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Can you entangle two already existing particles? or do you have to "create" entangled particles from "scratch"? if so, how would one do this? (say, entangle two already existing electrons).

Edit: Better example of what I'm asking, lets say I had an electron in a block of gold and then an electron in a block of silver I could entangle them and use each electron to manipulate each other?

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  • $\begingroup$ Whenever you let an environment interact with your system, the two get entangled, but nothing is (necessarily) getting created. A special case of this is 'measuring' your system. $\endgroup$ Jun 20 at 15:29
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All scattering experiments of high energy physics start with the scattering of two particles. At the point of interacion there exists a quantum mechanical wavefunction describing mathematically the interaction, and all particles coming out of the interaction are by definition "entangled" by conservations laws and quantum number conservations. See this answer of mine..

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  • $\begingroup$ Yes but technically aren’t you just trying to correlate the two? $\endgroup$ Jun 20 at 15:18
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    $\begingroup$ @BillAlsept In scattering experiments one is studying the quantum mechanical interaction between particles, that is the whole point of the experiment. What is called entanglement, the usual spin correlations are a tiny part of the knowledge of the interactions gained by the experiment. $\endgroup$
    – anna v
    Jun 20 at 15:33
  • $\begingroup$ Scattering two particles is a physical action. In the controlled experiment what are you physically doing to each particle. $\endgroup$ Jun 20 at 15:52
  • $\begingroup$ @BillAlsept making them meet at a "point" $\endgroup$
    – anna v
    Jun 20 at 15:56
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    $\begingroup$ @BillAlsept, no, I am talking of a single electron on a single electron-scattering experiment with known polarization of the incoming electrons. then the two outgoing electrons will have a polarization, and measuring one gives you also the polarization of the other, spin entanglement. $\endgroup$
    – anna v
    Jun 20 at 18:18
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Any interaction between two systems will typically lead to entanglement. If $\vert \phi_1\rangle$ and $\vert \psi_2\rangle$ are eigenstates of $H_1$ and $H_2$ respectively, then an interaction term $V_{12}$ will usually be so that the eigenstates of \begin{align} H=H_1+H_2+V_{12} \end{align} are not of the form $\vert\chi_{12}\rangle =\vert\phi_1\rangle\vert\psi_2\rangle$ but rather of the form \begin{align} \vert\chi_{12}\rangle =\sum_{ij}c_{ij}\vert\phi_i\rangle\vert\psi_i\rangle\, . \end{align} Loosely speaking, suppose \begin{align} H\vert\phi_1\rangle\vert\psi_2\rangle &=(H_1+H_2+V_{12})\vert\phi_1\rangle\vert\psi_2\rangle\, ,\\ &=(E_1+E_2)\vert\phi_1\rangle\vert\psi_2\rangle +\sum_{ij}\vert\phi_i\rangle \vert\psi_j\rangle\langle \phi_i;\psi_j\vert V_{12}\vert\phi_1;\psi_2\rangle\, . \end{align} Unless $\langle \phi_i;\psi_j\vert V_{12}\vert\phi_1;\psi_2\rangle$ is proportional to $\left(\sum_{i}c_i\vert\phi_i\rangle\right)\vert\psi_2\rangle$, the result will not be a separable state. But if $V_{12}$ gives such a linear combination, there is no interaction with the 2nd system as it has remained undisturbed by $V_{12}$.

This is not a rigorous argument but it does give insight into how entanglement typically results from interactions.

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  • $\begingroup$ But it still comes down to simply correlating the movement of two objects so that after sending them on their way you could later measure one and know something about the other. It takes precise effort to do this. You cannot simply smash two particles together and hope they become correlated. $\endgroup$ Jun 20 at 15:26
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    $\begingroup$ @BillAlsept that is incorrect, it is hard to avoid correlation, it is not hard to create correlation. Of course, it is hard to create correlation of a very particular nature; maybe that is what you are thinking of? $\endgroup$ Jun 20 at 15:31
  • $\begingroup$ @RubenVerresen if the two objects are not perfectly correlated then what kind of measurement can you make after that? $\endgroup$ Jun 20 at 15:56
  • $\begingroup$ @BillAlsept I'm not sure I understand the trust of your comment. Entanglement is synonymous with no-separable: that's all that's needed. There is no notion of perfect correlation, or post-measurements here: those are issues separate to entanglement itself. Maybe you can expand on what you have in mind when you bring the idea of "perfect correlation" into the debate? $\endgroup$ Jun 20 at 16:35
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    $\begingroup$ @ZeroTheHero i’m not sure what that means. What I know is they claim the predictions of quantum mechanics cannot be classically reproduced. Therefore something special (and incomplete) called entanglement is needed to try and explain what’s not understood. You don’t need the complicated things mentioned above, just stick with a simple experiment like multiple polarizers. cos2theta can be explained physically without something called entanglement. $\endgroup$ Jun 20 at 17:28

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