Comparison between quantum entanglement and a classical version

[Nugatory]

https://en.wikipedia.org/wiki/EPR_paradox

Yes, you will often hear the word "paradox" applied to the EPR argument. However, EPR themselves do not use that word and it doesn't even appear in the paper. Instead, they argue a syllogism: All complete theories have certain properties; QM does not have one of those properties; therefore QM is not a complete theory. It turns out that their major premise is mistaken (whatever complete theory we end up, whether QM or something else, will not have all of these properties), but this was not apparent at the time and didn't become apparent until decades later.

So in your opinion, how to explain the outcomes of Alice Bob spins-experiment and quantum correlation based on causality or based on anything?

You pick an interpretation of quantum mechanics and use whatever explanation that interpretation suggests. There's an interesting discussion of some of these issues at http://arxiv.org/abs/quant-ph/0212140.

 

[Adel Makram]

What non-locality?

Could you explain precisely what it is about QM that you see as non-local?

There is nothing Alice can do locally that in any way affects, superluminally, the results of a distant measurement - this is true in QM too. So what is it you're seeing as being 'non-local'?

Non locality means Bob particle changed to a definite spin state when Alice particle measured instantaneously according g to the collapse interpretation.

 

[Adel Makram]

Suppose there are two experiments set ups, one allows Bob to measure the expected outcome Of his measurment of spin direction from a singlet state and the other from an entangled state.
Knowing nothing about whether Alice, Bob measures his spin direction, up-down, and take the average. If the experiment set up is a singlet one, then Bob would have a 50% chance to have a spin up. If, instead the set up is entangled and Alice exists to the other side of the world pointing her detector to a direction antiparallel to Bob' one then Bob measures his spin up in 100% of times. Now Bob comes to know that not only the spin state s entangled but he knows the direction where Alice points her detector. They can even communicate using different angles as an agreed alphabet although no physical signal is used.

 

[Adel Makram]

I know that saying Bob measures his spin up in 100% of times sound contradictory with what have been discussed here when we say Alice has a 50% chance to measure her spin because there should nor be any preference of Alice on Bob. But I don't know how to clear this so.

 

[Simon Phoenix]

Bob to measure the expected outcome Of his measurment of spin direction from a singlet state and the other from an entangled state.

A singlet state is just the name given to a particular kind of entangled state - usually the state that looks like ~ |01> + |10>

If, instead the set up is entangled and Alice exists to the other side of the world pointing her detector to a direction antiparallel to Bob' one then Bob measures his spin up in 100% of times

No. That's completely incorrect. Sure Bob's result will be correlated with Alice's, but Alice's result is random.

 

[Simon Phoenix]

Non locality means Bob particle changed to a definite spin state when Alice particle measured instantaneously according g to the collapse interpretation.

As others have pointed out - there's a problem with using some notion of collapse here. Is it Alice or Bob who collapses the state? The answer is that it could be either depending on the frame of reference. So there's something a bit wonky with this idea that measurement causes some state change (collapse) in this scenario.

But even so it's important to realize that the issue only occurs because of a particular interpretation we've placed on things and not really a feature of QM.

Strip away the interpretative fun and games and QM is a theory that links preparations with measurement results - that's all. It doesn't tell us what's "really" happening in between these two things in any definite sense. Some people advocate thinking of the state, or wavefunction, as merely an abstract mathematical device that possesses no physical reality. It just allows us to calculate correct measurement probabilities. When we make a measurement we gain new information and so our probabilities instantly change - nothing at all non-local or strange about that process.

The point is that we don't actually need collapse as a concept in QM and nor do we need to think of the state as being some real physical entity that evolves.

Having said that, my own personal preference is to think in terms of measurement-induced collapses of a real physical entity, because that's the easiest and most pragmatic way of thinking about QM for me. But even so, there is no real issue with non-locality since no physical observables are being changed or linked by anything FTL - in other words there are no non-local experimental consequences. Only this nebulous wavefunction or state (whatever this might be) changes instantaneously in this perspective.

 

[Nugatory]

Non locality means Bob particle changed to a definite spin state when Alice particle measured instantaneously according to the collapse interpretation.

That is not what "non-locality" means.

Speaking loosely, non-locality refers to the fact that to calculate the result at either detector you need to know the settings of both detectors - any mathematical equation that correctly predicts the experimental results will include the angle between the two detectors somewhere. This non-locality does not necessarily conflict with relativity (although it will if you also assume wave function collapse, which is a good reason not to make that assumption).

Note that I did say "speaking loosely"... There are some subtleties here, and the Timson/Brown paper I linked above goes into them in more depth.

 

[stevendaryl]

Also if probability of Bob measuring spin down = sin² (β - Φ)/2)
And this formula is equated to 1/2 ( 1 - β ⋅ α ) above then can someone show this numerically also ?

It's just a trigonometric identity: sin^2(\frac{X}{2}) = \frac{1}{2}(1 - cos(X))

 

[stevendaryl]

Ok. So how SR explains the non locality then?

Well, that's what's tricky about QM. Although some interpretations (the collapse interpretation, the Bohm interpretation) involve faster than light interactions, there is no way to use QM to send a FTL signal, and so there is no way to use QM to demonstrate a violation of SR.

 

[Simon Phoenix]

Ok. So how SR explains the non locality then?

