Reading about QFT diagrams, I've seen examples like Bhabha scattering where the channel $u$ wasn't necessary due to the final states are distinguisable for being made of the different particles and having $t$ was enough (see answer to: Why does not Bhabha scattering contain u-channel diagram?).

But my question is: when I have to decide between $t$ or $u$ diagram, which one should I pick? Would I obtain the same thing?

  • $\begingroup$ The rule is that you have to draw external legs before drawing the rest of the diagram. Then fill in the internals in all possible ways. You can see that in Bhabha example there are only two ways of doing it at tree level. The "u-channel diagram" has different order of external legs, so it is actually a mirage that it is a new diagram. $\endgroup$ – Peter Kravchuk Dec 4 '18 at 0:12

If you have:

$$ k + p \rightarrow k' + p' $$

where I'm just labeling their particles by the momenta, then the Mandelstam variables are:

$$ s = (k_{\mu} + p_{\mu})^2 $$

$$ t = (k_{\mu} - k'_{\mu})^2 = (p_{\mu} - p'_{\mu})^2$$

$$ u = (k_{\mu} - p'_{\mu})^2 = (p_{\mu} - k'_{\mu})^2$$

If $k$ and $p$ annihilate, as in:

$$ e^++e^-\rightarrow \mu^++\mu^-$$

then $s$ is invariant mass of the intermediate state.

This is also true if the process is:

$$ e^++e^-\rightarrow e^++e^-$$

However, you don't really know if the particles annihilated. They may have just scattered, in which case $t$ is the invariant mass of the exchanged virtual quanta. Note the symmetry between $k$ and $p$ here: you can't really say the positron emitted a virtual photon and the electron absorbed it, because the electron may have emitted it and the positron absorbed it. The Feynman diagram covers both cases (note: there are separate diagrams for $\gamma$ exchange and $Z^0$ exchange, but $t$ doesn't know the difference and you have to add the amplitudes of all the diagrams).

But I digress: because all positrons are indistinguishable, you don't know if you detected the same one from your beam, or a "new" one that was created from the decay of the $s$-channel intermediate state. Hence: you must add the s and t channel amplitudes.

There is also scattering without the possibility of annihilation, e.g. inelastic electron-proton scattering:

$$ e^- + p \rightarrow e^- + X $$

Here the only channel is the t-channel (ignoring the case where $X$ contains an electron--which I've never heard of). Note that now it is really tempting to say the electron emitted a virtual photon that was absorbed by the proton causing to to break up, but that is not correct. The virtual quanta was exchanged and that's it. (That $t < 0$ means it's space-like, so time-ordering its existence doesn't really make sense).

Finally, consider scattering identical particles, say a beam of ultra-relativistic electrons with momentum $k_{\mu} = (E/c, 0,0, E/c) $ off stationary electrons in a target piece of foil, with $p_{\mu} = (mc, 0, 0, 0)$:

$$ e^- + e^- \rightarrow e^- + e^- $$

Let's say you put a spectrometer at 90 degrees and tune it to $k'$. When you detect a scattered electron you do not now where it came from. Was it from the beam:

$$ t = (E/c-k',-k', 0, E/c)^2 $$

or from the target:

$$ u = (mc-k',-k', 0, 0)^2 $$

The correct answer is both. Any electron you detect is part from the beam and part from the target: this is the double-slit experiment, where the particle goes through both slits.

Hence you have to add the amplitudes for the t and u channel diagrams.

  • $\begingroup$ What do you mean by 'mass invariant' when you talk about diagrams? $\endgroup$ – Vicky Dec 4 '18 at 8:15
  • $\begingroup$ The magnitude of a 4-momenta: $p_{\mu}p^{mu} = E^2-p^2 = m^2$ is a Lorentz invariant. For real particles it's the rest mass, for virtual particles it also looks like a squared-mass, except that is negative in the t-channel. $\endgroup$ – JEB Dec 4 '18 at 20:38

It depends on the expression of the 4-momentum squared for the mediator particle, depending on how the Mandelstam variables are defined.

See: https://en.wikipedia.org/wiki/Mandelstam_variables

  • $\begingroup$ I have the same question. Would you please consider give more explanation concerning your answer? $\endgroup$ – gamebm Jan 9 at 13:39

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