# Why can a coupled wavefunction of 2 nucleons seemingly arbitrarily give two different expressions?

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.

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 $1$.
• @Joshua |proton>|neutron>$\pm$ |neutron>|proton> Oct 26, 2017 at 17:47