# Breaking symmetries with a complex scalar field

I'm going through an example with symmetry breaking. It starts with the Lagrangian of a complex scalar field, $\Phi = \frac{1}{\sqrt2}(\Phi_1 + i\Phi_2)$,

$$\mathscr{L} = (\partial_u\Phi)^{*}(\partial^u\Phi) - m^2\Phi^{*}\Phi - \lambda(\Phi^{*}\Phi)^2.$$

Which is symmetric under the transformation $\Phi \rightarrow e^{i\alpha}\Phi$. I then select on of the minima of the potential using a change of variables $\Phi(x) = \frac{1}{\sqrt2}(\Phi_0 + \Phi_1'(x) + \Phi_2'(x))$ which yields

$$\mathscr{L'} = \frac{1}{2}\partial_{\mu}\Phi_1'\partial^{\mu}\Phi_1' + \frac{1}{2}\partial_{\mu}\Phi_2'\partial^{\mu}\Phi_2' + m^2{\Phi_2'} ^2.$$

My first question is how to get from the first Lagrangian to the second one using that change of variables. If someone could point me in the right direction that would be fantastic. Secondly, the whole idea is that the first Lagrangian is symmetric under the phase transformation but the second one isn't. It is easy for me to see how the first is symmetric under $\Phi \rightarrow e^{i\alpha}\Phi$ but what is the equivalent of this for the second? It is confusing me because now I have $\Phi_1'$ and $\Phi_2'$.

Sorry if I haven't explained this particularly well. Any information would be great. Thanks.

First of all, your initial Lagrangian is incorrect as it has a stable solution at $\Phi=0$ such that there will be no symmetry breaking. Below I present the correct Lagrangian and work out the symmmetry breaking, after reading this you should be able to recognize the errors in your Lagrangian.

I work with the "typical" experimental physics metric convention (+---) and L = T-V.

Question 1

My first question is how to get from the first Lagrangian to the second one using that change of variables?

$$\mathcal{L} = \partial_\mu \Phi \partial^\mu\Phi^* + m^2 \Phi\Phi^* - \lambda(\Phi\Phi^*)^2$$

such that $V(\Phi) = -m^2 \Phi\Phi^* + \lambda(\Phi\Phi^*)^2$ a plot is provided in this link: Mexican hat potential $m^2 = \lambda = 1$. This potential has two extrema at points where: $\partial V/\partial\Phi = -m^2\Phi^* + 2\lambda\Phi^2\Phi^* = 0 \rightarrow |\Phi| = 0 \text{ or } |\Phi|=\sqrt{m^2/2\lambda} = \Phi_0/\sqrt{2}$.

One way to break the symmetry is by writing $\Phi = \frac{1}{\sqrt{2}}(\Phi_1 + i\Phi_2 + \Phi_0)$ so that we are expanding around some value in the stable minimum of the potential in the real direction. Plugging this into the original Lagrangian gives us:

$$\mathcal{L} = \frac{1}{2}(\partial_\mu\Phi_1 +i\partial_\mu\Phi_2)(\partial^\mu \Phi_1 - i\partial^\mu\Phi_2) + \frac{1}{2}m^2(\Phi_1 + i\Phi_2 + \Phi_0)(\Phi_1 - i\Phi_2 + \Phi_0) - \frac{1}{4}\lambda((\Phi_1 + i\Phi_2 + \Phi_0)(\Phi_1 - i\Phi_2 + \Phi_0))^2$$

Working this out is a real hassle, luckily we can use the fact that $\partial V\partial\Phi = 0$ around the point where we are expanding. This tells us that there will be no linear terms in the final result such that we can just ignore them.

We will also ignore all terms that are third order in fields (to simplify things), doing all this gives us: Last term using Wolfrahm alpha

$$\mathcal{L} = \frac{1}{2}\partial \Phi_1 \partial \Phi_1 + \frac{1}{2}\partial\Phi_2\partial\Phi_2 + \frac{1}{2}m^2(\Phi_1\Phi_1 + \Phi_2\Phi_2) - \frac{\lambda}{4}(6\Phi_0^2 \Phi_1^2 + 2\Phi_0^2 \Phi_2^2) + trilinear\ terms$$

But now we know that $\Phi_0^2 = m^2/\lambda$ such that we get:

$$\mathcal{L} = \frac{1}{2}\partial \Phi_1 \partial \Phi_1 + \frac{1}{2}\partial\Phi_2\partial\Phi_2 + \frac{1}{2}m^2(\Phi_1\Phi_1 + \Phi_2\Phi_2) - \frac{\lambda}{4}(6\frac{m^2}{\lambda} \Phi_1^2 + 2\frac{m^2}{\lambda} \Phi_2^2) + trilinear\ terms$$

The mass term for the $\Phi_2$ field drops out and we get:

$$\mathcal{L} = \frac{1}{2}\partial \Phi_1 \partial \Phi_1 + \frac{1}{2}\partial\Phi_2\partial\Phi_2 - \frac{1}{2}m^2\Phi_1\Phi_1 + trilinear\ terms$$

Which is the Lagrangian for a massive and a massless scalar.

