# Lagrangians for Non-Interacting Scalar Fields in QFT

I am currently taking a QFT class and we are using both canonical and path integral quantization to solve non-interacting scalar fields. We have seen the real scalar field with Lagrangian $$\mathcal{L}_1=-\frac12\partial^\mu\phi\partial_\mu\phi-\frac12m^2\phi^2$$ and the complex scalar field with Lagrangian $$\mathcal{L}_2=-\partial^\mu\phi^\dagger\partial_\mu\phi-m^2\phi^\dagger\phi$$ which, if I am not mistaken, is equivalent to two real scalar fields $$\phi_1$$ and $$\phi_2$$ such that: $$\phi=\frac1{\sqrt{2}}[\phi_1+i\phi_2] \qquad\text{and}\qquad \phi^\dagger=\frac1{\sqrt{2}}[\phi_1-i\phi_2]$$ So the Lagrangian simply becomes the sum of the Lagrangians for the two real scalar fields: $$\mathcal{L}_2=-\frac12\partial^\mu\phi_1\partial_\mu\phi_1-\frac12m^2\phi_1^2-\frac12\partial^\mu\phi_2\partial_\mu\phi_2-\frac12m^2\phi_2^2$$ What are some other Lagrangians for non-interacting scalar fields? Does, for instance, $$\mathcal{L}_3=-\frac12\partial^\mu\phi_1\partial_\mu\phi_1-\frac12m^2\phi_1^2-\frac12\partial^\mu\phi_2\partial_\mu\phi_2-\frac12m^2\phi_2^2+g\phi_1\phi_2$$ describe a non-interacting case? Is there a rule for spotting such theories? For example, I think I read somewhere that they have to be at most quadratic.

• Your breakdown of of the complex field into a real and imaginary part is obviously incorrect. As regards your question, can you you diagonalise the fields? Nov 8, 2021 at 12:41
• A lagrangian describes a free field if there are no interaction terms (other than the mass term if one considers it as an interaction term). Nov 8, 2021 at 13:08
• It is always understood that quadratic terms are diagonalized right from the start of the discussion; this is a universal convention, leaving little bonafide leeway. Hardly deserving a "rule". Nov 8, 2021 at 14:00
• @Oбжорoв Sorry, but how is it obvious? Nov 8, 2021 at 19:21
• @GeorgeSmyridis you use the same symbols on both sides of the equation ... Nov 8, 2021 at 19:24

It is possible to describe the non-kinetic terms by a matrix

$$\left(\begin{array}{cc} \phi_1 & \phi_2\end{array}\right) \left( \begin{array}{cc} m^2 & -0.5 g \\ -0.5 g & m^2 \end{array} \right) \left(\begin{array}{cc} \phi_1 \\ \phi_2\end{array}\right) = \left( \begin{array}{c} \phi & \varphi\end{array}\right) \left( \begin{array}{cc} m^2 -0.5g & 0 \\ 0 & m^2 + 0.5g \end{array} \right) \left(\begin{array}{cc} \phi \\ \varphi\end{array}\right)$$ where

$$\phi = \frac{\phi_1 + \phi_2}{\sqrt{2}}\quad \text{and}\quad \varphi = \frac{\phi_1 - \phi_2}{\sqrt{2}}$$

As demonstrated the matrix can be diagonalized and the 2 modes can be decoupled. With decoupled modes it would indeed be an non-interacting scalar (quantum) field theory.
The rule you ask for is that as long as completely decoupled modes can be found it would be an non-interacting theory.

• Could you maybe demonstrate how I would go about it? Nov 8, 2021 at 20:23
• Find the transformation that diagonalizes the matrix, that same transformation will tell you the appropriate linear combos of fields for the non-interacting theory. For example, in the example given in this answer the appropriate linear combos (unnormalized) are $\phi_1 + \phi_2$ and $\phi_1 - \phi_2$. You can multiple by $1/\sqrt{2}$ to get nicer normalized fields.
– hft
Nov 8, 2021 at 21:46