fermions vs bosons

mpv_plate

A photon produced in annihilation is not just a packet of energy. Photon is a full-fledged particle with physical properties (like the spin, momentum). There is no reason to treat photons as inferior to other particles.

DaleSpam

Ah, I misunderstood, to me "transform" doesn't mean the same as "produce".

As far as I am aware there is no particle reaction which has as inputs only fermions and as outputs only bosons other than matter-antimatter annihilation. However, I am not a particle physicist, so there may be some of which I am not aware.

nouveau_riche

so i would come at the same question again...
an example where a fermion can transform into boson giving him charge, momentum and energy?

DaleSpam

Why don't matter-antimatter anhilation reactions count?

nouveau_riche

because i want to see a process where a fermion can modulate its intrinsic physical property that not just involves energy/momentum conservation but its field behavior

Dickfore

But the statistics an elementary or composite particle obeys is intimately connected to the intrinsic spin of the particle, as given by the Spin-Statistics Theorem.

So if you want to change a fermion into a boson, you would have to change its spin, thus making it a new particle.

There is a kind of a "loophole" with composite particles. Take He-4, for example. It has 2 protons, 2 electrons and 2 neutrons. All of these are fermions. But, when paired, they have integer spin, thus making the nucleus, and the atom of this isotope a boson. In fact, He-4 is the first substance that exhibited superfluid transition.

nouveau_riche

it is not just the pairing, they have to be cooled so as to make them superfluid

Dickfore

Of course. However, being a boson is a necessary condition. He-3, when cooled to the same conditions does not become superfluid.

DaleSpam

What do you mean by field behavior?

nouveau_riche

is it necessary that the bosons will always play the role of force carrier?

Dickfore

Why do you answer questions with questions?

nouveau_riche

i haven't answer the question yet but i need the answer before i can answer else you will be standing upright with your point

DaleSpam

Yes. More specifically, the forces of the standard model are described by gauge fields, and the quanta of those gauge fields are the gauge bosons. Each gauge field's gauge bosons are the carriers of the respective forces. Note that there are other non-gauge bosons which are not excitations of any of the gauge fields and therefore are not carriers of any of the corresponding forces.

Ilmrak

By definition fermion fields creation and destruction operators satisfies

[itex]
\{a_\alpha^\dagger, a_\beta^\dagger\}= 0\; , \\
\{a_\alpha,a_\beta\}= 0\; ,\\
\{a_\alpha^\dagger, a_\beta\}= \delta_{\alpha,\beta} \; ,
[/itex]

while boson fields creation and destruction operators satisfies

[itex]
\left[a_\alpha^\dagger, a_\beta^\dagger\right]= 0\; ,\\
\left[a_\alpha,a_\beta\right]= 0\; ,\\
\left[ a_\alpha^\dagger, a_\beta\right]= \delta_{\alpha,\beta}\; .
[/itex]

These (anti-)commutation rules implies that two fermions can't be in the same state ([itex]a_\alpha^\dagger a_\alpha^\dagger= 0[/itex]) while two bosons can ([itex]a_\alpha^\dagger a_\alpha^\dagger \neq 0[/itex]).

It can be shown that either a particle is a boson or it's a fermion, there isn't a third option.

Furthermore the Spin-Statistic theorem states that a particle with integer spin has to be a boson, while a semi-odd spin particle has to be a fermion.

Macroscopically a system of bosons can show superfluidity, while maybe the most known examples of the Fermi statistic are neutron stars.

Neutrons are fermions!!

There's really no need to invoke the Higgs field to explain how particles gains their statistical properties.
Actually your sentence sounds as a non-sense, as the Higgs is a boson (spin zero) and so statistics is needed to describe even the Higgs field.
Note that Fermi and Bose statistics reflects symmetries, not asymmetries.

All bosons has physical properties.
However, a process you may like is

[itex]e^- \;\bar{\nu}_e \rightarrow W^- \; ,[/itex]

where [itex]e^-[/itex] and [itex]\bar{\nu}_e[/itex] are fermions and [itex]W^-[/itex] is a boson.

No it's not necessary.
In the standard model there is exactly one boson that is not a gauge boson and so doesn't carry a force: the Higgs boson.

I hope this could help a bit.

Ilm

mpv_plate

Two bosons can scatter of each other thanks to intermediate fermions. That is the case of photon-photon scattering which happens due to interaction via the virtual electron-positron pair. However I am not sure if this is what you asked for.

Simon Bridge

No. It is the other way around: in the standard model, the gauge fields give rise to force carriers which are bosons.
Not all interactions need be mediated by bosons.
Not all spin 0 objects are force carriers.

Now please answer the question posed by dalespam: "what do you mean by field behaviour?"

We cannot help you if you will not answer questions.

nouveau_riche

what i meant to say was that if a fermion is changing into a boson that is a carrier of force, the field generating fermion has itself transformed into the carrier under some conditions, this could help to figure the fermion-boson relationship.