5
$\begingroup$

Why is the number of bosons not conserved while the number of fermions is conserved? Does it have something to do with the Pauli exclusion principle?

$\endgroup$
2
  • $\begingroup$ In electron-positron annihilation number of fermions is not conserved. $\endgroup$
    – Ruslan
    Jan 6, 2015 at 18:48
  • 1
    $\begingroup$ @Ruslan: In electron-positron annihilation fermion number is conserved. I think you are taking an overly literal interpretation of Roshan's question. $\endgroup$ Jan 7, 2015 at 11:12

3 Answers 3

7
$\begingroup$

If what you mean by conservation of "number of fermions" is that the number of fermions in the initial state must be equal to the number of fermions in the final state, well, it's wrong. Consider for example this nuclear reaction (which exists):

$$p + p \to \pi^+ + d$$

where $p\equiv$ proton (spin 1/2), $\pi^+$ pion (spin 0) and $d\equiv$ deuterium (spin 1), you see that there are 2 fermions in the initial state and none in the final. So no conservation of both the number of fermions and the number of bosons!

What is conserved by all reactions (so far…) are:

  1. Total angular momentum (and not only the spin)
  2. Energy/Momentum
  3. Total lepton number
  4. Baryon number
  5. electric charge

(plus the symmetry CPT but it's a bit different the context of this question).

Now, one can define the fermion number as $B+L$ where $B$ is the baryon number and $L$ the total lepton number. Its conservation is a consequence of the conservation of both $B$ and $L$ individually. But beware that fermion number doesn't mean "number of fermions".

$\endgroup$
7
  • 4
    $\begingroup$ I think your answer is disingenuous. Fermion number (as opposed to number of fermions) is conserved as far as we know. Why this should be is an interesting question that you have ignored. $\endgroup$ Jan 7, 2015 at 9:57
  • $\begingroup$ @John Rennie: can you define what you call fermion number? $\endgroup$
    – Paganini
    Jan 7, 2015 at 11:21
  • 1
    $\begingroup$ I suspect that you're referring to $B+L$ where $B$ is the baryon number and $L$ the total lepton number. Its conservation is a consequence of the conservation of both $B$ and $L$ individually. I have edited my answer accordingly. $\endgroup$
    – Paganini
    Jan 7, 2015 at 11:31
  • 1
    $\begingroup$ And you don't think this area could have been usefully explored in your answer? $\endgroup$ Jan 7, 2015 at 11:33
  • 1
    $\begingroup$ Cool, though if I were the OP I would now be asking why baryon and lepton number are conserved. $\endgroup$ Jan 7, 2015 at 11:37
0
$\begingroup$

I was just only making a comment, let me expand.The global charges of Fermion (f), baryon B) and Lepton (L) number are conserved in the standard model. But Global charges are not related to any gauge symmetry so there no strong reason for their conservation. In theories beyond the standard model they are predicted to be NOT conserved, though in some models , B-L charge Is conserved. Global charge conservation numbers must be violated to explain why our universe is mostly matter and not equal amounts of matter and anti matter which would give us a universe of mostly photons. Some models gauge B-L charge, these are left-right models, but these seem to be ruled out now by CERN. Beyond the standard models there is proton decay in SO(10)/SUSY SU(5) unification models and double beta neutrino-less decays which violate these global charges but none of this have been observed

$\endgroup$
1
  • $\begingroup$ "Some models gauge B-L charge... but these seem to be ruled out now by CERN." Can you please provide references on the "ruled out"? $\endgroup$
    – MadMax
    Mar 10, 2020 at 15:07
-3
$\begingroup$

With regard to Fermion number it gets strange in grand unification , because superforce bosons seem to carry the global charge of fermion number

$\endgroup$
2
  • $\begingroup$ This seems like a really incomplete answer in need of lots more detail. $\endgroup$ Dec 3, 2015 at 3:02
  • $\begingroup$ This is not an answer, more a comment. $\endgroup$
    – Gert
    Dec 3, 2015 at 3:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.