Novice's Question on Spontaneous SUSY Breaking & Goldstinos

[SmalltownBoy]

I'm a novice in SUSY and I'v got a question concerning spontaneous supersymmetry breaking and goldstinos. In Martin's review on page 68 there is a proof of a statement about existence of massless particle when one of $F_i$'s or $D_a$'s VEV is not zero. The thing I don't get is why vector $\tilde{G}$ is proportional to goldstino wave function. I guess that it means that after diagonalization of $m_F$ there is a field $\Psi = \sum F_i \, \Psi_i + \sum D_a \, \lambda_a$, that corresponds to zero eigenvalue of $m_F$. Is it correct? If so, could you please give any hints of proof of this statement.

 

[ChrisVer]

Well, if I understand your question, the vector $G$ has by definition zero mass eigenvalues, so it's a massless field.
This can be written as $m_F G = 0 \times G$.
As he states the $G$ is non-trivial only if the vevs are non-zero (so you have the symmetry breaking and thus the Nambu Goldstone field).

 

[SmalltownBoy]

Well, if I understand your question, the vector $G$ has by definition zero mass eigenvalues, so it's a massless field.
This can be written as $m_F G = 0 \times G$.
As he states the $G$ is non-trivial only if the vevs are non-zero (so you have the symmetry breaking and thus the Nambu Goldstone field).

Yes, basically I can prove that $G$ vector has zero eigenvalue. But obviously it's not goldstino field,because it's a constant. I assume that goldstino field is $\sum_i \, F_i \, \psi_i + \sum_a \, D_a \, \lambda_a$ but I cannot prove it.

 

[ChrisVer]

it doesn't have to be exactly this kind of combination. The components of the Weyl spinors and gauginos (your set of basis) just have to be proportional to the auxiliary fields' vevs (G), so that the Goldstino will keep being massless.

 

[SmalltownBoy]

it doesn't have to be exactly this kind of combination. The components of the Weyl spinors and gauginos (your set of basis) just have to be proportional to the auxiliary fields' vevs (G), so that the Goldstino will keep being massless.

Could you please prove this statement? In fact this is the point that I don't get... :(

 

[ChrisVer]

Because you want to get zero mass eigenvalue for the goldstino...
The goldstino is going to take away some degrees of freedom from your Weyl spinor or gauginos...

 

[SmalltownBoy]

Because you want to get zero mass eigenvalue for the goldstino...
The goldstino is going to take away some degrees of freedom from your Weyl spinor or gauginos...

Ok, let me explain how I see it...For simplcity let's consider theory wtih only chiral superfields. We know that in some mininum of scalar potential $V(\phi,\phi^*)$ VEV's of auxillary fields $F_i$ are nonzero and vector $(F_1,...,F_N)$ is anihillated by fermion mass matrix $m_F$. Than I would try to diagonalize this matrix by introducing new fermion fields $\tilde{\psi}_i$. My statement is as follows: one of this fields $\tilde{\psi}_k = \sum_{i} F_i \, \psi_i$ corresponds to zero eigenvalue. I guess that it's a true statement. Could you please prove it (algebraically, using formulas)? As for me, I have no idea how to do it :(

 

[ChrisVer]

I think it's pretty clear, the operator $m_F G$ acting on the space of $\left \{ \psi_i , \lambda_a \right\}$ is equivalent as an operator to the $0$, and that's the definition of a massless goldstino.

 

[SmalltownBoy]

I think it's pretty clear, the operator $m_F G$ acting on the space of $\left \{ \psi_i , \lambda_a \right\}$ is equivalent as an operator to the $0$, and that's the definition of a massless goldstino.

Unfortunately, it's not pretty clear for me because $m_F G$ is not even a matrix...

 

[ChrisVer]

it's zero...

 

[SmalltownBoy]

it's zero...

Ok, then arbitrary vector is goldstino field...?