High School How to differentiate b/w two collapsed values in Bell States

[randomuser3210]

Given the Bell State in Z basis $\psi = (a\uparrow \downarrow - b\downarrow \uparrow )$ where $a^2+b^2=1$.

Now Alice has one particle and sends the other to Bob. Suppose Alice decides to measure her particle in Z direction. Thus entanglement collapses.

Query:

1. How do we know which state the entanglement has collapsed to (if measured in Z axis). I mean both states give one $\uparrow$ and one $\downarrow$. So how do we know the collapse was the state $\uparrow \downarrow$ or $\downarrow \uparrow$?

2. Changing the value of a and b also seems to have no effect as long as we limit ourselves to Z axis a we always get one $\uparrow$ and one $\downarrow$?

P.S. We are limiting ourselves to Z-axis only.

What am I missing ?

 

[Nugatory]

1. How do we know which state the entanglement has collapsed to (if measured in Z axis). I mean both states give one $\uparrow$ and one $\downarrow$. So how do we know the collapse was the state $\uparrow \downarrow$ or $\downarrow \uparrow$?

If the measurement result at Alice's detector is spin-up, then the collapse was to $|\uparrow\downarrow\rangle$; if the measurement result at Alice's detector is spin=down, then the collapse was to $|\downarrow\uparrow\rangle$.

2. Changing the value of a and b also seems to have no effect as long as we limit ourselves to Z axis a we always get one $\uparrow$ and one $\downarrow$?

Yes, but if we perform the experiment many times (necessary anyways, just to confirm that Alice and Bob always get opposite results) we will find that Alice gets on average a spin up result $a^2$ times and a spin-down result $b^2$ times.

(This might be a good time to mention that preparing the two-particle system in the state $a|\uparrow\downarrow\rangle+b|\downarrow\uparrow\rangle$ with $a$ and $b$ not equal and both non-zero is only feasible in thought experiments)

 

[randomuser3210]

If the measurement result at Alice's detector is spin-up, then the collapse was to $|\uparrow\downarrow\rangle$; if the measurement result at Alice's detector is spin=down, then the collapse was to $|\downarrow\uparrow\rangle$.
Yes, but if we perform the experiment many times (necessary anyways, just to confirm that Alice and Bob always get opposite results) we will find that Alice gets on average a spin up result $a^2$ times and a spin-down result $b^2$ times.

Thank you. And this is true assuming we always decide to give which particle to whom randomly (say by a coin toss) ?

and I didn't get the thought experiment part. I mean is it due to limitation of our experimental setup or engineering capabilities as of today or something else?

 

[Nugatory]

Thank you. And this is true assuming we always decide to give which particle to whom randomly (say by a coin toss) ?

Yes, to the extent that even makes sense - which isn't much. There is no "which particle" here because they are indistinguishable.

You really want to break yourself of the classical habit of thinking of this as problem involving two particles. It is a single quantum system set up in such a way that some observables are measured at different locations. When we're talking about it using natural language (which is irretrievably biased by our lifetime of experience with classical objects) it's convenient to refer to things that might be measured at one location as properties of one particle and things that might be measured at the other detector as properties of the other particle. But that's not right - we have a single quantum system whose wave function allows the calculation of all the results of any sequence of measurements we carry out on it.

 

[PeroK]

Given the Bell State in Z basis $\psi = (a\uparrow \downarrow - b\downarrow \uparrow )$ where $a^2+b^2=1$.

Now Alice has one particle and sends the other to Bob. Suppose Alice decides to measure her particle in Z direction. Thus entanglement collapses.

Query:

1. How do we know which state the entanglement has collapsed to (if measured in Z axis). I mean both states give one $\uparrow$ and one $\downarrow$. So how do we know the collapse was the state $\uparrow \downarrow$ or $\downarrow \uparrow$?

2. Changing the value of a and b also seems to have no effect as long as we limit ourselves to Z axis a we always get one $\uparrow$ and one $\downarrow$?

