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In the lecture, the professor said something along the lines of:

"After a suitable gauge transformation, the standard model higgs field can be expanded as

$$\phi =\left(\begin{array}{c} 0 \\ v+H(x) \end{array}\right)$$ ".

Now, the argument I have been able to scramble from different sources goes along these lines:

  1. We can write small higgs field excitations as $$\tilde\phi =\left(\begin{array}{c} i \theta_1(x) + \theta_2(x) \\ v+H(x) - i \theta_3(x) \end{array}\right)$$

  2. An appropriately chosen local $SU(2)_L$ transformation transforms this into the above form where all $\theta=0$.

  3. Thus, by applying an appropriate local $SU(2)_L$ transformation to all elements of the Lagrangian, we can use this form without loss of generality.

I have the following two issues with this argumentation:

First, how do I know that the first statement is true?

Secondly and most importantly, wouldn't any symmetriy transformation that corrects the higgs weak spinor into the above from , given a random but fixed excitation of the higgs field, also mess up the spinors of the left handed leptons? In other words, how is it possible to find a local gauge transformation on the Lagrangian that corrects any $\tilde \theta$ into $\theta$, but does not change the form of $ \bar L =\left(\begin{array}{c} \bar\nu \\ \bar e \end{array}\right)$ ??

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1 Answer 1

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The first assumption is that whatever vev the higgs picks up is constant in space, because this has less energy than one that increases the kinetic term in the Lagrangian. So we can do one global transformation to make the vev be in the second component only. You can imagine doing this prior to symmetry breaking, if you know what it is going to be ahead of time, and since the other fields are invariant, bob's yer uncle. Stated differently, the pre-symmetry-breaking electrons and neutrinos are not the ones we observe, so we just label whatever remains as electrons and neutrinos.

"Without loss of generality", we work in an electron-neutrino (global) basis in which the higgs starts out with only the second component of the vev being nonzero and real.

If you buy that part, then it is just a matter of showing that you can perform a gauge transformation that gauges away all the other components of the Higgs except the real part of the second component. This gauge transformation will of course mix $\nu$ and $e$ spatially, but you can say that when we perform the path integral we have a gauge redundancy, and so we only integrate along a slice that obeys some gauge fixing condition. The components of $L$ might as well be labeled $L_1$ and $L_2$. It's only after we've chosen a gauge that we decide, hey, let's name them $\nu$ and $e$.

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ok, I absolutely buy the first part, still chewing on the second one... –  Konstantin Nov 29 '13 at 0:56
    
Think of $L_1$ and $L_2$ as variables in the path integral to be integrated over, and the gauge transformation mixes them up, but we still integrate over the entire domain, so it doesn't really matter. It might be instructive to think about what happens to the electron in QED when we go to say, the Lorentz gauge. Yes, the electron field changes by a spacetime dependent (and A-field dependent!) phase, but nobody seems to care :) –  lionelbrits Nov 29 '13 at 0:59
    
That means we look at the higgs excitation "of the world" and choose a local gauge transformation that gauges away all components but the real part in the second component, write down the Lagrangian and then we define that the resulting $L_1$ shall be called a neutrino? I think my problem with this was that a changing higgs excitation would require a redefinition of what we define as a neutrino and an electron. Maybe the answer to that is that we are working in spacetime, and hence a single gauge redefinition already works for all times and places? –  Konstantin Nov 29 '13 at 1:16
    
Yes. Now it wouldn't hurt to run that first sentence by your prof, since I'm just some guy on the internet, and on the internet, nobody knows your a dog. –  lionelbrits Nov 29 '13 at 1:23
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I think you can just postpone labeling $L_1$ and $L_2$ until the scalar field is in the form you want, because before symmetry breaking, there's no real difference between a left-handed neutrino and a left-handed electron. –  lionelbrits Nov 29 '13 at 11:08

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