Why do bosons tend to occupy the same state?
$begingroup$
It is often said that, while many fermions cannot occupy the same state, bosons have the tendency to do that. Sometimes this is expressed figuratively by saying, for example, that "bosons are sociable" or that "bosons want to stay as close as possible".
I understand that the symmetry of the wavefunction allows many bosons to be in the same one-particle state, but I can't see why they should prefer to do that rather than occupying different states.
Anyway, according to many science writers, bosons not only can be in the same state, but they also tend to do that. Why is it like that?
quantum-mechanics bosons
asked 12 hours ago
HicHaecHocHicHaecHoc
957
$endgroup$
add a comment |
$begingroup$
It is often said that, while many fermions cannot occupy the same state, bosons have the tendency to do that. Sometimes this is expressed figuratively by saying, for example, that "bosons are sociable" or that "bosons want to stay as close as possible".
I understand that the symmetry of the wavefunction allows many bosons to be in the same one-particle state, but I can't see why they should prefer to do that rather than occupying different states.
Anyway, according to many science writers, bosons not only can be in the same state, but they also tend to do that. Why is it like that?
quantum-mechanics bosons
asked 12 hours ago
HicHaecHocHicHaecHoc
957
$endgroup$
add a comment |
$begingroup$
It is often said that, while many fermions cannot occupy the same state, bosons have the tendency to do that. Sometimes this is expressed figuratively by saying, for example, that "bosons are sociable" or that "bosons want to stay as close as possible".
I understand that the symmetry of the wavefunction allows many bosons to be in the same one-particle state, but I can't see why they should prefer to do that rather than occupying different states.
Anyway, according to many science writers, bosons not only can be in the same state, but they also tend to do that. Why is it like that?
quantum-mechanics bosons
asked 12 hours ago
HicHaecHocHicHaecHoc
957
$endgroup$
It is often said that, while many fermions cannot occupy the same state, bosons have the tendency to do that. Sometimes this is expressed figuratively by saying, for example, that "bosons are sociable" or that "bosons want to stay as close as possible".
I understand that the symmetry of the wavefunction allows many bosons to be in the same one-particle state, but I can't see why they should prefer to do that rather than occupying different states.
Anyway, according to many science writers, bosons not only can be in the same state, but they also tend to do that. Why is it like that?
quantum-mechanics bosons
quantum-mechanics bosons
asked 12 hours ago
HicHaecHocHicHaecHoc
957
asked 12 hours ago
HicHaecHocHicHaecHoc
957
asked 12 hours ago
HicHaecHocHicHaecHoc
957
asked 12 hours ago
HicHaecHocHicHaecHoc
957
asked 12 hours ago
HicHaecHocHicHaecHoc
957
957
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Suppose you have two distinguishable coins that can either come up heads or tails. Then there are four equally likely possibilities,
$$text{HH}, text{HT}, text{TH}, text{TT}.$$
There is a 50% chance for the two coins to have the same result.
If the coins were fermions and "heads/tails" were quantum modes, the $text{HH}$ and $text{TT}$ states wouldn't be allowed, so there is a 0% chance for the two coins to have the same result.
If the coins were bosons, then all states are allowed. But there's a twist: bosons are identical particles. The states $text{HT}$ and $text{TH}$ are precisely the same state, namely the one with one particle in each of the two modes. So there are three possibilities,
$$text{two heads}, text{two tails}, text{one each}$$
and hence in the microcanonical ensemble (where each distinct quantum state is equally probable) there is a $2/3$ chance of double occupancy, not $1/2$. That's what people mean when they say bosons "clump up", though it's not really a consequence of bosonic statistics, just a consequence of the particles being identical. Whenever a system of bosonic particles is in thermal equilibrium, there exist fewer states with the bosons apart than you would naively expect, if you treated them as distinguishable particles, so you are more likely to see them together.
answered 11 hours ago
knzhouknzhou
45.9k11123222
$endgroup$
1
$begingroup$
@HicHaecHoc Yes, but compare this to the naive non-identical particle reasoning: in that case there are $k$ states where they are the same, but $k(k-1)$ when they aren't. For the bosons, this latter number is cut in half.
$endgroup$
– knzhou
11 hours ago
1
$begingroup$
@HicHaecHoc The statement is not that any two bosons are more than 50% likely to be in the same state, it's that you get more clumping with bosons than with distinguishable particles when you average over all states.
$endgroup$
– knzhou
11 hours ago
1
$begingroup$
Nevertheless, now I think that the common statements as "bosons want to stay close" and the like are a little bit misleading without some explanation.
