Mixing
Why can a particle decay into two photons but not one?
$begingroup$
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
edited Jan 13 at 0:36
Ben Crowell
50.4k6155298
asked Jan 12 at 3:42
user6760user6760
2,80111940
$endgroup$
add a comment |
$begingroup$
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
edited Jan 13 at 0:36
Ben Crowell
50.4k6155298
asked Jan 12 at 3:42
user6760user6760
2,80111940
$endgroup$
$begingroup$
Related: physics.stackexchange.com/q/427466/2451 and links therein.
$endgroup$
– Qmechanic♦
Jan 13 at 0:05
add a comment |
$begingroup$
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
edited Jan 13 at 0:36
Ben Crowell
50.4k6155298
asked Jan 12 at 3:42
user6760user6760
2,80111940
$endgroup$
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
edited Jan 13 at 0:36
Ben Crowell
50.4k6155298
asked Jan 12 at 3:42
user6760user6760
2,80111940
edited Jan 13 at 0:36
Ben Crowell
50.4k6155298
asked Jan 12 at 3:42
user6760user6760
2,80111940
edited Jan 13 at 0:36
Ben Crowell
50.4k6155298
edited Jan 13 at 0:36
Ben Crowell
50.4k6155298
edited Jan 13 at 0:36
Ben Crowell
50.4k6155298
50.4k6155298
asked Jan 12 at 3:42
user6760user6760
2,80111940
asked Jan 12 at 3:42
user6760user6760
2,80111940
asked Jan 12 at 3:42
user6760user6760
2,80111940
2,80111940
$begingroup$
Related: physics.stackexchange.com/q/427466/2451 and links therein.
$endgroup$
– Qmechanic♦
Jan 13 at 0:05
add a comment |
$begingroup$
Related: physics.stackexchange.com/q/427466/2451 and links therein.
$endgroup$
– Qmechanic♦
Jan 13 at 0:05
$begingroup$
Related: physics.stackexchange.com/q/427466/2451 and links therein.
$endgroup$
– Qmechanic♦
Jan 13 at 0:05
$begingroup$
Related: physics.stackexchange.com/q/427466/2451 and links therein.
$endgroup$
– Qmechanic♦
Jan 13 at 0:05
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
answered Jan 12 at 5:37
rob♦rob
40.7k972166
$endgroup$
$begingroup$
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♦
Jan 12 at 7:12
add a comment |
$begingroup$
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.
answered Jan 12 at 3:52
Dan YandDan Yand
10.2k21538
$endgroup$
1
$begingroup$
I just look up spin so spin can be negative
$endgroup$
– user6760
Jan 12 at 4:14
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "151"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f453657%2fwhy-can-a-particle-decay-into-two-photons-but-not-one%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered Jan 12 at 5:37
rob♦rob
40.7k972166
$endgroup$
$begingroup$
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♦
Jan 12 at 7:12
add a comment |
$begingroup$
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.
answered Jan 12 at 5:37
rob♦rob
40.7k972166
$endgroup$
$begingroup$
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♦
Jan 12 at 7:12
add a comment |
$begingroup$
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.
answered Jan 12 at 5:37
rob♦rob
40.7k972166
$endgroup$
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.
answered Jan 12 at 5:37
rob♦rob
40.7k972166
answered Jan 12 at 5:37
rob♦rob
40.7k972166
answered Jan 12 at 5:37
rob♦rob
40.7k972166
answered Jan 12 at 5:37
rob♦rob
40.7k972166
40.7k972166
$begingroup$
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♦
Jan 12 at 7:12
add a comment |
$begingroup$
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♦
Jan 12 at 7:12
$begingroup$
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♦
Jan 12 at 7:12
$begingroup$
@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♦
Jan 12 at 7:12
add a comment |
$begingroup$
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.
answered Jan 12 at 3:52
Dan YandDan Yand
10.2k21538
$endgroup$
1
$begingroup$
I just look up spin so spin can be negative
$endgroup$
– user6760
Jan 12 at 4:14
add a comment |
$begingroup$
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.
answered Jan 12 at 3:52
Dan YandDan Yand
10.2k21538
$endgroup$
1
$begingroup$
I just look up spin so spin can be negative
$endgroup$
– user6760
Jan 12 at 4:14
add a comment |
$begingroup$
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.
answered Jan 12 at 3:52
Dan YandDan Yand
10.2k21538
$endgroup$
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.
answered Jan 12 at 3:52
Dan YandDan Yand
10.2k21538
edited Jan 12 at 14:33
answered Jan 12 at 3:52
Dan YandDan Yand
10.2k21538
answered Jan 12 at 3:52
Dan YandDan Yand
10.2k21538
answered Jan 12 at 3:52
Dan YandDan Yand
10.2k21538
10.2k21538
1
$begingroup$
I just look up spin so spin can be negative
$endgroup$
– user6760
Jan 12 at 4:14
add a comment |
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
add a comment |
Thanks for contributing an answer to Physics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f453657%2fwhy-can-a-particle-decay-into-two-photons-but-not-one%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Related: physics.stackexchange.com/q/427466/2451 and links therein.
$endgroup$
– Qmechanic♦
Jan 13 at 0:05