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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 Two Stern-Gerlach apparatuses directed along z-direction

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2 Answers

up vote 2 down vote accepted

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.

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Frédéric, I like your answer, but my own reading of his question was a bit different. Here's a different version of the same problem, stated perhaps a bit more bluntly: If you detect an entangled net-zero-spin 1/2 spin particle pair using SGs at right angles to each other, spin conservation appears to be violated no matter what results you get, simply because there is no way to add up the orthogonal spin axis results to get zero total angular momentum. That's why I lean towards saying it stays quantum -- there is still entanglement in the system -- even after both detections have been made. –  Terry Bollinger Mar 10 '12 at 13:29
@Frédéric Grosshans Your calculations show that the singlet state is directionally independent, there is no doubt. Assume that SG{+z} measurement of the 1st particle spin direction finds it in +z axis directed spin. What is the result of measurement, we should expect from SG{+x}? –  Sergio Mar 10 '12 at 13:39
OK, I didn't read the question properly. If the first measurement is $+\hbar/2$ alon $z$, the second spin is in state $\left|\downarrow\right>$. Measuring this second spin in the $x$ direction would give a random result. This measurement does not allow to measure the total spin of the system, hence the apparent spin non-conservation. –  Frédéric Grosshans Mar 11 '12 at 15:44
@ Frédéric Grosshan It is strange to me, that spin does not conservate. What does it happen for total system, which consist of SG{+z}+entangled pair+SG{+x}? Does it mean the total system spin will not concerve? –  Sergio Mar 11 '12 at 17:14
1st: you don't need to prepare an entangled pair to have this effect. It appears also if you have a single spin polarized along z and measured along x. 2nd: I don't think the spin is conserved in this case. If it were conserved, it would be possible to measure simultaneously the spin both along x and along z. The non-conservating operation is the "wave-packet" reduction (in Copenhagen interpretation), which transforms $|\uparrow\rangle=\frac1{\sqrt2}(|\rightarrow\rangle+|\leftarrow\rangle)$ into a mixture of $|\rightarrow\rangle$ and $|\leftarrow\rangle$. But if the observer is included ... –  Frédéric Grosshans Mar 11 '12 at 17:49
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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.

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tnks for clarifing my question. –  Sergio Mar 11 '12 at 17:04
Sergio, it was a good question and you are most welcome. And to address your original question more directly: Your second option of 50/50 is definitely what an SG will do experimentally (see Feynman Lectures, Vol III). Also, all possible readings will nominally violate conservation of angular momentum, not just some, as Frédéric also noted above. A quick graphical way to verify that something is amiss is to draw the four possible outcomes of the SGs and try to add them like classical angular momentum vectors. It doesn't work. The resulting vectors are non-zero and also have invalid lengths. –  Terry Bollinger Mar 12 '12 at 5:17
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