Non-Identical Bosons

[Rade]

I have a question about bosons and how they relate to the Pauli Exclusion Principle. I think I understand the fermion situation.

 

Bosons come in many different sizes, from photon without rest mass, the gluon, the two quark pion, heavy bosons such as W-boson, and even clusters of nucleons such as the deuteron [NP] and helium-4 [NNPP]. All bosons by definition have integral quantum spin (0,1,2, ...). There are many bosons in the universe based on the photon abundance alone. Fermions on the other hand have factional spin (1/2, etc.).

 

As I understand the bosonic situation, the Pauli exclusion concludes that "two quantum identical bosons can be in the same quantum state". This is the opposite of the fermion situation where the Pauli exclusion concludes "two quantum identical fermions cannot be in the same quantum state". Of course then, logically, "two quantum non-identical fermions can be in the same quantum state".

 

Thus, the wavefunctions for a pair of bosons (or any number) are symmetric under the exchange of quantum properties such as spin. So, when applying the Pauli exclusion we regard two boson states as being identical when all good quantum numbers also are identical, including the spin quantum number. Thus, two (or many) spin identical (<<,>>) photons (as bosons) can occupy the same quantum state, as suggested by Hanbury Brown and Twiss (1956, Nature, Correlation between photons in 2 coherent beams of light, 177).

 

OK, no problem so far.

 

But my question is, what does the Pauli exclusion conclude about "two non-identical bosons" ? I have not seen this situation discussed (my ignorance).

 

It seems logical to me that the bosonic situation would be the opposite situation of "two non-identical fermions" ? For example, quantum theory predicts that two non-identical electrons (e-) can occupy the same orbital in an atom. Chemical bonds (covalent) allow for a pair of non-identical electrons to share orbitals. These situations result because the spin quantum numbers of the two electrons are in opposite directions e- (> spin) and e- (< spin). So, this is a situation where the Pauli exclusion is not violated--the two "fermions" with 1/2 spin (e-) can occupy the same quantum state for the simple reason they are not quantum identical fermions (the spins are opposite, < and > directions).

 

Thus, I suggest that for two non-identical bosons, the Pauli exclusion rule should be that two quantum non-identical bosons cannot be in the same quantum state. Would this be correct ?

 

As for electrons, one way for two "bosons" to be non-identical is for their spin states to be in opposite directions (<,>). If this is correct, then two non-identical photons (bosons with spin =1) cannot be in the same quantum state if they have opposite spin (<,>), but only if they have identical spin directions (>> or <<).

 

Another observation is that no two non-identical boson pairs of any combinations would be possible. Thus it would not be possible to see within the same quantum state a photon + pion, photon + W-boson, photon + gluon, gluon + pion, gluon + W-boson, etc. etc. etc.

 

OK, comments would be appreciated. If I have this incorrect, please explain why. In summary, here are the 4 Pauli exclusion rules that I have presented, it is rare to see them all presented together:

 

two quantum identical fermions cannot be in the same quantum state

two quantum non-identical fermions can be in the same quantum state

two quantum identical bosons can be in the same quantum state

two quantum non-identical bosons cannot be in the same quantum state

[ronthepon]

As I understand the bosonic situation, the Pauli exclusion concludes that "two quantum identical bosons can be in the same quantum state". This is the opposite of the fermion situation where the Pauli exclusion concludes "two quantum identical fermions cannot be in the same quantum state". Of course then, logically, "two quantum non-identical fermions can be in the same quantum state".

Erm... I may be mistaken, but I'm pretty sure the Pauli exclusion principle says nothing about bosons.

 

Furthermore, the principle says that two quantum identical fermions cannot exist.

 

I'd like to draw you attention to the difference with the sentence you've used. 'Quantum identical' things are, well, identical in all quantum numbers to begin with.

 

If, by chance you mean different fermions (say electron vs muon), the principle still says they can't have the same quantum state.

 

I'd say that since the Pauli Exclusion does not apply to bosons at all; and if you assume the different bosons appear to behave identically, they will properly follow Bose-Einstein Statistics; thus be able to have the same quantum state.

