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I was wondering if someone could provide me the name of the following equation

$$\square \varphi - \tilde{m}^2 \varphi = 0,$$

where $\square := \partial_t^2 - \nabla^2$. I am specifically seeking information on the physical interpretation of this equation. This comes from the minimisation of the following action

$$S = \int \frac{1}{2} (\partial_t \varphi)^{2} - \frac{1}{2} (\delta^{ij} \partial_i \varphi \partial_j \varphi) + \frac{1}{2} \tilde{m}^{2} \varphi^{2} \, dx \, dt$$

Clearly, it is similar in form to the Klein-Gordon equation, but with an imaginary mass $m \to i \tilde{m}$. I have read that the quartic potential for the Klein-Gordon equation with this imaginary mass

$$V(\varphi) = -\frac{1}{2} \tilde{m}^2 \varphi^2 + \frac{1}{2} \lambda \varphi^4$$

leads to spontaneous symmetry breaking, as referenced in this PBS spacetime video (starting at 10:29), but am not sure what is supposed to be happening when $\lambda = 0$.

I apologise if this question is not clear enough. I don't know enough about the equation above to know where to look for such information, nor the kinds of questions that I really want to ask about it, and this makes it rather difficult to ask a good question as per the site requirements.

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    $\begingroup$ Always fun when questions end up being about the site logo $\endgroup$
    – jacob1729
    Jun 12, 2022 at 22:24

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For what it's worth, a scalar field with an imaginary mass is called a tachyonic field. It should be stressed that it is not superluminous, cf. e.g. this Phys.SE post, but since the "wrong-sign" quadratic potential by itself is unstable, it must be accompanied with a higher-order stable potential, typically a $\phi^4$ potential. The Higgs field is an example.

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    $\begingroup$ Thanks for the response, I really appreciate it. I believe this makes a lot of sense in the context the equation is found; briefly, the field $\varphi$ is the solution to a particular system in a marginally stable state, and hence the 'wrong sign' quadratic potential is necessary to describe the spontaneous symmetry breaking from marginal stability to instability (this is how I have interpreted it anyway). Thanks again for your help. $\endgroup$ Jun 12, 2022 at 10:54

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