1
$\begingroup$

A video I watched from PBS spacetime talked about how the pilot wave theory provides an explanation for why quantum entanglement occurs, where it is a result of having global hidden variables. I just want to know if there is an explanation like this in the standard interpretation.

How does quantum mechanics predict quantum entanglement? Does the concept of entanglement come from any mathematical equations?

$\endgroup$
1
  • 4
    $\begingroup$ Once you accept that the state space of a composite system is the tensor product of the state spaces of the components, it's obvious on dimensional grounds that almost all states are entangled. $\endgroup$ – WillO Feb 27 at 6:51
-2
$\begingroup$

Entanglement arises from the mathematical structure of quantum mechanical states. It is not exclusive to quantum mechanics - here is a non-quantum example of entanglement.

Suppose I have a bag containing $40$ balls. An experiment consists of drawing a ball at random from the bag and then putting it back again. After doing many experiments I know that half the balls in the bag are red and half are blue. I also know that half the balls in the bag are large and half are small. I draw a ball from the bag and it is a red ball. What is the probability that it is also a large ball ?

It is tempting to say $50\%$ since half the balls in the bag are large and half are small. And indeed this is possible; if the bag contains $10$ large red balls, $10$ large blue balls, $10$ small red balls and $10$ small blue balls, then the probability of drawing a red or a blue ball is independent of the probability of drawing a large or a small ball. We could say the bag is a "tensor" product of two separate probability distributions or "states":

$(0.5 \text{ red} + 0.5 \text{ blue}) \otimes (0.5 \text{ large} + 0.5 \text{ small})$

But suppose the bag contains $20$ large red balls and $20$ small blue balls. The probability of drawing a red or blue ball and the probability of drawing a large or a small ball are now interdependent - we could say they are "entangled". There are now no values $p$ and $q$ that will let us write the contents of the bag as a tensor product

$(p \text{ red} + (1-p) \text{ blue}) \otimes (q \text{ large} + (1-q) \text{ small})$

Quantum entanglement arises in a similar way, although there are several important differences that mean the "bag of balls" analogy is not exact:

  1. The amplitudes in a quantum mechanical state are complex numbers, not real numbers.
  2. We may not be able to determine the colour and the size of a ball at the same time, because of Heisenberg's uncertainty principle.
  3. We think there is no exact equivalent of the bag of balls in quantum mechanics, because this would make quantum mechanics a hidden variable theory.
$\endgroup$
4
  • 4
    $\begingroup$ Not only is the analogy not exact, it is for the very least highly misleading, by suggesting that quantum entanglement in essence is nothing but classical correlations. However, the point is that quantum entanglement goes fundamentally beyond classical correlations, as we know latest since Bell's works. I think it is crucial that such a point is made very clear, and not declares as a "not exact" analogy. $\endgroup$ – Norbert Schuch Feb 27 at 11:58
  • $\begingroup$ @NorbertSchuch You are, of course, entitled to your opinion. But it seems clear to me that quantum entanglement is an extension of classical correlation (with some important differences which I have highlighted in my answer) but is not a fundamentally different phenomena. $\endgroup$ – gandalf61 Feb 27 at 12:33
  • 1
    $\begingroup$ Well, first of all, I wanted to explain my downvote. Second, and more importantly, I wanted to make it very clear that I find this answer highly misleading - I think this is important especially in the SE scheme. I agree it is an extension of classical correlations, just like classical correlations are an extension of uncorrelated systems, or just the way quantum entanglement is an extension of unentangled quantum systems, which by themselves are already much more rich than classical uncorrelated systems. However, unlike you I think there is a key difference to classical correlations ... $\endgroup$ – Norbert Schuch Feb 27 at 13:15
  • 2
    $\begingroup$ ... and, more importantly: Classical correlations are intuitive to everyone. My impression is that these classical analogies lead people to think that quantum entanglement is sth. which can be understood just as much from classical intuition. I think it is a major mistake to lure people into that belief: They think they understand quantum entanglement, while they don't. (A senior colleague who has a state named after him considers it a hallmark of bad explanations of quantum entanglement if they equally work with classical objects.) $\endgroup$ – Norbert Schuch Feb 27 at 13:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.