- #1

Markus Kahn

- 112

- 14

- Homework Statement
- Let ##A_\mu^a(x)## be the gluon field and ##G_{\mu\nu}^{ab}(x)## the gluon propagator. Show that

$$\left\langle 0\left|A_{\mu}^{a}\left(x_{1}\right) A_{\nu}^{b}\left(x_{2}\right) A_{\rho}^{c}\left(x_{3}\right)\right| 0\right\rangle=-i g f^{a^{\prime} b^{\prime} c^{\prime}} \int d x \partial_{\alpha}^{x}\left[G_{\mu \beta}^{a a^{\prime}}\left(x_{1}-x\right)\right] G_{\nu \alpha}^{b b^{\prime}}\left(x_{2}-x\right) G_{\rho \beta}^{c c^{\prime}}\left(x_{3}-x\right)+\text { perm.}$$

is in momentum-space

$$

\left\langle 0\left|A_{\mu}^{a}\left(p_{1}\right) A_{\nu}^{b}\left(p_{2}\right) A_{\rho}^{c}\left(p_{3}\right)\right| 0\right\rangle=(2 \pi)^{4} \delta^{(4)}\left(p_{1}+p_{2}+p_{3}\right) g f^{a^{\prime} b^{\prime} c^{\prime}} p_{1}^{\beta} g^{\alpha \gamma} G_{\mu \alpha}^{a a^{\prime}}\left(p_{1}\right) G_{\nu \beta}^{b b^{\prime}}\left(p_{2}\right) G_{\rho \gamma}^{c c^{\prime}}\left(p_{3}\right)+\text { perm. }$$

- Relevant Equations
- All given above.

This seems rather straight forward, but I can't figure out the details... Generally speaking and ignoring prefactors, the Fourier transformation of a (nicely behaved) function ##f## is given by

$$f(x)= \int_{\mathbb{R}^{d+1}} d^{d+1}p\, \hat{f}(p) e^{ip\cdot x} \quad\Longleftrightarrow \quad \hat{f}(p)= \int_{\mathbb{R}^{d+1}} d^{d+1}x\, {f}(x) e^{-ip\cdot x}, $$

where ##p\cdot x := \eta_{\mu\nu}x^\mu p^\nu##. We also notice that

$$\int_{\mathbb{R}^{d+1}} d^{d+1}x\, \left(\partial_\mu{f}(x)\right) e^{-ip\cdot x} =-\int_{\mathbb{R}^{d+1}} d^{d+1}x\, {f}(x)(\partial_\mu e^{-ip\cdot x}) =ip_\mu\int_{\mathbb{R}^{d+1}} d^{d+1}x\, {f}(x)e^{-ip\cdot x} = ip_\mu \hat{f}(p).$$

I think these are all the ingredients we need to show the above statement. I would bscly. substitute the following into the given eq. above (assuming ##d=3##)

$$G_{\mu\nu}^{ab}(x_k-x)= \int_{\mathbb{R}^{4}} d^{4}p_k\, \hat{G}_{\mu\nu}^{ab}(p_k) e^{ip_k\cdot (x_k-x)},$$

which gives

$$\begin{align*}

\langle 0|A_{\mu}^{a}(p_{1}) &A_{\nu}^{b}(p_{2}) A_{\rho}^{c}(p_{3})| 0\rangle\\

&= -i g f^{a^{\prime} b^{\prime} c^{\prime}} \int d x

\left[\partial_{\alpha}^{x}\int_{\mathbb{R}^{4}} d^{4}p_1\, \hat{G}_{\mu \beta}^{a a^{\prime}}\left(p_1\right)e^{ip_1(x_1-x)}\right]

\int_{\mathbb{R}^{4}} d^{4}p_2\, \hat{G}_{\nu \alpha}^{b b^{\prime}}\left(p_{2}\right) e^{ip_2(x_2-x)}

\int_{\mathbb{R}^{4}} d^{4}p_3\, \hat{G}_{\rho \beta}^{c c^{\prime}}\left(p_{3}\right)e^{ip_3(x_3-x)}

+\text { perm.}\\

&=

-i g f^{a^{\prime} b^{\prime} c^{\prime}} \int d x\, d^{4}p_1\,d^{4}p_2\, d^{4}p_3\,

(-i(p_1)_\alpha)\hat{G}_{\mu \beta}^{a a^{\prime}}\left(p_1\right)e^{ip_1(x_1-x)}

