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When we perform a double slit experiment we receive interference pattern. when we measure in a specific slit the exact location of the photon (or electron) the interference pattern disappears and we receive only 2 light spots with no interference pattern.

Now let's assume that we perform the double slit experiment on two entangled photon (or electrons) far away from each other (Alice & Bob).

If Alice measures the location of the photon in the slits, she doesn't receive interference pattern, but because of entanglement the location of Bob's photon is now known and he also doesn't get an interference pattern. This way he knows instantaneously that Alice performed a measurement even when she is far away.

If she doesn't measure the location of the photon in the slits, they will both receive interference pattern and Bob will know instantaneously that she didn't measure the photons location.

This ideas based on the EPR famous paper enables faster than light communication. What did I miss here?

Emilio Pisanty
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Eran Sinbar
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  • “the location of Bob's photon is now known” is a wrong assumption. – HolgerFiedler Dec 21 '18 at 06:11
  • Sorry I misread your question at first. This is perfectly legitimate, you don’t even need to use double slit experiment, usually this is shown with different polarizations of light. The point is that this does not send information from Alice to Bob. – lcv Dec 21 '18 at 07:48
  • BTW why the downvotes? Although the answer is out there in many places, this is a legitimate question. And at some point in history the answer was probably not so clear to many. – lcv Dec 21 '18 at 11:39
  • You never get two spots. You will still get a single slit pattern. – Lambda Jun 09 '19 at 15:11

2 Answers2

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This is actually the usual context in which EPR is discussed and the apparent paradox that it may be used to send information faster than light. First note that you don’t need the double slit part, the usual EPR situation works the same way. If Alice measure the polarization in the vertical direction she knows immediately that Bob has horizontal (assuming the system was prepared in the usual singlet state). But this cannot be used to send information from Alice to Bob.

Added comment

The situation is the same as the following. Imagine there are two balls in an urn, a white one and a black one. Alice picks up one without looking and leave the other to Bob. Then travels to Alpha Centauri. When she looks at the ball, in Alpha Centauri, she immediately knows that Bob has the opposite color. No information has been transmitted in this process and obviously no entanglement was present.

lcv
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  • if interference is 0 and non interference is 1 they can communicate between them . – Eran Sinbar Dec 21 '18 at 10:00
  • @EranSinbar This is perfectly clear. As I said, it works the same way in the original EPR experiment (say with photons). The wavefunction is prepared as $|\Psi\rangle = (|V\rangle_A |H\rangle_B - |H\rangle_A |V\rangle_B )/\sqrt{2}$. When Alice measures polarization and gets, say, $V$ she knows that Bob has $H$. See my added comment. – lcv Dec 21 '18 at 11:34
  • @EranSinbar they can't communicate, neither of them knows the order in which the 0's and 1's are going to come, so they can't arrange a message. It is clear that it is not possible to construct a message, because before I measure mine, I don't know what I am going to get, so even thought after I measure I will know what the other person has, I will not be able to send any information. The best thing you could use entanglement for is encryption – Hugo V Dec 21 '18 at 11:41
  • They can communicate in this manner: – Eran Sinbar Dec 22 '18 at 08:03
  • they can communicate in this manner: before leaving earth to Alpha Centauri Alice tells Bob that when she wants to say hello she will measure the location of the entangled photon in the double slit , and Bob will not get interference pattern with its entangled partner. when she is not communicating with Bob he will keep getting interference pattern because she will not measure the location of the entangled photon in her slits. why can't they communicate this way? – Eran Sinbar Dec 22 '18 at 08:09
  • @EranSinbar Because your scheme does not represent what Bob will actually measure. See my answer for the details. – Emilio Pisanty Dec 31 '18 at 23:31
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You've misunderstood how entanglement works. None of your predictions for the experimental measurements of Bob are correct.

From your text, it appears that you're considering an entangled state of the form $$ |\Psi⟩ = |\text{slit }1⟩_A |\text{slit }1⟩_B + |\text{slit }2⟩_A |\text{slit }2⟩_B. $$ Under these conditions neither particle will show an interference pattern, regardless of how you measure.

In quantum mechanics it is not enough to simply refrain from doing an explicit measurement (say, of the which-way information in Alice's slit 1 vs slit 2); if the information is available even in principle then it will still kill any interference patterns that depend on it.

As such, the only way for Bob to measure an interference pattern on his particle is for Alice to perform a quantum eraser experiment on her particle and to communicate those results to Bob.

Specifically, the protocol would require Alice to measure in the basis of states $$ \bigg\{ |{+}⟩_A = |\text{slit }1⟩_A + |\text{slit }2⟩_A, \ |{-}⟩_A = |\text{slit }1⟩_A - |\text{slit }2⟩_A \bigg\}. $$ If Alice gets the result $|{+}⟩_A$, then that will collapse Bob's particle to the state $$ |{+}⟩_B = |\text{slit }1⟩_B + |\text{slit }2⟩_B, $$ and the electron will produce an interference pattern. Similarly, if Alice gets the result $|{-}⟩_A$, then that will collapse Bob's particle to the state $$ |{-}⟩_B = |\text{slit }1⟩_B - |\text{slit }2⟩_B, $$ and that will also produce an interference pattern on Bob's side, but that interference pattern is complementary to that produced by the state $|{+}⟩_B$. Unless Bob knows how to split up the results between Alice's $|{+}⟩_A$s and $|{-}⟩_A$s, he can only ever observe the sum of those two complementary patterns, which adds up to a absolutely-no-interference shapeless blob:

Mathematica graphics

(Results from $|{+}⟩_B$ in red, results from $|{-}⟩_B$ in blue, the sum of the two in black.)

To resolve the two, he needs to wait for Alice to perform her measurements and send the results on a classical channel, which must necessarily go, at most, at the speed of light.

To put it concisely: Your scheme doesn't work. You cannot use entanglement to communicate faster than light. Ever.

Emilio Pisanty
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