Entangled particle Superposition(Simple question)

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Consider two virtual entangled particles (+ve & -ve particles) which emerged out of nothing.
We keep +ve and -ve in two different boxes. If the box containing +ve particle is closed and we do not observe the particle, then it is said to be in a superposition of +ve and -ve, Right? After some time, we open the box and observe the particle. We then again keep the particle back in the box and close it i.e. stop observing it. Will it again be back into a state of superposition of +ve and -ve?

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Staff Emeritus
2019 Award
Consider two virtual entangled particles (+ve & -ve particles) which emerged out of nothing.
There is no such thing, so it's hard to consider it.

It doesn't need to "return" to it. Whatever its state after the observation, it will be a superposition in some base.

PeroK
Homework Helper
Gold Member
You seem to be confusing entanglement and superposition. Do you understand what these terms mean?

Demystifier
Gold Member
Consider two virtual entangled particles (+ve & -ve particles) which emerged out of nothing.
We keep +ve and -ve in two different boxes. If the box containing +ve particle is closed and we do not observe the particle, then it is said to be in a superposition of +ve and -ve, Right? After some time, we open the box and observe the particle. We then again keep the particle back in the box and close it i.e. stop observing it. Will it again be back into a state of superposition of +ve and -ve?
Measurement is not reversible. Once the superposition is destroyed by measurement, stopping observation will not restore the superposition.

Derek P and Allen_Wolf
Staff Emeritus
2019 Award
PeroK is right. It seems like you don't understand what a lot of the terms mean: at least virtual, superposition, entanglement. It might make more sense for you to re-pose your question, carefully checking to see that the words you use are the words you mean.

To answer the question you asked - which I suspect is not the question you intended to ask - if I measure a partcle's angular momentum along the z-direction, it's in a superposition of states along the x-direction.

Allen_Wolf
PeterDonis
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2019 Award
Consider two virtual entangled particles (+ve & -ve particles) which emerged out of nothing.
We keep +ve and -ve in two different boxes.
You can't keep virtual particles in boxes.

Consider two virtual entangled particles (+ve & -ve particles) which emerged out of nothing.
We keep +ve and -ve in two different boxes. If the box containing +ve particle is closed and we do not observe the particle, then it is said to be in a superposition of +ve and -ve, Right? After some time, we open the box and observe the particle. We then again keep the particle back in the box and close it i.e. stop observing it. Will it again be back into a state of superposition of +ve and -ve?
If you prepare a particle as a member of an entangled pair you cannot ascribe a state to it at all, you have to specify the other particle and talk about the state of the two together. Of course if you cleanly ignore the other particle you can talk about the one you still have but then it is simply a probability distribution - it might be +ve or it might be -ve.

I can see what you are trying to do. You want to specify a scenario where the particle is unambiguously in superposition. Far better to prepare your superposition by taking a single particle, putting it into a known state and then, if necessary, forcing it into a superposition of the two states you want. Doing this with charge is next to impossible, but with spin or polarisation it is quite easy.

Also, quantum states are nothing to do with opening boxes - Schroedinger's cat has a lot to answer for!

And observation isn't a continuous process that we start and stop. It's an interaction.