Physics Asked by suissidle on November 28, 2020
I was watching Leonard Susskind’s video series on quantum entanglement, where he looks at the spins of two electrons. In particular, there are entangled states of the form $$alphaleft|uparrow downarrow rightrangle + betaleft|downarrow uparrowrightrangletag {*})$$
A special case is the singlet state $left|uparrow downarrow rightrangle – left|downarrow uparrowrightrangle$, whose spin can be measured to be 0 in any direction.
In order to be 100% sure to have an entangled spin state, one would have to measure it, but can entangled states be eigenvectors of Hermitian operators (= results of measurements) other than the trivial one? Can these operators be expressed as tensor products of measurements on each electron (“sigmas” and “taus” in the lectures)?
[I imagine the following method. Electron pairs shot through a Stern-Gerlach-type apparatus split into three families:
$$left|uparrow uparrow rightrangle$$
$$left|uparrow downarrow rightrangle, left|downarrow uparrowrightrangle$$
$$left|downarrow downarrow rightrangle$$
(careful not to split the pairs!)
The $0$-spin fraction then contains the entangled pairs, with the coefficients of the linear combination (*) such that the spin is with 100% probability equal to 0 on the measured axis. One can then refine this fraction again along the remaining directions and end up with the singlet state.]
I'm not sure your question is as well posed as you think it is.
In order to be 100% sure to have an entangled spin state, one would have to measure it, but can entangled states be eigenvectors of Hermitian operators (= results of measurements) other than the trivial one?
If you know something is in one of various orthogonal states then in principle the official answer is that you can find out which one. But otherwise in general you can find out of something was in a particular state unless you have many copies (and by no cloning you can't make the copies you have to already have them). And doing a measurement doesn't help. He word measurement is seriously misleading. When you so the thing called a measurement you make something become an eigenvector of that measurement. But you can do it in such a way that if it already were an eigenvector of that operator that it doesn't change.
When you force it into becoming an eigenvector you force it to be in one of various subspaces. But the set of entangled states ... is ... not ... a ... subspace.
For instance $left|uparrow downarrow rightrangle-left|downarrow uparrowrightrangle$ is an entangled state as is $left|uparrow downarrow rightrangle+left|downarrow uparrowrightrangle$ but their sum is $left|uparrow downarrow rightrangle$ which is not. So there is not a subspace of entangled states onto which you could project.
You could project onto a specific entangled state.
Can these operators be expressed as tensor products of measurements on each electron ("sigmas" and "taus" in the lectures)?
No. And as for the pair thing, that's vague and a Stern-Gerlach is just going to separate states. If you want to measure the z component of the total spin you could break it into three subspaces the span of $left|uparrow uparrowrightrangle,$ the span of $left|downarrow downarrowrightrangle,$ and the span of ${left|uparrow downarrow rightrangle,left|downarrow uparrowrightrangle}.$ But this would not give you entangled states if they weren't already entangled. And the result would not tell you if they were entangled.
Answered by Timaeus on November 28, 2020
Get help from others!
Recent Answers
Recent Questions
© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP