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?

  • $\begingroup$ 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. $\endgroup$ Dec 23, 2019 at 20:34
  • $\begingroup$ Is $\lvert 0 \rangle$ in your first equation the vacuum state? $\endgroup$
    – yellon
    Dec 25, 2019 at 10:32
  • $\begingroup$ No it's not, thats state has zero three momentum. $\endgroup$ Dec 25, 2019 at 12:39
  • $\begingroup$ 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...) $\endgroup$
    – yellon
    Dec 26, 2019 at 21:44

1 Answer 1


I got this. Since $\phi'$ is a scalar field we have \begin{equation} U(\Lambda)\phi'(x)U(\Lambda)^{-1}=\phi'(\Lambda^{-1}x), \end{equation} for any element $\Lambda$ of the Lorentz group, therefore \begin{equation} \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 \end{equation} 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.


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