# What properties are entangled in quantum entanglement?

When two particles are entangled, one can measure the properties of one of the particles and instantaneously know the properties of the other. This is because the two particles possess the same properties when entangled, no matter how far apart.

Which properties are entangled?

Is it just fundamental properties, such as spin, or can it be instantaneous/changing properties, such as the speed of the particles?

It may seem very odd to suggest that speed could be entangled, but I think it's a reasonable question to ask as quantum entangles is plenty weird anyway.

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In principle, you can entangle any observable properties of a given pair of particles that you wish. All you have to do is to form the right entangled state between the two particles given the observable you care about.

Consider, for example, two identical particles 1 and 2 whose states are both described by the Hilbert space $\mathcal H$, and let $\Omega$ be an observable on $\mathcal H$, two of whose normalized eigenvectors are $|\omega_1\rangle$ and $|\omega_2\rangle$ with corresponding eigenvalues $\omega_1$ and $\omega_2$. This observable can in principle be anything you wish such as spin, momentum, angular momentum, etc. Now consider the state $$|\psi\rangle = \frac{1}{\sqrt 2}\left(|\omega_1\rangle|\omega_2\rangle +|\omega_2\rangle|\omega_1\rangle\right)$$ of the combined, two-particle system. This state has the property that if the observable $\Omega$ is measured on a given particle with result $\omega_1$, then we know with certainty that a simultaneous measurement of $\Omega$ on the other particle will yield the result $\omega_2$ with certainty, and vice versa.

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Not only you can entangle any property of the two particles, you can also, in principle, entangle one property of one particle (say its spin), with a different property of the second particle (say its position). – becko Mar 13 '13 at 0:49
Or entangle one property of a particle with another property of the same particle. But then maybe people don't call that entanglement any more. – Dan Piponi Mar 13 '13 at 1:17

$\vert \psi(p_{1},p_{2})\rangle=\vert \psi(p_{1})\rangle\otimes\vert \psi(p_{2})\rangle - \vert \psi(p_{2})\rangle\otimes\vert \psi(p_{1})\rangle$ $(p_{1},p_{2})$are the momentum's of particle 1 & 2 Infact Cooper pair(arising in superconductivity) is one such state of two electrons which remain entangled in momentum space.

In heisenberg picture you can define velocity operator

$\hat{v}=\frac{d\hat{x}}{dt}=\frac{i}{\hbar}[\hat{x},H]+\frac{\partial \hat{x}}{\partial t}$. And then define eigenstates of velocity operator and write down the entangled state.

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