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Here are 2 doubts:

  1. If we change the sign of the mass term in the free massive KG Lagrangian to get:

    $L = \frac{1}{2}\partial^\mu\phi\partial_\mu\phi + \frac{1}{2}m^2\phi^2$,

    What would be the $physical$ implications of this change? (aside from on shell condition not being satisfied)?

  2. Let $\phi^{(1)}$ and $\phi^{(2)}$ be 2 real scalar fields with the Lagrangian:

    $L = \frac{1}{2}\sum_{i}\partial^\mu\phi^{(i)}\partial_\mu\phi^{(i)} - \frac{1}{2}m^2 \sum_{i,j,k} \phi^{(i)} M_{ij}M_{jk} \phi^{(k)}$ .

    where $M_{11} = \lambda$, $M_{12} = 1$, $M_{21} = 1$, $M_{22} = 0$ AND $\lambda >> 1$.

    What is the mass ratio of the 2 particles in the theory?

    EDIT: As for the 2nd question, I found that $L$ should simplify to $\frac{1}{2}\sum_{i}\partial^\mu\phi^{(i)}\partial_\mu\phi^{(i)} - \frac{1}{2} m^2(\lambda \phi^{(1)} + \phi^{(2)})^2$. I tried continuing from the viewpoint that we can consider the superposition $\phi^{(3)} = \lambda \phi^{(1)} + \phi^{(2)}$ and try to eliminate cross-terms, thereby getting a simple sum of non-interacting Lagrangians, but that didn't work. (Now I'm out of my depths)

Thanks in advance

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2  
Please do not remove the content of your question like that. –  dmckee Mar 11 '13 at 19:03

2 Answers 2

up vote 3 down vote accepted

Here we will only address the first question. A tachyonic negative mass-square term corresponds to a potential term $V$ that is unbounded from below. This drives the value $|\phi| \to \infty$. In other words, a physical instability.

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If possible could you give a reference which explains how the potential term is unbounded and hence $\phi$ goes to infinity? I have a vague idea that ordinarily particles are small oscillations of the field about the minimum of the potential, but I can't put any of that in rigorous terms. –  1989189198 Mar 10 '13 at 19:52
2  
$V(\phi) =- \frac{1}{2} m^2 \phi^2$ (with $m^2>0$) implies that $V$ as a function of $\phi$ is unbounded from below, no need for anything fancy. –  alexarvanitakis Mar 10 '13 at 20:00
    
Ah sorry! My bad, I'm considering everything too complicated... –  1989189198 Mar 10 '13 at 20:09

For the second part of the question, note that the kinetic term for $N$ scalar fields $\phi_i$ is invariant under and O(N) rotation of the fields $\phi_i \to O^j_i \phi_j$:

$$ (\partial \phi_i) (\partial \phi_i) \to (\partial \phi_j O^j_i) (\partial \phi_k O^k_i) = O^j_i O^k_i (\partial \phi_j)(\partial \phi_k),$$

and using $O^j_i O^k_i = \delta^{jk}$. But the mass term changes:

$$ \phi_i (M^2)_{ij} \phi_j \to \phi_k (O^k_i (M^2)_{ij} O^l_j) \phi_l.$$

Since the mass matrix is symmetric you can choose an $O^k_i$ to diagonalise it:

$$ O^k_i (M^2)_{ij} O^l_j = m_k^2 \delta_{kl}, $$

(no sum on $k$ on the right hand side) and the eigenvalues $m_i^2$ are the physical masses and the eigenvectors are the physical mass eigenstates.

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Thanks and apologies for the late acknowledgement! –  1989189198 May 25 '13 at 4:37

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