Definition of indistinct in pauli exclusion principle

[RedX]

definition of "indistinct" in pauli exclusion principle

I'm a little confused about what constitutes a distinct particle.

For example, a muon is not an electron as they've got different masses. So the wavefunction for the electron/muon system does not have to be antisymmetric (although it can be).

However, is a u-quark distinct from a d-quark?

The idea is that there might be just one type of quark, but it can have different quantum numbers.

An electron has different quantum numbers like spin up and spin down, and if you have two electrons, you have to antisymmetrize them:

|\uparrow \downarrow>-|\downarrow \uparrow>If the u,d states of the quark are just different states like spin up and spin down, then you have to antisymmetrize a state of two quarks:

|ud>-|du>

But if u and d are taken as distinct particles (as opposed to distinct states of a single particle), then it is okay to have just |ud>.

The reason I ask is because in various books I've often seen it argued that u and d are different states of SU(3), rather than different particles. Moreover, the Pauli exclusion principle applies to these different states when you have more than one particle, so that you have to choose something like:

|ud>-|du>

rather than |ud>.

The justification for this is (at least I think) that they say that SU(3) is a good symmetry, and the u d and s quarks to a good approximation have the same mass, so that they are all the same particle and u d and s just means different states like spin up and spin down.

So is there an approximate Pauli exclusion principle, that says if 2 particles are almost the same such as the u and d quarks, then their wavefunction must be antisymmetric under exchange of particles to a good approximation? Are they intrinsically the same, but something else (electromagnetic field, Higgs) makes them different? Is that why there is an approximate Pauli exclusion principle that applies to them, even though they're distinct?

Note that this could apply to not just quarks. Is a neutrino different from an electron? Are they different states of SU(2)? What if you turn off Higgs and electromagnetism? Even if you don't turn them off, if you have a neutrino and an electron, are their wavefunctions approximately antisymmetric? Because the argument for quarks seems to be that they should be antisymmetric, even though quarks are distinct - they're close enough that they say that the hadron wavefunctions are antisymmetric.

 

[A. Neumaier]

I'm a little confused about what constitutes a distinct particle.

For example, a muon is not an electron as they've got different masses. So the wavefunction for the electron/muon system does not have to be antisymmetric (although it can be).

However, is a u-quark distinct from a d-quark?

The idea is that there might be just one type of quark, but it can have different quantum numbers.

Yes, and particle known to have distinct quantum numbers are distinct. If you don't know then they are indistinguishable. How they are modeled depends therefore on the context.

An electron has different quantum numbers like spin up and spin down, and if you have two electrons, you have to antisymmetrize them:

|\uparrow \downarrow>-|\downarrow \uparrow>

Yes, but you wouldn't antisymmetrize two bound electrons in an atomic lattice, bound at different sites. Here the position distinguishes the two.

If the u,d states of the quark are just different states like spin up and spin down, then you have to antisymmetrize a state of two quarks:

|ud>-|du>

Yes, this is the usual situation.

But if u and d are taken as distinct particles (as opposed to distinct states of a single particle), then it is okay to have just |ud>.

The reason I ask is because in various books I've often seen it argued that u and d are different states of SU(3), rather than different particles.

This is no contradiction, since ''particles'' are in some sense just labels for the components of a wave vector.

 

[RedX]

Yes, but you wouldn't antisymmetrize two bound electrons in an atomic lattice, bound at different sites. Here the position distinguishes the two.

What makes bound electrons special?

Say I know that a spin up electron is at x=2, and a spin down electron is at x=7. Then would I have to antisymmetrize the wavefunction to be:

|2,\uparrow \rangle \otimes |7,\downarrow \rangle -<br /> |7,\downarrow \rangle \otimes |2,\uparrow \rangle<br />

Or could I just have |2,\uparrow \rangle \otimes |7,\downarrow \rangle?

An electron might have several quantum numbers such as position and spin. In this example position serves to distinguish the two electrons?

If both electrons had spin up, then position would be totally uncorrelated with spin (knowing either one doesn't give you further information about the other), and the wavefunction could be written as a direct product:

|2,7 \rangle \otimes |\uparrow \uparrow \rangle

or would I have to antisymmetrize this direct product to:(|2,7 \rangle - |7,2 \rangle ) \otimes |\uparrow \uparrow \rangle ?

If you don't know then they are indistinguishable. How they are modeled depends therefore on the context.

Creation operators for fermions anti-commute. But don't creation operators for distinct fermions commute? So the question as to whether a u quark is distinct from a d quark or if there is one quark but u and d are different states of a flavor symmetry should depend on if there are two separate creation operators or just one.

So for example the electron has creation operator:

a^{\dagger}(\vec{k},s)

and the momentum and spin are distinct states of the electron. Electrons with different momentum or spin are still the same particle and must be antisymmetrized.

But say your quark creation operator has flavor f (which can take on values u or d):

a^{\dagger}(\vec{k},s,f)

Then a u quark and a d quark must be antisymmetrized, because of anticommutation of the creation operator.

But say the creation operator for u and d are different:

a^{\dagger}(\vec{k},s) for u, and
b^{\dagger}(\vec{k},s) for d.

Then you would not have to antisymmetrize of symmetrize a state containing a u and d quark, because they two operators commute.

So there should be no ambiguity of context: the creation operators fundamentally determine symmetrization of antisymmetrization.

 

[A. Neumaier]

What makes bound electrons special?

They are effective electrons distinguished by sitting at a particular place. This labels them and makes them distinguished. While the electrons in the conduction band of a conducting material float and cannot be labelled. The two kinds of electrons live in different effective Hilbert spaces!

Creation operators for fermions anti-commute. But don't creation operators for distinct fermions commute?

No; they still anticommute.

So the question as to whether a u quark is distinct from a d quark or if there is one quark but u and d are different states of a flavor symmetry should depend on if there are two separate creation operators or just one.

Normally (i.e., in the standard QCD calculations) they are indistinguishable and need to be antisymmetrized.

 

[RedX]

Normally (i.e., in the standard QCD calculations) they are indistinguishable and need to be antisymmetrized.

The u and d quarks can be transformed into each other under flavor isospin SU(2).

Are there separate creation operators for u and d quarks? Or is there just one creation operator, with isospin a label just like spin or momentum:

a^{\dagger}(\vec{k},s,I_3)

I.e., the creation operator creates a quark with momentum k, spin s, and isospin I₃?

 

[A. Neumaier]

The u and d quarks can be transformed into each other under flavor isospin SU(2).

Are there separate creation operators for u and d quarks? Or is there just one creation operator, with isospin a label just like spin or momentum:

a^{\dagger}(\vec{k},s,I_3)

I.e., the creation operator creates a quark with momentum k, spin s, and isospin I₃?

It doesn't matter which notations one uses. Mathematically, both your choices give the same results. Particle kinds are just labels for field components, and their naming is historical accident.