Question 2

Secondly, the whole idea is that the first Lagrangian is symmetric under the phase transformation but the second one isn't. It is easy for me to see how the first is symmetric under $\Phi \rightarrow e^{i\alpha} \Phi$ but what is the equivalent of this for the second? It is confusing me because now I have $Φ^{'}_1$ and $Φ^{'}_2$.

This is actually not so hard, your $\Phi$ field is still a complex field so nothing really changes:

$$\Phi = \frac{1}{\sqrt{2}}(\Phi_1 + i\Phi_2 + \Phi_0) \rightarrow \frac{e^{i\alpha}}{\sqrt{2}}(\Phi_1 + i\Phi_2 + \Phi_0)$$

And obviously your Lagrangian is still invariant under this shift. As to "what does it mean" go back to the plot of the mexican hat potential. A complex phase just rotates the fields into each other, it sort of corresponds to rolling a ball down in the valley.

This picture also explains why one field is massive and one it massless. The massless field corresponds to an excitation that is staying in the valley and is basically a pure gauge field. The massive field is oscillating up and down the valley which gives it some mass.

The fact that one field is pure gauge becomes immediately obvious if we choose the expansion $\Phi = (\Phi_0 + \Phi_1)e^{i\Phi_2}$ which is another fully general expansion for every complex field $\Phi$, in this representation it is immediately obvious that $\Phi_2$ is a pure gauge artifact.

I hope this helped, if you have any more questions (as is typical for this kind of subject) please tell me so in the comments :)

Edit To clear of something, the orginal theory had two symmetries, one for each component of the complex field $\Phi$. The broken theory has only one symmetry left corresponding to rotations in the valley of the mexican hat (see plot).

This is consistent with the Nambu goldstone theorem that tells us that every broken symmetry will generate a massive particle. Which is exactly as observed !

• Hi, I really appreciate the thorough answer. I do have a few questions as you probably expected. 1) All of the places I look have the Lagrangian with the negative sign infront of the mass, rather than your positive sign. When I compute the second derivative I get $\Phi = 0$ as unstable? 2) How do you have $\Phi_1*$ and $\Phi_2*$? I thought they were defined to be real since it's the $\Phi(x)$ that is complex. 3) I thought the point of this was that the new Lagrangian isn't symmetric anymore, hence the symmetry is breaking. But you're saying this new Lagrangian still has the symmetry? – Horro Mar 20 '17 at 22:13
• 1)People might be working with a different definition for their metric trace or lagrangian. You should always look this up in your reference and than check that $V = -m^2 \Phi \Phi^*+ \lambda (\Phi \Phi^*)^2$ such that it becomes unstable at $\Phi = 0$, just look at the plot. 2) Indeed I should not have written $\Phi^*_{1,2}$ this was a mistake I edited it. 3)Originally there were two symmetries one for each component of the complex field $\Phi$. After the breaking only one symmetry is left namely the rotations in the valley of the mexican hat. – gertian Mar 20 '17 at 22:31
• Hi, I think I'm starting to understand but there are just a few things I'm struggling with. I understand that "rolling" around the brim of the "hat" in the potential corresponds to a massless field whereas "rolling" up the hill in the hat corresponds to a massive field. However, we see this potential BEFORE we break the symmety - doesn't this suggest that our original Lagrangian has a massless and massive field? Should I be visualising both potentials as the mexican hat but in the first instance the "ball" is at the top, then after the breaking it is in the brim? – Horro Mar 21 '17 at 10:38
• Yes you should. The "ball" in this visual picture is at the point around which you are expanding your theory. So before breaking it is at the top of the hat in an unstable situation and both direction are unstable with negative $m^2$, this is what we call a tachyon. After breaking the ball is down in the valley one as you said one excitation is massless and the other will be massive. – gertian Mar 21 '17 at 11:41