can particles that have two entangled properties remain entangled if only 1 of the properties is measured at a time?

Question

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?

Answer 1

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.