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When calculating a simple diagram I came across an ambiguity in the conservation of momentum, i.e. it seems to me that the particle could come out of the process with opposite momentum with respect to its initial state. This might be a complete triviality, in that case I apologize, but I can't find my mistake. Take the following Feynman diagram in $\lambda \phi^4$ theoryenter image description here

Call $x$ and $y$ the two external points, $z_1$ and $z_2$ the two internal points. Using Feynman's rules in position space this is proportional to the integral

$$\int dz_1 dz_2 D_F(x-z_1)D_F(z_1-z_2)^2D_F(z_2-z_2)D_F(z_1-y) $$

Where $D_F$ is the Feynman propagator of the Klein Gordon field

$$ D_F(x-y)=\int\frac{d^4p}{(2\pi^4)}\frac{i}{p^2-m^2+i\varepsilon}e^{-ip(x-y)}$$

Inserting this into the diagram, and writing $dq=\frac{d^4p_x}{(2\pi)^4}\frac{d^4p_y}{(2\pi)^4}\frac{d^4p_1}{(2\pi)^4}\frac{d^4p_2}{(2\pi)^4}\frac{d^4p_3}{(2\pi)^4}$ for the sake of brevity

$$ \int dz_1 dz_2 D_F(x-z_1)D_F(z_1-z_2)^2D_F(z_2-z_2)D_F(z_2-y)=\\=\int dz_1 dz_2 \int dq \frac{e^{-ip_x(x-z_1)}}{p_x^2-m^2+i\varepsilon}\frac{e^{-ip_1(z_1-z_2)}}{p_1^2-m^2+i\varepsilon}\frac{e^{-ip_2(z_1-z_2)}}{p_2^2-m^2+i\varepsilon}\frac{1}{p_3^2-m^2+i\varepsilon}\frac{e^{-ip_y(z_1-y)}}{p_y^2-m^2+i\varepsilon}=\\=\int dz_1 dz_2 \int dq \frac{e^{-iz_1(-p_x+p_1+p_2+p_y)}}{p_x^2-m^2+i\varepsilon}\frac{e^{-iz_2(-p_1-p_2)}}{p_1^2-m^2+i\varepsilon}\frac{e^{-i(p_x x-p_y y)}}{p_2^2-m^2+i\varepsilon}\frac{1}{p_3^2-m^2+i\varepsilon}\frac{1}{p_y^2-m^2+i\varepsilon} $$

The momenta in the exponentials are gonna give $$ \delta^{(4)}(p_x-p_1-p_2-p_y)\delta^{(4)}(p_1+p_2)$$

Which implies conservation of momentum, and everyone is happy. But let's get back to the first expression of the integral, where I wrote

$$\int dz_1 dz_2 D_F(x-z_1)D_F(z_1-z_2)^2D_F(z_2-z_2)D_F(z_1-y) $$

I could have written

$$\int dz_1 dz_2 D_F(x-z_1)D_F(z_1-z_2)^2D_F(z_2-z_2)D_F(\color{red}{y-z_1}) $$

because the propagator is symmetric. This seemingly harmless change changes the exponential that yields the delta in

$$ e^{-iz_1(-p_x+p_1+p_2-p_y)}$$

and the conservation of momentum will be $$\delta^{(4)}(p_x-p_1-p_2 \color{red}{+p_y} )\delta^{(4)}(p_1+p_2)$$

which in the end implies $p_x=-p_y$, or, the particle comes out with opposite momentum, which seems weird to me. Have I made a mistake or is the sign of the momentum not important in some way?