$endgroup$
– HicHaecHoc
10 hours ago
3
$begingroup$
@HicHaecHoc Popularization is always like that. I think virtual particles are even more misleading, but when my well-intentioned little sister (with zero physics background) wants to know what I'm studying, you better bet I'm weaving a fantastic tale of how virtual particles crash into each other.
$endgroup$
– knzhou
10 hours ago
1
$begingroup$
In coming up with the 2/3 value you make a logical leap in assuming equiprobability of events. It's sort of along the same lines as "Either Bill Gates will give me a million dollars tonight or he won't. Therefore, there's a 50% chance I'll be a millionaire tomorrow!" Now, in this particular case equiprobability of the three states may be correct (versus "one each" being twice as likely as either of the others), but the answer could be improved by more directly addressing it, rather treating it in a fashion which normally would be a probabilistic fallacy.
$endgroup$
– R.M.
4 hours ago
|
show 6 more comments
$begingroup$
Anyway, according to many science writers, bosons not only can be in the same state, but they also tend to do that. Why is it like that?
That's simply wrong. It's one of many clichés dear to popular science writers but with little physical content. What's true, as you said, is that bosons can occupy the same state, as opposed to fermions. As to tendency, it has the same half-truth as when one says "a system tends
to stay in its ground state".
In fact assume the latter statement is inconditionally true and you have an ensemble of non-interacting bosons. Then each boson will tend to occupy its ground state and - since nothing forbids that - all particles will sit there. In presence of interaction things may go different or not depending on the interaction.
But the real issue is if there is the tendency of a system to stay in its ground state or go into it if initially placed in a state of higher energy. However this is a question not easy to deal with in few words.
answered 11 hours ago
Elio FabriElio Fabri
3,4001214
$endgroup$
1
$begingroup$
No, bosons do actually 'clump' relative to classical Maxwell-Boltzmann particles, at least under some circumstances. A nice example is the Hong-Ou-Mandel effect: en.wikipedia.org/wiki/Hong%E2%80%93Ou%E2%80%93Mandel_effect
$endgroup$
– Rococo
11 hours ago
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%2f470316%2fwhy-do-bosons-tend-to-occupy-the-same-state%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
$begingroup$
Suppose you have two distinguishable coins that can either come up heads or tails. Then there are four equally likely possibilities,
$$text{HH}, text{HT}, text{TH}, text{TT}.$$
There is a 50% chance for the two coins to have the same result.
If the coins were fermions and "heads/tails" were quantum modes, the $text{HH}$ and $text{TT}$ states wouldn't be allowed, so there is a 0% chance for the two coins to have the same result.
If the coins were bosons, then all states are allowed. But there's a twist: bosons are identical particles. The states $text{HT}$ and $text{TH}$ are precisely the same state, namely the one with one particle in each of the two modes. So there are three possibilities,
$$text{two heads}, text{two tails}, text{one each}$$
and hence in the microcanonical ensemble (where each distinct quantum state is equally probable) there is a $2/3$ chance of double occupancy, not $1/2$. That's what people mean when they say bosons "clump up", though it's not really a consequence of bosonic statistics, just a consequence of the particles being identical. Whenever a system of bosonic particles is in thermal equilibrium, there exist fewer states with the bosons apart than you would naively expect, if you treated them as distinguishable particles, so you are more likely to see them together.
answered 11 hours ago
knzhouknzhou
45.9k11123222
$endgroup$
1
$begingroup$
@HicHaecHoc Yes, but compare this to the naive non-identical particle reasoning: in that case there are $k$ states where they are the same, but $k(k-1)$ when they aren't. For the bosons, this latter number is cut in half.
$endgroup$
– knzhou
11 hours ago
1
$begingroup$
@HicHaecHoc The statement is not that any two bosons are more than 50% likely to be in the same state, it's that you get more clumping with bosons than with distinguishable particles when you average over all states.
$endgroup$
– knzhou
11 hours ago
1
$begingroup$
Nevertheless, now I think that the common statements as "bosons want to stay close" and the like are a little bit misleading without some explanation.
$endgroup$
– HicHaecHoc
10 hours ago
3
$begingroup$
@HicHaecHoc Popularization is always like that. I think virtual particles are even more misleading, but when my well-intentioned little sister (with zero physics background) wants to know what I'm studying, you better bet I'm weaving a fantastic tale of how virtual particles crash into each other.
