# Constantness of the field renormalisation constant (according to Sidney Coleman)

I encountered the following passage in Quantum Field Theory: Lectures of Sidney Coleman, page 280:

Since Lorentz transformations don't change $$\phi'(0)$$, or change any one-meson state to any other one-meson state, the coefficient $$\langle k\rvert\phi'(0)\lvert 0\rangle$$ of $$e^{ik\cdot x}$$ must be Lorentz invariant, and so can depend only on $$k^2$$. Then $$k^2=\mu^2$$, and $$\langle k\rvert\phi'(0)\lvert 0\rangle$$ is a constant.

Then the author goes on defining the field renormalisation constant $$Z_3$$ from this quantity.

My problem is, I thought that one-meson states $$\lvert k\rangle$$ were not invariant, since they should transform to $$\lvert \Lambda k\rangle$$, $$\Lambda$$ being the Lorentz transform. The other quantity, $$\phi'(0)$$, is surely constant since $$\phi$$ is a scalar field and $$\Lambda 0=0$$. I checked in the previous chapters on scalar fields, and indeed I found this transformation law. What am I missing here? Is that quantity really Lorentz-invariant, and if so, how can it be explained?

• Write |k⟩ as U(Λ)|0⟩, where Lambda is the Lorentz transformation that takes k to (μ,0,0,0). Make the same U act on the vacuum (it is invariant under Lorentz transformations). Then you get something like $U^{\dagger}\phi(0)U$ which will again give you back $\phi(0)$. Since μ is a constant, the field strength renormalization will also be a constant. Dec 23, 2019 at 20:34
• Is $\lvert 0 \rangle$ in your first equation the vacuum state? Dec 25, 2019 at 10:32
• No it's not, thats state has zero three momentum. Dec 25, 2019 at 12:39
• So... by your first comment, I should get (I'll call $\lvert\emptyset\rangle$ the vacuum state) that $\langle k\rvert\phi'(0)\lvert\emptyset\rangle=\langle 0\rvert U(\Lambda)^{-1}\phi'(0)\lvert\emptyset\rangle$... then since $U(\Lambda)^{-1}\phi'(0)U(\Lambda)=\phi'(0)$ I obtain $U(\Lambda)^{-1}\phi'(0)=\phi'(0)U(\Lambda)^{-1}$, and then $\phi'(0)U(\Lambda)^{-1}\lvert\emptyset\rangle=\phi'(0)\lvert\emptyset\rangle$ following from the invariance of the vacuum state. Is this what you mean? (Even then, this is not the quantity I started from...) Dec 26, 2019 at 21:44

I got this. Since $$\phi'$$ is a scalar field we have $$$$U(\Lambda)\phi'(x)U(\Lambda)^{-1}=\phi'(\Lambda^{-1}x),$$$$ for any element $$\Lambda$$ of the Lorentz group, therefore $$$$\langle k\rvert\phi'(0)\lvert 0\rangle= \langle k\rvert U(\Lambda)^{-1}\phi'(0)U(\Lambda)\lvert 0\rangle= \langle \Lambda k\rvert\phi'(0)\lvert 0\rangle$$$$ which means that $$\langle k\rvert\phi'(0)\lvert 0\rangle$$ is invariant with respect to Lorentz transformations of $$k$$. This implies that the function
1. doesn't depend on $$k$$, or
2. is a function of $$k^2$$ only, since this is the only invariant quantity we can build from $$k$$.
Since every particle associated to the field $$\phi'$$ has the same mass $$\mu$$, and $$k^2=\mu^2$$, the function $$\langle k\rvert\phi'(0)\lvert 0\rangle$$ actually doesn't depend on $$k$$ in any of the two cases.