A query regarding Spin Entanglement Measurement

[Student149]

Given a pair of Spin $1/2$ entangled particles created in the $z^→$ direction according to the following equation $Ψ=1/√2(\uparrow\uparrow+\downarrow\downarrow)$. One entangled particle is sent to Alice and another to Bob.

Now if Alice measures her particle in the $z^→$ direction she gets her particle $\uparrow or \downarrow$ each with $(1/√2)^2=1/2=50$% probability. When Bob measures his particle in $z^→$ direction he always gets his particle in the same direction as Alice 100% of the times.

Moreover when Alice decides to measure her particle in any random direction (say $k^→$) as long as Bob measures his in same $k^→$ direction both still get either $\uparrow\uparrow or \downarrow\downarrow$ with the same 50% each probability.

Query 1: Similar to above, is it possible to create an entangled pair in the $z^→$ direction such that no matter which random direction (say $k^→$) Alice measures her particle as long as Bob measures in the same $k^→$ direction the following is true:

1. Both always get either $\uparrow\uparrow or \downarrow\downarrow$ with 100% probability.
2. The probability of getting $\uparrow\uparrow$ is $x$% and $\downarrow\downarrow$ is $(1-x)$% where $x≠50$.

Note: When we talk about direction, we are talking in context of the 3 dimensional Euclidean plane.
P.S. My background is not in physics thus if there is something trivially incorrect apologies beforehand.

 

[PeroK]

Moreover when Alice decides to measure her particle in any random direction (say $k^→$) as long as Bob measures his in same $k^→$ direction both still get either $\uparrow\uparrow or \downarrow\downarrow$ with the same 50% each probability.

That can't be correct. In any other direction they may get opposite spins.

 

[Student149]

That can't be correct. In any other direction they may get opposite spins.

what I meant is that 'in that chosen random direction both would be spin up or down with same 50:50 probability. This is my source:

 

[PeroK]

I haven't time to watch a nine-minute video to figure out what precisely you have misunderstood. He's pretty reliable, so I assume he doesn't say what you think he says.

If you are able to calculate the probabilities for $\vec k = \hat y$ for the state you gave that might be interesting.

PS are you sure he doesn't use the state $\psi = \frac 1 {\sqrt 2}(\uparrow \downarrow - \downarrow \uparrow)$?

 

[Student149]

He doesn't use any eqn. for any state at at all but conceptual description so I am clueless. I have no other source for laymen too thus I came here.

What is the difference b/w two entangled states. I mean if there is any of the 4 possible entangled states that satisfies the original Q I would be glad to know.

And I don't know the generic formula that let us know with what probability the original eqn. (whichever of the 4 we choose) is satisfied in random directions..

Sorry but I am struggling thus ended up here..

 

[PeroK]

He doesn't use any eqn. for any state at at all but conceptual description so I am clueless. I have no other source for laymen too thus I came here.

What is the difference b/w two entangled states. I mean if there is any of the 4 possible entangled states that satisfies the original Q I would be glad to know.

And I don't know the generic formula that let us know with what probability the original eqn. (whichever of the 4 we choose) is satisfied in random directions..

Sorry but I am struggling thus ended up here..

The state $\psi = \frac 1 {\sqrt 2}(\uparrow \downarrow - \downarrow \uparrow)$ is almost certainly the one used in the video. This is the "singlet" entangled state. It's effectively spinless, in the sense that it has a total spin of zero. And, therefore, if you measure spin about any axis you will get a total of zero, hence you must get an up and a down.

The state you quoted $\psi = \frac 1 {\sqrt 2}(\uparrow \uparrow + \downarrow \downarrow)$ is an entirely different state. It's still an entangled state and if you measure spin about the z-axis you will get either up-up or down down. But, it doesn't have the property you claim that this is true about any axis. If you measure spin about the y-axis, you will get either up-down or down-up.

PS the state $\psi = \frac 1 {\sqrt 2}(\uparrow \downarrow + \downarrow \uparrow)$ is also interesting. This is a "triplet" state and although it has zero spin about the z-axis, it has a total spin of $1$. The spins are, therefore, correlated (not anti-correlated) about the x and y axes.

 

[PeroK]

what I meant is that 'in that chosen random direction both would be spin up or down with same 50:50 probability. This is my source:

All you can do from a video like this is try to understand what he's saying. To start doing your own analysis you need to learn the mathematical formalism (of a spin-1/2 system) from a proper academic source. That's how you learn to do physics. Pop-science videos, however good they are, are more like a spectator sport. They don't actually teach you how to play the game.

