Do Standard Model Yukawa couplings depend on the gauge choice?

In the standard model and the Unitary gauge, we write the Higgs field as

$\phi = \frac{1}{\sqrt{2}} \begin{pmatrix} 0 \\ v + H \end{pmatrix}$

and the Yukawa couplings (leaving out the neutrino terms) are

$\mathscr{L}_{\mathrm{Yuk}} = - \Gamma^u_{mn} \bar{q}_{\mathrm{L} m} \epsilon \phi u_{\mathrm{R}n} - \Gamma^d_{mn} \bar{q}_{\mathrm{L} m} \phi d_{\mathrm{R}n} - \Gamma^e_{mn} \bar{\ell}_{\mathrm{L} m} \phi e_{\mathrm{R}n} + \text{h.c.}$

where $\epsilon$ is the antisymmetric tensor and the $\Gamma$ matrices are related to the mass matrices. I assume these terms are independent of the gauge choice (maybe I'm wrong to think this). Continuing on, we can write the Yukawa couplings in the Unitary gauge

$\mathscr{L}_{\mathrm{Yuk}}^{\mathrm{U}} = - \Gamma^u_{mn} \bar{u}_{\mathrm{L} m} (H + v) u_{\mathrm{R}n} - \Gamma^d_{mn} \bar{d}_{\mathrm{L} m} (H + v) d_{\mathrm{R}n} - \Gamma^e_{mn} \bar{e}_{\mathrm{L} m} (H + v) e_{\mathrm{R}n} + \text{h.c.}$

So far so good.

But I'm confused by how the choice in gauge seems to dictate the Yukawa couplings. I've also seen the unitary gauge written (specifically in Srednicki's Quantum Field Theory) as

$\phi' = \frac{1}{\sqrt{2}} \begin{pmatrix} v + H \\ 0 \end{pmatrix}$

which seems equally valid since it's also an expansion around the vacuum expectation value of $\phi$. But this would do very weird things to the Yukawa couplings:

$\mathscr{L}_{\mathrm{Yuk}}^{\text{?}} = - \Gamma^u_{mn} \bar{d}_{\mathrm{L} m} (H + v) u_{\mathrm{R}n} - \Gamma^d_{mn} \bar{u}_{\mathrm{L} m} (H + v) d_{\mathrm{R}n} - \Gamma^e_{mn} \bar{e}_{\mathrm{L} m} (0) e_{\mathrm{R}n} + \text{h.c.}$

i.e. the charged leptons become massless and the quark mass terms don't make a lot of sense. Of course this would all be fine if I'd removed the $\epsilon$ in the first term of $\mathscr{L}_{\mathrm{Yuk}}$ and added it to the second two terms, but it seems weird to have this depend on the choice of gauge.

So what am I doing wrong here? Does the choice of gauge ($\phi$ or $\phi'$) depend on the author's choice of how the Yukawa couplings are defined? The way I see the unitary gauge presented in most cases doesn't make any reference to the eventual choice in Yukawa couplings.

Very weird things are happening indeed. Unless you take a step back and think...

By choosing the upper component to carry the vacuum expectation value, you will find that the electric charge $Q=T^3+Y$ appears to no longer be conserved by the vacuum (try it out by making the appropriate transformation). Instead, the conserved charge is now $Q_\text{new} = -T^3 + Y$.

This means you have to reinterpret the meaning of electric charge as $Q_\text{new}$. So what then are the electric charge assignments of the fermions?

The upper components of quarks have a new electric charge of $-1/3$ and the lower components have a new charge of $+2/3$. Clearly, the down quark field must be occupying the upper component, and the up quark field must be occupying the lower component in this new scheme.

Let's look at the leptons: the lower components are what you had previously called the "charged leptons"... not anymore in this new scheme -- the lower components have $Q_\text{new}=0$, and the upper components have charge "$Q_\text{new}=-1$"... they got flipped around too.

Having correctly relabeled all the fields, you should find by inspecting the Yukawa couplings, that the fermions have their correct masses.

So what happened? By putting the vacuum expectation value in the upper component, you have essentially done an isospin gauge rotation by $\pi/2$. But when you make an isospin gauge rotation, you must remember to make the rotation for all fields that transform under isospin, and this includes the fermions too. Since the $T^3$ in $Q=T^3+Y$ also transforms, the electric charge got transformed too. That is why all the relabeling occurred, and the masses of fermions are unchanged.

• Unfortunately, I don't follow your second paragraph; I was going through Srednikcki's Quantum Field Theory and noticed that he puts the VEV in the upper component, but uses the same charge definition as other sources that put it in the lower component. Is he doing something wrong? – Shep Feb 15 '15 at 21:33
• @Shep Srednicki assigns the Higgs a hypercharge of $Y=-1/2$ (which is opposite from most everyone else's definition). As a result, Srednicki's Higgs field is rotated by $\pi/2$ around the isospin-$y$ axis, followed by $C$-conjugation with respect to everyone else Higgs's field [analog of the $G$-parity in hadron physics]. – QuantumDot Feb 15 '15 at 21:57
• OK, that makes a lot more sense. I should have mentioned Srednicki in the original post (added now). I was seeing both versions and assuming the choice of upper vs lower component was arbitrary, since the unitary gauge is usually justified as a way to get to the $V(\phi)$ minimum without much discussion of vacuum charge conservation. But it turns out there's a bit more subtlety to the unitary gauge than I realized. So it's also chosen to make charge conservation work in the vacuum? – Shep Feb 15 '15 at 22:09
• @Shep Yes, you can think of it that way. Otherwise you have to do a little more work to identify the conserved charge, and identify that as $Q_\text{new}$. – QuantumDot Feb 16 '15 at 9:17

The actual values of the Yukawa couplings are arbitrary, and put in by hand (depending on the mass) and the matrices chosen are labelled however you like.

That said, you are always free to choose which component of your scalar field gets a vev, by a global rotation. The scalar fields as you have written them are related by such a rotation $$\phi' = \sigma_1\phi$$

Usually, choosing the unitary gauge in this case means fixing the value of $\chi(x)$ in $$\phi' = e^{\frac iv\chi(x)}\phi$$ This restricts the scalar degrees of freedom so that there are no Goldstone boson interactions in your theory (making it physical) and has nothing to do with the field you choose to gain a vev - the same gauge choice has been made in both situations.

If you choose the scalar field vev to be $\phi'$ instead of $\phi$, it amounts to doing a global SU(2) transformation to the lagrangian (most importantly, to the doublet fields - applying $\sigma_1$). Under such a transformation, the upper and lower components of each doublet switch and you should again derive the same form of your lagrangian.

• I'm slightly confused, because the theory should be equally physical in any gauge, shouldn't it? I'd further expected that the values of the Yukawa couplings would be independent of the gauge. – Shep Feb 14 '15 at 21:44
• It's the relabeling that confuses me: the first version of $\mathscr{L}_{\mathrm{Yuk}}$ that I give should be gauge invariant, so I shouldn't have to relabel it for different gauge choices. But the lepton masses are zero if I choose the $\phi'$ gauge... – Shep Feb 14 '15 at 21:52