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# 物理代写|量子力学代写Quantum mechanics代考|TFYS7030 Calculation of the propagator

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## 物理代写|量子力学代写Quantum mechanics代考|Calculation of the propagator

Let us now calculate the 2-point function, or propagator, $G_{E}^{(0)}\left(x-x^{\prime}\right)$ for infinite Euclidean space. This is the case of interest in QFT at $T=0$. In section $5.8$, we will calculate the propagator at finite temperature.

Eq. (5.139) tells us that $G_{E}^{(0)}\left(x-x^{\prime}\right)$ is the Green function of the operator $-\partial^{2}+m^{2}$. We will use Fourier transform methods and write $G_{E}^{(0)}\left(x-x^{\prime}\right)$ in the form
$$G_{E}^{(0)}\left(x-x^{\prime}\right)=\int \frac{d^{D} p}{(2 \pi)^{D}} G_{0}^{E}(p) e^{i p_{\mu}\left(x_{\mu}-x_{\mu}^{\prime}\right)}$$
which is a solution of eq. (5.139) if
$$G_{E}^{(0)}(p)=\frac{1}{p^{2}+m^{2}}$$
Therefore, the correlation function in real (Euclidean!) space is the integral
$$G_{E}^{(0)}\left(x-x^{\prime}\right)=\int \frac{d^{D} p}{(2 \pi)^{D}} \frac{e^{i p_{\mu}\left(x_{\mu}-x_{\mu}^{\prime}\right)}}{p^{2}+m^{2}}$$
We will often encounter integrals of this type and for that reason, let us evaluate this one in some detail. We begin by using the identity
$$\frac{1}{A}=\frac{1}{2} \int_{0}^{\infty} d \alpha e^{-\frac{A}{2} \alpha}$$
where $A>0$ is a positive real number. The variable $\alpha$ is called a “Feynman-Schwinger parameter.”

Now choose $A=p^{2}+m^{2}$, and substitute this expression back in eq. (5.160), which takes the form
$$G_{E}^{(0)}\left(x-x^{\prime}\right)=\frac{1}{2} \int_{0}^{\infty} d \alpha \int \frac{d^{D} p}{(2 \pi)^{D}} e^{-\frac{\alpha}{2}\left(p^{2}+m^{2}\right)+i p_{\mu}\left(x_{\mu}-x_{\mu}^{\prime}\right)}$$

## 物理代写|量子力学代写Quantum mechanics代考|Behavior of the propagator in Minkowski spacetime

Let us find the behavior of the propagator in real time. Now we must do the analytic continuation back to real time.

Recall that in going from Minkowski to Euclidean space, we continued $x_{0} \rightarrow-i x_{4}$. In addition, there is also a factor of $i$ difference in the definition of the propagator. Thus, the propagator in Minkowski spacetime, $G^{(0)}\left(x-x^{\prime}\right)$, is the expression that results from the analytic continuation:
$$G^{(0)}\left(x-x^{\prime}\right)=\left.i G_{E}^{(0)}\left(x-x^{\prime}\right)\right|{x{4} \rightarrow i x_{0}}$$
We can also obtain this result from the path integral formulation in Minkowski spacetime. Indeed, the generating functional for a free real massive scalar field $Z[J]$ in $D=(d+1)$-dimensional Minkowski spacetime is
$$Z[J]=\int \mathcal{D} \phi e^{i \int d^{D} x\left[\frac{1}{2}(\partial \phi)^{2}-\frac{m^{2}}{2} \phi^{2}+J \phi\right]}$$
Hence, the expectation value of the time-ordered product of two fields is
$$\langle 0|T \phi(x) \phi(y)| 0\rangle=-\left.\frac{1}{Z[J]} \frac{\delta Z[J]}{\delta J(x) \delta J(y)}\right|{J=0}$$ In contrast, for a free field, the generating function is given by (up to a normalization constant $\mathcal{N})$ $$Z[J]=\mathcal{N}\left[\operatorname{Det}\left(\partial^{2}+m^{2}\right)\right]^{-1 / 2} e^{\frac{i}{2}} \int d^{D} x \int d^{D} y J(x) G^{(0)}(x-y) J(y)$$ where $G{0}(x-y)$ is the Green function of the Klein-Gordon operator and satisfies
$$\left(\partial^{2}+m^{2}\right) G^{(0)}(x-y)=\delta^{D}(x-y)$$

## 物理代写|量子力学代写Quantum mechanics代考|Calculation of the propagator

$$G_{E}^{(0)}\left(x-x^{\prime}\right)=\int \frac{d^{D} p}{(2 \pi)^{D}} G_{0}^{E}(p) e^{i p_{\mu}\left(x_{\mu}-x_{\mu}^{\prime}\right)}$$

$$G_{E}^{(0)}(p)=\frac{1}{p^{2}+m^{2}}$$

$$G_{E}^{(0)}\left(x-x^{\prime}\right)=\int \frac{d^{D} p}{(2 \pi)^{D}} \frac{e^{i p_{\mu}\left(x_{\mu}-x_{\mu}^{\prime}\right)}}{p^{2}+m^{2}}$$

$$\frac{1}{A}=\frac{1}{2} \int_{0}^{\infty} d \alpha e^{-\frac{A}{2} \alpha}$$

$$G_{E}^{(0)}\left(x-x^{\prime}\right)=\frac{1}{2} \int_{0}^{\infty} d \alpha \int \frac{d^{D} p}{(2 \pi)^{D}} e^{-\frac{\alpha}{2}\left(p^{2}+m^{2}\right)+i p_{\mu}\left(x_{\mu}-x_{\mu}^{\prime}\right)}$$

## 物理代写|量子力学代写Quantum mechanics代考|Behavior of the propagator in Minkowski spacetime

$$G^{(0)}\left(x-x^{\prime}\right)=i G_{E}^{(0)}\left(x-x^{\prime}\right) \mid x 4 \rightarrow i x_{0}$$

$$Z[J]=\int \mathcal{D} \phi e^{i \int d^{D}\left[\frac{1}{2}(\partial \phi)^{2}-\frac{\pi^{2}}{2} \phi^{2}+J \phi\right]}$$

$$\langle 0|T \phi(x) \phi(y)| 0\rangle=-\frac{1}{Z[J]} \frac{\delta Z[J]}{\delta J(x) \delta J(y)} \mid J=0$$

$$Z[J]=\mathcal{N}\left[\operatorname{Det}\left(\partial^{2}+m^{2}\right)\right]^{-1 / 2} e^{\frac{i}{2}} \int d^{D} x \int d^{D} y J(x) G^{(0)}(x-y) J(y)$$

$$\left(\partial^{2}+m^{2}\right) G^{(0)}(x-y)=\delta^{D}(x-y)$$

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