Quantum entanglement of spin along multiple orthogonal axes

Question

Picture an entangled pair of spin 1/2-spin particles with total spin 0. In the diagram, particle 1 of the pair is moving to the left (-y), and particle 2 to the right (+y).

If a z-oriented SG$^*$ is used to detect the spin direction of particle 1 on the left, then the spin direction of particle 2 can be predicted with 100% certainty by using another z-oriented SG on the right. For example, if the left SG finds particle 1 to have spin $\frac{\hbar}{2}$, there is be a 100% probability that a z-directed SG on the right will detect particle 2 as having spin $-\frac{\hbar}{2}$.

Now consider leaving the left SG unchanged (pointing to +z), and rotating the right SG so it will point to +x. If particle 1 is detected on the left to have spin $\frac{\hbar}{2}$, two possibilities can be considered for what will happen when particle 2 reaches the +x-directed SG:

• Particle 2 is detected with 100% probability as having spin $-\frac{\hbar}{2}$ in the +x-direction, or

• Particle 2 is detected with a 50% probability of having spin $-\frac{\hbar}{2}$ in the +x direction, and 50% of having spin $+\frac{\hbar}{2}$ in the -x direction.

In the second case, obviously, it may turn out that the total spin of the system is not equal to zero.

^{*SG: Stern-Gerlach apparatus}

Answer 1

Edited to add : This is an answer to a different question. The real answer is in the comments

If both particle are measures along the same direction, the obtained results would be opposite whatever the direction is. The 0 totalt spin of the singlet state means that there is no specific direction associated to it.

More formally for the $x$ direction, the EPR (singlet) state is $$\begin{align} \left|\Psi^-\right>=\frac1{\sqrt2}\left(\left|\uparrow\downarrow\right>-\left|\downarrow\uparrow\right>\right) \end{align}$$ To rewrite it in the $x$ basis, one use $$\begin{align} \left|\rightarrow\right>&=\frac1{\sqrt2}\left(\left|\uparrow\right>+\left|\downarrow\right>\right)& \left|\leftarrow\right>&=\frac1{\sqrt2}\left(\left|\uparrow\right>-\left|\downarrow\right>\right) \\ \left|\uparrow\right>&=\frac1{\sqrt2}\left(\left|\leftarrow\right>+\left|\rightarrow\right>\right)& \left|\downarrow\right>&=\frac1{\sqrt2}\left(\left|\leftarrow\right>-\left|\rightarrow\right>\right) \end{align}$$ $$\begin{align} \left|\uparrow\downarrow\right>&=\frac12\left(\left|\leftarrow\leftarrow\right>-\left|\leftarrow\rightarrow\right>+\left|\rightarrow\leftarrow\right>+\left|\rightarrow\rightarrow\right>\right) \\ \left|\downarrow\uparrow\right>&=\frac12\left(\left|\leftarrow\leftarrow\right>+\left|\leftarrow\rightarrow\right>-\left|\leftarrow\rightarrow\right>+\left|\rightarrow\rightarrow\right>\right)& \end{align}$$ Therefore $$ \left|\Psi^-\right>=\frac1{\sqrt2}\left(\left|\uparrow\downarrow\right>-\left|\downarrow\uparrow\right>\right) =\frac1{\sqrt2}\left(\left|\rightarrow\leftarrow\right>-\left|\leftarrow\rightarrow\right>\right), $$ which is the result you predicted from the null total spin.

Answer 2

Excellent question.

A quick answer is that spin entanglement can involve large-scale objects also. When a particle spin is detected by the Stern-Gerlach apparatus, the entire apparatus becomes entangled with the other (still isolated) particle. This allows a seemingly sold answer for the particle spin that contradicts conservation, but the answer is an illusion in the sense that spin state of the apparatus as a whole will now be "ready" to balance out whatever result the other particle gives.