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To phrase my question I will use the example, which we used as an exercise in my introducory QFT lecture. We considered a theory of a real scalar field $\Phi$ and a complex scalar field $\phi$. The Lagrangian of the theory was given as $$ \mathcal L = \frac 1 2\partial_\mu \Phi\partial^\mu\Phi - \frac 1 2M^2\Phi^2 + (\partial_\mu \phi^\dagger)(\partial^\mu \phi) - m^2 \phi^\dagger\phi + \lambda\phi^\dagger\phi\Phi . $$ So now we had to expand the $S$-operator up to quadratic order in the coupling $\lambda$. Hence I calculated the following $$ \begin{align} S &= T\left\{\exp\left(i\int \text d^4x\ \ \mathbf{: }\,\lambda\phi^\dagger\phi\Phi\,\mathbf{:}\right)\right\}\\ &= \mathbf 1 + i\lambda \int\text d^4x\ \ T\left\{\mathbf{:}\phi^\dagger(x)\phi(x)\Phi(x)\mathbf{:}\right\} \\ & \phantom{=1i}+ \frac{(i\lambda)^2}{2}\int\int\text d^4x_1 \text d^4x_2 T\left\{\mathbf{:}\phi^\dagger(x_1)\phi(x_1)\Phi(x_1)\mathbf{:}\ \mathbf{:}\phi^\dagger(x_2)\phi(x_2)\Phi(x_2)\mathbf{:}\right\} + \mathcal O(\lambda^3),\quad(1) \end{align} $$ where $T\{\circ\}$ denotes the time ordering of operators and $\mathbf{:}\circ\mathbf{:}$ denotes the normal ordering of operators. This expression I further "simplified" using Wicks Theorem. But in the solution to the exercise there was no normal ordering right at the start, so the solution wrote $$ \begin{align} S &= T\left\{\exp\left(i\int \text d^4x\ \lambda\phi^\dagger\phi\Phi\right)\right\}\\ &= \mathbf 1 + i\lambda \int\text d^4x\ \ T\left\{\phi^\dagger(x)\phi(x)\Phi(x)\right\} \\ & \phantom{=1i}+ \frac{(i\lambda)^2}{2}\int\int\text d^4x_1 \text d^4x_2 T\left\{\phi^\dagger(x_1)\phi(x_1)\Phi(x_1)\phi^\dagger(x_2)\phi(x_2)\Phi(x_2)\right\} + \mathcal O(\lambda^3),\quad (2) \end{align} $$ and then applied Wicks theorem. So one can see that these two approaches do not lead to the same answer, since by applying Wicks theorem on approach $(1)$ you won't get any contraction between fields evaluated at the same point in space-time (for terms quadratic in the coupling). Whereas in approach $(2)$ you will get contractions between fields evaluated at the same point in space-time.

So to my understanding we need the normal ordering of the field operators to make "sense" out of products of "distribution-valued" operators. Therefore I used approach (1), since I thought that equation (2) would be even less sensible.

But the tutor said that if we would not normal order in the first place the terms we would get additionally would cancel with their counter terms in renormalization. So actually I am pretty confused, is eq. (1) or eq. (2) right or both? And if both are "correct" depending on how the result is treated in further calculations could someone shed some light on why eq. (2) makes any sense in the first place (since I thought we need normal ordering due to the distributional nature of the field operators)?

Edit: As discussed in the comments this might boil down to the fact that the quantization procedure itself does not contain the "need" or rather require normal ordering, but we need to make sense of the whole calculation and it does not matter when to do so. (Which is actually not really satisfying)

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  • $\begingroup$ @Jakob No, I have no problem applying Wicks Theorem. The Problem is even before we apply Wicks theorem. Is the interaction Lagrangian always normal ordered, and if not why? $\endgroup$ Commented Jun 17, 2021 at 8:19
  • $\begingroup$ I know, but I am fully aware of this fact, actually this leads to the problem I am having, which I adress in: "So one can see that these two approaches do not lead to the same answer, since using Wicks theorem on approach (1) you won't get any contraction between fields evaluated at the same point in space-time (for terms quadratic in the coupling). Whereas in approach (2) you will get contractions between fields evaluated at the same point in space-time." $\endgroup$ Commented Jun 17, 2021 at 8:22
  • $\begingroup$ Okay, I see. Could you provide a reference for equation (1)? Or better, the definition of the S-matrix in the first line. $\endgroup$ Commented Jun 17, 2021 at 8:23
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    $\begingroup$ The contraction of two fields taken at the same point is divergent. These divergent terms can be eliminated in calculations by the appropriate counterterms. So which approach you use, makes no difference in the end. I believe that is what your tutor may mean. $\endgroup$ Commented Jun 17, 2021 at 12:05
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    $\begingroup$ Formally you are entirely correct. Pragmatically this is not important. Realistically maybe your tutor was just being careless by not using NO but didn’t want to admit it and found a way out - or just wanted you to think about it… $\endgroup$ Commented Jun 17, 2021 at 19:42

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