# Commutator of normal ordered squared scalar field

The teacher left us prove the following statement.

Let $$\phi(x)$$ be a scalar field $$\phi(x) = \int \frac{d^3p}{2\pi \sqrt{2\omega_{\boldsymbol{p}}}} \left[ a_{\boldsymbol p} e^{-ip \cdot x} + a^+_{\boldsymbol p} e^{ip \cdot x} \right]$$ with $$p = (p^0, \boldsymbol p)$$, $$\omega_{\boldsymbol p} = p^0 = \sqrt{\boldsymbol{p}^2 + m^2}$$.

Show that the below commutator is not well defined

$$\left[ : \phi^2(x) :\, , \, :\phi^2(x): \right]$$

Where with $$:\phi^2(x):$$ is understood the normal ordering of the creation and annihilation orperators, namely, creation operators at the left and the annihilation ones at the right.

I find commutators of the form $$\left[ a_{\boldsymbol p} a_{\boldsymbol q}, a_{\boldsymbol k} a_{\boldsymbol s} \right], \quad \left[ a^+_{\boldsymbol p} a_{\boldsymbol q}, a_{\boldsymbol k} a_{\boldsymbol s} \right], \,\dots \, , \left[ a^+_{\boldsymbol p} a^+_{\boldsymbol q}, a^+_{\boldsymbol k} a^+_{\boldsymbol s} \right]$$

why they are not defined?

The Klein-Gordon field is: $$\phi(x)=\int Dp (a_p e^{-ip\cdot x}+a_p^\dagger e^{ip\cdot x})$$ with: $$\int Dp=\int \frac{d^4p}{(2\pi)^3 2\omega_\vec{p}}\theta(p_0)=\int\frac{d\vec{p}}{(2\pi)^{3}2\omega_\vec{p}}$$ and: $$[a_p,a_q^\dagger]=(2\pi)^32\omega_\vec{p}\delta(\vec{p}-\vec{q})$$ Then: $$:\phi(x)^2:\,\,\,=\, \int Dp\int Dq \,:(a_p e^{-ip\cdot x}+a_p^\dagger e^{ip\cdot x})(a_q e^{-iq\cdot x}+a_q^\dagger e^{iq\cdot x}):\,\,\,=\int Dp\int Dq :(a_pa_qe^{-ipx-iqx}+a_pa_q^\dagger e^{-ipx+iqx}+a_p^\dagger a_qe^{ipx-iqx}+a_p^\dagger a_q^\dagger e^{ipx+iqx}):\,\,\,=\int Dp\int Dq (a_pa_qe^{-ipx-iqx}+a_q^\dagger a_p e^{-ipx+iqx}+a_p^\dagger a_qe^{ipx-iqx}+a_p^\dagger a_q^\dagger e^{ipx+iqx})$$ If we want now to calculate: $$[:\phi^2(x):\,,:\phi^2(x)]=\int Dp\int Dq \int Dl \int Ds\,\left[(a_pa_qe^{-ipx-iqx}+a_q^\dagger a_p e^{-ipx+iqx}+a_p^\dagger a_qe^{ipx-iqx}+a_p^\dagger a_q^\dagger e^{ipx+iqx}),(a_la_se^{-ilx-isx}+a_s^\dagger a_l e^{-ilx+isx}+a_l^\dagger a_se^{ilx-isx}+a_l^\dagger a_s^\dagger e^{ilx+isx})\right]$$ we can stop here and look at these commutators, it is clear that there will be terms which will not be normal ordered, and so there will eventually come up divergences. So this is not a well defined operator in the context of QFT.