
#1
Jun811, 01:24 AM

P: 525

Is it possible to put a Boson and Fermion in superposition? If not possible, why not?




#2
Jun811, 09:27 AM

Sci Advisor
P: 1,395

[tex]\psi=(B1>\otimesF1> + B2>\otimesF2>)[/tex] Where B1 and B2 are boson states and F1 and F2 are fermion states. However, if you are asking if a wavefunction can be created for a superposition where a fermion exchanges with a boson, then I am almost 100% positive that the answer is no. The only possible exception I can think of is some highenergy particle physics process where fermions and bosons can be interconverted, but I am not even sure that is theoretically possible. Exchange is a property of indistinguishable particles, and fermions are always distinguishable from bosons. 



#3
Jun811, 09:52 AM

P: 160

This is a very good question indeed. I take it to mean "if there is a state of a single boson [itex]B\rangle[/itex] and a state of a single fermion [itex]F\rangle[/itex], it the linear combination [itex]B\rangle+F\rangle[/itex] a 'good' state?". And the answer, as far as we know, is no!
This has to do with the topology of the Poincare group of spacetime symmetries (the rotations, boosts and translations). We can think just about the rotations for this. If you rotate by [itex]2\pi[/itex], a boson goes back to the same thing but a fermion picks up a minus sign. Each of these behaviours are fine on their own, because all we need when the states transform is what is called a projective representation, which means we don't care about a change in overall phase. But if you have the linear combination, what you end up with is not just a phase, so the combination is denied by the symmetry. This is what is called a 'superselection rule'. The mathsy bit: The reason these 'projective representations' exists is because the rotation group is not 'simply connected', which means it has a hole: there are paths through the group which can't be contracted continuously to a point. If you want more on this, I'll be happy to expand a little; there is much more to be said! 



#4
Jun811, 09:56 AM

Sci Advisor
Thanks
P: 2,133

Superposition betweeen Boson and Fermion
That's not a superposition of a boson and a fermion state, since the twoparticle system, consisting of a fermion and a boson is, of course, a fermion again. So you superpose two fermion states, and it's indeed "nothing wrong" with that.
Whether or not you allow certain superpositions or not is of course a question of observations, as any fundamental question in physics. In theoretical physics you make models, often based on symmetry assumptions. Now, if you assume that rotational symmetry is realized in nature, this leads you to the mathematical assumption that you can realize the symmetry by a unitary or antiunitary ray representation of the according group. Thanks to Bargmann and Wigner's famous theorem any such symmetry can be lifted to an unitary or antiunitary representation of the covering group in Hilbert space. For the rotation group, SO(3), this means you investigate all unitary representations of its covering group, SU(2), which leads to the halfinteger and integerspin representations of rotations. Now a rotation around [tex]2 \pi[/tex] should represent the identity. Now for particles with halfinteger (integer) spin all their state vectors get multiplied by 1 (1). As long as you don't superpose states which belong to halfinteger and integerspin representations, there's no trouble with these phase factors since they are not observable. Thus, if you assume that nature respects rotational invariance and thus the Hilbert space admits a representation of the rotation group, one must not superimpose states, where one belongs to halfinteger and the other to integerspin representations. This is what is called a superselection rule: Certain linear combinations are forbidden since this would violate some symmetry principle. Now, the spinstatistics theorem tells us that halfintegerspin (integerspin) representations belong necessarily to fermions (bosons). Thus the rotational superselection rule forbids the superposition of a fermion and a boson state. 



#5
Jun811, 05:33 PM

P: 525

I was made aware of this superposition question between boson and fermion by Vlatko Vedral amazon book page (was referred to it after reading his June Scientific American cover story "Living in a Quantum World"). What do you think of the following? Is he talking about whether wavefunction can be created for a superposition where a fermion exchanges with a boson, which SpectraCat believes is not possible categorically? Vedral is a physicist. He is not sure the answer is no. Why? Anyone has same position as him or do all of you physicists share SpectraCat "almost 100% positive the answer is no"?
http://www.amazon.com/DecodingReali...7571796&sr=81 "Vlatko Vedral: It is indeed depressing that quantum physics has been so consistently accurate over the past hundred years. There is really no obvious deviation from experiments (we physicists would get really excited if there were). The main issue I think is how general the quantum superposition principle is: Can any property really be superposed? Roger Penrose, for instance, believes that gravity will prevent superposing a massive object in two different places. Along with many other physicists, I think that this is a technological (not fundamental) problem. On top of this, we are far away from being able to experiment with time and space on scales relevant for quantum gravity. A more interesting issue for me (as well as being more readily accessible to experiments) is the existence of two different types of particles, fermions and bosons. It seems that every particle we observe is either a fermion (electrons, for example) or a boson (photons, for example). But can it be that we can have a particle in a superposition between a fermion and a boson? We are now in a position to be able to attempt to superpose these two properties in practice. If we show that this cannot be done, however, it is not clear what this means for quantum physics. Some of us like to think of everything in the universe as being quantum and finding limitations even in one aspect would tell us that there might be more out there…" 



#6
Jun811, 07:23 PM

P: 160

BUT this is physically completely indistinguishable from a slightly different situation: We could have a case where the symmetry of nature is in fact a bigger group, the universal covering group (SU(2) in the case of rotations). This group is essentially several copies of the original group glued together. For example, SU(2) basically contains two copies of the rotation group and a 'rotation by 360 degrees' makes you jump between the two. This covering group is designed so that we don't need projective representations, so the superselection rule must then be imposed as an extra constraint: it really is a rule rather than a derived consequence. But then if we found that nature chose to break the rule, that would be fine: we would just conclude that it's not actually rotations that are the fundamental symmetry of nature, but the bigger group, and we throw out the superselection rules. So AFAIK, while it would be interesting to find out that we could make these superpositions, I don't think it really has any serious consequences from a theoretical point of view. As to the practical consequences, I'm afraid you'll have to ask someone else, as I really don't know! 



#7
Jun911, 03:06 AM

Sci Advisor
P: 1,722

Can a boson state and a fermion state be superposed in a physically
sensible nontrivial way? I'm 100% certain the answer is "no", if local Poincare symmetry holds. Actually, all we need is local rotational invariance under boring old SO(3) so that boson and fermion state spaces correspond to unitary irreducible representations of that group. The argument is given in Ballentine's textbook "QM  A Modern Development", p183. Here's a quick summary. Consider a hypothetical Hilbert space which is a direct sum of the boson and fermion (1particle) spaces. Denote two states from the boson and fermions subspaces as [tex] +\rangle ~~~ \mbox{and}~~~ \rangle [/tex] respectively. Denote by "R" the operator corresponding to rotation by 2pi. Let A be any physical observable (which Ballentine has already showed to commute with R). Then [tex] \langle+ R A \rangle ~=~ \langle+ A R \rangle [/tex] implying [tex] \langle+ A \rangle ~=~  \, \langle+ A \rangle [/tex] and therefore [tex] \langle+ A \rangle ~=~ 0. [/tex] So, (in Ballentine's words), or (local) GR doesn't change the argument because one still has local SO(3) symmetry. 



#8
Jun911, 03:22 AM

Sci Advisor
P: 3,369

The superselection rule in question is called univalence superselection. It was historically the first superselection rule to be discovered.
A nice account can be found here: http://www.springerlink.com/content/n24439h86601x418/ 


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