Heisenberg exchange term in phenomenological nuclear potentials

In phenomenological nucleon-nucleon potentials there are three exchange forces terms:

1. $V_M=V_M(r) P_{x}$ : Majorana exchange (exchange of position coordinates, $P_{x}$ is the operator of exchange of coordinates)
2. $V_B=V_B(r) P_{s}$ : Bartlett exchange (exchange of spin, $P_{s}$ is the operator of exchange of spins)
3. $V_H=-V_H(r) P_{t}$ : Heisenberg exchange (exchange of isospin, $P_{t}$ is the operator of exchange of isospins)

(The minus sign comes from Pauli principle)

I can't undersand the following about term 3.: The nucleon nucleon interaction should not be dependent on the isospin orientation, but if I consider:

• a singlet of isospin (proton-neutron), which is antisymmetric, then the eigenvalue of $P_t$ is $-1$
• a state of isospin triplet proton-proton, which is symmetric, then the eigenvalue of $P_t$ is $+1$

The $V_H$ changes its sign in the two cases, i.e. the interaction is different if the two nucleons have parallel or antiparallel isospin, or, in other words, the force is different in the systems neutron-proton and proton-proton (or neutron-neutron), but this cannot be true.

Am I missing something about how the term $V_H$ actually work?

• There is also a proton-neutron isospin triplet state ($|\uparrow\downarrow\rangle+|\downarrow\uparrow\rangle$), so it's not about whether the isospin is parallel or antiparallel. – probably_someone Jun 30 '18 at 22:25
• "but this cannot be true." why do you say that? As far as I can remember this is actually a true statement. – Quantum spaghettification Jul 1 '18 at 3:50