In chapter 12.1 of Peskin and Schroeder we derive the propagator of a high momenta shell of a $\phi^4$ theory. Following the derivation, we ignore quartic terms as well as mass terms.

The original action is $$ \int{d^dx\left(\frac{1}{2}(\partial_\mu\hat{\phi})^2+\frac{1}{2}m^2\hat{\phi}^2+\lambda\left(\frac{1}{6}\phi^3\hat{\phi}+\frac{1}{4}\phi^2\hat{\phi}^2+\frac{1}{6}\phi\hat{\phi}^3+\frac{1}{4!}\hat{\phi}^4\right)\right)}\tag{12.5} $$ then we only keep the first term giving us $$ \mathcal{S}=\int{d^dx\frac{1}{2}(\partial_\mu\hat{\phi})^2}=\int{d^dx\frac{1}{2}\hat{\phi}\partial^2\hat{\phi}}=\frac{1}{2}\int{\frac{d^dk}{(2\pi)^d}\hat{\phi}^*(k)k^2\hat{\phi}(k)}. $$ First of all, my question is the following. Shouldn't the second $\phi$ be $\phi^*$? I was under the impression that the '$^*$' is used because the fields are Fourier transformed.

Furthermore, how do we get the following propagator(how can we derive it?)? $$ \frac{\int{\mathcal{D}\hat{\phi}}\exp(-\mathcal{S})\hat{\phi}(k)\hat{\phi}(p)}{\int{\mathcal{D}\hat{\phi}\exp(-\mathcal{S})}}=\frac{1}{k^2}(2\pi)^d\delta^{(d)}(k+p)\Theta(k)\tag{12.8} $$ where $\Theta(k)$ is defined in eq. (12.9).

Note: the momentum integrals are down between $b\Lambda \leq k < \Lambda$ for which $\Lambda$ is the momenta cutoff and $b$ a parameter between 0 and 1.

From these I understand the use of the step-function $\Theta$ but I'm not sure how we rederive the propagator. This is essentially the same as the klein-gordon propagator but in Euclidean space and in the limit of low mass ($m^2\ll \Lambda^2$).

In chapter 12.1 of Peskin and Schroeder we derive the propagator of a high momenta shell of a $\phi^4$ theory. Following the derivation, we ignore quartic terms as well as mass terms.

The original action is $$ \int{d^dx\left(\frac{1}{2}(\partial_\mu\hat{\phi})^2+\frac{1}{2}m^2\hat{\phi}^2+\lambda\left(\frac{1}{6}\phi^3\hat{\phi}+\frac{1}{4}\phi^2\hat{\phi}^2+\frac{1}{6}\phi\hat{\phi}^3+\frac{1}{4!}\hat{\phi}^4\right)\right)}\tag{12.5} $$ then we only keep the first term giving us $$ \mathcal{S}=\int{d^dx\frac{1}{2}(\partial_\mu\hat{\phi})^2}=\int{d^dx\frac{1}{2}\hat{\phi}\partial^2\hat{\phi}}=\frac{1}{2}\int{\frac{d^dk}{(2\pi)^d}\hat{\phi}^*(k)k^2\hat{\phi}(k)}. $$ First of all, my question is the following. Shouldn't the second $\phi$ be $\phi^*$? I was under the impression that the '$^*$' is used because the fields are Fourier transformed.

Furthermore, how do we get the following propagator  (how can we derive it?)? $$ \frac{\int{\mathcal{D}\hat{\phi}}\exp(-\mathcal{S})\hat{\phi}(k)\hat{\phi}(p)}{\int{\mathcal{D}\hat{\phi}\exp(-\mathcal{S})}}=\frac{1}{k^2}(2\pi)^d\delta^{(d)}(k+p)\Theta(k)\tag{12.8} $$ where $\Theta(k)$ is defined in eq. (12.9).

Note: the momentum integrals are down between $b\Lambda \leq k < \Lambda$ for which $\Lambda$ is the momenta cutoff and $b$ a parameter between 0 and 1.

From these I understand the use of the step-function $\Theta$ but I'm not sure how we rederive the propagator. This is essentially the same as the klein-gordon propagator but in Euclidean space and in the limit of low mass ($m^2\ll \Lambda^2$).

