1
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ 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? $\endgroup$ Commented Nov 8, 2021 at 12:41
  • 1
    $\begingroup$ 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). $\endgroup$ Commented Nov 8, 2021 at 13:08
  • $\begingroup$ 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". $\endgroup$ Commented Nov 8, 2021 at 14:00
  • $\begingroup$ @Oбжорoв Sorry, but how is it obvious? $\endgroup$ Commented Nov 8, 2021 at 19:21
  • $\begingroup$ @GeorgeSmyridis you use the same symbols on both sides of the equation ... $\endgroup$ Commented Nov 8, 2021 at 19:24

1 Answer 1

2
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Could you maybe demonstrate how I would go about it? $\endgroup$ Commented Nov 8, 2021 at 20:23
  • $\begingroup$ 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. $\endgroup$
    – hft
    Commented Nov 8, 2021 at 21:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.