$endgroup$
– knzhou
10 hours ago
1
$begingroup$
In coming up with the 2/3 value you make a logical leap in assuming equiprobability of events. It's sort of along the same lines as "Either Bill Gates will give me a million dollars tonight or he won't. Therefore, there's a 50% chance I'll be a millionaire tomorrow!" Now, in this particular case equiprobability of the three states may be correct (versus "one each" being twice as likely as either of the others), but the answer could be improved by more directly addressing it, rather treating it in a fashion which normally would be a probabilistic fallacy.
$endgroup$
– R.M.
4 hours ago
|
show 6 more comments
$begingroup$
Suppose you have two distinguishable coins that can either come up heads or tails. Then there are four equally likely possibilities,
$$text{HH}, text{HT}, text{TH}, text{TT}.$$
There is a 50% chance for the two coins to have the same result.
If the coins were fermions and "heads/tails" were quantum modes, the $text{HH}$ and $text{TT}$ states wouldn't be allowed, so there is a 0% chance for the two coins to have the same result.
If the coins were bosons, then all states are allowed. But there's a twist: bosons are identical particles. The states $text{HT}$ and $text{TH}$ are precisely the same state, namely the one with one particle in each of the two modes. So there are three possibilities,
$$text{two heads}, text{two tails}, text{one each}$$
and hence in the microcanonical ensemble (where each distinct quantum state is equally probable) there is a $2/3$ chance of double occupancy, not $1/2$. That's what people mean when they say bosons "clump up", though it's not really a consequence of bosonic statistics, just a consequence of the particles being identical. Whenever a system of bosonic particles is in thermal equilibrium, there exist fewer states with the bosons apart than you would naively expect, if you treated them as distinguishable particles, so you are more likely to see them together.
answered 11 hours ago
knzhouknzhou
45.9k11123222
$endgroup$
1
$begingroup$
@HicHaecHoc Yes, but compare this to the naive non-identical particle reasoning: in that case there are $k$ states where they are the same, but $k(k-1)$ when they aren't. For the bosons, this latter number is cut in half.
$endgroup$
– knzhou
11 hours ago
1
$begingroup$
@HicHaecHoc The statement is not that any two bosons are more than 50% likely to be in the same state, it's that you get more clumping with bosons than with distinguishable particles when you average over all states.
$endgroup$
– knzhou
11 hours ago
1
$begingroup$
Nevertheless, now I think that the common statements as "bosons want to stay close" and the like are a little bit misleading without some explanation.
$endgroup$
– HicHaecHoc
10 hours ago
3
$begingroup$
@HicHaecHoc Popularization is always like that. I think virtual particles are even more misleading, but when my well-intentioned little sister (with zero physics background) wants to know what I'm studying, you better bet I'm weaving a fantastic tale of how virtual particles crash into each other.
$endgroup$
– knzhou
10 hours ago
1
$begingroup$
In coming up with the 2/3 value you make a logical leap in assuming equiprobability of events. It's sort of along the same lines as "Either Bill Gates will give me a million dollars tonight or he won't. Therefore, there's a 50% chance I'll be a millionaire tomorrow!" Now, in this particular case equiprobability of the three states may be correct (versus "one each" being twice as likely as either of the others), but the answer could be improved by more directly addressing it, rather treating it in a fashion which normally would be a probabilistic fallacy.
$endgroup$
– R.M.
4 hours ago
|
show 6 more comments
$begingroup$
Suppose you have two distinguishable coins that can either come up heads or tails. Then there are four equally likely possibilities,
$$text{HH}, text{HT}, text{TH}, text{TT}.$$
There is a 50% chance for the two coins to have the same result.
If the coins were fermions and "heads/tails" were quantum modes, the $text{HH}$ and $text{TT}$ states wouldn't be allowed, so there is a 0% chance for the two coins to have the same result.
If the coins were bosons, then all states are allowed. But there's a twist: bosons are identical particles. The states $text{HT}$ and $text{TH}$ are precisely the same state, namely the one with one particle in each of the two modes. So there are three possibilities,
$$text{two heads}, text{two tails}, text{one each}$$
and hence in the microcanonical ensemble (where each distinct quantum state is equally probable) there is a $2/3$ chance of double occupancy, not $1/2$. That's what people mean when they say bosons "clump up", though it's not really a consequence of bosonic statistics, just a consequence of the particles being identical. Whenever a system of bosonic particles is in thermal equilibrium, there exist fewer states with the bosons apart than you would naively expect, if you treated them as distinguishable particles, so you are more likely to see them together.
answered 11 hours ago
knzhouknzhou
45.9k11123222
$endgroup$
Suppose you have two distinguishable coins that can either come up heads or tails. Then there are four equally likely possibilities,
$$text{HH}, text{HT}, text{TH}, text{TT}.$$
There is a 50% chance for the two coins to have the same result.
