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Let's suppose you have two entangled particles that each have two entangled properties. The first property can be either A or B and the second can be either 0 or 1. According to this video, measuring the first property will lock its state, but then measuring the second property will unlock the first property's state. Does this mean that if you only measure one property at a time of two entangled particles that each have two entangled properties, the two particles will continue to remain entangled? So for example if first measure the 0 1 property on one particle and then measure the A B property on the same particle then the two particles will then continue to be entangled for the 0 1 property?

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It depends on what those properties are, and whether simultaneous observation of the two properties is possible, and how, specifically, the entanglement is set up. I will assume in this answer that measurement of one property does not change the other in any way (formally, I will assume that the operators for these properties commute). If the two properties are incompatible with simultaneous measurement, then measuring one property essentially randomizes the other property.

Each particle has four possible measurable states: $|A0\rangle$, $|A1\rangle$, $|B0\rangle$, and $|B1\rangle$. Therefore, the system of two particles has sixteen possible measurable states: $|A0A0\rangle$, $|B0A0\rangle$, $|A1A0\rangle$, etc. Any possible state of the two-particle system can be written as a superposition of these sixteen states.

Suppose the two-particle system was in the state $\frac{1}{\sqrt{2}}(|A0A0\rangle+|B1B1\rangle)$. This is an entangled state because it cannot be separated into a product of single-particle states. In this state, an observer has a 50 percent chance of measuring both the first particle and the second particle to have properties A0, and an observer has a 50 percent chance of measuring both particles to have properties B1. In this state, the properties of both particles are correlated; measuring one particle gives you knowledge about the properties of the other. More than that, though, in this particular state, the properties of an individual particle are also correlated; measuring the first property of the first particle to be A, for example, tells you that the second property is 0. Therefore, measuring one property of one particle means that the system is no longer entangled.

This is not true of all states, however; for example, take the state $\frac{1}{\sqrt{3}}(|A0A0\rangle+|B0A0\rangle+|A1B0\rangle)$. This state is still entangled; if you measure the first particle's properties and find them to be A1, then the second particle is in the state B0. Otherwise, the second particle is in state A0. If you measure the second particle's properties and find them to be A0, then the first particle is in an equal superposition of states A0 and B0. Otherwise, the first particle is in state A1. But now the properties are not necessarily fully correlated. If you measure the first property of the first particle to be A, then the system is in the state $\frac{1}{\sqrt{2}}(|A0A0\rangle+|A1B0\rangle)$. So the system is still entangled, and in particular, the second property of the first particle and the first property of the second particle are correlated (measuring 0 for the first particle means that the second particle will have A, and measuring 1 for the first particle means that the second particle will have B).

Different entangled states do different things when you measure them, depending on the particular way in which they are entangled.

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  • $\begingroup$ Thanks for you excellent answer. If the case is that simultaneous measurements are not possible and the two properties are not correlated in any way, then would it be possible to first measure the first property on a particle and get A, then measure the second property and get either 0 or 1, then measure the first property again on the same particle and get B, and then measure the second property and get either 0 or 1? If this is possible, would this act of these series of measurements cause the second particle not to be entangled anymore to the first particle? $\endgroup$ – kishdude Dec 7 '18 at 16:22
  • $\begingroup$ @kishdude (1/2) If simultaneous measurements of both properties are not possible, then measurement of one property affects the other property. After measuring the first property, the second property is effectively randomized. So once you measure the first property, you don't really get any new information from the measurement of the second property, and so on with the subsequent measurements. So every measurement past the first one is not helpful. $\endgroup$ – probably_someone Dec 7 '18 at 16:33
  • $\begingroup$ @kishdude (2/2) In any case, once you measure the first property of the first particle, that affects the second property of the first particle, and also lets you know what the state of the first property of the second particle is, assuming that you measure the first property of the second particle. If instead you measure the second property of the second particle, then that affects the first property of the second particle, which means that the information you got about this property from your earlier measurement of the first particle will be incorrect. $\endgroup$ – probably_someone Dec 7 '18 at 16:36
  • $\begingroup$ @kishdude So, with a system like you have described, the only way you can retrieve any information from entanglement is if you measure one property of the first particle, then measure the same property of the second particle, without interacting with either of them in between. Doing such a pair of measurements will return the system to an unentangled state. (Now you see why I assumed that simultaneous measurement of both properties was possible...) $\endgroup$ – probably_someone Dec 7 '18 at 16:37
  • $\begingroup$ So in essence what you are saying is that measuring the first property of the first particle and then measuring the first property of the second particle will completely unentangle the particles such that even the second property of the particles will have no correlation? $\endgroup$ – kishdude Dec 7 '18 at 17:41

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