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http://www.damtp.cam.ac.uk/user/tong/qft/qft.pdf

Why can we not have a u-channel diagram in Figure 13 on p64?

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- Thread starter latentcorpse
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http://www.damtp.cam.ac.uk/user/tong/qft/qft.pdf

Why can we not have a u-channel diagram in Figure 13 on p64?

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fzero

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I think I see why there is no s channel in the nucleon nucleon case. Is this because, if there were an s channel diagram, we would have 2 [itex]\psi[/itex] particles on the left of the 1st vertex and 0 [itex]\psi[/itex] particles on the right? Since the number of psi particles is a quantum number, it must be conserved and so this diagram is unphysical.

I tried applying a similar argument to the u channel diagram in the nucleon-antinucleon case but I just cannot see why it won't work! Can you elaborate please?

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fzero

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I think I see why there is no s channel in the nucleon nucleon case. Is this because, if there were an s channel diagram, we would have 2 [itex]\psi[/itex] particles on the left of the 1st vertex and 0 [itex]\psi[/itex] particles on the right? Since the number of psi particles is a quantum number, it must be conserved and so this diagram is unphysical.

Yes the interaction vertex conserves the nucleon number.

I tried applying a similar argument to the u channel diagram in the nucleon-antinucleon case but I just cannot see why it won't work! Can you elaborate please?

To form a u-channel diagram, you would need to have a vertex that turned a nucleon into an antinucleon, but this violates nucleon number.

We can also note that for the n-n interaction in Fig 9, the u-channel is just a permutation of the final state momenta when compared to the t-channel. This is possible because the nucleons are indistinguishable. In the nucleon-antinucleon case, we can't just swap the momenta because the states are distinguishable.

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