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

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: $$\left\{|\!\uparrow\uparrow\rangle, |\!\uparrow\downarrow\rangle, |\!\downarrow\uparrow\rangle, |\!\downarrow\downarrow\rangle\right\}.$$

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.

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.

• Thanks for the answer. I have two thoughts on this: 1. I think I got it initially wrong because I found out that $S^2$ actually only covers the magnitude of the spin operator, I think I was really looking for $\hat{S}$ but I don't really know how to get this for two spin half particles. 2. Regarding your answer: This means I need to show that every component $S_x$, $S_y$ and $S_z$ does not commute with $P$? Does this need to be true for all of them or is it enough that only one component does not commute with $P$? Commented Aug 13, 2023 at 8:58
• @DisposableGuy See my edit.
– Buzz
Commented Aug 13, 2023 at 22:07
• So I found out that $S^2$ and $S_z$ do commute with $P$, but $S_x$ and $S_y$ do not commute with $P$, which should mean that the operator $P$ does not conserve total angular momentum. So it is sufficient for only one of the components of $S$ to not commute with an operator $O$ to say that the total angular momentum is not conserved? Commented Aug 14, 2023 at 2:22
• Could you confirm that this is right what I said? Commented Aug 16, 2023 at 4:01
• @DisposableGuy Yes, that is correct.
– Buzz
Commented Aug 16, 2023 at 14:17