- #1
Plantation
- 14
- 1
- Homework Statement
- $$G_0(x,y) = \frac{\int \mathcal{D}\phi \phi(x) \phi(y) e^{i\int d^4 x \mathcal{L}_0[ \phi]}}{\int \mathcal{D}\phi e^{i \int d^4x \mathcal{L}_0[\phi]}}$$
- Relevant Equations
- $$ G^{(n)}_0 ( x_1, \dots ,x_n) = \frac{1}{i^n} \frac{\delta^n \bar{Z}_0 [J]}{\delta J(x_1) \dots \delta J(x_n)}|_{J=0}
= \frac{1}{i^n}\frac{1}{Z_0[J=0]}\frac{1}{i^n} \frac{\delta^n Z_0 [J]}{\delta J(x_1) \dots \delta J(x_n)}|_{J=0}$$
I am reading the Lancaster & Blundell, Quantum field theory for gifted amateur, p.225 and stuck at understanding some derivations.
We will calculate a generating functional for the free scalar field. The free Lagrangian is given by
$$ \mathcal{L}_0 = \frac{1}{2}(\partial _\mu \phi)^2 - \frac{m^2}{2}\phi^2. \tag{24.9}$$
And in the p.224, he get expression for normalized generating functional for the free scalar field as
$$ \bar{Z}_0[J] = \frac{ \int \mathcal{D} \phi e^{\frac{i}{2} \int d^4 x \phi \{ - ( \partial^2 + m^2) \} \phi + i \int d^4x J \phi }}{\int \mathcal{D} \phi e^{\frac{i}{2} \int d^4 x \phi \{ - ( \partial^2 + m^2)\} \phi} } = e^{- \frac{1}{2} \int d^4x d^4 y J(x) \Delta(x-y)J(y)} \tag{24.17} $$
Here, ##\Delta(x,y)=\Delta(x-y)## is the free Feynman propagator (C.f. their book (17.24) (p.159) )
In the page 225, he saids that " Specifically we have for free fields that the propagator is given, in terms of the normalized generating functional, by (C.f. their book (22.8) )
$$ G^{(n)}_0 ( x_1, \dots ,x_n) = \frac{1}{i^n} \frac{\delta^n \bar{Z}_0 [J]}{\delta J(x_1) \dots \delta J(x_n)}|_{J=0}
= \frac{1}{i^n}\frac{1}{Z_0[J=0]}\frac{1}{i^n} \frac{\delta^n Z_0 [J]}{\delta J(x_1) \dots \delta J(x_n)}|_{J=0} \tag{24.20}$$"
, where ##G^(n)(x_1, \dots x_n)## is the Green's function.
And next, he saids
"We'll evaluate this in two different ways for the single-particle propagator ##G_0(x,y)##. Differentiating the expression for the functional integral ##\bar{Z}_0[J]## with respect to the ##J##'s gives us
$$G_0(x,y) = \frac{\int \mathcal{D}\phi \phi(x) \phi(y) e^{i\int d^4 x \mathcal{L}_0[ \phi]}}{\int \mathcal{D}\phi e^{i \int d^4x \mathcal{L}_0[\phi]}} \tag{24.21}$$
while differentiating the expression for the normalized generating functional ##\bar{Z}_0[J] = e^{- \frac{1}{2} \int d^4x d^4 y J(x) \Delta(x,y)J(y)} ## ( C.f. their book (24.17) )gives us the expected answer ##G_0(x,y) = \Delta(x,y) ##."
And why these two statements are true? I've been trying to calculate these formulas continuously by brutal force differentiation but I don't know how to perform differentiation exactly at all. What should I note to make calculations easier? Can anyone give me a hint or helps?
We will calculate a generating functional for the free scalar field. The free Lagrangian is given by
$$ \mathcal{L}_0 = \frac{1}{2}(\partial _\mu \phi)^2 - \frac{m^2}{2}\phi^2. \tag{24.9}$$
And in the p.224, he get expression for normalized generating functional for the free scalar field as
$$ \bar{Z}_0[J] = \frac{ \int \mathcal{D} \phi e^{\frac{i}{2} \int d^4 x \phi \{ - ( \partial^2 + m^2) \} \phi + i \int d^4x J \phi }}{\int \mathcal{D} \phi e^{\frac{i}{2} \int d^4 x \phi \{ - ( \partial^2 + m^2)\} \phi} } = e^{- \frac{1}{2} \int d^4x d^4 y J(x) \Delta(x-y)J(y)} \tag{24.17} $$
Here, ##\Delta(x,y)=\Delta(x-y)## is the free Feynman propagator (C.f. their book (17.24) (p.159) )
In the page 225, he saids that " Specifically we have for free fields that the propagator is given, in terms of the normalized generating functional, by (C.f. their book (22.8) )
$$ G^{(n)}_0 ( x_1, \dots ,x_n) = \frac{1}{i^n} \frac{\delta^n \bar{Z}_0 [J]}{\delta J(x_1) \dots \delta J(x_n)}|_{J=0}
= \frac{1}{i^n}\frac{1}{Z_0[J=0]}\frac{1}{i^n} \frac{\delta^n Z_0 [J]}{\delta J(x_1) \dots \delta J(x_n)}|_{J=0} \tag{24.20}$$"
, where ##G^(n)(x_1, \dots x_n)## is the Green's function.
And next, he saids
"We'll evaluate this in two different ways for the single-particle propagator ##G_0(x,y)##. Differentiating the expression for the functional integral ##\bar{Z}_0[J]## with respect to the ##J##'s gives us
$$G_0(x,y) = \frac{\int \mathcal{D}\phi \phi(x) \phi(y) e^{i\int d^4 x \mathcal{L}_0[ \phi]}}{\int \mathcal{D}\phi e^{i \int d^4x \mathcal{L}_0[\phi]}} \tag{24.21}$$
while differentiating the expression for the normalized generating functional ##\bar{Z}_0[J] = e^{- \frac{1}{2} \int d^4x d^4 y J(x) \Delta(x,y)J(y)} ## ( C.f. their book (24.17) )gives us the expected answer ##G_0(x,y) = \Delta(x,y) ##."
And why these two statements are true? I've been trying to calculate these formulas continuously by brutal force differentiation but I don't know how to perform differentiation exactly at all. What should I note to make calculations easier? Can anyone give me a hint or helps?