# What makes particles lose entanglement or do they ever lose it?

Once we generate a pair of entangled electrons or photons, will they ever lose their entanglement?

If yes, then what causes them to lose entanglement?

If measuring them causes them to lose entanglement then how can we be sure of the EPR experiment? As measuring them anyway breaks entanglement and the result generated may be totally unrelated?

So do the particles remain entangled?

• physics.stackexchange.com/questions/218085/… – pathintegral Jun 15 '19 at 3:52
• check out the monogamy property of quantum entanglement. according to this, it seems that if spin A and spin B were initially entangled, and after a measurement, B and the apparatus becomes entangled, the A and B will no longer be entangled. – pathintegral Jun 15 '19 at 3:55

If lighting a firework causes it to burn all of its gunpowder, then how can we be sure it contained gunpowder in the first place? The answer: we look at the explosion of the firework, which indicates that it contained gunpowder (and, with precise enough observations, can even tell us how much gunpowder it contained). Similarly, measurements of an entangled system carry information about the entanglement. In particular, classically-impossible correlations between measurements of different parts of the system indicate that the system was entangled (and, with precise enough observations, can even tell us how/in what way the system was entangled), even though the parts of the system are not entangled afterwards.

• OK , so to be clear, 1.The entanglement is lost after measurement and particles are not entangled after measurement? 2.We know of entanglement only from "classically-impossible correlations"? – VARUN.N RAO Jun 12 '19 at 13:10
• @VARUN.NRAO 1) When you measure a part of an entangled system, it is no longer entangled with other parts of the system (in systems consisting of more than two parts, there still may be entanglement between two or more unmeasured parts). The other parts of the system will be in a state that is related in a specific way with the state of the measured part. 2) That's the main evidence that favors entanglement over alternatives like local hidden-variable theories. The experimental violation of Bell's inequality leaves as possibilities either entanglement or a nonlocal hidden-variable theory. – probably_someone Jun 12 '19 at 13:48
• @VARUN.NRAO In the above, I'm assuming that by "measurement" you mean the usual definition, namely an interaction whose result is a pure state not in any kind of superposition. There is significant interest in "weak measurements" for which this is not the case (but from which very little information can be gained), and the above doesn't necessarily apply to those. – probably_someone Jun 12 '19 at 13:52
• Thank you so much. – VARUN.N RAO Jun 12 '19 at 16:23
• I'm assuming that by "measurement" you mean the usual definition, namely an interaction whose result is a pure state not in any kind of superposition. A pure state isn't in contradistinction to a superposition. Every pure state can be considered to be either a superposition or not a superposition, depending on what basis you choose. Presumably what you mean is that a measurement results in decoherence between outcomes in which the measurement is a different eigenvalue of the operator being measured. But that wouldn't be relevant to your answer. Your answer is simply asserting the [...] – user4552 Jun 12 '19 at 17:48

If measuring them causes them to lose entanglement [...]

Measurement doesn't destroy entanglement. Quantum mechanics works through the unitary evolution of the wavefunction. Suppose that a pion decays into an electron and an antielectron with opposite spins. The spins are entangled. If the correlation between their spins were to somehow be erased, it would violate conservation of angular momentum. If Ed measures the electron's spin, and Alice measures the antielectron's, then, through a unitary process, Ed and Alice have become entangled as well. Ed and Alice are in a mixture of the states $$\Psi_1=|\uparrow\downarrow\rangle$$ and $$\Psi_2=|\downarrow\uparrow\rangle$$.

Now for practical reasons, we're not going to be able to observe interference between $$\Psi_1$$ and $$\Psi_2$$, because it's too hard to observe interference involving objects like human bodies. However, that doesn't mean that the state has become separable. Suppose that it did, and the result was that the system ended up in state 1. Then what would the time-evolution of the system look like in the basis consisting of the states 1 and 2? Since it takes any mixture of the electron-antielectron states 1 and 2 and sends it to a pure state 1, we could imagine that it would be represented by a matrix something like

$$\left(\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}\right).$$

But this is a nonunitary matrix, which is impossible.

• Could the downvoter explain their reason for downvoting? If I'm wrong, I'd like to understand what my mistake was. – user4552 Jun 12 '19 at 17:41
• I'm not the downvoter, but this is very confusing as the other person is disagreeing with you. I made some extra research and found that entanglement is destroyed the moment you measure the concerned quantity. – VARUN.N RAO Jun 13 '19 at 1:56
• Oh I get it, what you're saying is way profound. You are saying the entanglement property of the electron once measured is not limited to electrons but now even the measuring devices are entangled. Right? The system on the whole remains connected. But my main question is for all the practical purposes of Bell's experiment, the two electrons lose all statistical weirdness once measured. They are not locally entangled anymore, they are globally entangled. If I understand correctly. This is a beautiful world view and answer. I'm upvoting – VARUN.N RAO Jun 13 '19 at 2:02