0
$\begingroup$

We know that if we measure the spin of electron along X and it turns out to be up and then we measure along Y and then again along X, the probability of spin being up along X is 50% and down is 50%

What happens if you do this on an entangled pair of electron. Say A and B. Say A was up and B was down along X, now we measure A along Y and then measure again along X, let's say it turns out to be down, has the spin of B now changed to up?

$\endgroup$
2
$\begingroup$

I'm going to write this out in excruciating detail in hopes that this will be useful if you're just getting used to this kind of calculation.

I'm also going to indulge my habit of not distinguishing between a quantum state and the vector that represents that state, and not requiring vectors to be normalized, so that I'll cheerfully write equations like $A=2A$.

You haven't completely specified the problem because (among other things) you haven't specified the state spaces or the initial entangled state, but let's assume the state spaces are two-dimensional and you're starting with a generic entangled state $$\Phi=\alpha U\otimes U+\beta U\otimes D+\gamma D\otimes U+\delta D\otimes D$$ where $U$ and $D$ are (orthogonal) eigenstates of your "$X$-direction" observable.

"Measuring Particle $A$ in the $X$-direction" means applying an observable with eigenspaces $U\otimes {\cal H}_Y$ and $V\otimes{\cal H}_Y$. You've assumed that the outcome of your measurement picks out the first of these subspaces, so you want to project $\Phi$ onto that subspace to get your new state $$\Phi'=\alpha U\otimes U+\beta U\otimes D$$

Then you measure particle $B$ in the $X$-direction, which means applying an observable with eigenspaces ${\cal H}_X\otimes U$ and ${\cal H}_X\otimes D$. You've assumed that the outcome of your measurement picks out the second of these subspaces, so you want to project $\Phi'$ onto that subspace to get your new state $$\Phi''=U\otimes D$$

Now you're going to "measure $A$ along $Y$", but haven't told us what specific operator you've got in mind (or what it's eigenstates are) but I'll assume it's an operator $E$ with eigenspaces $(U+D)\otimes {\cal H}_Y$, $(U-D)\otimes {\cal H}_Y$ (if you want some other eigenspaces you can modify what follows accordingly).

Thus your measurement along $Y$ picks out one of these subspaces. I'll assume it picks out the first, and you can mimic the calculation to see what happens if it picks out the second. We now want to project $\Phi''$ onto $(U+D)\otimes{\cal H}_Y$, giving a new state $\Phi'''=(U+D)\otimes D$. (Check it out: $\Phi''=U\otimes D=(U+D)\otimes D+(U-D)\otimes D$).

Now if you "measure $B$ along $X$" again, y your eigenspaces are ${\cal H}_X\otimes U$ and ${\cal H}_X\otimes D$ --- and your state $\Phi'''$ is already in the second of these eigenspaces. So the only possible outcome for your measurement is $D$.

$\endgroup$
  • $\begingroup$ OH MY GOD!!!!!!! $\endgroup$ – VARUN.N RAO Dec 26 '18 at 19:12
  • $\begingroup$ Ok to paraphrase and dummify your answer, it means, no B is still going to have spin down along X axis. $\endgroup$ – VARUN.N RAO Dec 26 '18 at 19:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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