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  Goldstones as matrices acting on the vacuum

+ 2 like - 0 dislike

In this physicsforums post (post #10), @samalkhaiat states that the Goldstone bosons can be parameterized by a matrix acting on the vacuum state (which in general can be represented by some vector, $\vec{v}$).

Explicitly the user claims that we can write,

\begin{equation}\phi_i = R ( g (x)) _{i,j} v _j\end{equation}

where $g (x) $, if I understand correctly, are the group generators of the unbroken group. How come this is justified?

asked Apr 16, 2014 in Theoretical Physics by JeffDror (650 points) [ revision history ]

1 Answer

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Let $H\subset G$ be the subgroup that acts trivially on $\vec{v}$ i.e., $R(h)_{i,j} v_j = v_i$. You can choose the most general $g(x)\in G$.  However, any two elements $g(x)$ and $g'(x)=g(x)h(x)$ give rise to the same $\phi_i(x) = R(g(x))_{i,j} v_j$. This freedom shows that the fluctuations about a vacuum solution are parametrized by elements of the coset $G/H$. We can write (for a Lie group) $g(x) =\exp(i \sum_a \theta_a(x) T_a)$, where the sum runs of the generators of $G$ that are not in $H$ (somewhat loose terminology) and $\theta_a(x)$ parametrize the Goldstone modes.

answered Apr 19, 2014 by suresh (1,545 points) [ revision history ]

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