In chapter 12.1 of Peskin and Schroeder we derive the propagator of a high momenta shell of a $\phi^4$ theory. Following the derivation, we ignore quartic terms as well as mass terms.

The original action is $$ \int{d^dx\left(\frac{1}{2}(\partial_\mu\hat{\phi})^2+\frac{1}{2}m^2\hat{\phi}^2+\lambda\left(\frac{1}{6}\phi^3\hat{\phi}+\frac{1}{4}\phi^2\hat{\phi}^2+\frac{1}{6}\phi\hat{\phi}^3+\frac{1}{4!}\hat{\phi}^4\right)\right)} $$ then we only keep the first term giving us $$ \mathcal{S}=\int{d^dx\frac{1}{2}(\partial_\mu\hat{\phi})^2}=\int{d^dx\frac{1}{2}\hat{\phi}\partial^2\hat{\phi}}=\int{\frac{d^dk}{(2\pi)^d}\hat{\phi}^*(k)k^2\hat{\phi}(k)} $$ First of all, my question is the following. Shouldn't the second $\phi$ be $\phi^*$? I was under the impression that the '$^*$' is used because the fields are Fourier transformed.

Furthermore, how do we get the following propagator(how can we derive it?)? $$ \frac{\int{\mathcal{D}\hat{\phi}}\exp(-\mathcal{S})\hat{\phi}(k)\hat{\phi}(p)}{\int{\mathcal{D}\hat{\phi}\exp(-\mathcal{S})}}=\frac{1}{k^2}(2\pi)^d\delta^{(d)}(k+p)\Theta(k) $$

Note: the momentum integrals are down between $b\Lambda \leq k < \Lambda$ for which $\Lambda$ is the momenta cutoff and $b$ a parameter between 0 and 1.

From these I understand the use of the step-function $\Theta$ but I'm not sure how we rederive the propagator. This is essentially the same as the klein-gordon propagator but in Euclidean space and in the limit of low mass ($m^2\ll \Lambda^2$).

In chapter 12.1 of Peskin and Schroeder we derive the propagator of a high momenta shell of a $\phi^4$ theory. Following the derivation, we ignore quartic terms as well as mass terms.

The original action is $$ \int{d^dx\left(\frac{1}{2}(\partial_\mu\hat{\phi})^2+\frac{1}{2}m^2\hat{\phi}^2+\lambda\left(\frac{1}{6}\phi^3\hat{\phi}+\frac{1}{4}\phi^2\hat{\phi}^2+\frac{1}{6}\phi\hat{\phi}^3+\frac{1}{4!}\hat{\phi}^4\right)\right)}\tag{12.5} $$ then we only keep the first term giving us $$ \mathcal{S}=\int{d^dx\frac{1}{2}(\partial_\mu\hat{\phi})^2}=\int{d^dx\frac{1}{2}\hat{\phi}\partial^2\hat{\phi}}=\frac{1}{2}\int{\frac{d^dk}{(2\pi)^d}\hat{\phi}^*(k)k^2\hat{\phi}(k)}. $$ First of all, my question is the following. Shouldn't the second $\phi$ be $\phi^*$? I was under the impression that the '$^*$' is used because the fields are Fourier transformed.

Furthermore, how do we get the following propagator(how can we derive it?)? $$ \frac{\int{\mathcal{D}\hat{\phi}}\exp(-\mathcal{S})\hat{\phi}(k)\hat{\phi}(p)}{\int{\mathcal{D}\hat{\phi}\exp(-\mathcal{S})}}=\frac{1}{k^2}(2\pi)^d\delta^{(d)}(k+p)\Theta(k)\tag{12.8} $$ where $\Theta(k)$ is defined in eq. (12.9).

Note: the momentum integrals are down between $b\Lambda \leq k < \Lambda$ for which $\Lambda$ is the momenta cutoff and $b$ a parameter between 0 and 1.

From these I understand the use of the step-function $\Theta$ but I'm not sure how we rederive the propagator. This is essentially the same as the klein-gordon propagator but in Euclidean space and in the limit of low mass ($m^2\ll \Lambda^2$).