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Is Quantum Entanglement the one-dimensional spatial relationship between two particles as described by mathematics?

By one dimensional I mean, when they travel at opposite directions their spin is also opposite (without any spooky communication between the two, just the plain fact of their direction as described by our observations).

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  • $\begingroup$ I am a noob in Quantum Computing so if you thing this question is somewhat irrelevant (more of a physics question), offsetting, already put and answered or wrongly put, please reply as soon as you can. $\endgroup$
    – petpet
    Mar 15 '20 at 10:56
  • $\begingroup$ Maybe only when the direction of the movement of each particle is opposite can we speak of entanglement or, maybe better formulated, ideal entanglement. $\endgroup$
    – petpet
    Mar 15 '20 at 13:08
  • $\begingroup$ Hi @petpet and welcome to the Quantum Computing SE :). In most of the quantum hardware systems, the entanglement is not related to the qubit movement direction just because they are always fixed in their positions. $\endgroup$ Mar 15 '20 at 13:17
  • $\begingroup$ or the aproximation of an 180 degrees angle $\endgroup$
    – petpet
    Mar 15 '20 at 13:18
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    $\begingroup$ @petpet note that you need to tag people in a comment for them to be notified by your comment. A system being "bipartite" refers to it having more than one degree of freedom. The polarisation of spatially separated particles is a common example of this, but you can have a bipartite state composed of different degrees of freedom of a single particle, e.g. position and polarisation of a single photon. E.g. a single photon's state $|H,\text{left}\rangle+|V,\text{right}\rangle$ is entangled. As I was saying, spatial separation of particles is usually involved, but it's not a fundamental feature $\endgroup$
    – glS
    Mar 16 '20 at 10:13
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A qubit is a two-level quantum system and these two-levels in some hardware can be implemented by spins of the electron, but it's not the only option (can be just two "stabile" quantum states not essentially associated with spins). In this sense it is not only about the spin (directional) correlation between qubits, it's about state correlation between them. But suppose that we have spin qubits: two quantum levels are two spin directions (up or down) in some chosen direction. An example of an entangled state for two qubits:

$$| \psi \rangle = \frac{1}{\sqrt{2}}\left( | \downarrow_z \rangle| \downarrow_z \rangle + | \uparrow_z \rangle| \uparrow_z \rangle \right)$$

This means that if we will do a measurement of spin in the z-direction we will always measure both qubits down or up (nice explanation of the entanglement 2:50 - 4:15 of Anton Zeilinger's presentation). But not only this. A valid question is what is our $|\psi\rangle$ state expressed by spin up and down states in x-direction. In other words what will be the measurement results, if we measure the spin in the x-direction. Note that (video lecture by Allan Adams about spin: especially from 47:45-52:45):

\begin{align*} | \uparrow_z \rangle = \frac{1}{\sqrt{2}}\left( | \uparrow_x \rangle + | \downarrow_x \rangle \right) \qquad | \downarrow_z \rangle = \frac{1}{\sqrt{2}}\left( | \uparrow_x \rangle - | \downarrow_x \rangle \right) \end{align*}

That's why:

$$| \psi \rangle = \frac{1}{\sqrt{2}}\left( | \downarrow_x \rangle| \downarrow_x \rangle + | \uparrow_x \rangle| \uparrow_x \rangle \right)$$

So, if we will measure in x-direction we will still have a correlation between the $| \uparrow_x \rangle$ and $| \downarrow_x \rangle$ measurement results of two entangled qubits. The same is true for y-direction. Thus, this is not only about a one-dimensional correlation. And, it is not about where are physically our qubits: they can be in a fixed position, they can be at some distance and no matter what the measurement results should be the same.

For more, I highly recommend this lecture by Allan Adams (about separable states and entangled states from 28:33-36:02) and this video from Veritasium youtube channel.

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