 

I'm neither qualified, nor do I know for sure, so these are just like educated guesses. Maybe a mix of different bosons won't follow vanilla Bose Einstein Statistics, maybe they don't get the same quantum numbers... I'd guess they do.

 

:rolleyes:

 

It seems logical to me that the bosonic situation would be the opposite situation of "two non-identical fermions" ?

One last thing; (again, i confess this is just me guessing...) I'd say that it would be wrong to assert that bosons and fermions are 'opposites', and apply that reasoning here. They're opposites with respect to adherence to Pauli's exclusion, but that might just be all.

[Erasmus00]

Quantum statistics tells you two identical fermions (spin up electrons, say) cannot be in the same state, ever. They can be in different states.

 

Bosons can be in any state- whats more, they are tend to clump in states. For a low temperature systems, bosons will bose-condense. This means all the bosons will drop to the lowest energy state possible.

-Will

[Qfwfq]

You seem to be putting many carts before their horses.

 

By definition, bosons are particles described according to symmetrized states, fermions those described according to antisymmetrized states. These are the two options when the buggers are indistinguishable.

 

But my question is, what does the Pauli exclusion conclude about "two non-identical bosons" ? I have not seen this situation discussed (my ignorance).

Pauli's exclusion principle is an elementary consequence of antisymmetrized states. It concludes nothing about bosons. When the are identical, as Will says, they have somewhat the opposite behaviour (a fact which follows less trivially, I still remember getting terribly bungled when I was told to show it in an oral :rolleyes:).

 

two quantum identical fermions cannot be in the same quantum state

two quantum non-identical fermions can be in the same quantum state

two quantum identical bosons can be in the same quantum state

two quantum non-identical bosons cannot be in the same quantum state

Cute :phones: but, alas, it is non-sequitur.

 

If one is an apple and the other is a pear, there is simply no constraint of symmetrized nor of antisymmetrized state; the whole tensor product of the Hilbert spaces is fine.

[Rade]

Cute but, alas, it is non-sequitur.

..bosons can be in any state...

 

Thanks for the comments. Can you provide an experimental example of two quantum non-identical bosons observed in the same quantum state ? So, for example, two photons (each by definition a boson), one with spin = 1 in (<) spin direction and another with spin = 1 in (>) direction. Or any other such example.

 

I suppose what I am saying is that the Pauli Exclusion (as stated by Pauli) really is a sub-set of a more comprehensive "exclusion rule" concerning all logical ways that fermions and bosons behave--thus the Pauli Exclusion is only one of these 4 more general "exclusion rules" (i.e., the first one):

 

Originally Posted by Rade:

two quantum identical fermions cannot be in the same quantum state

two quantum non-identical fermions can be in the same quantum state

two quantum identical bosons can be in the same quantum state

two quantum non-identical bosons cannot be in the same quantum state

 

The first three of these "exclusion rules" are well supported by experimental data--but it is the last one that I know nothing about experimentally. So, it would be great if someone could provide a link to a published paper that falsifies the last rule above. I appreciate the help.

[Erasmus00]

A circularly polarized laser is an equal combination of transverse linear polarized photons. So you can have two photons with different polarizations in the same spatial state.

[Rade]

A circularly polarized laser is an equal combination of transverse linear polarized photons. So you can have two photons with different polarizations in the same spatial state.

Thanks, but is this really the quantum situation ?

 

I have Roger Penrose book "The road to reality"--a good reference. On p. 557 Penrose makes this statement about the polarized state of the photon:

 

"...the general polarized state of a [individual] photon is a complex linear combination of the states of a positive helicity |+> and negative helicity |->" , thus he shows that the individual photon wavefunction state is:

 

photon state = {(w|+>) + (z|->)}

 

He then says.."the physical interpretation of such a state is called elliptical polarization, which generally has two particular cases, linear plane and circular polarization"

 

Thus, a circularly polarized laser is not a combination of two different photons (w|+>) and (z|->) within the same state, it is a particular case of the more general elliptical condition where both of the helicity states are superposed within a single photon.

 

Please me know if this is incorrect.

 

==

 

Also, on p. 595, Penrose clarifies my OP question...as everyone here has responded, I was incorrect, two non-identical bosons can be in the same state.

