Why is a deuteron an antisymmetric singlet in

[RedX]

Why is a deuteron an antisymmetric singlet in isospin:

|\uparrow\downarrow>-|\downarrow\uparrow>=|0,0>

whereas a proton and neutron that are separated are a combination of an antisymmetric singlet and a symmetric triplet:

|\uparrow\downarrow>=|0,0>+|1,0>

I don't understand the difference. Isn't a deuteron a neutron and a proton? (the spin up arrow stands for a proton state, and the spin down arrow a neutron state).

I read somewhere that a deuteron is antisymmetric in isospin because it is composed of two identical fermions so that the wavefunction needs to be antisymmetric! Since when was a proton identical to a neutron? And why doesn't that logic hold for a proton and a neutron that are separated?

 

[Bill_K]

The deuteron is a neutron and a proton in an S-state with parallel spins. The idea of isospin is that the neutron and proton are both the "same" particle in different isospin states. Being fermions, their overall wavefunction must be antisymmetric. But the wavefunction now consists of three parts: space, spin and isospin.

For the deuteron the space and spin parts are symmetric, hence the isospin part must be antisymmetric. The state in which the isospin part is symmetric would correspond to an antisymmetric spin part (antiparallel spins) but for a neutron and proton this state happens to be unbound.

 

[RedX]

The deuteron is a neutron and a proton in an S-state with parallel spins. The idea of isospin is that the neutron and proton are both the "same" particle in different isospin states. Being fermions, their overall wavefunction must be antisymmetric. But the wavefunction now consists of three parts: space, spin and isospin.

For the deuteron the space and spin parts are symmetric, hence the isospin part must be antisymmetric. The state in which the isospin part is symmetric would correspond to an antisymmetric spin part (antiparallel spins) but for a neutron and proton this state happens to be unbound.

Thanks.

I was under the impression that the Pauli exclusion principle and antisymmetry only applies to indistinct particles. If one of the particles is slightly different than the other - however slight the difference - then they are distinct particles. So I don't think antisymmetry is a requirement. Yet that is what the notes I have say.

Now there is nothing wrong with having an antisymmetric wave function - even distinct particles can have an antisymmetric wave function - but it's just not required.

The neutron and the proton are distinct. If you have to apply antisymmetry and Pauli exclusion to them, then I think you have to show that particles that are very similar to each other, obey the Pauli exclusion principle too and antisymmetry. As far as I know, these are only requirements for particles that are exactly indistinct.

Also I'm not sure why this logic doesn't hold for a neutron and a proton that are not bound, but are both free in space. I'd think the wavefunction would be:

\psi(x_1,x_2)|\mbox{singlet or triplet}>|\uparrow \downarrow>

where the spatial part doesn't necessarily have any type of symmetry on swapping x₁ and x₂, the spin state is some linear combination of the singlet and triplet state, and the isospin state has no type of symmetry on swapping the two arrows - the first particle (at x₁) is a proton, and the second (at x₂) is a neutron.

Now what is the difference between a proton and neutron that are widely separated, and a deuteron? I think the answer is in the spatial wavefunction:


\psi(x_1,x_2)|\mbox{singlet or triplet}>|\uparrow \downarrow>

For a deuteron \psi(x_1,x_2) vanishes when x₁ is far from x₂. But changing the spatial wavefunction doesn't change the spin or isospin states when you have a direct product. So by this logic, the deuteron should be the same wavefunction:

\psi(x_1,x_2)|\mbox{singlet or triplet}>|\uparrow \downarrow>

except \psi(x_1,x_2) will change to vanish outside (x_1-x_2)^2 < r^2, where r is the spatial size of deuteron.

But obviously it's not, and deuteron is actually:

\psi_S(x_1,x_2)|\mbox{triplet}>|\uparrow \downarrow-\downarrow\uparrow >

but I can't figure out why.

 

[Bill_K]

I was under the impression that the Pauli exclusion principle and antisymmetry only applies to indistinct particles. If one of the particles is slightly different than the other - however slight the difference - then they are distinct particles. So I don't think antisymmetry is a requirement. Yet that is what the notes I have say.

Most certainly, the exclusion principle applies only to identical particles, and the proton and neutron are not identical. Never mind "slightly different", they ARE different. Isospin is a formalism for treating nonidentical particles. It's convenient, but not in any way fundamental. In place of Ψ = (space)(spin) you add an extra degree of freedom and consider Ψ = (space)(spin)(isospin). For a state which contains both a proton and a neutron, the actual wavefunction (space)(spin) can have either symmetry. So you tack on an isospin state which makes up for that, so that the overall (space)(spin)(isospin) is antisymmetric. Again, this is just formalism.

Also I'm not sure why this logic doesn't hold for a neutron and a proton that are not bound, but are both free in space.

It does. Isospin is used in scattering problems too.

 

[Bill_K]

Note there are two relevant questions

1) When can you do it?
2) When is it useful?

It's important to keep the two issues separate. Only after you realize the answer to (1) is "always" -- only then should you consider (2)!