QFT and the SM by Schwartz, checking

[Genady]

Checking with the physicists here:

On the p.17 it says,

Shouldn't it say force rather than potential? Isn't any potential rather quadratic close to equilibrium?

[joigus]

1 hour ago, Genady said:

Checking with the physicists here:

On the p.17 it says,

Shouldn't it say force rather than potential? Isn't any potential rather quadratic close to equilibrium?

Absolutely. It's an erratum. The potential is quadratic, so it's the force that's linear.

Close to equilibrium the Taylor expansion of the potential must be quadratic, as at the equilibrium position, the gradient (the force) must be zero. So the next-to-zero^th-order term for the force is proportional to V''(x₀).

V(x)=V(x₀)+(1/2)V''(x₀)(x-x₀)²+...