 

However, the boson situation does differ between the case where there are two interacting bosons that Penrose calls "not identical" vs "identical". Thus, not all boson pairs behave the same.

 

The "identical boson pair" state is more complex than the situation of having two "non identical" bosons. So, there is a different "exclusion" rule for non-identical bosons as compared to identical bosons, and these two rules differ from the fermion state possibilities (i.e., identical fermions vs non identical fermions).

 

So, here are my new "exclusion" rules for all fermion and boson possibilities--I think I now have them correct ? Someone let me know if they are incorrect. I find it useful to have these all together for reference--only the first is the Pauli Exclusion Principle, the others do not have names as far as I know:

 

two quantum identical fermions cannot be in the same state

two quantum non-identical fermions can be in the same state

two quantum non-identical bosons can be in the same state, and the state is a tensor product under exchange {A B}

two quantum identical bosons can be in the same state, and the state is symmetric under exchange { (A :naughty: + (B' A') }

 

Thanks everyone for the assistance.

[Qfwfq]

Thanks for the comments. Can you provide an experimental example of two quantum non-identical bosons observed in the same quantum state ? So, for example, two photons (each by definition a boson), one with spin = 1 in (<) spin direction and another with spin = 1 in (>) direction.

You get into some sticky semantic issues here. If their spins are opposite, are they in the same state? If they are both photons, are they distinguishable? If they are of different species, does the question even have any meaning?

 

I suppose what I am saying is that the Pauli Exclusion (as stated by Pauli) really is a sub-set of a more comprehensive "exclusion rule" concerning all logical ways that fermions and bosons behave--thus the Pauli Exclusion is only one of these 4 more general "exclusion rules" (i.e., the first one):

I believe I had already said what Pauli's exclusion principle is a specific case of, but it seems you need it said in more detail.

 

If both particles are of the same species and there can't be a way of distinguishing between them (the first and the second one, so to speak), then any description that would allow this can't be correct, right?

 

If you agree with this, then in oder to descreibe two-particle states in terms of the single-particle states, one must at the very least restrict the tensor product to elements which are either symmetric or antisymmetric, discarding all others. Linearity, however, further imposes the aut-aut, hence the fact that a given species is either a boson or a fermion.

 

Consider the single-particle sates [imath]\varphi(\xi)[/imath] and [imath]\psi(\xi)[/imath] (with [imath]\xi[/imath] standing for whatever set of parameters) for which the antisymmetric combination is [imath]\varphi(\xi_1)\psi(\xi_2)-\varphi(\xi_2)\psi(\xi_1)[/imath]. Pauli's exclusion principle is due to the fact that this expression reduces to 0 whenever [imath]\varphi[/imath] and [imath]\psi[/imath] are the same.

 

Now, if one is an apple and the other is a lemon, is there any reason for the tight restriction on the states? How could the lack of this restriction be grounds on which to discard the ones in which [imath]\varphi[/imath] and [imath]\psi[/imath] are the same? Or even any of them? If both are bosons but of different species, how is it a reason to antisymmetrize instead of symmetrizing?

 

This said, I don't see the point of trying to experimentally falsify your conjecture.

[Rade]

You get into some sticky semantic issues here. If their spins are opposite, are they in the same state?

Logically, no, they would not be in the same quantum state, given that "spin" is a good quantum number. I assume this is what you mean by "state" ?

 

If they are both photons, are they distinguishable?

Logically yes, If "spin" is a good quantum number, then the label "photon" (whatever it represents) comes in a 1+ and 1- possibility quantum spin state. Thus it must be true that one can distinguish a set of spin (1+) photons from another set of spin (1-) photons after the measurement event, if they wish to observe what is called spin.

 

If they are of different species, does the question even have any meaning?

OK, the answer here really depends on what you mean by the word "species" ? (I never used this term). In philosophy (from Aristotle) and biology (see Mayr) the concept "species" has great meaning--since you use it here in discussion of quantum theory I assume it also has meaning for it. Now, if by "species" you mean a set of quantum entities such as photons (bosons) all with spin = (1+), and then you want to compare the physical interaction of the "species=1+ spin set" with another set of photons that you label "species=1- spin set"--then, yes, any quantum mechanical "question" you ask about how these two species "sets" interact physically has great "meaning".

