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 quantum-statistics
edited Apr 4 at 15:01
knzhou
46.7k11126224
asked Apr 3 at 14:45
HicHaecHocHicHaecHoc
1338
$endgroup$
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 quantum-statistics
edited Apr 4 at 15:01
knzhou
46.7k11126224
asked Apr 3 at 14:45
HicHaecHocHicHaecHoc
1338
$endgroup$
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 quantum-statistics
edited Apr 4 at 15:01
knzhou
46.7k11126224
asked Apr 3 at 14:45
HicHaecHocHicHaecHoc
1338
$endgroup$
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-statistics
quantum-mechanics bosons quantum-statistics
edited Apr 4 at 15:01
knzhou
46.7k11126224
asked Apr 3 at 14:45
HicHaecHocHicHaecHoc
1338
edited Apr 4 at 15:01
knzhou
46.7k11126224
asked Apr 3 at 14:45
HicHaecHocHicHaecHoc
1338
edited Apr 4 at 15:01
knzhou
46.7k11126224
edited Apr 4 at 15:01
knzhou
46.7k11126224
edited Apr 4 at 15:01
knzhou
46.7k11126224
46.7k11126224
asked Apr 3 at 14:45
HicHaecHocHicHaecHoc
1338
asked Apr 3 at 14:45
HicHaecHocHicHaecHoc
1338
asked Apr 3 at 14:45
HicHaecHocHicHaecHoc
1338
1338
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 Apr 3 at 15:08
knzhouknzhou
46.7k11126224
$endgroup$
2
$begingroup$
@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
Apr 3 at 15:27
5
$begingroup$
@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
Apr 3 at 16:41
3
$begingroup$
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.
Apr 3 at 22:16
2
$begingroup$
@R.M. I said "in the microcanonical ensemble", which is precisely the assumption that the states are equally likely. (This is what happens when you give the system a fixed energy and let it come to thermal equilibrium.) But yeah, it's true I didn't justify why that happens -- it's kind of the fundamental postulate that makes all of our reasoning about thermal physics work, so it would be a whole book to say anything more! You could phrase it other ways, but I preferred talking about a thermodynamic example because it makes it easier to connect to the probabilities I had above.
$endgroup$
– knzhou
Apr 3 at 22:41
3
$begingroup$
@gented No, the Hilbert space is 3-dimensional and not 4-dimensional. It's the symmetrized tensor product, not the ordinary tensor product. "Indistinguishable" has a deeper meaning than just being indistinguishable in practice.
$endgroup$
– knzhou
Apr 4 at 11:59
|
show 14 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 Apr 3 at 15:15
Elio FabriElio Fabri
3,5851214
$endgroup$
2
$begingroup$
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
Apr 3 at 15:49
add a comment |
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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 Apr 3 at 15:08
knzhouknzhou
46.7k11126224
$endgroup$
2
$begingroup$
@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
Apr 3 at 15:27
5
$begingroup$
@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
Apr 3 at 16:41
3
$begingroup$
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.
Apr 3 at 22:16
2
$begingroup$
@R.M. I said "in the microcanonical ensemble", which is precisely the assumption that the states are equally likely. (This is what happens when you give the system a fixed energy and let it come to thermal equilibrium.) But yeah, it's true I didn't justify why that happens -- it's kind of the fundamental postulate that makes all of our reasoning about thermal physics work, so it would be a whole book to say anything more! You could phrase it other ways, but I preferred talking about a thermodynamic example because it makes it easier to connect to the probabilities I had above.
$endgroup$
– knzhou
Apr 3 at 22:41
3
$begingroup$
@gented No, the Hilbert space is 3-dimensional and not 4-dimensional. It's the symmetrized tensor product, not the ordinary tensor product. "Indistinguishable" has a deeper meaning than just being indistinguishable in practice.
$endgroup$
– knzhou
Apr 4 at 11:59
|
show 14 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 Apr 3 at 15:08
knzhouknzhou
46.7k11126224
$endgroup$
2
$begingroup$
@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
Apr 3 at 15:27
5
$begingroup$
@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
Apr 3 at 16:41
3
$begingroup$
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.
Apr 3 at 22:16
2
$begingroup$
@R.M. I said "in the microcanonical ensemble", which is precisely the assumption that the states are equally likely. (This is what happens when you give the system a fixed energy and let it come to thermal equilibrium.) But yeah, it's true I didn't justify why that happens -- it's kind of the fundamental postulate that makes all of our reasoning about thermal physics work, so it would be a whole book to say anything more! You could phrase it other ways, but I preferred talking about a thermodynamic example because it makes it easier to connect to the probabilities I had above.
$endgroup$
– knzhou
Apr 3 at 22:41
3
$begingroup$
@gented No, the Hilbert space is 3-dimensional and not 4-dimensional. It's the symmetrized tensor product, not the ordinary tensor product. "Indistinguishable" has a deeper meaning than just being indistinguishable in practice.
$endgroup$
– knzhou
Apr 4 at 11:59
|
show 14 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 Apr 3 at 15:08
knzhouknzhou
46.7k11126224
$endgroup$
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 Apr 3 at 15:08
knzhouknzhou
46.7k11126224
edited Apr 3 at 22:43
answered Apr 3 at 15:08
knzhouknzhou
46.7k11126224
answered Apr 3 at 15:08
knzhouknzhou
46.7k11126224
answered Apr 3 at 15:08
knzhouknzhou
46.7k11126224
46.7k11126224
2
$begingroup$
@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
Apr 3 at 15:27
5
$begingroup$
@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
Apr 3 at 16:41
3
$begingroup$
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.
