# Proof of U(1) gauge invariance

1. Aug 26, 2009

### Astrofiend

1. The problem statement, all variables and given/known data

I want to show explicitly that the Lagrangian...

$$L_\Phi = (D_\mu \Phi)^\dagger (D^\mu \Phi) - \frac{m^2}{2\phi_0 ^2} [\Phi^\dagger \Phi - \phi_0 ^2]^2$$

where $$\Phi$$ is a complex doublet of scalar fields, and

$$D_\mu = (\partial_u + i \frac{g_1}{2} B_\mu)$$

is the covariant derivative, with $$B_\mu$$ the introduced gauge field

...is invariant under the U(1) gauge transformation

$$\Phi' = e^{-i(g_1 /2)\chi\sigma^0} \Phi$$

$$B'_\mu = B_\mu + \partial_\mu \Chi$$

where $$\chi = \chi (x)$$ - i.e an arbitrary function of spacetime, $$\sigma_0$$ is the 2x2 identity matrix.

3. The attempt at a solution

I've had a couple of flaccid cracks at it, but I'm afraid that I'm pretty lost with this one. The problem is, most textbooks or internet resources that I can find just basically state that the Lagrangian is gauge invariant under such U(1) transformations, and that this can be shown quite easily. Not exactly helpful for someone like me trying to work out how the maths works in the first place! Is anybody able to help out, or point me to any resources where this sort of gauge invariance is demonstrated explicitly, so I can get a feel for how the maths works? Help would be greatly appreciated!

Last edited: Aug 26, 2009
2. Aug 26, 2009

### javierR

Write the "new" Lagrangian L' by putting primes on everything that changes: D', B', Phi'. Expand everything out using the expressions for the primed objects in terms of unprimed ones. Keep in mind that operators obey $$(AB)^{\dagger}=B^{\dagger}A^{\dagger}$$ and $$\partial_{\mu}^{\dagger}=\partial_{\mu}$$. The main thing to look for is that, because the transformation is local, there is a spacetime-dependence of the function $$\chi$$, so that you get terms proportional to $$\partial_{\mu}\chi\partial^{\mu}\chi$$ in your expansion, which are to be canceled by the new terms from the the transformed gauge field B'. Be careful with signs, including those due to products of "i" and you will end up with the original L (in terms of unprimed quantities), thus showing that L'=L.

By the way, showing gauge invariance of the interaction terms is the easy part since exp(+iA)exp(-iA)=1...the kinetic (derivative) term is the messy part.

Last edited: Aug 26, 2009
3. Aug 26, 2009

### Ben Niehoff

Also, you have a typo in one of the formulas in your post...make sure you don't have the same mistake in your work. The gauge field should transform as

$$B'_{\mu} = B_{\mu} + \partial_{\mu} \chi$$

4. Aug 26, 2009

### Astrofiend

Thanks guys. I'll try to nut it out this afternoon...

Thanks for picking up that typo...

5. Aug 27, 2009

### Astrofiend

OK - I think I'm really not getting how the maths goes on this. If anyone can have even a brief look at this and tell me what I'm doing wrong, that'd be great.

I've tried to expand out the first (kinetic) part of the primed Lagrangian as described above, and got (using $$g = \frac{g1}{2}$$ which I used in the original post):

$$(\partial_\mu \Phi^\dagger - igB_\mu\Phi^\dagger - ig\partial_\mu\chi\Phi^\dagger)(\partial^\mu \Phi^\dagger - igB^\mu\Phi^\dagger - ig\partial^\mu\chi\Phi^\dagger) =$$

$$\partial_\mu \Phi^\dagger\partial^\mu\chi\Phi + ig\partial_\mu \Phi^\dagger B^\mu\Phi + ig\partial_\mu\Phi^\dagger\partial^\mu\chi\Phi -$$

$$ig\partial^\mu\Phi B_\mu\Phi^\dagger + g^2B_\mu\Phi^\dagger\\B^\mu\Phi + g^2B_\mu\partial^\mu\chi\Phi^\dagger\Phi -$$

$$ig\partial^\mu\Phi\partial_\mu\chi\Phi^\dagger + g^2\partial_\mu \chi B^\mu\Phi^\dagger\Phi + g^2\partial_\mu\chi\partial^\mu\chi\Phi^\dagger\Phi$$

....and then I made the Phi-primed substitution, using

$$\partial_\mu \Phi ' = \partial_\mu e^{-ig \chi \sigma^0} \Phi = -ig \partial_\mu \chi \Phi '$$

...to get

$$g^2 \partial_\mu \chi \partial^\mu \chi \Phi^\dagger \Phi - g^2 \partial_\mu \chi \Phi^\dagger \Phi B^\mu - g^2 \partial_\mu \chi \partial^\mu \chi \Phi^\dagger \Phi +$$

$$g^2\partial^\mu \chi \Phi^\dagger \Phi B_\mu + g^2 B_\mu B^\mu \Phi^\dagger \Phi + g^2 \partial^\mu \chi B_\mu \Phi^\dagger \Phi +$$

$$g^2 \partial^\mu \chi \partial_\mu \chi \Phi^\dagger \Phi + g^2 \partial_\mu \chi B^\mu \Phi^\dagger \Phi +g^2 \partial_\mu \chi \partial^\mu \chi \Phi^\dagger \Phi$$

So I'm trying to show that the full lagrangian is invariant under U(1) transformations, and I can't see how this has gotten me anywhere. Have I completely misunderstood how the maths works? Done something completely stupid? Or am I sort of on the right track but just can't see it?

Once again - any help much appreciated.

6. Aug 28, 2009

### Astrofiend

Done! (only took 5 hours of staring at it waiting for my slow brain to tick over...) I didn't actually need to go through all of that ungodly expanding.

Thanks to those who helped! Much appreciated.

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