- #1

- 970

- 3

[tex]V_{eff}(\phi)=\frac{1}{4!}\lambda(M)\phi^4+\frac{\lambda(M)^2}{(16 \pi)^2}\phi^4[log\frac{\phi^2}{M^2}-\frac{25}{6}]+O(\lambda(M)^3)[/tex]

The effective potential is basically the potential with quantum corrections built in and is basically what an experimentalist would measure.

I should note that there is no mass term proportional to [tex]\phi^2 [/tex] in [tex]V_{eff}(\phi) [/tex] because Zee is enforcing that the mass term is zero by hand by choice of renormalization constants - the model he is considering really does have its physical mass equal to zero.

From the equation, symmetry is broken because of the [tex]\phi^4 log(\phi^2) [/tex] term, so that although the original potential is simply [tex]V(\phi)=\lambda \frac{\phi^4}{4!}[/tex] and hence has a minimum at [tex]\phi=0[/tex], the effective potential has a potential not at 0 because the log term becomes infinitely negative at small [tex]\phi [/tex] so the minimum of the potential occurs at small values of [tex]\phi [/tex] not quite zero, so symmetry is broken and all the consequences of symmetry-breaking such as particles getting mass hold.

My question is how is symmetry restored at high energies from this equation for the effective potential? If the effective potential is always of that form, then isn't there no way of escaping symmetry breaking? The only variable in the equation that can be adjusted is the mass scale M, so at high energies M should increase, but I don't see how an increase of M (and the accompanying increase in [tex]\lambda [/tex] by the beta function) changes the fact that the log term will cause symmetry breaking.

In Kaku's book on QFT there is a homework problem that asks you to prove that symmetry restoration occurs at a temperature [tex]T^2=-\frac{24m^2}{\lambda} [/tex]. This takes a bit of thermal QFT and the imaginary-time formalism (Kaku's book is sometimes ridiculous in what it expects you to know) and the calculation is not very simple, but conceptually you can see how symmetry is restored at that temperature. But I can't see how varying the energy restores symmetry.