Embedding of $SU(2)_L \times U(1)_Y$ into $SU(2)_L \times SU(2)_R$ in electroweak chiral effective theories

Question

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 1

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.

Answer 2

We can write $Y=2(Q-T_3)$ and observe that in the standard model only left-handed quarks carry weak isospin. These papers simply don't write the QED part.

Answer 3

Geeky footnote for the above answer

I realized from your conceptual comment request that a mere reassuring demonstration of fact (as in my answer above) was not enough, without the explicit technical details of the Longhitano magic-hat trick in your implicit hidden question, which I understand as

How does the hypercharge transformation $e^{i\beta /2}$ on a complex Higgs doublet $\Phi$ morph into the $e^{-i\beta \tau_3/2}$ acting on the right of the Goldstone boson matrix picture?

Referring you to Longhitano's thesis paper of 1981, as I did, again glosses over the routine, but still esoteric, reparameterization to the exponential Gürsey realization that has been the trusty side-knife of many. So, I'll archive the explicit details here for possible utility to future nitpickers.

Longhitano starts from the standard Higgs weak isodoublet and weak hypercharge 1 (as in WP) $$ \Phi = \begin{pmatrix} \phi^+ \\ \phi^0 \end{pmatrix}\equiv \frac{1}{\sqrt 2} \begin{pmatrix} \varphi_1-i\varphi_2 \\ \sigma +i\chi \end{pmatrix}. $$ The remnant physical Higgs is $\sigma$, soon to be frozen to decoupling rigidity by taking its mass to infinity, thus leaving only the goldstons behind.

The conjugate doublet is also a left isotriplet, but, naturally, with the opposite value (-1) of weak hypercharge, $$ \tilde \Phi =i\tau_2 \Phi^*= \begin{pmatrix} \phi^{0~~*} \\ -\phi^- \end{pmatrix} , $$ so that $$ \Phi \mapsto e^{i(\beta +\vec{\alpha}\cdot \vec{\tau})/2} \Phi ~,$$ hence $$ \tilde \Phi \mapsto e^{i(-\beta +\vec{\alpha}\cdot \vec{\tau})/2}\tilde \Phi ~.$$

Now, the celebrated Higgs matrix is defined as a side-by-side juxtaposition of these two left-doublets serving as columns, $$ M\equiv \sqrt{2}(\tilde\Phi, \Phi)= \sqrt {2} \begin{pmatrix} \phi^{0~~*} &\phi^+ \\ -\phi^- & \phi^0 \end{pmatrix}. $$

It is then evident that its transform is $$ \bbox[yellow]{ e^{i\vec{\alpha}\cdot \vec{\tau}/2} \sqrt{2}(\tilde\Phi e^{-i\beta/2}, \Phi e^{i\beta/2})= e^{i\vec{\alpha} \cdot \vec{\tau}/2}\sqrt {2} \begin{pmatrix} \phi^{0~~*}e^{-i\beta/2} &\phi^+e^{i\beta/2} \\ -\phi^- e^{-i\beta/2} & \phi^0 e^{i\beta/2} \end{pmatrix}= e^{i\vec{\alpha}\cdot \vec{\tau}/2} M e^{-i\beta \tau_3/2} }. $$ This is the core group theory trick.

All one needs now is to send the mass of the Higgs to infinity, so $\sigma \to v\sqrt {1-\chi^2/v^2}$, the standard linear $\sigma$-model limit to the non-linear one, orthogonally rotate the definition of the three (adjoint) Goldstone variables a bit, $$ \Phi\to \frac{1}{2}\begin{pmatrix} \varpi_2+i\varpi_1 \\ v\sqrt{1-\varpi^2/v^2}-i\varpi_3 \end{pmatrix}, $$ and normalize $M$ to a unitary matrix, $$ M/v\to U= \begin{pmatrix}v\sqrt{1-\varpi^2/v^2}+i\varpi_3& \varpi_2+i\varpi_1 \\ -\varpi_2+i\varpi_1& v\sqrt{1-\varpi^2/v^2}-i\varpi_3 \end{pmatrix}\frac{1}{v} \\ = 1\!\! 1 ~\sqrt{1-\varpi^2/v^2} + i\frac{\vec{\varpi}}{v}\cdot \vec \tau ~ . $$

Finally, to ward off loss of mind, change goldston variables to parallel ones, in the more elegant/sensible chiral model parameterization of a unitary matrix, $$ \vec \varpi /v\equiv \hat \pi \sin \frac{|\vec \pi|}{v} ~, $$ so that $$ U= \mathbb {1} \cos \frac{|\vec \pi|}{v} + i\hat \pi \cdot \vec \tau \sin \frac{|\vec \pi|}{v}= e^{i\vec \pi \cdot \vec \tau/v} ~, $$ the standard chiral Goldstone boson matrix with the originally striking weak hypercharge property in question.