# Scalar field mass alteration due to interaction term $\mathcal{L_{int}} = \mu^{2}\phi_1\phi_2$

With non-interactive (free) Action of two scalar fields $$\phi_1$$ and $$\phi_2$$ I add an interaction term $$\mathcal{L_{int}} = \mu^{2}\phi_1\phi_2 \ \$$ i.e.:

$$S = S^{free}[\phi_1, \phi_2] + S_{int} [\phi_1, \phi_2 ]= \int d^4 x ( \partial_{\mu}\phi_1 \partial^{\mu}\phi_1 +\partial_{\nu}\phi_2 \partial^{\nu}\phi_2 - m^{2}_1\phi^{2}_1 - m^{2}_2\phi^{2}_2 + \mu^2 \phi_1\phi_2)$$

It is clear to me how $$\mu=0$$ gives me Klein-Gordon Equations for both $$\phi_1$$ and $$\phi_2$$, but I am supposed to obtain the aforementioned Action as a non-interacting one, with the masses $$m_1$$ and $$m_2$$ only changed. What's the way here, or am I missing something?

• Well try a field transformation that completes the square :) Commented Jun 9, 2022 at 20:31

\begin{align} m_1^2 \phi_1^2 +m_2^2\phi_2^2 - \mu^2 \phi_1 \phi_2 =& \ m_1^2 \phi_1^2 - \mu^2 \phi_1 \phi_2 + a^2\phi_2^2 + (m_2^2-a^2) \phi_2^2 = \\ = & \ (m_1 \phi_1 - a \phi_2) ^2 + (m_2^2-a^2) \phi_2^2 \end{align}
where you chose $$a$$ such that it fullfills $$-2m_1a=-\mu^2$$.
• What is the mass of the altered field configuration? I see $m^2_2 -a^2 = {m'}^2$ the mass term for $\phi_2$ but what is it for that superposition term of $m_1 \phi_2 - a\phi_1$? Commented Jun 9, 2022 at 21:16
• The new field term $m_1\phi_1 - a\phi_2$ will have a modulus $\sqrt{|m_1|^2+|a^2|}$, so the mass term will be that squared, which gives a final Lagrangian term: $(|m_1|^2+|a^2|) \phi_1'^2$ Commented Jun 9, 2022 at 21:26
• But will the kinetic energy term change accordingly? I don't think so, since mixed terms like $\partial^{\mu}{\phi_1}\partial_{\mu}{\phi_2}$ will come along. How do we deal with that? Commented Jun 10, 2022 at 6:52