 

[Student149]

All you can do from a video like this is try to understand what he's saying. To start doing your own analysis you need to learn the mathematical formalism (of a spin-1/2 system) from a proper academic source. That's how you learn to do physics. Pop-science videos, however good they are, are more like a spectator sport. They don't actually teach you how to play the game.

thank you. my background is CS but I wan't to understand from the scratch up (especially and specifically entanglement). any book pointers for a novice like me (from scratch up and not too dense) ?

 

[PeroK]

thank you. my background is CS but I wan't to understand from the scratch up (especially and specifically entanglement). any book pointers for a novice like me (from scratch up and not too dense) ?

The most accessible introduction to QM I know is:

https://physics.mq.edu.au/~jcresser/Phys304/Handouts/QuantumPhysicsNotes.pdf

There's also Susskind's theoretical minimum:

https://theoreticalminimum.com/courses/quantum-mechanics/2012/winter

There is a book as well as these lectures.

 

[PeterDonis]

Now if Alice measures her particle in the $z^→$ direction she gets her particle $\uparrow or \downarrow$ each with $(1/√2)^2=1/2=50$% probability. When Bob measures his particle in $z^→$ direction he always gets his particle in the same direction as Alice 100% of the times.

The video you referenced does not appear to say this. It appears to say that Bob always measures his particle to spin in the opposite direction from Alice if they both measure spin along the same axis.

For a pair of fermions, such as electrons, I don't think it's possible to put them in an entangled state such that their spins will always be measured the same along the same axis. AFAIK you can only do that with bosons, such as photon polarizations.

 

[vanhees71]

Of course you can prepare two electrons either in a triplet or a singulet-spin state. Only the total state must be antisymmetric under particle exchange not the spin state for itself.

 

[PeroK]

For a pair of fermions, such as electrons, I don't think it's possible to put them in an entangled state such that their spins will always be measured the same along the same axis.

That is indeed the case. In order to achieve correlation we would need the state to be (in the usual z-basis):
$$|\psi \rangle = \alpha \uparrow_z \uparrow_z + \beta \downarrow_z \downarrow_z$$
If we express this in the x-basis we get:
$$|\psi \rangle = \frac 1 2 (\alpha + \beta)(\uparrow_x \uparrow_x + \downarrow_x \downarrow_x) - \frac 1 2(\alpha - \beta)(\uparrow_x \downarrow_x + \downarrow_x \uparrow_x)$$
To have correlation wrt the x-axis we would need $\alpha - \beta = 0$. So we do actually get correlation in the x direction as well for the original state, where $\alpha = \beta = \frac 1 {\sqrt 2}$.

But, if we express the state in the y-basis we get:
$$|\psi \rangle = \frac 1 2 (\alpha - \beta)(\uparrow_y \uparrow_y + \downarrow_y \downarrow_y) - \frac 1 2(\alpha + \beta)(\uparrow_y \downarrow_y + \downarrow_y \uparrow_y)$$
And in this case we would need $\alpha = -\beta$. We cannot, therefore, have correlation wrt all three axes. We may have correlation wrt at most two axes, in which case we have anti-correlation wrt the third axis.

 

[vanhees71]

That's of course true, because the spin state is only rotation invariant if it's in the $S=0$ state, i.e., in the singulet (antisymmetric in the single-particle) spins. So I misunderstood @PeterDonis 's remark. So forget about my posting #11.

Nevertheless in the discussion about the entangled-spin measurements with indistinguishable particles it can be important to put the full states, i.e., including also the "orbital part". This is most easily done, of course, in the 2nd-quantization formalism with the appropriate fermionic or bosonic field operators, taking the necessary (anti-)symmetrization into account automatically.

Take, e.g., electrons and let's stick to the non-relativistic case. Then define annihilation operators $\hat{\psi}_j(\vec{x},\sigma_z)$, where $j$ labels two (e.g., Gaussian) wave packets peaking at different positions (maybe very far apart) with $\sigma_z \in \{\pm 1/2 \}$. Then the full state of the kind discussed here is described as
$$|\Psi \rangle=[\alpha \hat{\psi}_1^{\dagger} (\vec{x}_1,1/2) \hat{\psi_2}^{\dagger}(x_2,1/2) + \beta \hat{\psi}_1^{\dagger}(x_1,-1/2) \hat{\psi}_2^{\dagger}(x_2,-1/2)]|\Omega \rangle,$$
with $|\Omega \rangle$ the vacuum state.