Why can a particle decay into two photons but not one?












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I recently read an old physics news about the Higgs boson where it was observed to decay into 2 photons and I was wondering why it wouldn't have decayed into a single photon with the combined energy of 2 photons?






particle-physics photons momentum conservation-laws higgs







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  • $begingroup$
    Related: physics.stackexchange.com/q/427466/2451 and links therein.
    $endgroup$
    – Qmechanic♦
    2 days ago


















10












$begingroup$


I recently read an old physics news about the Higgs boson where it was observed to decay into 2 photons and I was wondering why it wouldn't have decayed into a single photon with the combined energy of 2 photons?






particle-physics photons momentum conservation-laws higgs







share|cite|improve this question











$endgroup$












  • $begingroup$
    Related: physics.stackexchange.com/q/427466/2451 and links therein.
    $endgroup$
    – Qmechanic♦
    2 days ago
















10












10








10


2



$begingroup$


I recently read an old physics news about the Higgs boson where it was observed to decay into 2 photons and I was wondering why it wouldn't have decayed into a single photon with the combined energy of 2 photons?






particle-physics photons momentum conservation-laws higgs







share|cite|improve this question











$endgroup$




I recently read an old physics news about the Higgs boson where it was observed to decay into 2 photons and I was wondering why it wouldn't have decayed into a single photon with the combined energy of 2 photons?






particle-physics photons momentum conservation-laws higgs




particle-physics photons momentum conservation-laws higgs






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share|cite|improve this question













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share|cite|improve this question








edited 2 days ago









Ben Crowell

49.2k4153294




49.2k4153294










asked Jan 12 at 3:42









user6760user6760

2,53611738




2,53611738












  • $begingroup$
    Related: physics.stackexchange.com/q/427466/2451 and links therein.
    $endgroup$
    – Qmechanic♦
    2 days ago




















  • $begingroup$
    Related: physics.stackexchange.com/q/427466/2451 and links therein.
    $endgroup$
    – Qmechanic♦
    2 days ago


















$begingroup$
Related: physics.stackexchange.com/q/427466/2451 and links therein.
$endgroup$
– Qmechanic♦
2 days ago






$begingroup$
Related: physics.stackexchange.com/q/427466/2451 and links therein.
$endgroup$
– Qmechanic♦
2 days ago












2 Answers
2






active

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34












$begingroup$

No massive particle can decay into a single photon.



In its rest frame, a particle with mass $M$ has momentum $p=0$. If it decayed to a single photon, conservation of energy would require the photon energy to be $E=Mc^2$, while conservation of momentum would require the photon to maintain $p=0$. However, photons obey $E=pc$ (which is the special case of $E^2 = (pc)^2 + (mc^2)^2$ for massless particles). It's not possible to satisfy all these constraints at once. Composite particles may emit single photons, but no massive particle may decay to a photon.






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  • $begingroup$
    Massive particle as in fermion with half integer spin right, so it have to decay into some other particles on top of a photon to conserve energy and spin momentum is this what you are saying?
    $endgroup$
    – user6760
    Jan 12 at 6:04








  • 3




    $begingroup$
    @user6760 This argument is about linear, not angular, momentum. And there are plenty of massive particles which obey Bose-Einstein statistics and have integer spins.
    $endgroup$
    – rob♦
    2 days ago





















17












$begingroup$

The Higgs boson has spin $0$. A photon has spin $1$. The total angular momentum cannot change in the decay, so a Higgs boson cannot decay into a single photon, regardless of the energy. But the total angular momentum of two photons can be zero (because their spins can be oriented in opposite directions), so this decay mode can conserve angular momentum.



As emphasized in a comment, conservation of angular momentum is only a necessary condition, not a sufficient one. Please see rob's answer for clarification about this.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    I just look up spin so spin can be negative
    $endgroup$
    – user6760
    Jan 12 at 4:14











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









34












$begingroup$

No massive particle can decay into a single photon.



In its rest frame, a particle with mass $M$ has momentum $p=0$. If it decayed to a single photon, conservation of energy would require the photon energy to be $E=Mc^2$, while conservation of momentum would require the photon to maintain $p=0$. However, photons obey $E=pc$ (which is the special case of $E^2 = (pc)^2 + (mc^2)^2$ for massless particles). It's not possible to satisfy all these constraints at once. Composite particles may emit single photons, but no massive particle may decay to a photon.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Massive particle as in fermion with half integer spin right, so it have to decay into some other particles on top of a photon to conserve energy and spin momentum is this what you are saying?
    $endgroup$
    – user6760
    Jan 12 at 6:04








  • 3




    $begingroup$
    @user6760 This argument is about linear, not angular, momentum. And there are plenty of massive particles which obey Bose-Einstein statistics and have integer spins.
    $endgroup$
    – rob♦
    2 days ago


















34












$begingroup$

No massive particle can decay into a single photon.



In its rest frame, a particle with mass $M$ has momentum $p=0$. If it decayed to a single photon, conservation of energy would require the photon energy to be $E=Mc^2$, while conservation of momentum would require the photon to maintain $p=0$. However, photons obey $E=pc$ (which is the special case of $E^2 = (pc)^2 + (mc^2)^2$ for massless particles). It's not possible to satisfy all these constraints at once. Composite particles may emit single photons, but no massive particle may decay to a photon.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Massive particle as in fermion with half integer spin right, so it have to decay into some other particles on top of a photon to conserve energy and spin momentum is this what you are saying?
    $endgroup$
    – user6760
    Jan 12 at 6:04








  • 3




    $begingroup$
    @user6760 This argument is about linear, not angular, momentum. And there are plenty of massive particles which obey Bose-Einstein statistics and have integer spins.
    $endgroup$
    – rob♦
    2 days ago
















34












34








34





$begingroup$

No massive particle can decay into a single photon.



