Entanglement swapping and Bohmian mechanics

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Since the recent thread on Bohmian mechanics (BM) is now closed, I open a new thread in which I want to explain something that bothers @DrChinese for a long time. How can BM explain the entanglement swapping? But I want to make this explanation simple and intuitive, rather than technical. I want to explain what, in my opinion, is the main conceptual misunderstanding behind DrChinese's impression that BM can't explain entanglement swapping. For that purpose I will start with something completely fictional, not directly related to actual science, because this fiction, I believe, will help much to convey the basic intuitive idea that I want to convey.

Suppose that I have the paranormal powers of clairvoyance and telekinesis. By clairvoyance, I can see things in my mind without seeing them by my eyes. By telekinesis, I can move things without touching them. Clearly, I am talking about such fictional paranormal powers because they are analogous to nonlocal action at a distance in BM.

Now suppose that there is a card which I don't see by my eyes and can't touch. Nevertheless, I can see it in my mind by clairvoyance and move at will by telekinesis. And suppose that, for some reason, I decide to follow the following rule: It the card is red, move it to the left. If the card is blue, don't do anything.

Now the crucial conceptual question that I want to address is the following: When the card is blue, do I really have paranormal powers at all? Someone looking from the side might think that I don't, because I don't move the card and don't show any sign that I know the color of the card. And yet, from my own point of view, I do have paranormal powers because I do know that the color is blue and it is precisely this knowledge why I don't move the card. From my perspective, the fact that I don't move the card is a demonstration that I do have the power of clairvoyance. I could even move it if I wanted, but I don't do it because I have chosen so. I just follow the rule that I decide to follow by my own. In other words, my paranormal powers do not cease when the card is blue. They in fact never cease, they just don't manifest themselves explicitly under certain conditions so it looks as if they ceased, but in reality they didn't.

Now we can move to physics. BM is very similar to these paranormal powers. According to BM, particles always have the power of clairvoyance. Each particle instantaneously knows the positions of all other particles in the universe. However, it doesn't mean that they always use this power for telekinesis. Instead, the particles follow a rule, and the rule itself changes. The rule is encoded in the pilot wave (the wave function of the universe), and this pilot wave changes in time, according to the Schrodinger equation. One of the rules encoded in the pilot wave says the following: If the wave function is not entangled, then don't apply the telekinesis powers. For instance, if the wave function of two particles A and B, with positions ##x_A## and ##x_B##, has the product form ##\psi(x_A,x_B)=\psi_A(x_A)\psi_B(x_B)##, then the rule says that velocity of A does not depend on the position of B, and vice versa. In other words, when the particles are not entangled it looks as if all nonlocal powers of BM vanish. But that's an illusion, they don't vanish, they just don't manifest themselves explicitly.

Let us illustrate this by a simple example. Suppose that, at a certain time ##t##, ##\psi(x_A,x_B)## has a form
$$\psi(x_A,x_B)=\psi_A(x_A)\psi_B(x_B) \;\;\; {\rm for} \;\;\; x_B>0$$
$$\psi(x_A,x_B) \neq \psi_A(x_A)\psi_B(x_B) \;\;\; {\rm for} \;\;\; x_B\leq 0$$
This should be considered as one wave function, written separately for ##x_B>0## and ##x_B\leq 0##. The full wave function, valid for all ##x_A,x_B##, is not a product. It is an entangled wave function. Any yet, for ##x_B>0## it looks like a product wave function without entanglement. So if the B particle happens to have the Bohmian position ##X_B>0##, then the rule says that the particle A has to move by velocity that does not depend on the position of B. But in order to obey this rule, the particle A has to know that ##X_B>0##. So, the motion of A "independently" on B actually depends on B. This demonstrates that motion always depends on positions of all particles, even when this dependence is not manifest.

And now we can finally discuss the entanglement swapping. It involves 4 particles, A, B, C and D. Initially the wave function has the form ##\psi_{AB}\psi_{CD}##, so A is entangled with B, and C is entangled with D, but there is no entanglement between A and D. Nevertheless, each particle knows positions of all other particles. In particular, A knows the position of D, and vice versa, but this knowledge does not have a direct manifestation. In other words, the motion of A depends on the position of D, and vice versa, but this dependence is not manifest. The entanglement swapping is a way to make this dependence manifest.

More specifically, one brings the waves of B and C into an interaction, making B entangled with C. The details can be found in the standard literature and they are not important here. The point is that this changes the full wave function of all 4 particles and the result is that A and D become mutually entangled. In other words, now the Bohmian motion of A depends on the position of D, and vice versa. What seems to be confusing to DrChinese is how can suddenly the motion of A may start to depend on the position D, if there was no such dependence from the beginning? How can such a dependence be created without the interaction between A and D? And what exactly creates such a dependence? The answer is that the dependence was there from the start, it was never really created. It is only that certain changes in the system changed the rules of the game (encoded in the change of the pilot wave), so that, at the time when B and C interacted, the dependence of A on D became manifest.
 
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What about the case where the measuremtns on A and D are done, the particles no loger exist and only then B and C are measured? This a point @DrChinese makes often.