2
$\begingroup$

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.

$\endgroup$

1 Answer 1

1
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ 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 $P$? $\endgroup$ Aug 13, 2023 at 8:58
  • $\begingroup$ @DisposableGuy See my edit. $\endgroup$
    – Buzz
    Aug 13, 2023 at 22:07
  • $\begingroup$ 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? $\endgroup$ Aug 14, 2023 at 2:22
  • $\begingroup$ Could you confirm that this is right what I said? $\endgroup$ Aug 16, 2023 at 4:01
  • $\begingroup$ @DisposableGuy Yes, that is correct. $\endgroup$
    – Buzz
    Aug 16, 2023 at 14:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.