0
$\begingroup$

I know that an adjoint field $\Phi$ spontaneously breaks an SU(N) symmetry when it gets a VEV in the diagonal form

$$ \langle\Phi\rangle = diag(v_1, v_1,...,v_1, v_2,.., v_2,...) $$ where the number of $v_1$ is $n_1$, $v_2$ is $n_2$, etc.

The unbroken symmetry is SU($n_1$)$\times$SU($n_2$)$\times$U(1). The last U(1) comes from the combination of all the unbroken subgroup generators, i.e. proportional to $\langle\Phi\rangle$ itself. I have a couple of questions

  1. How to relate this to the fact the unbroken symmetry generator should vanish the vacuum, $Q|0\rangle=0$? I don't see, say, the SU($n_1$) generator vanish the diagonal VEV matrix.
  2. What if in the VEV matrix, there are some entries which only have one copy. For example, consider an SU(5) adjoint with a VEV $$ \langle\Phi\rangle \propto diag(-1,-1,-1,-1,4) $$ Does the last entry (which only has one copy) generate an unbroken U(1)? But apparently the whole $\langle\Phi\rangle$ generates an unbroken U(1). So I'm not sure whether the unbroken symmetry group is SU(4)$\times$U(1)$\times$U(1) or SU(4)$\times$U(1)?
  3. The most extremal situation is that all the entries in VEV only have one copy. Let's say the SU(2) adjoint with a VEV $$ \langle\Phi\rangle \propto diag(-1,1) $$ If the two elements separately generate an unbroken U(1), then $\langle\Phi\rangle$ itself apparently does not generate an independent U(1), otherwise originally SU(2) have three generators whereas after SSB there are still three generators in three U(1)'s...But it's also possible that the two elements do not generate unbroken U(1) while $\langle\Phi\rangle$ does. So, in this case, is the unbroken symmetry U(1)$\times$U(1) or U(1)?
$\endgroup$
3
$\begingroup$

The adjoint representation is the same as the Lie algebra $\mathfrak{su}(n)$ itself, with the action given by the Lie bracket/commutator. Since you are apparently presenting $\Phi$ as a matrix, note that $\mathfrak{su}(n)$ are the traceless Hermitian matrices, so you should have $\sum_i n_i v_i = 0$ for the VEV to actually be possible (your examples fulfill this, but I'm spelling it out for posterity). The diagonal traceless Hermitian matrices are precisely the Cartan subalgebra of $\mathfrak{su}(n)$. The reason we may assume that the VEV lies in the Cartan algebra is that one can show that every element of a Lie algebra can be conjugated to an element of the Cartan algebra.

  1. The action of a generator $T^a$ on $\Phi$ is, as I said, the commutator $[T^a,\Phi]$. It should be obvious that this commutator vanishes if $T^a$ is a block matrix $$ \begin{pmatrix}A & 0 \\ 0 & B\end{pmatrix},\tag{1}$$ where $A$ is an $n_1\times n_1$ matrix and $B$ an $n_2\times n_2$-matrix. The generators of this form are precisely those which generate the $\mathrm{SU}(n_1)\times\mathrm{SU}(n_2)\times\mathrm{U}(1)$ - the ones with $B=0$ generate $\mathrm{SU}(n_1)$, the ones with $A=0$ generate $\mathrm{SU}(n_2)$, and the one with $\mathrm{tr}(A) = -\mathrm{tr}(B)$ generates the $\mathrm{U}(1)$.

  2. The statement has no special case for $n_i = 1$. $\mathrm{SU}(1)$ is the trivial group, and so in the case $n_1= 4,n_2=1$ you have $\mathrm{SU}(4)\times\mathrm{U}(1)$ as the symmetry group.

  3. See 2.

Basically, it all just comes down to looking which generators are of the form (1). To exemplify this for case 3.: If $A$ and $B$ are both just one-dimensional, then you should see that there cannot be a generator where one of them is zero but the other is not since that matrix cannot be traceless. Furthermore, there is only one non-zero traceless diagonal matrix of dimension two up to scalar multiplication, so there is only one generator that commutes with $\mathrm{diag}(v_1,-v_1)$.

$\endgroup$
  • 1
    $\begingroup$ You could add the following corollary: Since the VEV is (can be chosen to be) in the Cartan subalgebra, the other Cartan generators commute with it, and so a breaking by an adjoint VEV preserves the rank. Hence, if you start with e.g. $SU(5)$ and break to $SU(4)\times\text{something}$, that something can basically only be a single $U(1)$ (since it's rank one and one-dimensional, hence Abelian). $\endgroup$ – Toffomat May 11 '17 at 10:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.