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I understand that if I have an action $$S=\int \phi(\Box + m^2 )\phi$$ Then the field $\phi$ has mass $m$ since this is the pole of the propagator of $\phi$. Now If I have an action $$S=\int \phi_1 \Box \phi_2 + m^2 \phi_1^2$$ Then how do I interpret the mass of the field $\phi_1$ or $\phi_2$?

My thinking was that If I find the equations of motion for $\phi_2$ then I have $$\Box \phi_1 =0$$ Which is a Klein-Gordon equation with no-mass so we interpret the field $\phi_1$ to be massless??

Thank you

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  • $\begingroup$ Minor comment to the post (v2). 1. Where are the integration measures? 2. The mass terms has the wrong signs. 3. The kinetic term in the 2-boson theory is not positive definite. $\endgroup$ – Qmechanic♦ Aug 15 '19 at 15:54
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In order to calculate the mass you need to go to an eigenbasis. Unfortunately, your theory is ill-defined: the kinetic term has a negative eigenvalue: $$ \begin{pmatrix}0&1/2\\1/2&0\end{pmatrix}\sim \begin{pmatrix}1/2&0\\0&-1/2\end{pmatrix} $$

Therefore, the Hamiltonian is unbounded from below and the model is simply inconsistent. If your kinetic term was positive definite, then you could simultaneously diagonalise it together with the mass term, and easily read off the masses.

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    $\begingroup$ I think your answer would benefit from being explicit about how you extracted these matrices from the OP's action. $\endgroup$ – Martin C. Aug 15 '19 at 12:48
  • $\begingroup$ @MartinC. It seems OP was able to figure it out by themselves :-) $\endgroup$ – AccidentalFourierTransform Aug 15 '19 at 14:16
  • $\begingroup$ Thank you @AccidentalFourierTransform . Just to make sure I understood it correctly. So you packaged phi_1 and phi_2 into a vector V={{phi_1,phi_2}} and rewrote the kinetic in terms of matrix V^T {{0,1/2},{1/2,0}} \Box V? I'm curicus if we can rewrite it as V^T {{0,1},{0,0}} \Box V, then eigenvalue of this would be {0,0}. which I guess would make the theory inconsistant anyway becase of zero kinetic term? $\endgroup$ – user239970 Aug 15 '19 at 14:57
  • $\begingroup$ @user239970 Yes, basically. But recall that the kinetic term is a quadratic form, so its matrix representation is always the symmetric one. (The rest of representations are not canonical; their eigenvalues do not represent the principal axes of the transformation). $\endgroup$ – AccidentalFourierTransform Aug 15 '19 at 15:12
  • $\begingroup$ @AccidentalFourierTransform Ok thanks really helpful. $\endgroup$ – user239970 Aug 15 '19 at 16:46

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