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## Homework Statement

Take the action

[tex] S = \int d^4x \frac{1}{2} \left( \partial_{\mu}\phi(x)\partial^{\mu}\phi(x) - m^2\phi^2(x) - g\phi(x)^p \right) , [/tex]

and consider the following transformations:

[tex] x^{\mu} \rightarrow x^{'\mu} = \lambda x^{\mu} [/tex]

[tex] \phi(x) \rightarrow \phi^{'}(x) = \lambda^{-D}\phi(\lambda^{-1} x) . [/tex]

For what values of ##m## and ##p## is this a symmetry?

## Homework Equations

N/A

## The Attempt at a Solution

This seems like an elementary calculus problem, but I'm having trouble. Under the condition that the derivative terms remain invariant, we want

[tex] \frac{\partial}{\partial x^{\mu}} \phi(x) \rightarrow \frac{\partial}{\partial(\lambda^{-1}x^{\mu})}\phi(\lambda^{-1}x), [/tex]

which is satisfied when ##D = -2##. From the action we calculate the equation of motion and get

[tex] \partial_{\mu}\partial^{\mu} \phi(x) + m^2\phi(x) + gp\phi^{p-1}(x) = 0. [/tex]

In the primed frame, the Euler-Lagrange equations are taken with respect to the primed coordinates and the equation of motion reads

[tex] \frac{\partial}{\partial(\lambda x^{\mu})} \left( \partial^{'}_{\mu} \phi^{'}(x) \right) + m^2\phi^{'}(x) + gp\phi^{'}(x)^{p-1} = 0. [/tex]

Invoking the condition that ##D = -2##, the ##\partial^{'}_{\mu}\phi^{'}(x)## simply becomes ##\frac{\partial}{\partial\alpha}\phi(\alpha)##, where ##\alpha = \lambda^{-1}x##. The remaining derivative in the first time requires an extra factor of ##\lambda^{-2}## for it to be evaluated over ##\alpha## as well, and the entire equation becomes

[tex] \frac{\partial}{\partial\alpha_{\mu}}\frac{\partial}{\partial\alpha^{\mu}}\phi(\alpha) + \lambda^{4}m^2\phi(\alpha) + \lambda^{2p}gp\phi(\alpha)^{p-1} = 0. [/tex]

In order for the transformation to be a symmetry, we want ##\lambda## to equal unity, and we find that ##m^2 = m^2/\lambda^4##. But this is some sort of recursion relation and doesn't really make any sense.

I'm also skeptical about whether or not the equation I have written for the EOM in the primed frame is actually a symmetry when ##\lambda = 1## because the derivatives aren't even taken with respect to the primed frame's coordinates. If it is, though, I'm kind of lost. (Incidentally it's also possible that there was a typo in the problem; I don't see any reason to define the transformation such that ##x## transforms by a different factor than the argument of ##\phi(x)##...?)

Help appreciated, thanks!