If the coins were fermions and "heads/tails" were quantum modes, the $text{HH}$ and $text{TT}$ states wouldn't be allowed, so there is a 0% chance for the two coins to have the same result.
If the coins were bosons, then all states are allowed. But there's a twist: bosons are identical particles. The states $text{HT}$ and $text{TH}$ are precisely the same state, namely the one with one particle in each of the two modes. So there are three possibilities,
$$text{two heads}, text{two tails}, text{one each}$$
and hence in the microcanonical ensemble (where each distinct quantum state is equally probable) there is a $2/3$ chance of double occupancy, not $1/2$. That's what people mean when they say bosons "clump up", though it's not really a consequence of bosonic statistics, just a consequence of the particles being identical. Whenever a system of bosonic particles is in thermal equilibrium, there exist fewer states with the bosons apart than you would naively expect, if you treated them as distinguishable particles, so you are more likely to see them together.
answered 11 hours ago
knzhouknzhou
45.9k11123222
edited 4 hours ago
answered 11 hours ago
knzhouknzhou
45.9k11123222
answered 11 hours ago
knzhouknzhou
45.9k11123222
answered 11 hours ago
knzhouknzhou
45.9k11123222
45.9k11123222
1
$begingroup$
@HicHaecHoc Yes, but compare this to the naive non-identical particle reasoning: in that case there are $k$ states where they are the same, but $k(k-1)$ when they aren't. For the bosons, this latter number is cut in half.
$endgroup$
– knzhou
11 hours ago
1
$begingroup$
@HicHaecHoc The statement is not that any two bosons are more than 50% likely to be in the same state, it's that you get more clumping with bosons than with distinguishable particles when you average over all states.
$endgroup$
– knzhou
11 hours ago
1
$begingroup$
Nevertheless, now I think that the common statements as "bosons want to stay close" and the like are a little bit misleading without some explanation.
$endgroup$
– HicHaecHoc
10 hours ago
3
$begingroup$
@HicHaecHoc Popularization is always like that. I think virtual particles are even more misleading, but when my well-intentioned little sister (with zero physics background) wants to know what I'm studying, you better bet I'm weaving a fantastic tale of how virtual particles crash into each other.
$endgroup$
– knzhou
10 hours ago
1
$begingroup$
In coming up with the 2/3 value you make a logical leap in assuming equiprobability of events. It's sort of along the same lines as "Either Bill Gates will give me a million dollars tonight or he won't. Therefore, there's a 50% chance I'll be a millionaire tomorrow!" Now, in this particular case equiprobability of the three states may be correct (versus "one each" being twice as likely as either of the others), but the answer could be improved by more directly addressing it, rather treating it in a fashion which normally would be a probabilistic fallacy.
$endgroup$
– R.M.
4 hours ago
|
show 6 more comments
1
$begingroup$
@HicHaecHoc Yes, but compare this to the naive non-identical particle reasoning: in that case there are $k$ states where they are the same, but $k(k-1)$ when they aren't. For the bosons, this latter number is cut in half.
$endgroup$
– knzhou
11 hours ago
1
$begingroup$
@HicHaecHoc The statement is not that any two bosons are more than 50% likely to be in the same state, it's that you get more clumping with bosons than with distinguishable particles when you average over all states.
$endgroup$
– knzhou
11 hours ago
1
$begingroup$
Nevertheless, now I think that the common statements as "bosons want to stay close" and the like are a little bit misleading without some explanation.
$endgroup$
– HicHaecHoc
10 hours ago
3
$begingroup$
@HicHaecHoc Popularization is always like that. I think virtual particles are even more misleading, but when my well-intentioned little sister (with zero physics background) wants to know what I'm studying, you better bet I'm weaving a fantastic tale of how virtual particles crash into each other.
$endgroup$
– knzhou
10 hours ago
1
$begingroup$
In coming up with the 2/3 value you make a logical leap in assuming equiprobability of events. It's sort of along the same lines as "Either Bill Gates will give me a million dollars tonight or he won't. Therefore, there's a 50% chance I'll be a millionaire tomorrow!" Now, in this particular case equiprobability of the three states may be correct (versus "one each" being twice as likely as either of the others), but the answer could be improved by more directly addressing it, rather treating it in a fashion which normally would be a probabilistic fallacy.
