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Entangled states

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So i understand that two particles are said to be entangled if their joint wavefunction cannot be written as a product of their separate wavefunctions. So [math]\psi(\underline{r}_1, \underline{r}_2) = \psi_1(\underline{r}_1) \psi_2(\underline{r}_2) [/math] is not entangled whereas [math] \psi(\underline{r}_1, \underline{r}_2) = \frac{1}{\sqrt{2}}\Big(\psi_1(\underline{r}_1) \psi_2 (\underline{r}_2) - \psi_2(\underline{r}_2) \psi_1 (\underline{r}_1) \Big) [/math] is entangled.

 

Does this stem from the fact that if two random variables in a joint probability distribution are independent of each other then their respective marginal distributions can also be written as a simple product: and if they cannot be expressed as products then they are not independent?

I think that is a valid observation - although it should be borne in mind that if observations/measurements are obtained from sets of two entangled particles then the observed distributions can be shown not have come from a single, (even if hidden), joint distribution.

  • 3 weeks later...

Agreed with the probabilistic approach, but just keep in mind that the wave function is much more than a way to compute probability densities. Especially, it is a complex number with a phase.

 

The definition "not a product of separate wave functions" shows how general and common entanglement is. For instance, two electrons in a helium atom are entangled. If by some means you observe one electron in a small subvolume of the atom, then the other electron probably isn't there, because of their electric repulsion. The Schrö equation for both electrons includes this repulsion which would be seen in the solution if only we knew an algebraic solution.

 

QM gets a little bit abstract at that point, at least to me, because the wave function for these two electrons in the fundamental state is still stationary. That is, no time involved here, except the term exp(iEt/hbar). The probability density is strictly static, so it's not when one electron is in a small subvolume, but if, that the other electron probably isn't. Even our language lacks simple expressions for that.

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