Exchange particles in two reactions

astenroo
Messages
47
Reaction score
0

Homework Statement



Hi all!

I was wondering in two Feynman diagrams (1 and 2)
1) in an electron positron annihilation two photons are created. Now, is the virtual electron the exchanged particle? I'm having problems understanding the diagrams.

2) A proton decays to a neutron as a mu-neutrino collide and a muon (u-) is emitted. As far as I know the exchange particle should be a Boson, but still I'm clueless as it comes to which. I'm ruling out the Z boson, and guessing it should be the W-.

Homework Equations



1. e+ + e- -> y (photons) exchange particle?


2. v(u) + n -> p + u-
Here a down quark decays to an up quark and the exchange particle should be the W- boson. Am I on the right track here at all?

The Attempt at a Solution



Some help would be much appreciated.

-Alex
 
Physics news on Phys.org
Hi Alex! :smile:
astenroo said:
1) in an electron positron annihilation two photons are created. Now, is the virtual electron the exchanged particle? I'm having problems understanding the diagrams.

the lowest-order feynman diagram (there are infinitely many higher order ones, of course) is the "H" diagram, with an electron and positron at the bottom, a line across the middle, and two photons at the top

yes, that line is a virtual electron (or positron, same thing) … that gives each vertex an electron in, an electron out (or positron in), and a photon :wink:
2) A proton decays to a neutron as a mu-neutrino collide and a muon (u-) is emitted. As far as I know the exchange particle should be a Boson, but still I'm clueless as it comes to which. I'm ruling out the Z boson, and guessing it should be the W-.

let's see …

yes, µ doesn't feel the strong interaction, so it has to be a Z or a W

and there's a change in charge, so it can't be a Z or W0 :smile:
 
Then to "conserve" charge it should be the W- boson :)
 
yup! :biggrin:
 
tiny-tim said:
yup! :biggrin:

Thank you for the help :biggrin:
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top