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  How to calculate the mass spectrum in Lagrangian with SO(3) and SU(2)?

+ 2 like - 0 dislike

Consider a Lagrangian:

$\mathcal{L}=\frac{1}{2}\partial_\mu\phi_i\partial^\mu\phi_i - V(\phi);\\V(\phi)=-\frac{1}{2}\mu^2\phi_i\phi_i+\frac{1}{4}\lambda\phi_i\phi_i\phi_j\phi_j.$

I understand that the mass spectrum of the particles in the theory can be obtained by:


The minimum of the potential is given by:

$\frac{\partial V}{\partial \phi_k}= -\mu^2\phi_k+\lambda\phi_k\phi_j\phi_j =0\implies\phi_k=0 $ or $\phi_j\phi_j=\frac{\mu^2}{\lambda}$

Now to obtain the mass spectrum:

$\frac{\partial^2V}{\phi_k\phi_l}=\frac{\partial}{\partial \phi_l}(-\mu^2\phi_k+\lambda\phi_k\phi_j\phi_j)=-\mu^2\delta_{kl}+\lambda\phi_j\phi_j\delta_{kl}+2\lambda\phi_k\phi_l$

Substituting the minimum obtained above:

$\frac{\partial^2V}{\phi_k\phi_l}|_{\phi=\phi_0}=-\mu^2\delta_{kl}+\lambda\delta_{kl} \frac{\mu^2}{\lambda}+2\lambda\frac{\mu^2}{\lambda}=2\mu^2.$

I was expecting to obtain a matrix with the masses of the particles, but I just have one mass. What did I do wrong?

asked Sep 1, 2018 in Theoretical Physics by NewTheorist [ no revision ]

1 Answer

+ 2 like - 0 dislike

The following expression is correct:

$\frac{\partial^2V}{\phi_k\phi_l}=\frac{\partial}{\partial \phi_l}(-\mu^2\phi_k+\lambda\phi_k\phi_j\phi_j)=-\mu^2\delta_{kl}+\lambda\phi_j\phi_j\delta_{kl}+2\lambda\phi_k\phi_l$

but the following expression is incorrect:

$\frac{\partial^2V}{\phi_k\phi_l}|_{\phi=\phi_0}=-\mu^2\delta_{kl}+\lambda\delta_{kl} \frac{\mu^2}{\lambda}+2\lambda\frac{\mu^2}{\lambda}=2\mu^2.$

The correct expression is

$\frac{\partial^2V}{\phi_k\phi_l}|_{\phi=\phi_0}=-\mu^2\delta_{kl}+\lambda\delta_{kl} \frac{\mu^2}{\lambda}+2\lambda\phi_k\phi_l=2\lambda\phi_k\phi_l.$

Now you have the correct matrix.  Do you agree?

answered Sep 2, 2018 by juancho (1,130 points) [ no revision ]

well corrected :)

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