Showing that measurement of spin parity does not conserve total angular momentum

Question

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 $$ |0\rangle = \frac{1}{\sqrt{2}}|\!\uparrow\downarrow\rangle + |\!\downarrow\uparrow\rangle $$ $$ |1\rangle = \frac{1}{\sqrt{2}}|\!\uparrow\downarrow\rangle - |\!\downarrow\uparrow\rangle.$$ I have the positive operator $P$ to measure spin parity: $$P = (1-s)(|\!\uparrow\uparrow\rangle\langle\uparrow\uparrow\!| + |\!\downarrow\downarrow\rangle\langle\downarrow\downarrow\!|) + s(|\!\uparrow\downarrow\rangle\langle\uparrow\downarrow\!| + |\!\downarrow\uparrow\rangle\langle\downarrow\uparrow\!|) $$ where $s=0\rightarrow$ spins are same and $s=1\rightarrow$ 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: $$\left\{|\!\uparrow\uparrow\rangle, |\!\uparrow\downarrow\rangle, |\!\downarrow\uparrow\rangle, |\!\downarrow\downarrow\rangle\right\}.$$

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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 I thought the outer product of e.g. $|\!\uparrow\uparrow\rangle\langle\uparrow\uparrow\!|$, where $|\!\uparrow\uparrow\rangle$ has the form $(1,0,0,0)^{T}$ 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 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: $$ \begin{equation*} S^2 = \hbar^{2} \begin{pmatrix} 2 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 2 \end{pmatrix}. \end{equation*} $$ 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.

Answer 1

For angular momentum to be conserved under the action of an operator $O$, it is not enough that $[O,J^{2}]=0$. Instead, each of the three commutators $[O,J_{k}]$ must vanish; all three angular momentum components $J_{k}$ need to be included. For a composite angular momentum $\vec{J}=\vec{S}\quad\!\!\!\!\!^{(1)}+\vec{S}\quad\!\!\!\!\!^{(2)}$, the components $J_{k}$ are simply the sums $S^{(0)}_{k}+S^{(1)}_{k}$, all of which must commute with the Hamiltonian if the angular momentum (vector) is to be conserved. (Since angular momentum is a vector, it is time dependent if any of its Cartesian components is.)

To see how that this must be the case, consider a single spin-$\frac{1}{2}$ particle. In that case the magnitude of the angular momentum never changes; $S^{2}=\frac{3}{4}\hbar^{2}\mathbf{1}$ is proportional to the identity and so commutes with anything. However, the components of $\vec{S}$ do not need to be conserved in the presence of an external torque. For example, with a magnetic dipole Hamiltonian $H=-\gamma\vec{S}\cdot\vec{B}$, the commutators $[H,S_{x}]$ and $[H,S_{y}]$ are nonzero, so those two components of the spin angular momentum are not conserved. However, since $[H,S_{z}]=0$, the $z$-component of the spin is actually conserved.