P.S. We are limiting ourselves to Z-axis only.

What am I missing ?

First, the $\psi$ you specify is not anti-symmetric in the two particles. Unless $a = b$, of course. It's not a valid spin state unless you have an anti-symmetric spatial wave-function. It's simpler, perhaps, to consider these as distinguishable particles, in which case Alice and Bob know which ones they've got. You might have an electron and a proton, for example.

PS and it's not a Bell state.

 

[randomuser3210]

Nugatory

Yes, to the extent that even makes sense - which isn't much. There is no "which particle" here because they are indistinguishable.

You really want to break yourself of the classical habit of thinking of this as problem involving two particles. It is a single quantum system set up in such a way that some observables are measured at different locations. When we're talking about it using natural language (which is irretrievably biased by our lifetime of experience with classical objects) it's convenient to refer to things that might be measured at one location as properties of one particle and things that might be measured at the other detector as properties of the other particle. But that's not right - we have a single quantum system whose wave function allows the calculation of all the results of any sequence of measurements we carry out on it.

Thank you. I understand. Of course the limitation of classical language. I would take care.And I am curious about the 'only in thought experiment' as I asked. WhI mean is it due to limitation of our experimental setup or engineering capabilities as of today or something else?

 

[PeroK]

And I am curious about the 'only in thought experiment' as I asked. WhI mean is it due to limitation of our experimental setup or engineering capabilities as of today or something else?

Before you get too deeply into what happens if we measure a two-particle system in this state, you might ask how you prepare two particles in this state to begin with.

 

[randomuser3210]

First, the $\psi$ you specify is not anti-symmetric in the two particles. Unless $a = b$, of course. It's not a valid spin state unless you have an anti-symmetric spatial wave-function. It's simpler, perhaps, to consider these as distinguishable particles, in which case Alice and Bob know which ones they've got. You might have an electron and a proton, for example.

PS and it's not a Bell state.

I should have stated 'Generic Version of Bell Equation' or something to that effect. But even if we have two different particles after collapse so we can distinguish them as you suggest, the rest of the original wave function, random coin toss distribution of the wave function (some part with Alice and other with Bob), and the resultant probabilities and everything explained above still stands, correct ?

 

[randomuser3210]

Before you get too deeply into what happens if we measure a two-particle system in this state, you might ask how you prepare two particles in this state to begin with.

I assumed they could be prepared as the theory (with my naive knowledge) permits it. I could be wrong though?

 

[PeroK]

I should have stated 'Generic Version of Bell Equation' or something to that effect. But even if we have two different particles after collapse so we can distinguish them as you suggest, the rest of the original wave function, random coin toss distribution of the wave function (some part with Alice and other with Bob), and the resultant probabilities and everything explained above still stands, correct ?

I don't think there's any change to the calculations, but with distinguishable particles there's no mystery which one is which. All a coin toss is doing is determining whether Alice gets the proton or the electron to measure.

 

[PeroK]

I assumed they could be prepared as the theory (with my naive knowledge) permits it. I could be wrong though?

Yes, in theory that is a valid state. But, if you want to talk about an experiment involving such a state, then you may not have a free hand in specifying the measurement set-up without also thinking about the preparation procedure.

 

[randomuser3210]

Yes, in theory that is a valid state. But, if you want to talk about an experiment involving such a state, then you may not have a free hand in specifying the measurement set-up without also thinking about the preparation procedure.

So if I understand correctly, preparation procedure is the main limit.

But, maybe it would be possible to create this in the future with relevant tech. (to translate any relevant thought experiment with it) into an actual one right (assuming a much smarter civilisation and so on) ? :)

 

[PeroK]

But, maybe it would be possible to create this in the future with relevant tech. (to translate any relevant thought experiment with it) into an actual one right (assuming a much smarter civilisation and so on) ? :)

This sort of state preparation might be relevant in quantum computing - you could start looking at the issues there.