau^3 B_\mu)(...)^\dagger \to \frac{v^2}{8} (gW^3_\mu-g'B_\mu )^2\\ =\frac{g^2 v^2}{8} (W^3_\mu-\tan \theta_W~~B_\mu )^2 = \frac{g^2 v^2}{8\cos^2\theta_W} Z_\mu^2. $$ In the first line, one performs the trace after going to the unitary gauge $U=\mathbb 1$, where all goldstons are transformed away. That's it: the familiar mass matrix of neutral bosons.
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