Apr 3 at 22:16
2
$begingroup$
@R.M. I said "in the microcanonical ensemble", which is precisely the assumption that the states are equally likely. (This is what happens when you give the system a fixed energy and let it come to thermal equilibrium.) But yeah, it's true I didn't justify why that happens -- it's kind of the fundamental postulate that makes all of our reasoning about thermal physics work, so it would be a whole book to say anything more! You could phrase it other ways, but I preferred talking about a thermodynamic example because it makes it easier to connect to the probabilities I had above.
$endgroup$
– knzhou
Apr 3 at 22:41
3
$begingroup$
@gented No, the Hilbert space is 3-dimensional and not 4-dimensional. It's the symmetrized tensor product, not the ordinary tensor product. "Indistinguishable" has a deeper meaning than just being indistinguishable in practice.
$endgroup$
– knzhou
Apr 4 at 11:59
|
show 14 more comments
2
$begingroup$
@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
Apr 3 at 15:27
5
$begingroup$
@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
Apr 3 at 16:41
3
$begingroup$
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.
Apr 3 at 22:16
2
$begingroup$
@R.M. I said "in the microcanonical ensemble", which is precisely the assumption that the states are equally likely. (This is what happens when you give the system a fixed energy and let it come to thermal equilibrium.) But yeah, it's true I didn't justify why that happens -- it's kind of the fundamental postulate that makes all of our reasoning about thermal physics work, so it would be a whole book to say anything more! You could phrase it other ways, but I preferred talking about a thermodynamic example because it makes it easier to connect to the probabilities I had above.
$endgroup$
– knzhou
Apr 3 at 22:41
3
$begingroup$
@gented No, the Hilbert space is 3-dimensional and not 4-dimensional. It's the symmetrized tensor product, not the ordinary tensor product. "Indistinguishable" has a deeper meaning than just being indistinguishable in practice.
$endgroup$
– knzhou
Apr 4 at 11:59
2
2
$begingroup$
@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
Apr 3 at 15:27
$begingroup$
@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
Apr 3 at 15:27
5
5
$begingroup$
@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
Apr 3 at 16:41
$begingroup$
@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
Apr 3 at 16:41
3
3
$begingroup$
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.
Apr 3 at 22:16
$begingroup$
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.
Apr 3 at 22:16
2
2
$begingroup$
@R.M. I said "in the microcanonical ensemble", which is precisely the assumption that the states are equally likely. (This is what happens when you give the system a fixed energy and let it come to thermal equilibrium.) But yeah, it's true I didn't justify why that happens -- it's kind of the fundamental postulate that makes all of our reasoning about thermal physics work, so it would be a whole book to say anything more! You could phrase it other ways, but I preferred talking about a thermodynamic example because it makes it easier to connect to the probabilities I had above.
$endgroup$
– knzhou
Apr 3 at 22:41
$begingroup$
@R.M. I said "in the microcanonical ensemble", which is precisely the assumption that the states are equally likely. (This is what happens when you give the system a fixed energy and let it come to thermal equilibrium.) But yeah, it's true I didn't justify why that happens -- it's kind of the fundamental postulate that makes all of our reasoning about thermal physics work, so it would be a whole book to say anything more! You could phrase it other ways, but I preferred talking about a thermodynamic example because it makes it easier to connect to the probabilities I had above.
$endgroup$
– knzhou
Apr 3 at 22:41
3
3
$begingroup$
@gented No, the Hilbert space is 3-dimensional and not 4-dimensional. It's the symmetrized tensor product, not the ordinary tensor product. "Indistinguishable" has a deeper meaning than just being indistinguishable in practice.
$endgroup$
– knzhou
Apr 4 at 11:59
$begingroup$
@gented No, the Hilbert space is 3-dimensional and not 4-dimensional. It's the symmetrized tensor product, not the ordinary tensor product. "Indistinguishable" has a deeper meaning than just being indistinguishable in practice.
$endgroup$
– knzhou
Apr 4 at 11:59
|
show 14 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 Apr 3 at 15:15
Elio FabriElio Fabri
3,5851214
$endgroup$
2
$begingroup$
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
Apr 3 at 15:49
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 Apr 3 at 15:15
Elio FabriElio Fabri
3,5851214
$endgroup$
2
$begingroup$
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
Apr 3 at 15:49
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 Apr 3 at 15:15
Elio FabriElio Fabri
3,5851214
$endgroup$
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 Apr 3 at 15:15
Elio FabriElio Fabri
3,5851214
answered Apr 3 at 15:15
Elio FabriElio Fabri
3,5851214
answered Apr 3 at 15:15
Elio FabriElio Fabri
3,5851214
answered Apr 3 at 15:15
Elio FabriElio Fabri
3,5851214
3,5851214
2
$begingroup$
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
Apr 3 at 15:49
add a comment |
2
$begingroup$
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
Apr 3 at 15:49
2
2
$begingroup$
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
Apr 3 at 15:49
$begingroup$
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
Apr 3 at 15:49
add a comment |
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