$endgroup$
– R.M.
4 hours ago
1
1
$begingroup$
@HicHaecHoc Yes, but compare this to the naive non-identical particle reasoning: in that case there are $k$ states where they are the same, but $k(k-1)$ when they aren't. For the bosons, this latter number is cut in half.
$endgroup$
– knzhou
11 hours ago
$begingroup$
@HicHaecHoc Yes, but compare this to the naive non-identical particle reasoning: in that case there are $k$ states where they are the same, but $k(k-1)$ when they aren't. For the bosons, this latter number is cut in half.
$endgroup$
– knzhou
11 hours ago
1
1
$begingroup$
@HicHaecHoc The statement is not that any two bosons are more than 50% likely to be in the same state, it's that you get more clumping with bosons than with distinguishable particles when you average over all states.
$endgroup$
– knzhou
11 hours ago
$begingroup$
@HicHaecHoc The statement is not that any two bosons are more than 50% likely to be in the same state, it's that you get more clumping with bosons than with distinguishable particles when you average over all states.
$endgroup$
– knzhou
11 hours ago
1
1
$begingroup$
Nevertheless, now I think that the common statements as "bosons want to stay close" and the like are a little bit misleading without some explanation.
$endgroup$
– HicHaecHoc
10 hours ago
$begingroup$
Nevertheless, now I think that the common statements as "bosons want to stay close" and the like are a little bit misleading without some explanation.
$endgroup$
– HicHaecHoc
10 hours ago
3
3
$begingroup$
@HicHaecHoc Popularization is always like that. I think virtual particles are even more misleading, but when my well-intentioned little sister (with zero physics background) wants to know what I'm studying, you better bet I'm weaving a fantastic tale of how virtual particles crash into each other.
$endgroup$
– knzhou
10 hours ago
$begingroup$
@HicHaecHoc Popularization is always like that. I think virtual particles are even more misleading, but when my well-intentioned little sister (with zero physics background) wants to know what I'm studying, you better bet I'm weaving a fantastic tale of how virtual particles crash into each other.
$endgroup$
– knzhou
10 hours ago
1
1
$begingroup$
In coming up with the 2/3 value you make a logical leap in assuming equiprobability of events. It's sort of along the same lines as "Either Bill Gates will give me a million dollars tonight or he won't. Therefore, there's a 50% chance I'll be a millionaire tomorrow!" Now, in this particular case equiprobability of the three states may be correct (versus "one each" being twice as likely as either of the others), but the answer could be improved by more directly addressing it, rather treating it in a fashion which normally would be a probabilistic fallacy.
$endgroup$
– R.M.
4 hours ago
$begingroup$
In coming up with the 2/3 value you make a logical leap in assuming equiprobability of events. It's sort of along the same lines as "Either Bill Gates will give me a million dollars tonight or he won't. Therefore, there's a 50% chance I'll be a millionaire tomorrow!" Now, in this particular case equiprobability of the three states may be correct (versus "one each" being twice as likely as either of the others), but the answer could be improved by more directly addressing it, rather treating it in a fashion which normally would be a probabilistic fallacy.
$endgroup$
– R.M.
4 hours ago
|
show 6 more comments
$begingroup$
Anyway, according to many science writers, bosons not only can be in the same state, but they also tend to do that. Why is it like that?
That's simply wrong. It's one of many clichés dear to popular science writers but with little physical content. What's true, as you said, is that bosons can occupy the same state, as opposed to fermions. As to tendency, it has the same half-truth as when one says "a system tends
to stay in its ground state".
In fact assume the latter statement is inconditionally true and you have an ensemble of non-interacting bosons. Then each boson will tend to occupy its ground state and - since nothing forbids that - all particles will sit there. In presence of interaction things may go different or not depending on the interaction.
But the real issue is if there is the tendency of a system to stay in its ground state or go into it if initially placed in a state of higher energy. However this is a question not easy to deal with in few words.
answered 11 hours ago
Elio FabriElio Fabri
3,4001214
$endgroup$
1
$begingroup$
No, bosons do actually 'clump' relative to classical Maxwell-Boltzmann particles, at least under some circumstances. A nice example is the Hong-Ou-Mandel effect: en.wikipedia.org/wiki/Hong%E2%80%93Ou%E2%80%93Mandel_effect
$endgroup$
– Rococo
11 hours ago
add a comment |
$begingroup$
Anyway, according to many science writers, bosons not only can be in the same state, but they also tend to do that. Why is it like that?
