- #1
LCSphysicist
- 644
- 162
- Homework Statement
- Show that the Fourier transform of the triangle diagram in x space in Fig. 1.3b is the
star diagram in p space in Fig. 1.3c.
- Relevant Equations
- .
OBS: Ignore factors of ## (2 \pi) ##, interpret any differential ##dx,dp## as ##d^4x,d^4p##, ##\int = \int \int = \int ... \int##. I am using ##x,y,z## instead of ##x_i##.
Honestly, i am a little confused how to show this "triangle-star duality". Look, the propagators in positions space gives me ##\int \frac{e^{ip(x-y)}}{p^2+m^2} dp##
$$
\int d x d y d z dp_x dp_y dp_z \frac{1}{p_x^2+m^2} \frac{1}{p_y^2+m^2} \frac{1}{p_z^2+m^2} e^{i(p_x (x-y) + p_y (y-z) + p_z (z-x))} e^{-i(q_1 x + q_2 y + q_3 z)}
$$
$$
\int d y d z dp_x dp_y dp_z \frac{1}{p_x^2+m^2} \frac{1}{p_y^2+m^2} \frac{1}{p_z^2+m^2} e^{i(p_x (-y) + p_y (y-z) + p_z (z))} e^{-i( q_2 y + q_3 z)} \delta(p_x - p_z - q_1)
$$
$$
\int d y d z dp_y dp_z \frac{1}{(p_z+q_1)^2+m^2} \frac{1}{p_y^2+m^2} \frac{1}{p_z^2+m^2} e^{i((p_z + q_1) (-y) + p_y (y-z) + p_z (z))} e^{-i( q_2 y + q_3 z)}
$$
$$
\int d z dp_y dp_z \frac{1}{(q_2 - p_y)^2+m^2} \frac{1}{p_y^2+m^2} \frac{1}{p_z^2+m^2} e^{i( + p_y (-z) + p_z (z))} e^{-i( q_3 z)} \delta (-p_z - q_1 + p_y - q_2)
$$
$$
\int d z dp_z \frac{1}{(q_1+p_z)^2+m^2} \frac{1}{(q_2 + q_1 + p_z)^2+m^2} \frac{1}{p_z^2+m^2} e^{i( (q_2+q_1+p_z)(-z) + p_z (z))} e^{-i( q_3 z)}
$$
$$
\int dp_z \frac{1}{(q_1+p_z)^2+m^2} \frac{1}{(q_2 + q_1 + p_z)^2+m^2} \frac{1}{p_z^2+m^2} \delta(q_1+q_2+q_3)
$$
If there was no ##p_z## integral, i think the answer would be correct (the ##\delta## i got is an indication of it, i think). Where did i committed an error?