\hat{G}_{\nu \alpha}^{b b^{\prime}}\left(p_{2}\right) e^{ip_2(x_2-x)}

\hat{G}_{\rho \beta}^{c c^{\prime}}\left(p_{3}\right)e^{ip_3(x_3-x)}

+\text { perm.}\\

&=

-p_1^\lambda\eta_{\alpha\lambda}g f^{a^{\prime} b^{\prime} c^{\prime}} \int d^{4}p_1\,d^{4}p_2\, d^{4}p_3\,

\hat{G}_{\mu \beta}^{a a^{\prime}}\left(p_1\right)

\hat{G}_{\nu \alpha}^{b b^{\prime}}\left(p_{2}\right)

\hat{G}_{\rho \beta}^{c c^{\prime}}\left(p_{3}\right)e^{ip_1x_1+p_2x_2+p_3x_3} \int dx\, e^{-i(p_1+p_2+p_3)x}

+\text { perm.}\\

&=

- (2\pi)^4 \delta^{(4)}(p_1+p_2+p_3)p_1^\lambda\eta_{\alpha\lambda}g f^{a^{\prime} b^{\prime} c^{\prime}} \int d^{4}p_1\,d^{4}p_2\, d^{4}p_3\,

\hat{G}_{\mu \beta}^{a a^{\prime}}\left(p_1\right)

\hat{G}_{\nu \alpha}^{b b^{\prime}}\left(p_{2}\right)

\hat{G}_{\rho \beta}^{c c^{\prime}}\left(p_{3}\right)e^{ip_1x_1+p_2x_2+p_3x_3}

+\text { perm.}\\

\end{align*}$$

This is as far as I can go and comparing it to the solution we see some key differences...

$$\left\langle 0\left|A_{\mu}^{a}\left(p_{1}\right) A_{\nu}^{b}\left(p_{2}\right) A_{\rho}^{c}\left(p_{3}\right)\right| 0\right\rangle=(2 \pi)^{4} \delta^{(4)}\left(p_{1}+p_{2}+p_{3}\right) g f^{a^{\prime} b^{\prime} c^{\prime}} p_{1}^{\beta} g^{\alpha \gamma} G_{\mu \alpha}^{a a^{\prime}}\left(p_{1}\right) G_{\nu \beta}^{b b^{\prime}}\left(p_{2}\right) G_{\rho \gamma}^{c c^{\prime}}\left(p_{3}\right)+\text { perm. }$$

$$

\langle 0|A_{\mu}^{a}(p_{1}) A_{\nu}^{b}(p_{2}) A_{\rho}^{c}(p_{3})| 0\rangle=

- (2\pi)^4 \delta^{(4)}(p_1+p_2+p_3)p_1^\lambda\eta_{\alpha\lambda}g f^{a^{\prime} b^{\prime} c^{\prime}} \int d^{4}p_1\,d^{4}p_2\, d^{4}p_3\,

\hat{G}_{\mu \beta}^{a a^{\prime}}\left(p_1\right)

\hat{G}_{\nu \alpha}^{b b^{\prime}}\left(p_{2}\right)

\hat{G}_{\rho \beta}^{c c^{\prime}}\left(p_{3}\right)e^{ip_1x_1+p_2x_2+p_3x_3}

+\text { perm.}

$$

$$f(x)= \int_{\mathbb{R}^{d+1}} d^{d+1}p\, \hat{f}(p) e^{ip\cdot x} \quad\Longleftrightarrow \quad \hat{f}(p)= \int_{\mathbb{R}^{d+1}} d^{d+1}x\, {f}(x) e^{-ip\cdot x}, $$

where ##p\cdot x := \eta_{\mu\nu}x^\mu p^\nu##. We also notice that

$$\int_{\mathbb{R}^{d+1}} d^{d+1}x\, \left(\partial_\mu{f}(x)\right) e^{-ip\cdot x} =-\int_{\mathbb{R}^{d+1}} d^{d+1}x\, {f}(x)(\partial_\mu e^{-ip\cdot x}) =ip_\mu\int_{\mathbb{R}^{d+1}} d^{d+1}x\, {f}(x)e^{-ip\cdot x} = ip_\mu \hat{f}(p).$$

I think these are all the ingredients we need to show the above statement. I would bscly. substitute the following into the given eq. above (assuming ##d=3##)

$$G_{\mu\nu}^{ab}(x_k-x)= \int_{\mathbb{R}^{4}} d^{4}p_k\, \hat{G}_{\mu\nu}^{ab}(p_k) e^{ip_k\cdot (x_k-x)},$$

which gives

$$\begin{align*}

\langle 0|A_{\mu}^{a}(p_{1}) &A_{\nu}^{b}(p_{2}) A_{\rho}^{c}(p_{3})| 0\rangle\\

&= -i g f^{a^{\prime} b^{\prime} c^{\prime}} \int d x

\left[\partial_{\alpha}^{x}\int_{\mathbb{R}^{4}} d^{4}p_1\, \hat{G}_{\mu \beta}^{a a^{\prime}}\left(p_1\right)e^{ip_1(x_1-x)}\right]

