# 4th order feynman-diagrams for $e^- + \mu^+ \to \nu_e + \bar{\nu}_\mu$

The task is to draw the two 4th-order Feynman diagrams of:

$$e^- + \mu^+ \to \nu_e + \bar{\nu}_\mu$$

I drew the first one as (time left->right):

   mu^+    \bar{\nu}_\mu    \nu_\mu
--<---------<----------<-----
/         /
/W^-      /Z^0
/         /
/         /
--->-------<--------<-----
e^-    \nu_e        \nu_e


which is correct. I drew the second one with a $W^+$ instead of a $W^-$, with the $W^+$ going the other direction (from the $mu^+$ to the $e^-$). However, the solutions gives the second diagram differently: rather, the order (in time) of the $W^-$ and the $Z^0$ is interchanged. I agree that this should be a valid feynman diagram as well for the process, but I don't understand why my alternative with the $W^+$ would be illegal. As far as I can see, it obeys all conservation laws.

Anyone?

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Welcome, Carl! We have the MathJax rendering engine active on the site which means that you can write equations as LaTeX. I've marked up you question that way, ad you can see how I did it my clicking the "edit" link (you can always cancel out). – dmckee Jul 31 '12 at 17:14
Oh, thanks for the help. I know, I've actually used the LaTeX engine before when asking questions here, don't know why it didn't come to my mind this time. Thanks anyway for the edits. (The diagram doesn't render the LaTeX since I needed fixed-width font, but maybe the LaTeX syntax is rather readable if you're used to it as well.) – Carl Jul 31 '12 at 18:35

A $W^-$ in one direction and a $W^+$ in the opposite direction are exactly the same thing. (That's what it means to be an antiparticle.)