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I am reading P&S, specifically Chapter 12. I have trouble understanding why the propagator in momentum space (if the following is indeed the propagator in momentum space) in a $\phi^4$ theory in the Euclidean space of the path integral over the configuration of fields over a single momentum shell has the following form $$\hat{\phi}(k)\hat{\phi}(k)=\frac{1}{k^2}(2\pi)^d\delta^{(d)}(k+p)\Theta(k),\tag{12.8}$$ where despite the fact that I can not denote it, a Wick contraction is implied between the two fields on the left hand side. I understand that the $\Theta$ function must be there because we are integrating over a single momentum shell $b\Lambda<|k|<\Lambda$, cf. eq. (12.9), and hence its role is to ensure that the propagator vanishes for momentum values outside that shell. Moreover, I realise that the $\delta$ function is somehow related to the fact that the field takes only real values. Despite the fact that I realise (to a point) why these $\delta$ and $\Theta$ functions are for, I do not seem to comprehend how they appear in the propagator.

Any help will be appreciated.

P.S. The $\Theta$ function is vanishing besides the shell $b\Lambda<|k|<\Lambda$, where it is equal to one, cf. eq. (12.9).

P.S.#2 I have noticed that there may be a confusion on what I am asking. I am not asking how to derive the free Lagrangian in momentum space (as in Propagator in $\phi^4$ theory). What I am asking is how to read off the propagator from that expression in a way such that Eq. (12.8) from P&S occurs.

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  • $\begingroup$ Possible duplicate: physics.stackexchange.com/q/693565/2451 $\endgroup$
    – Qmechanic
    Feb 21, 2022 at 22:56
  • $\begingroup$ There is no overlap with the link you suggest. I have also derived the expression for the free Lagrangian in momentum space. I just do not know how to associate it with the propagator and how these extre $\Theta$ and $\delta$ functions occur. This is what I was asking for... So the answer in the link you provide is not something new to me and it certainly does not help me realise what I am asking for. Can you please re-open my question? $\endgroup$
    – schris38
    Feb 22, 2022 at 6:37
  • $\begingroup$ @ohneVal I have seen that... As I mentioned above, my problem was why do we conclude that this is the form of the propagator, given the Lagrangian has the form it has (in momentum space). Thanks though! $\endgroup$
    – schris38
    Feb 24, 2022 at 13:27

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Start with expression (12.7): $$\int\frac{d^dk}{(2\pi)^d}\hat{\phi}(-k)k^2\hat{\phi}(k)$$ and add sources to that expression, i.e. $$\int\frac{d^dk}{(2\pi)^d}\hat{\phi}(-k)k^2\hat{\phi}(k)+J(-k)\hat{\phi}(k)+J(k)\hat{\phi}(-k)$$ then do the usual shift in the field $\hat{\phi}(\pm k)=\hat{\phi}'(\pm k)-\int\frac{d^dp}{(2\pi)^d}D_F(\pm k,p)J(p)$. The result is merely a shift with respect to the functional integral. The only way of doing all the necessary steps (completing the square etc) and arriving the desired result (momentum space representation of the result in chapter 9.2 from P&S) $$\int\mathcal{D}\hat{\phi}'e^{\int\mathcal{L}_0(\hat{\phi}')} \exp\bigg[\int d^dk\int d^dp\ J(-k)D_F(\pm k,p)J(p)\bigg]$$ is to use that $k^2D_F(k,p)=(2\pi)^d\delta^{(d)}(k\pm p)$ in order for us to reach the final expression written above. In any way, one can identify that between the two sources lies the propagator in momentum space... Some signs may not be exact, please excuse me for that, but anyone willing to work out the details will derive the necessary result.

The final conclusion is that, in between the sources, the quantity that is located there is the propagator $$D_F(p,k)=\frac{(2\pi)^d}{k^2}\delta^d(k+p)\Theta(k)$$ where $\Theta$ must be included such that we integrate over all momenta that lie in the desired momentum shell $(b\Lambda<|k|<\Lambda)$

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