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When looking at the Isospin representation of a proton-neutron system (with the notation $|I,I_3\rangle$), you can go from an uncoupled to a coupled representation like this:

$$\textstyle|\frac{1}{2},+\frac{1}{2}\rangle|\frac{1}{2},-\frac{1}{2}\rangle=\sqrt{\frac{1}{2}}(|1,0\rangle+|0,0\rangle)$$

However, you can also find the following antisymmetric wavefunction when switching the first two kets around (or switching proton and neutron):

$$\textstyle|\frac{1}{2},-\frac{1}{2}\rangle|\frac{1}{2},+\frac{1}{2}\rangle=\sqrt{\frac{1}{2}}(|1,0\rangle-|0,0\rangle)$$

If you now want to calculate the probability of finding the proton-neutron system in a $|1,0\rangle$ state, you have to take the square of the inproduct of the above expressions with $\langle1,0|$, which gives $\frac{1}{2}$ for both expressions.

Do you now have to add these 2 probabilities together, since they should be equivalent, though one is symmetric and the other is asymmetric.
However, adding the probabilities means the probability to find the proton-neutron state in the $|1,0\rangle$ state is 1, but you could reason similarly for $|0,0\rangle$ and so find a total probability of 2 which doesn't seem right.

It seems weird that the wavefunction is different if you assume a proton-neutron vs. a neutron-proton system.

As a background for this question, I need to calculate the relative probabilities for 2 reactions, one of which is a pion and deuteron reacting to form a proton and neutron, so I'm writing left and right side in the coupled representation and taking the inproduct.

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It seems there is a bit of confusion here. A state like $$ \textstyle\vert \frac{1}{2},\frac{1}{2}\rangle_1\vert \frac{1}{2},-\frac{1}{2}\rangle_2 \tag{1} $$ does not transform into itself under permutation of the particle index so is neither symmetric nor antisymmetric. As written, (1) describes distinguishable nucleons: the first nucleon is certainly the proton and the second is certainly the neutron.

If your nucleons are indistinguishable, you need to work with properly symmetrized states $$ \vert\psi_\pm\rangle =\frac{1}{\sqrt{2}}\Bigl(\textstyle\vert \frac{1}{2},\frac{1}{2}\rangle_1\vert \frac{1}{2},-\frac{1}{2}\rangle_2\pm \vert \frac{1}{2},-\frac{1}{2}\rangle_1\vert \frac{1}{2},\frac{1}{2}\rangle_2\Bigr) $$ which are basically the $\vert 1,0\rangle$ and $\vert 0,0\rangle$ states you already have.

To compute the probability of finding the proton-neutron system (as a system of distinguishable particles) in $\vert 1,0\rangle$, you have $$ \textstyle \vert \langle 1,0\left[\vert \frac{1}{2},\frac{1}{2}\rangle_1\vert \frac{1}{2},-\frac{1}{2}\rangle\right]\vert^2 =\frac{1}{2}\, . $$ This is the probability of having the first nucleon a proton, and the second nucleon a neutron, in the isospin state $\vert 1,0\rangle$.

At this point you can ask about the probability of having the first nucleon a neutron, and the second a proton, and this must be $1/2$ since the probabilities must sum to 1. You can also ask about the probability of having the first nucleon a proton, and the second nucleon a neutron, in the isospin state $\vert 0,0\rangle$, which again must be $1/2$ for the same reason that the probabilities sum to $1$.

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  • $\begingroup$ So in the case of Isospin states, a proton and neutron system is always a set of distinguishable particles, right? I'm not really getting how nucleons can be undistinguishable, if they're the same then their isospin projection is identical and the problem above is not relevant any more. If I'm understanding correctly, the answer to my problem is just 1/2? Also I'm having trouble understanding your last paragraph, does the first sentence refer to the case of finding the nucleons in the |1,0> state? $\endgroup$
    – Joshua
    Oct 26, 2017 at 16:18
  • $\begingroup$ @Joshua Protons and neutrons are identical when it comes to strong interactions, which is the same between any pair of nucleons is the same. Also, isospin manifest itself in the near-degeneracy of some particle masses: the proton and the neutron have almost the same mass and the difference can be attributed to electromagnetic effects at the quark level rather than effects of the strong force. See en.wikipedia.org/wiki/Isospin#Isospin_symmetry $\endgroup$ Oct 26, 2017 at 16:36
  • $\begingroup$ I know about Isospin symmetry, the strong force doesn't see protons or neutrons, rather nucleons with isospin up or down. If "a proton-neutron" system is a distinguishable set, could you give an example of an undistinguishable set? Is the distinction made between "a proton and a neutron" and "two nucleons where one has isospin up and the other down"? And again referring to your last paragraph, I don't really understand your first sentence there, are you solely talking about the chance that nucleon 1 is a neutron and nucleon 2 is a proton? $\endgroup$
    – Joshua
    Oct 26, 2017 at 17:38
  • $\begingroup$ @Joshua |proton>|neutron>$\pm$ |neutron>|proton> $\endgroup$ Oct 26, 2017 at 17:47

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