 

If both particles are of the same species and there can't be a way of distinguishing between them (the first and the second one, so to speak), then any description that would allow this can't be correct, right?

No, false. The "if-then" argument you present is based on a contradiction in the use of terms. Consider how you use the term "species" in relation to what I just posted...

 

1.....if both particles are of the same species..... OK--the answer obviously depends if you are asking about (1) two boson particles or (2) two fermion particles. Clearly this must be logically true--correct ? So, I will present the two logical answers:

1. The boson situation. I will use photons as the example, but would apply for any two bosons. The problem you present is the situation where you have 2 photons, and they are of the same species--that is, they each have spin 1+ (or the opposite, does not matter). So, this is the quantum system {(photon #1 = spin 1)+ (photon #2 = spin 1+)}. In this situation, because you have "two identical bosons" the "exclusion rule" that is to be applied to the quantum situation is what I posted as "exclusion rule #3" in my last (modified) post--which is:

 

Exclusion Rule #3: two quantum identical bosons can be in the same state, and the state is symmetric under exchange { (A + (B' A') }

 

This rule is the answer to your question .....if both particles are of the same species.....and both particles are bosons.

 

2. Fermion situation. Here we can use two electrons with spin =1/2, but again, the answer would apply to any two types of fermions. The exclusion rule to apply to the problem ....if both particles are of the same species....is completely different if the two particles are fermions. In that case, the exclusion rule that applies (for two electrons) is this one from my list:

 

Exclusion Rule #1: two quantum identical fermions cannot be in the same state

 

...If you agree with this, then ....

Well, I do not agree with what you presented, for the reason just presented. So, I have no way to respond to anything else you wrote in that section of your post.

 

...Now, if one is an apple and the other is a lemon, is there any reason for the tight restriction on the states?

Yes, there is every reason--but again, it depends what you mean by "apple" and "lemon". I will assume a common sense distinction of common sense words, and if so, then apple vs lemon can be taken in two different logical ways in the strange reality of quantum theory (1) apple = boson <---> (2) lemon = fermion. And, (1') apple = quantum identical <---> (2') lemon = quantum non-identical.

 

Why is this important to consider the common sense words apple and lemon in two different logical ways? For the very reason that what is called the Pauli Exclusion Principle (as he described it) does not apply for two fermions (the term he used) that have non-identical quantum numbers [such as spin]. This is exactly why two electrons (e-) defined as "fermions" CAN be in the same quantum state if their respective quantum numbers spins are opposite, one (+) the other (-). I mean, all of Chemistry (electron shell theory) demands that the Pauli Exclusion not apply to this situation of two APPLES (fermions) that behave quantum mechanically as two LEMONS !

 

If both are bosons but of different species, how is it a reason to antisymmetrize instead of symmetrizing?

OK, to your question ....if both are bosons but of a different species....what you are asking about is the exclusion rule #3 I presented:

two quantum non-identical bosons can be in the same state, and the state is a tensor product under exchange {A B}

 

So, consider two photons (bosons) and are of different species (1) one set has spin 1+, (2) the other set has spin 1-. Now, clearly, the theory of quantum mechanics allows for this situation (the equation for it is provided by Roger Penrose in his book "The Road to Reality"). The solution is not to "antisymmetrize", the solution of the final state is a simple tensor product under exchange. Two completely differ operations--is this not correct ?

 

I don't see the point of trying to experimentally falsify your conjecture.

Forget the conjecture--that was a hypothesis for my initial (and I now understand) false set of exclusion rules. You are correct, there is no point.

 

But, this does not mean that my revised set of selection rules do not hold true--as presented again below--these are now what I asked to be "falsified":

 

Exclusion Rule #1: two quantum identical fermions cannot be in the same state

Exclusion Rule #2: two quantum non-identical fermions can be in the same state

Exclusion Rule#3: two quantum non-identical bosons can be in the same state, and the state is a tensor product under exchange {A B}

Exclusion Rule#4: two quantum identical bosons can be in the same state, and the state is symmetric under exchange { (A :clock: + (B' A') }

 

I greatly appreciate and respect your interest in this topic--I am here at this forum to learn.

