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I saw this answer:

Is quantum entanglement functionally equivalent to a measurement?.

I had follow up questions that seem to have been addressed. Here are just a few similar topics:

Why does observation collapse the wave function?.

Does the observer or the camera collapse the wave function in the double slit experiment?.

What constitutes an observation/measurement in QM?.

After reading these, this is my question: The light from a laser in a 2-slit experiment must travel through a medium (air, water, glass, etc..). Why would a diffraction pattern would still be seen if the light interacts with particles just before entering the slit. These particles leak information about which slit the light went through into the environment. Why doesn't this information getting out cause decoherence?

Consider the case where just one particle sits in front of one of the slits. There is 2-slit experiment set up with a single particle emitted (particle A), and there is a second particle (particle B) already sitting and waiting in front of one of the slits. In order for particle A to go through this slit, it must pass by particle B which would B's state to change. Then particle A continues. It passes through the slit. On the other hand, if particle A passed through the other slit, then it does not interact with particle B, and no state change occurs for B. Particle A eventually lands on the landing detector. After particle A has landed, we can check particle B. If particle B's state has changed since we placed it in front of the slit before starting the experiment, then we know particle A went through that slit. Otherwise, A went through the other slit. Therefore, the final outcomes of A and B are correlated.

According to other answers, the wave function should collapse because particle B acts as a detector. Some say particle B measures particle A. So when A interacted with B, it's wave function collapsed, so there should not be an interference pattern.

This isn't my main question, but this leaves me wondering why there would also not be an interference pattern if A went through the other slit and did not become entangled with B. We know if B's state is unchanged, A went through the other slit. So how does B's mere presence cause A's wave function to collapse? No interaction occurred. If there was an interaction, then B's state would have changed. But how could B's state change if A went through the other slit? It's paradoxical.

However, this is my main question. I've seen a laser in a 2-slit setup produce an interference pattern in a classroom. My point is that particle B could have been any particle floating in front of the slit. Most photons would dodge the dust particles, but with the soup of all kinds of cosmic particles floating around, does the light from the laser really make it to the slits without interacting with anything along the way? If the light does in fact interact with other particles along the way, then does that mean the measurement problem depends on whether or not the information leaked out is discernable? In other words, these particles A interacts with carry information about which slit it went through. However, since we are unable to capture any of these witnessing particles, A goes on to create an interference pattern. It is only in a controlled setting where we can discern particle B's state that we can derive information about which slit A went through. How would such a system know in advance whether or not we would be able to discern such a thing after the event, thus deciding whether or not A should result in a interference pattern or not?

PS: If you think the laser is just dodging particles in the air, suppose we make the room very dusty. Or suppose we conduct the experiment underwater.

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  • $\begingroup$ A laser in air makes a fine 2-slit (or 1-slit) diffraction pattern. Not clear why entanglement is needed at all. A very confused question. $\endgroup$ – Jon Custer Jun 15 '17 at 15:53
  • $\begingroup$ My question is asking why would it make a diffraction pattern since photons would be interacting with other particles just before going through any slit. Why don't these interactions count as measurement, thus preventing diffraction patterns? Entanglement is not needed but is a consequence of the photon interacting with one of the particles floating in front of the slit. In the simplified example with A and B, an entangled particle can function as a which-way detector. $\endgroup$ – Croolsby Jun 15 '17 at 16:06
  • $\begingroup$ I revised my question to say "correlated" instead of "entangled". This language is closer to the point that I am trying to make. $\endgroup$ – Croolsby Jun 15 '17 at 16:25
  • $\begingroup$ The point is that 'correlated' and 'entangled' mean precise things in physics. In particular, entanglement is generally not a consequence of a photon scattering off of some random particle. $\endgroup$ – Jon Custer Jun 15 '17 at 16:31
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    $\begingroup$ Why do you think that quora discussion support your ideas? It is specifically discussing about whether a photon retains its entanglement with another photon as it bounces off mirrors vs randomly polarising surfaces? What's the connection with your question? $\endgroup$ – user154997 Jun 16 '17 at 0:49
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I am afraid your question is conceptually very similar to Feynman classic discussions: for the interferences, you use photons where he used electrons; for the slit detection, you use air molecules where he used photons. I strongly encourage you to go and read it. This is in this chapter of lectures. The specific experiment I referred to is in section 1-4 "An experiment with electrons" but it is best to read the whole chapter.

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Let's do the math for this problem. We're dealing with a two particle system, if particle A moves through one of the two slits, then this will change the state of particle B. If the initial state is formally denoted by $\left|B\right>$, the final state will be some different state $\left|B'\right>$. But if particle A moves through the other slit then the state of particle B will be unchanged, it will still be $\left|B\right>$. Then let's consider the complete quantum state of the two particle system, just before particle A is going to hit the screen. We can formally denote this as:

$$\left|\psi\right> = \left|A_1\right>\left|B\right> + \left|A_2\right>\left|B'\right>$$

Here $\left|A_1\right>$ is the state of particle A if it moves through slit 1, and that then doesn't affect particle B, and $\left|A_2\right>$ is the state of particle A if it moves through the other slit, and then the state of particle B will have been affected. The probability $P(x)$ of particle A being detected at some position $x$ on the screen is given by the squared norm of the state obtained by projecting $\left|\psi\right>$ to the state of particle A being at position $x$. So, we have:

$$P(x) = \left|\left<x\right.\left|A_1\right>\left|B\right> + \left<x\right.\left|A_2\right>\left|B'\right>\right|^2$$

So, we have a linear combination of the states $\left|B\right>$ and $\left|B'\right>$ and we need to compute the squared norm of that state, this just the inner product of that linear combination with itself. Since $\left|B\right>$ and $\left|B'\right>$ are normalized to 1, we have:

$$P(x) = \left|\left<x\right.\left|A_1\right>\right|^2 + \left|\left<x\right.\left|A_2\right>\right|^2 + 2 \operatorname{Re}\left[\left<x\right.\left|A_1\right>^* \left<x\right.\left|A_2\right>\left<B\right.\left|B'\right>\right] $$

What we see here is that the last term is the interference term and this will vanish if the states $\left|B\right>$ and $\left|B'\right>$ are orthogonal. Such states are then physically distinct states while states that are not orthogonal cannot be perfectly distinguished from each other. If the states are not orthogonal then the interference pattern will still be visible, but it will be reduced by a factor given by the overlap between the two states. In that case the fact that you can't perfectly distinguish the two states means that you can't tell which slit A went through with certainty. But the more certain this is, the weaker the interference pattern will be.

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