How to translate expression into momentum-space correctly

• Markus Kahn
In summary: Apart from that, I don't see any major mistakes, but I'm not an expert in this area. In summary, the conversation discusses the Fourier transformation of a nicely behaved function, detailing the necessary ingredients and steps to show a given statement. Differences between the two versions of the solution are identified and addressed.
Markus Kahn
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...
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...
Could somebody maybe look over this and tell me (if) what I'm doing wrong?

One mistake (or typo, whatever you want to call it) is the exponent in the third line of yours should be ##\exp(i(p_1x_1+p_2x_2+p_3x_3))## and not ##\exp(ip_1x_1+p_2x_2+p_3x_3)##, so it seems you are missing paranatheses.

1. What is momentum-space and why is it important in scientific research?

Momentum-space is a mathematical representation of the momentum of particles in a system. It is important in scientific research because it allows for a more precise and accurate description of the behavior of particles, particularly in quantum mechanics.

2. How do I translate an expression into momentum-space correctly?

To translate an expression into momentum-space, you will need to use the Fourier transform. This involves converting the expression from position-space to momentum-space using a specific mathematical formula. It is important to follow the correct steps and use the appropriate variables to ensure accuracy.

3. Can I use any expression in momentum-space or are there limitations?

There are limitations to the types of expressions that can be translated into momentum-space. Generally, only expressions that are continuous and have a finite energy can be translated accurately. Additionally, certain mathematical operations, such as differentiation, may affect the translation and should be considered carefully.

4. What are some common mistakes to avoid when translating an expression into momentum-space?

One common mistake is not using the correct variables in the Fourier transform. It is important to use the correct variables for position and momentum to ensure an accurate translation. Another mistake is not taking into account the limitations mentioned earlier, which can result in incorrect calculations and interpretations.

5. How can I check if my translation into momentum-space is correct?

One way to check the accuracy of your translation is by performing an inverse Fourier transform. This will convert the expression back to position-space, allowing you to compare it to the original expression. Additionally, you can consult with other scientists or use mathematical software to verify your results.

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