So I've been sitting on the following question for days now and I really gave my best, but I just can't seem to get the right solution.
The initial problem was that we have a singlet-triplet qubit $$ |0\rangle = \frac{1}{\sqrt{2}}|\uparrow\downarrow\rangle + |\downarrow\uparrow\rangle $$$$ |0\rangle = \frac{1}{\sqrt{2}}|\!\uparrow\downarrow\rangle + |\!\downarrow\uparrow\rangle $$ $$ |1\rangle = \frac{1}{\sqrt{2}}|\uparrow\downarrow\rangle - |\downarrow\uparrow\rangle $$
I$$ |1\rangle = \frac{1}{\sqrt{2}}|\!\uparrow\downarrow\rangle - |\!\downarrow\uparrow\rangle.$$ I have the positive operator $P$ to measure spin parity:
$$P = (1-s)(|\uparrow\uparrow\rangle\langle\uparrow\uparrow| + |\downarrow\downarrow\rangle\langle\downarrow\downarrow|) \\ + s(|\uparrow\downarrow\rangle\langle\uparrow\downarrow| + |\downarrow\uparrow\rangle\langle\downarrow\uparrow|) $$ $$P = (1-s)(|\!\uparrow\uparrow\rangle\langle\uparrow\uparrow\!| + |\!\downarrow\downarrow\rangle\langle\downarrow\downarrow\!|) + s(|\!\uparrow\downarrow\rangle\langle\uparrow\downarrow\!| + |\!\downarrow\uparrow\rangle\langle\downarrow\uparrow\!|) $$ where $s=0$ ->$s=0\rightarrow$ spins are same and $s=1$ ->$s=1\rightarrow$ spins are different.
I need to explain that the measurement of the spin parity by $P$ does not conserve total angular momentum.
I know that I need to do this by showing that the total angular momentum operator and parity operator do not commute, so $[J,P] \neq 0$, and I can assume the form of the states being in an arbitrary superposition of the basis: $$|\uparrow\uparrow\rangle, |\uparrow\downarrow\rangle, |\downarrow\uparrow\rangle, |\downarrow\downarrow\rangle$$$$\left\{|\!\uparrow\uparrow\rangle, |\!\uparrow\downarrow\rangle, |\!\downarrow\uparrow\rangle, |\!\downarrow\downarrow\rangle\right\}.$$
What I did was to try to find the matrices for both operators and then calculate $PJ-JP$, but I think I got it wrong, because when I do it, they seem to commute.
I tried to express $P$ as: $$ \begin{equation*} P = \begin{pmatrix} (1-s) & 0 & 0 & 0 \\ 0 & s & 0 & 0 \\ 0 & 0 & s & 0 \\ 0 & 0 & 0 & (1-s) \end{pmatrix} \end{equation*} $$
because$$ \begin{equation*} P = \begin{pmatrix} (1-s) & 0 & 0 & 0 \\ 0 & s & 0 & 0 \\ 0 & 0 & s & 0 \\ 0 & 0 & 0 & (1-s) \end{pmatrix}, \end{equation*} $$ because I thought the outer product of e.g. $|\uparrow\uparrow\rangle\langle\uparrow\uparrow|$$|\!\uparrow\uparrow\rangle\langle\uparrow\uparrow\!|$, where $|\uparrow\uparrow\rangle$$|\!\uparrow\uparrow\rangle$ has the form $(1,0,0,0)$$(1,0,0,0)^{T}$ gives a matrix,
$$ \begin{equation*} A = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \end{equation*} $$
and $$ \begin{equation*} A = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}, \end{equation*} $$ and when I do this for all $P$, I end up with the matrix above.
For $J$, I thought just to use the spin operator $S^2$ for two spin-$1/2$ particles (because I assumed there is no orbital momentum), which I found here:
$$ \begin{equation*} S^2 = \hbar \begin{pmatrix} 2 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 2 \end{pmatrix} \end{equation*} $$
But $$ \begin{equation*} S^2 = \hbar^{2} \begin{pmatrix} 2 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 2 \end{pmatrix}. \end{equation*} $$ But when I do this, $[J,P]=0$, so they do commute.
I'm really struggling here, so would be open for any help. Am I on the right track or is it completely wrong?
I'm not looking to get the complete solution, just some hints what I may need to consider.