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    $\begingroup$ Here $p_y$ is a dummy variable you are integrating over (which is why you could do the change in $D_F$), so you can change $p_y$ to $-p_y$. To really check momentum conservation, you need to fourier transform the full diagram with respect to x and y. $\endgroup$ – Adam Jan 6 at 22:13
  • $\begingroup$ @Adam Hi, thanks for the answer. So you mean that, because I can change variable in the integral, in this case the sign of $p_y$ has no relevance, but this doesn't have any physical meaning, right? What do you mean by "fourier transform the full diagram with respect to $x$ and $y$"? I can't really integrate further as I'll run into a divergence when integrating the second time ($\delta^{(4)}(p_1+p_2)$ becomes $\delta^{(4)}(0)$ when I integrate either of the two) $\endgroup$ – user2723984 Jan 7 at 8:20
  • $\begingroup$ See my answer, it should be clear now. $\endgroup$ – Adam Jan 7 at 9:31
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Let's call the OP's diagram $\Sigma(x-y)$, which can be rewritten as $$ \Sigma(x-y)=A\int dp_xdp_y \delta(p_x-p_y) e^{-i (p_x x-p_y y)}D_F(p_x)D_F(p_y), $$ where $A$ is a constant which comes from the two loops contained in the diagram, and $D_F(p)=(p^2-m^2+i\epsilon)^{-1}$.

Playing the game of the second part of the question (i.e. replacing $D_F(z_1-y)$ by $D_F(y-z_1)$), we would end up with $$ \Sigma(x-y)=A\int dp_xdp_y \delta(p_x+p_y) e^{-i (p_x x+p_y y)}D_F(p_x)D_F(p_y), $$ which is obviously equivalent to the first expression, by the change of variable $p_y\to-p_y$. There is therefore no contradiction, and both way of doing the calculation are equivalent.

What confuses the OP is the question of momentum conservation, which looks strange in the second equation. This is because we should look at momentum conservation in momentum space (whereas we are still in real space here), i.e. we should compute $ \Sigma(p_1,p_2)=\int dx dy e^{ip_1 x-ip_2 y}\Sigma(x-y)$.

Starting from either expression, we obtain $$ \Sigma(p_1,p_2)=\delta(p_1-p_2) A D_F(p_1)D_F(p_2),$$ which obviously conserve momentum. One usually factors out the delta function and introduce the function $\Sigma(p)=A D_F(p)D_F(p)$ such that $\Sigma(p_x,p_y)=\delta(p_x-p_y)\Sigma(p_x)$.

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  • $\begingroup$ Thank you! Though, I'm still not sure. You say that starting from either expression we get that, but I tried starting with the expression with the plus and I got $\delta(p_1+p_2)$, which looks like the same problem. Plus you only get $\delta(p_1-p_2)$ in the minus case because you chose an appropriate convention for the Fourier transform, i.e. $ip_1 x-ip_2y$ but any other sign choice would have been equivalent, and would have yielded a different sign in the $\delta$. And another point, when you integrated on $dp_1 dp_2$, what happened to the resulting $\delta(0)$? $\endgroup$ – user2723984 Jan 7 at 11:42
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    $\begingroup$ @user2723984 The FT convention is unambiguous, it is the one consistent with your convention to go from $D_F(z)$ to $D_F(p)$. Do the FT starting from both expression carefully, you'll see that all is fine (be careful with all the signs in front of all $p_y$ !)). I don't integrate over $p_1$ or $p_2$, so I don't understand your problem with that. $\endgroup$ – Adam Jan 7 at 13:07
  • $\begingroup$ ah, you're right! They turn out to be equal. This is like magic! So the sign ambiguity in the position space integral is irrelevant, and actually all the signs of the momenta in the position space integral are irrelevant. I guess it's nevertheless good custom to arrange them so that they make physical sense. I'll accept your answer right away, but as a bonus point, does this "freedom" of choice of the momentum direction have any physical interpretation or consequence? I guess it boils down to the symmetry of the propagator in the momentum, but why is that? $\endgroup$ – user2723984 Jan 7 at 13:23
  • $\begingroup$ @user2723984 It is not that the signs are irrelevant, it is just that they have to be consistent. Start in real space (which is the physical one), pick a Fourier convention (whatever it is), and all is consistent. However, depending on the Fourier convention, conservation of momentum might read as $\delta(p_1-p_2)$ or $\delta(p_1+p_2)$. What really matters here (and is independent of the convention), is that the propagators and the interaction are translation invariant (as well as parity invariant $x\to-x$), which translates into conservation of momentum and invariance under $p\to-p$. $\endgroup$ – Adam Jan 7 at 13:51

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