In its rest frame, a particle with mass $M$ has momentum $p=0$. If it decayed to a single photon, conservation of energy would require the photon energy to be $E=Mc^2$, while conservation of momentum would require the photon to maintain $p=0$. However, photons obey $E=pc$ (which is the special case of $E^2 = (pc)^2 + (mc^2)^2$ for massless particles). It's not possible to satisfy all these constraints at once. Composite particles may emit single photons, but no massive particle may decay to a photon.






share|cite|improve this answer









$endgroup$



No massive particle can decay into a single photon.



In its rest frame, a particle with mass $M$ has momentum $p=0$. If it decayed to a single photon, conservation of energy would require the photon energy to be $E=Mc^2$, while conservation of momentum would require the photon to maintain $p=0$. However, photons obey $E=pc$ (which is the special case of $E^2 = (pc)^2 + (mc^2)^2$ for massless particles). It's not possible to satisfy all these constraints at once. Composite particles may emit single photons, but no massive particle may decay to a photon.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 12 at 5:37









rob♦rob

40k972165




40k972165












  • $begingroup$
    Massive particle as in fermion with half integer spin right, so it have to decay into some other particles on top of a photon to conserve energy and spin momentum is this what you are saying?
    $endgroup$
    – user6760
    Jan 12 at 6:04








  • 3




    $begingroup$
    @user6760 This argument is about linear, not angular, momentum. And there are plenty of massive particles which obey Bose-Einstein statistics and have integer spins.
    $endgroup$
    – rob♦
    2 days ago




















  • $begingroup$
    Massive particle as in fermion with half integer spin right, so it have to decay into some other particles on top of a photon to conserve energy and spin momentum is this what you are saying?
    $endgroup$
    – user6760
    Jan 12 at 6:04








  • 3




    $begingroup$
    @user6760 This argument is about linear, not angular, momentum. And there are plenty of massive particles which obey Bose-Einstein statistics and have integer spins.
    $endgroup$
    – rob♦
    2 days ago


















$begingroup$
Massive particle as in fermion with half integer spin right, so it have to decay into some other particles on top of a photon to conserve energy and spin momentum is this what you are saying?
$endgroup$
– user6760
Jan 12 at 6:04






$begingroup$
Massive particle as in fermion with half integer spin right, so it have to decay into some other particles on top of a photon to conserve energy and spin momentum is this what you are saying?
$endgroup$
– user6760
Jan 12 at 6:04






3




3




$begingroup$
@user6760 This argument is about linear, not angular, momentum. And there are plenty of massive particles which obey Bose-Einstein statistics and have integer spins.
$endgroup$
– rob♦
2 days ago






$begingroup$
@user6760 This argument is about linear, not angular, momentum. And there are plenty of massive particles which obey Bose-Einstein statistics and have integer spins.
$endgroup$
– rob♦
2 days ago













17












$begingroup$

The Higgs boson has spin $0$. A photon has spin $1$. The total angular momentum cannot change in the decay, so a Higgs boson cannot decay into a single photon, regardless of the energy. But the total angular momentum of two photons can be zero (because their spins can be oriented in opposite directions), so this decay mode can conserve angular momentum.



As emphasized in a comment, conservation of angular momentum is only a necessary condition, not a sufficient one. Please see rob's answer for clarification about this.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    I just look up spin so spin can be negative
    $endgroup$
    – user6760
    Jan 12 at 4:14
















17












$begingroup$

The Higgs boson has spin $0$. A photon has spin $1$. The total angular momentum cannot change in the decay, so a Higgs boson cannot decay into a single photon, regardless of the energy. But the total angular momentum of two photons can be zero (because their spins can be oriented in opposite directions), so this decay mode can conserve angular momentum.



As emphasized in a comment, conservation of angular momentum is only a necessary condition, not a sufficient one. Please see rob's answer for clarification about this.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    I just look up spin so spin can be negative
    $endgroup$
    – user6760
    Jan 12 at 4:14














17












17








17





$begingroup$

The Higgs boson has spin $0$. A photon has spin $1$. The total angular momentum cannot change in the decay, so a Higgs boson cannot decay into a single photon, regardless of the energy. But the total angular momentum of two photons can be zero (because their spins can be oriented in opposite directions), so this decay mode can conserve angular momentum.



As emphasized in a comment, conservation of angular momentum is only a necessary condition, not a sufficient one. Please see rob's answer for clarification about this.






share|cite|improve this answer











$endgroup$



The Higgs boson has spin $0$. A photon has spin $1$. The total angular momentum cannot change in the decay, so a Higgs boson cannot decay into a single photon, regardless of the energy. But the total angular momentum of two photons can be zero (because their spins can be oriented in opposite directions), so this decay mode can conserve angular momentum.



As emphasized in a comment, conservation of angular momentum is only a necessary condition, not a sufficient one. Please see rob's answer for clarification about this.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 2 days ago

























answered Jan 12 at 3:52









Dan YandDan Yand

8,33211234




8,33211234








  • 1




    $begingroup$
    I just look up spin so spin can be negative
    $endgroup$
    – user6760
    Jan 12 at 4:14














  • 1




    $begingroup$
    I just look up spin so spin can be negative
    $endgroup$
    – user6760
    Jan 12 at 4:14








1




1




$begingroup$
I just look up spin so spin can be negative
$endgroup$
– user6760
Jan 12 at 4:14




$begingroup$
I just look up spin so spin can be negative
$endgroup$
– user6760
Jan 12 at 4:14


















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