# Spontaneous symmetry breaking to subspace not giving massless bosons

I'm currently trying to understand spontaneously symmetries broken in general and have stumbled upon a weird result which doesn't seem to correspond to my knowledge about broken gauge symmetries.

Suppose we start with an SU(2) invariant theory with a double Higgs, $$\phi = \left( \begin{array}{c} \phi _1 \\ \phi _2 \end{array} \right)$$ and explore the possible breaking patterns. We now break SU(2) into a subgroup, presumably U(1). To do this I would like to find some combination of the generators that annihilate the vacuum, $$\alpha_a \frac{ \sigma _a }{ 2} \left( \begin{array}{c} v _1 \\ v _2 \end{array} \right) = 0$$ where $v _i$ are the VEVs of $\phi _i$. In order to have a nontrivial solution to the equation above we must have, $\det ( \alpha _a \sigma _a ) = 0$. One such solution is $\alpha = \left( \begin{array}{ccc}1 & -i & 0\end{array} \right)^T$. This annihilates the vacuum, $\left( \begin{array}{cc}0 & v\end{array} \right) ^T$ since, $$\frac{1}{2} ( \sigma _x - i \sigma _y ) \left( \begin{array}{c} 0 \\ 1 \end{array} \right) = 0$$

So far so good. However, if I go and find the masses of the gauge bosons in this theory I find that they are all massive: \begin{align} D _\mu \phi ^\dagger D ^\mu \phi & \rightarrow g ^2 v ^2 W _{ a ,\mu } \left( \begin{array}{cc}0 & 1\end{array} \right) \left( \sigma _a \sigma _b \right) \left( \begin{array}{c} 0 \\ 1 \end{array} \right) W ^\mu _b \\ & = g ^2 v ^2 W _{a, \mu } W _a ^\mu \end{align} where we have used, $\sigma _a \sigma _b = i \epsilon _{ a b c} \sigma _c + \delta _{ ab }$.

Since we can transform the vacuum with a combination of generators and leave the theory invariant I expected to have massless gauge bosons as well. Why aren't any of the gauge bosons massless?

Note: This is not a problem in the standard model since there we don't get the mass basis after spontaneous breaking. Then we use the Weinberg angle to rotate between bases.

• You should consider only hermitian operators annihilating the vacuum, that is linear combinations with real coefficients of the hermitian generators. The reason is that your group is generated by real parameters. Your example instead involves an imaginary linear combination. – TwoBs Nov 1 '14 at 17:54

There is no contradiction since you should not be doing complex linear combination of generators that are already hermitian (indeed, you want the group transformation to be unitary). Hence, your linear combination with a complex coefficient (that brings you away from the $SU(2)$ group) doesn't imply a massless excitation.

I guess instead of me trying to explain, it is way better if you read section "Non Abelian Examples" of Peskin and Schroeder, in chapter 20. It explains exactly what you are asking for. It's actually the predecessor of the Standard model.

Georgi and Glashow proposed this model before the SM, because they didn't know about the Z boson. So it is exactly what you need, 2 massive (W's) and 1 massless (foton). but it turned out you also need an extra massive (Z).

Anyway, the idea is simply that you are mixing bases. You chose a basis to see that a particular combination of your generators leave the vacuum invariant. But then you looked at the diagonal basis for the masses. In this diagonal basis, it seems like all 3 have acquired mass, but if you rotate 45 degrees, to your 1-i2 axis, you'll see that 1 of them is massless.

SU(2) is basically the 3D rotations, so you are rotating around the z-axis, leaving the 3rd generator massless. But the generators corresponding to rotations around the x-axis and y-axis acquire mass.

But again, read that section, it's way better explained!

• Thanks for your response. I think I understand now. I think you can't just do a basis rotation to pick up a massless boson because we are all ready in the mass basis. This would require a nonunitary transformation. I hope you don't mind I put together an answer from what I learned. If you have any comment on it I'd be interested in hearing your thoughts. – JeffDror Oct 29 '14 at 20:56

After reading the section put forth by @gcsantucci I think I kind of understand what was happening, but I'm eager to hear feedback on this.

If you break $SU(2)$ using a doublet you indeed break all the generators and get massive gauge bosons. What was confusing is that a linear combination of generators leaves the vacuum invariant (namely, $\sigma _1 + i \sigma _2$). In essence what I was suggesting was that we should first rotate the basis of the generators to, $$\bigg\{ \underbrace{ \frac{1}{2} ( \sigma _1 + i \sigma _2 )} _{ \sigma _+ } , \underbrace{ \frac{1}{2} ( \sigma _1 - i \sigma _2 )} _{ \sigma _- } , \sigma _3 \bigg\}$$ and then the first combination indeed annihilates the vacuum. The issue is that this is not a valid representation of $SU(2)$. This is because the matrices no longer obey a key property of two generator matrices: $$\mbox{Tr} \left[ t _r ^a t _r ^b \right] = 2 \delta ^{ ab }$$ since e.g., $$\mbox{Tr} \left[ \sigma _+ \sigma _+ \right] = 0$$

As mentioned by @gcsantucci you could rotate the $W _i$ basis such that one of the bosons is indeed massless. This is true, but would require a nonunitary transformation. I suspect this is somehow related to this basis change (but I can't quite figure out how).

• now I see your point. I'm sorry about that. It's actually a nice point. I'm afraid I don't know what is going on. – gcsantucci Oct 30 '14 at 4:14
• BTW (general comment), you would get 2 massive bosons and one massless if the Higgs were in the adjoint (triplet) representation of $SU(2)$. It's easy to see that from similarity to $SO(3)$ -- any direction is left invariant under a particular axis of rotation. Since experiments show 2+1 massive bosons and 1 massless, we need an extra U(1), but to give mass to 3 bosons, the Higgs should be in the fundamantal (doublet) representation of $SU(2)$. – Siva Nov 2 '14 at 19:00