\int_{\mathbb{R}^{4}} d^{4}p_2\, \hat{G}_{\nu \alpha}^{b b^{\prime}}\left(p_{2}\right) e^{ip_2(x_2-x)}

\int_{\mathbb{R}^{4}} d^{4}p_3\, \hat{G}_{\rho \beta}^{c c^{\prime}}\left(p_{3}\right)e^{ip_3(x_3-x)}

+\text { perm.}\\

&=

-i g f^{a^{\prime} b^{\prime} c^{\prime}} \int d x\, d^{4}p_1\,d^{4}p_2\, d^{4}p_3\,

(-i(p_1)_\alpha)\hat{G}_{\mu \beta}^{a a^{\prime}}\left(p_1\right)e^{ip_1(x_1-x)}

\hat{G}_{\nu \alpha}^{b b^{\prime}}\left(p_{2}\right) e^{ip_2(x_2-x)}

\hat{G}_{\rho \beta}^{c c^{\prime}}\left(p_{3}\right)e^{ip_3(x_3-x)}

+\text { perm.}\\

&=

-p_1^\lambda\eta_{\alpha\lambda}g f^{a^{\prime} b^{\prime} c^{\prime}} \int d^{4}p_1\,d^{4}p_2\, d^{4}p_3\,

\hat{G}_{\mu \beta}^{a a^{\prime}}\left(p_1\right)

\hat{G}_{\nu \alpha}^{b b^{\prime}}\left(p_{2}\right)

\hat{G}_{\rho \beta}^{c c^{\prime}}\left(p_{3}\right)e^{ip_1x_1+p_2x_2+p_3x_3} \int dx\, e^{-i(p_1+p_2+p_3)x}

+\text { perm.}\\

&=

- (2\pi)^4 \delta^{(4)}(p_1+p_2+p_3)p_1^\lambda\eta_{\alpha\lambda}g f^{a^{\prime} b^{\prime} c^{\prime}} \int d^{4}p_1\,d^{4}p_2\, d^{4}p_3\,

\hat{G}_{\mu \beta}^{a a^{\prime}}\left(p_1\right)

\hat{G}_{\nu \alpha}^{b b^{\prime}}\left(p_{2}\right)

\hat{G}_{\rho \beta}^{c c^{\prime}}\left(p_{3}\right)e^{ip_1x_1+p_2x_2+p_3x_3}

+\text { perm.}\\

\end{align*}$$

This is as far as I can go and comparing it to the solution we see some key differences...

**Solution:**$$\left\langle 0\left|A_{\mu}^{a}\left(p_{1}\right) A_{\nu}^{b}\left(p_{2}\right) A_{\rho}^{c}\left(p_{3}\right)\right| 0\right\rangle=(2 \pi)^{4} \delta^{(4)}\left(p_{1}+p_{2}+p_{3}\right) g f^{a^{\prime} b^{\prime} c^{\prime}} p_{1}^{\beta} g^{\alpha \gamma} G_{\mu \alpha}^{a a^{\prime}}\left(p_{1}\right) G_{\nu \beta}^{b b^{\prime}}\left(p_{2}\right) G_{\rho \gamma}^{c c^{\prime}}\left(p_{3}\right)+\text { perm. }$$

**My version:**$$

\langle 0|A_{\mu}^{a}(p_{1}) A_{\nu}^{b}(p_{2}) A_{\rho}^{c}(p_{3})| 0\rangle=

- (2\pi)^4 \delta^{(4)}(p_1+p_2+p_3)p_1^\lambda\eta_{\alpha\lambda}g f^{a^{\prime} b^{\prime} c^{\prime}} \int d^{4}p_1\,d^{4}p_2\, d^{4}p_3\,

\hat{G}_{\mu \beta}^{a a^{\prime}}\left(p_1\right)

\hat{G}_{\nu \alpha}^{b b^{\prime}}\left(p_{2}\right)

\hat{G}_{\rho \beta}^{c c^{\prime}}\left(p_{3}\right)e^{ip_1x_1+p_2x_2+p_3x_3}

+\text { perm.}

$$

**Differences:**- the momentum integrals are missing
- the ##e^{ip_kx_k}## factor is missing
- the sign is wrong
- pretty much all of the indices are different and I'm not sure if the solutions just relabeled everything or if I made a mistake...