# Does field redefinition make Klein Gordon Action unhealthy?

+ 6 like - 0 dislike
111 views
Consider the action $L=\chi\square^3\chi$

On one hand, Ostrogradski's theorem seems to indicate that it is not unitary.

On the other hand, it can be reached by replacing $\phi$ with $\square \chi$ in KleinGordon action.

Namely, KleinGordon Action is well known to be healthy. Now if we replace in it with $\phi=\square\chi$, we get the action $L=\chi\square^3\chi$.

Does this make the Klein Gordon Action unhealthy?
edited Dec 31, 2014
Hi Xavier, what exactly do you mean by healthy? Do you mean useful as an effective action? As I understand it, non-unitarity is always a bad thing ...
Sorry, my bad choice of title, I've reedited the question, please review it, I'll be waiting for comments

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOverflo$\varnothing$Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.