[Qfwfq]

Logically yes, If "spin" is a good quantum number, then the label "photon" (whatever it represents) comes in a 1+ and 1- possibility quantum spin state. Thus it must be true that one can distinguish a set of spin (1+) photons from another set of spin (1-) photons after the measurement event, if they wish to observe what is called spin.

You are saying that the different states are distinct. This does not imply it being possible to distinguish the photons. Can you slap a label on each of them, in order to say which one is in which of two states?

 

OK, the answer here really depends on what you mean by the word "species" ? (I never used this term).

That explains a lot! If you follow the sequence, the if clauses in those two consecutive questions might have suggested what I meant by that word. "If they are of different species" was an alternative to "If they are both photons", according to the use of the term in particle physics. Apples and lemons were simply placeholders for different species of particles, without choosing specific ones existing in particle physics, and I don't get how you can justify your affermative answer about there being a reason for the tight restriction on the states.

 

No, false. The "if-then" argument you present is based on a contradiction in the use of terms. Consider how you use the term "species" in relation to what I just posted...

So, where is the contradiction? If we can't slap a label on each particle, how can we accept a description that may predict different results according to which is which? Why should we, to the contrary, reject it when there is more than just a label on each particle? If one is an apple and one is a lemon, how can there be a reason to reject such descriptions?

 

Why is this important to consider the common sense words apple and lemon in two different logical ways? For the very reason that what is called the Pauli Exclusion Principle (as he described it) does not apply for two fermions (the term he used) that have non-identical quantum numbers [such as spin]. This is exactly why two electrons (e-) defined as "fermions" CAN be in the same quantum state if their respective quantum numbers spins are opposite, one (+) the other (-). I mean, all of Chemistry (electron shell theory) demands that the Pauli Exclusion not apply to this situation of two APPLES (fermions) that behave quantum mechanically as two LEMONS !

Here you contradict your own reason for a negative answer to my "If their spins are opposite, are they in the same state?" question. The opposite-spin pairs of electron orbitals are not the same state, since their spins differ. :)

 

I greatly appreciate and respect your interest in this topic--I am here at this forum to learn.

If this is so, you do well to ask what I meant by saying certain things because I couldn't guess what you do and don't understand, but I saw less point in lengthy discussion of logical consequences of wild guesses for what I might possibly have meant.

 

For brevity, I had tried to gloss over the matter of labelling when I posted yesterday, but it seems you lack the familiarity necessary to do without it. When I said "in oder to descreibe two-particle states in terms of the single-particle states" this implicitly requires a pseudo-labelling and the very reason for restricting to either symmetric or antisymmetric states is exactly to make these labels irrelevant. The only reason for this is in describing states of two particles of a same species.

[Rade]

Qfwfq...

 

I can see you are very upset--not my intent in asking and responding to questions here.

==

 

previous comment deleted...see correction

 

==

 

OK, now I see in looking at the equations for symmetric and anti-symmetric at Wiki (http://en.wikipedia.org/wiki/Identical_particles), that a concept of "non-identical bosons" is impossible based on quantum field theory. So, I must revise, hopefully for the last time, my selection rules and I am now down to 3:

 

Let Boson = A particle with integral spin quantum number (i.e., 0,1,2..)

Let Fermion = A particle with non-integral spin quantum number (i.e., 1/2, 2/3,..), then the following sections rules hold for the final state:

 

Selection Rule #1: two quantum identical fermions cannot be in the same state

Selection Rule #2: two quantum non-identical fermions can be in the same state

Selection Rule #3: two quantum bosons can be in the same state, and the state is symmetric under exchange { (A :phones: + (B' A') }

 

Example: The deuteron [NP] is a boson with spin = 1. Two boson particles (no matter if the parity of the spin is + or -) can be in the same state, and the state is symmetric under exchange. Thus, it is possible (in theory) that two [NP] could combine to form helium-4 {(NP]+[NP]} that would have a spin quantum number of = 0. This is allowed by Section Rule #3 above.

 

How does this look ? Again, of interest, only the first is given a name (Pauli Exclusion Principle).