Derivative interactions in the Wilsonian renormalisation Group

I am currently working through some basic renormalisation group problems, and have come to one about derivative interactions. It has been a while since I have studied QFT formally so bear with me please. The question is as follows:

Integrating out high energy degrees of freedom can generate derivative interactions in the low energy theory. For example, the expansion of an exponential term of the form $$e^{-\int d^d x \lambda \phi^2 \hat{\phi}^2},$$ where $$\phi$$ is the low-energy degree of freedom and $$\hat{\phi}$$ is the high-energy degree of freedom. Expanding in $$\lambda$$ we find $$e^{-\int d^d x \lambda \phi^2 \hat{\phi}^2}=1-\lambda \int d^d x \phi^2 \hat{\phi}^2+\frac{\lambda^2}{2 !} \int d^d x d^d y \phi^2(x) \hat{\phi}^2(x) \phi^2(y) \hat{\phi}^2(y)+\ldots$$ Show that integrating out $$\hat{\phi}$$ will generate a term of the form $$\phi^2\left(\partial_\mu \phi\right)^2$$. You will need to use $$\langle\hat{\phi}(k) \hat{\phi}(p)\rangle \sim \frac{\delta^d(k+p)}{k^2}$$.

So the only part of the exponential expansion which can contribute such a term is the quadratic. By using the inverse Fourier Transform to decompose the fields in position space to momentum space, we get an integral of the form, for the term $$O\left(\lambda^2\right)$$, as follows (according to the answer):

$$I=\int d x d y d k_1 d k_2 d q_1 d q_2 e^{i\left(k_1+k_2+q_1+q_2\right) \cdot x} \\ \Phi\left(k_1\right) \Phi\left(k_2\right) \hat{\phi}\left(q_1\right) \phi\left(q_2\right) \\ e^{i\left(k_3+k_4+q_3+q_4\right) \cdot y}
d k_3 d k_4 d q_3 d q_4 \\ \phi\left(k_3\right) \phi\left(k_4\right) \hat{\phi}\left(q_3\right) \hat{\phi}\left(q_4\right)$$
ignoring coefficients and using $$d x$$ instead of $$d^d x$$. \begin{aligned} & \Rightarrow \\ & I=\int d k_1 d k_2 d k_3 d k_4 d q_1 d q_2 d q_3 d q_4 \\ & \delta \left(k_1+k_2+q_1+q_2\right) \delta \left(k_3+k_4+q_3+q_4\right) \\ & \phi\left(K_1\right) \phi\left(K_2\right) \phi\left(K_3\right) \phi\left(K_4\right) \\ & \hat{\phi}(q_1) \hat{\phi}(q_2) \hat{\phi}(q_3) \hat{\phi}(q_4) \\ & \end{aligned}

As is my understanding, the task is to integrate over the momentums of the high energy field. My confusion is that the next part of the answer seems to take the expectation value of I, which then seemingly is only applied to the high energy field as such:

$$< \hat{\phi}(q_1) \hat{\phi}(q_2) \hat{\phi}(q_3) \hat{\phi}(q_4) >$$ and yields:

$$\frac{\delta^d(q_1 + q_3)}{q_1^2} \\\ \frac{\delta^d(q_2 + q_4)}{q_2^2}$$

So I have 4 questions:

1. Why do we take the expectation value when integrating out?
2. Why is the expectation value of the entire integral only performed over the high-energy field?
3. Why does this only consider the combination sof 1,3 and 2,4 instead of say 1,4 etc?
4. Is there a particular reason why it would be $$q_1$$ instead of $$q_3$$ in the denominator or is this just an arbitrary choice?

Any help is appreciate as always.

I will try to answer your questions one by one. Please comment if you need any clarifications. Before I do that, let me briefly review the idea of RG flow which might make the answers to (1) and (2) clearer.

The idea of RG is to start from a theory that includes both low- and high energy modes, and somehow get rid of the high-energy modes while keeping track of the relevant physics. In the path integral language, the partition function is defined as $$Z = \int D\varphi \exp(-S[\varphi]).$$ We can formally separate $$\varphi = \phi + \tilde{\phi}$$, which are the low and high energy (momentum) modes respectively. Keeping track of these separately, we have \begin{align} Z &= \int D\phi D\tilde{\phi} \exp(-S[\phi,\tilde{\phi}]) \\ &= \int D\phi D\tilde{\phi} \exp(-S_0[\phi] - S_0[\tilde{\phi}] - S_\text{int}[\phi,\tilde{\phi}]). \end{align} Here, $$S_0$$ denotes the quadratic part of the action which separates for $$\phi$$ and $$\tilde{\phi}$$ because of momentum conservation.

Next is the key idea of RG. We want a new action containing only the slow modes such that all correlations functions of the slow variables are preserved. It is easy to see that if we define \begin{align} e^{-S_{\text{eff}}[\phi]} &= \int D\tilde{\phi} \exp(-S[\phi,\tilde{\phi}]) \\ &= e^{-S_0[\phi]}\int D\tilde{\phi} \exp( - S_0[\tilde{\phi}] - S_\text{int}[\phi,\tilde{\phi}]) \\ &= e^{-S_0[\phi]}\int D\tilde{\phi} \exp( - S_0[\tilde{\phi}]) \left[1 - S_\text{int}[\phi,\tilde{\phi}] + \frac{1}{2}S_\text{int}[\phi,\tilde{\phi}]^2 + \dots\right] , \end{align} we preserve all correlation functions while throwing away the fast modes. Going from $$S$$ to $$S_\text{eff}$$ is the RG flow. Note that we are integrating over only the high energy modes to get the effective action, this is just because these are the degrees of freedom we don't care about anymore, and we want an effective action for the slow modes.

1. You have an expression that looks like \begin{align} &\quad \int D\tilde{\phi} e^{-S_0[\tilde\phi]}\phi \phi \phi \phi \tilde\phi \tilde\phi \tilde\phi \tilde\phi \\ &= \phi \phi \phi \phi \frac{\int D\tilde{\phi} e^{-S_0[\tilde\phi]}\tilde\phi \tilde\phi \tilde\phi \tilde\phi}{\int D\tilde{\phi} e^{-S_0[\tilde\phi]}} \tilde Z \\ &=\phi \phi \phi \phi \tilde{Z}\langle \tilde\phi \tilde\phi \tilde\phi \tilde\phi\rangle \end{align} where $$\tilde Z \equiv \int D\tilde{\phi} e^{-S_0[\tilde\phi]}$$ and the fraction in the second expression is just the formula for expectation values in the path integral formalism. That is why you take expectation values of the high-energy modes.

2. See earlier point. You integrate over the high energy modes to get the effective action in terms of the low-energy mode. In the process of integrating out, you naturally recover these expectation values.

3. Your idea is correct. But, because you integrate over all possible '$$q$$'s, the answer is independent of your choice of contraction. Therefore, you just need to keep track of the number of possible contractions (2 in this case: the 1-2, 3-4 contraction can be ignored because of the linked cluster theorem when we take a log later to get the correction to the action) and multiply it by the result for any one of the contractions.

4. The $$\delta$$-function in the numerator enforces that $$q_1^2 = q_3^2$$, so it does not matter which one you put.