That's simply wrong. It's one of many clichés dear to popular science writers but with little physical content. What's true, as you said, is that bosons can occupy the same state, as opposed to fermions. As to tendency, it has the same half-truth as when one says "a system tends
to stay in its ground state".
In fact assume the latter statement is inconditionally true and you have an ensemble of non-interacting bosons. Then each boson will tend to occupy its ground state and - since nothing forbids that - all particles will sit there. In presence of interaction things may go different or not depending on the interaction.
But the real issue is if there is the tendency of a system to stay in its ground state or go into it if initially placed in a state of higher energy. However this is a question not easy to deal with in few words.
answered 11 hours ago
Elio FabriElio Fabri
3,4001214
$endgroup$
1
$begingroup$
No, bosons do actually 'clump' relative to classical Maxwell-Boltzmann particles, at least under some circumstances. A nice example is the Hong-Ou-Mandel effect: en.wikipedia.org/wiki/Hong%E2%80%93Ou%E2%80%93Mandel_effect
$endgroup$
– Rococo
11 hours ago
add a comment |
$begingroup$
Anyway, according to many science writers, bosons not only can be in the same state, but they also tend to do that. Why is it like that?
That's simply wrong. It's one of many clichés dear to popular science writers but with little physical content. What's true, as you said, is that bosons can occupy the same state, as opposed to fermions. As to tendency, it has the same half-truth as when one says "a system tends
to stay in its ground state".
In fact assume the latter statement is inconditionally true and you have an ensemble of non-interacting bosons. Then each boson will tend to occupy its ground state and - since nothing forbids that - all particles will sit there. In presence of interaction things may go different or not depending on the interaction.
But the real issue is if there is the tendency of a system to stay in its ground state or go into it if initially placed in a state of higher energy. However this is a question not easy to deal with in few words.
answered 11 hours ago
Elio FabriElio Fabri
3,4001214
$endgroup$
Anyway, according to many science writers, bosons not only can be in the same state, but they also tend to do that. Why is it like that?
That's simply wrong. It's one of many clichés dear to popular science writers but with little physical content. What's true, as you said, is that bosons can occupy the same state, as opposed to fermions. As to tendency, it has the same half-truth as when one says "a system tends
to stay in its ground state".
In fact assume the latter statement is inconditionally true and you have an ensemble of non-interacting bosons. Then each boson will tend to occupy its ground state and - since nothing forbids that - all particles will sit there. In presence of interaction things may go different or not depending on the interaction.
But the real issue is if there is the tendency of a system to stay in its ground state or go into it if initially placed in a state of higher energy. However this is a question not easy to deal with in few words.
answered 11 hours ago
Elio FabriElio Fabri
3,4001214
answered 11 hours ago
Elio FabriElio Fabri
3,4001214
answered 11 hours ago
Elio FabriElio Fabri
3,4001214
answered 11 hours ago
Elio FabriElio Fabri
3,4001214
3,4001214
1
$begingroup$
No, bosons do actually 'clump' relative to classical Maxwell-Boltzmann particles, at least under some circumstances. A nice example is the Hong-Ou-Mandel effect: en.wikipedia.org/wiki/Hong%E2%80%93Ou%E2%80%93Mandel_effect
$endgroup$
– Rococo
11 hours ago
add a comment |
1
$begingroup$
No, bosons do actually 'clump' relative to classical Maxwell-Boltzmann particles, at least under some circumstances. A nice example is the Hong-Ou-Mandel effect: en.wikipedia.org/wiki/Hong%E2%80%93Ou%E2%80%93Mandel_effect
$endgroup$
– Rococo
11 hours ago
1
1
$begingroup$
No, bosons do actually 'clump' relative to classical Maxwell-Boltzmann particles, at least under some circumstances. A nice example is the Hong-Ou-Mandel effect: en.wikipedia.org/wiki/Hong%E2%80%93Ou%E2%80%93Mandel_effect
$endgroup$
– Rococo
11 hours ago
$begingroup$
No, bosons do actually 'clump' relative to classical Maxwell-Boltzmann particles, at least under some circumstances. A nice example is the Hong-Ou-Mandel effect: en.wikipedia.org/wiki/Hong%E2%80%93Ou%E2%80%93Mandel_effect
$endgroup$
– Rococo
11 hours ago
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%2f470316%2fwhy-do-bosons-tend-to-occupy-the-same-state%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
