standard model - Why $U(1)_Y$ hypercharge rather than $U(1)_text{em}$ electromagnetism?

In the Standard Model we have $SU(2)_I\times U(1)_Y$, where $U(1)_Y$ is weak hypercharge and $SU(2)_I$ is the symmetry group of weak isospin. Why do we introduce $U(1)_Y$ of weak hypercharge rather than $U(1)_\text{em}$ of electromagnetism?

Answer

Short answer: to accurately model reality.

Long answer: The weak interaction has several peculiar properties:

1. The $W$ bosons are vector bosons (so the weak theory is likely a gauge theory)


2. The $W$ bosons have electric charge


3. The $W$ bosons have mass. (The $Z$ boson hadn't been observed experimentally; it was a prediction of the SM)


4. The $W$ bosons couple chirally, meaning left-handed and right-handed fermions belong to different representations of the gauge group. In particular, left-handed fermions transform in the doublet representation, right-handed in the singlet (trivial) rep.


5. EM is a gauge interaction that couples vectorially.

This leads to two problems that weren't understood before the 60s. The first problem is how to have massive gauge bosons (items 1 and 3). Naively a vector boson mass term violates gauge symmetry and unitarity. The second problem is how to get massive fermions (item 4). A fermion mass term couples left- and right-handed fermions, but those belong to different weak representations so the fermion mass term also breaks gauge invariance.

The realization of Higgs et al. and then later Glashow and Weinberg was that both of these problems can be solved by spontaneous symmetry breaking. The Higgs mechanism says that spontaneous symmetry breaking of a gauge symmetry leads to massive gauge bosons. And if the operator that gets a vacuum expectation value has the right quantum numbers you it can couple to the fermions in just the right way to make an effective fermion mass term.

The questions are, what pattern of spontaneous symmetry breaking corresponds to reality, and what kind of operator should get a vev?

Because of item 5, we need an SSB that leaves a $U(1)_{em}$ unbroken. Since there is only one Lie group with doublet irreps we also know the original gauge group should include an $SU(2)$ factor. Moreover, since the $W$ bosons have charge (item 2), the generator of $U(1)_{em}$ must not commute with some $SU(2)$ generators!

To fix item 4 and make a gauge-invariant mass term we need an operator to get a vev which transforms in the doublet of $SU(2)$. This rules out the symmetry breaking pattern $SU(2)\rightarrow U(1)_{em}$ (there is no generator of $SU(2)$ that leaves a non-trivial doublet invariant, meaning a doublet always breaks $SU(2)\rightarrow {1}$ which doesn't leave room for a photon).

The next simplest pattern to try is $SU(2)\times U(1)_Y\rightarrow U(1)_{em}$. We could achieve this with an uncharged doublet and $U(1)_Y = U(1)_{em}$, but then $[U(1)_{em},SU(2)]=0$ and the $W$ bosons wouldn't be charged. The only other way is to have the Standard Model pattern where the $U(1)_{em}$ generator is a linear combination of $U(1)_Y$ and some generator of $SU(2)$, which is indeed a possible pattern.

So, the symmetry breaking pattern $SU(2)\times U(1)_Y \rightarrow U(1)_{em}$ with $Q = Y + T_3$ is the minimal SSB pattern that matches general weak phenomenology. It's not the only possible mechanism, so it's really very nice that nature chose this one. It's also nice because it predicts the $Z$ boson.