I think it's also useful to realize that we don't actually need non-locality to see a violation of the statistical inequality we call Bell's inequality, and nor do we need entanglement. If that seems counter to everything you've read on Bell's inequality then consider the following :

In the usual scenario we have Alice <---------------- Source ----------------> Bob
Let's bring Alice closer to the source so that we have Alice <- Source ---------------->Bob

In both these cases when statistics are compared we see a violation of the BI (for appropriately chosen measurement angles)

Now let's put the source in Alice's lab. She doesn't allow any of the entangled particles to reach Bob, but she makes a measurement of one of the entangled particles as she would in the normal set-up. Now she uses the result she obtains to prepare another spin-1/2 particle and sends this off to Bob

So we have || Alice <- Source -> || ......Bob
then Alice ---------------->Bob

So in fact we don't need the entangled source at all - Alice can just randomly prepare spin-1/2 particles in the appropriate states.

Now when we compare the statistics there's a violation of the Bell inequality but it's between Alice's state preparation and Bob's measurement.

No problems with FTL signals here - and we don't need entangled particles - yet we're still seeing a violation of a mathematical inequality (the BI)

Of course this is a different experimental set-up which doesn't carry with it the same implications for 'reality' - and it was the brilliance of EPR and subsequently Bell to focus on a simple physical system that could bring out these implications for 'reality' - but it's not the 'non-local' bit of QM (or the entanglement) that's actually giving us mathematics capable of violating the mathematical inequality. A bit hard to state precisely, sorry, but I hope you get my drift here.

 

[DrChinese]

Now let's put the source in Alice's lab. She doesn't allow any of the entangled particles to reach Bob, but she makes a measurement of one of the entangled particles as she would in the normal set-up. Now she uses the result she obtains to prepare another spin-1/2 particle and sends this off to Bob

So we have || Alice <- Source -> || ......Bob
then Alice ---------------->Bob

So in fact we don't need the entangled source at all - Alice can just randomly prepare spin-1/2 particles in the appropriate states.

Now when we compare the statistics there's a violation of the Bell inequality but it's between Alice's state preparation and Bob's measurement.

So you are saying exactly what? Bell inequalities demonstrate that local hidden variable explanations are not viable for some experiments. Yours isn't one of them.

 

[Simon Phoenix]

So you are saying exactly what? Bell inequalities demonstrate that local hidden variable explanations are not viable for some experiments. Yours isn't one of them.

Agreed :-)

I'm sorry I was hoping that I made it clear that this different experimental set-up doesn't have implications for reality in the same way that the usual Bell set-up does. My fault.

What I was trying to get at was that in both cases the maths is identical - just that we interpret it differently in each case. For me I think it brings out that the non-commutatitivity is driving the violation in QM - and when we apply this to the usual Bell set up we also get all these interesting implications for hidden variable theories.

In a nutshell there's nothing Bob can do even when given Alice's data to distinguish whether the results are from genuine entangled particles or Alice's state preparations. This has practical implications in that we can use this in a quantum key distribution using the Bell inequality protocol, but now with single particles instead of having to use entangled sources which are practically a bit trickier.

 

[Simon Phoenix]

I think there's also a sense in which the issue of non-locality is almost a red herring. That's a little bit of an overstatement and I hope the following will make it clearer what I mean.

If we impose the condition that results 'here' do not depend on settings 'there' even if here and there are only timelike separated, then a hidden variable theory of QM of this kind would satisfy the Bell inequality. We can't rule out a hidden variable theory of physics which did allow this. So even for timelike here and there we can rule out a certain class of local hidden variable theories.

So the possible candidate hidden variable theories are those in which information about settings 'there' is transferred to 'here' and affects results 'here'. I've not seen any of these constructions which look like anything like a 'sensible' theory of physics. They are possible, but somewhat artificial (at least the ones I've seen) - even in the local case where 'here' and 'there' are timelike separated. The Bohm theory, which is non-local anyway, has some attractive features and is perhaps the most 'natural' I've seen, but even there we have to introduce this complex guiding potential.

If anyone does know of a 'natural' local hidden variable theory that works (i.e. violates the BI) in this more restricted timelike case - then I'd love to see it if you have a link.

So in my view 'classical' physics is struggling even before we hit the issue of non-locality. In this case then I think it's useful to figure out what it is about QM that's causing the violation - even in the local case which only rules out a more restricted class of local hidden variable theories.

Of course it's critical that we go that last step and consider spacelike separated 'here' and 'there' - and then we put the final nail in the coffin of local hidden variable theories.

 

[Adel Makram]

If we impose the condition that results 'here' do not depend on settings 'there' even if here and there are only timelike separated, then a hidden variable theory of QM of this kind would satisfy the Bell inequality. We can't rule out a hidden variable theory of physics which did allow this. So even for timelike here and there we can rule out a certain class of local hidden variable theories.

So what class of local hidden variables that satisfies BI in a time-like experiment?

 

[Nugatory]

So what class of local hidden variables that satisfies BI in a time-like experiment?

Any theory in which the result at either detector depends on the initial state (hidden and unhidden variables) of the particle pair and on the state (hidden and unhidden variables) of that detector at the time detection will satisfy the inequality - that IS the theorem, restated in layman's language, as the theorem derives the inequality from the premise that the theory is as I just said. The timelike or spacelike separation of the measurement events is irrelevant to this proof.

 

[Adel Makram]

Will the quantum correlation be reproduced by putting 2 polarizers with opposite direction between the source and the detectors (one polarizer at each side of the source that pass only spin up along its direction) and repeating the experiments at different angles of both polarizers near the source and the polarizers at Alice and Bob?

Let`s put 2 polarizers, one on each side of the source, P<sub>toward Alice</sub> ( between the source and Alice) and P<sub>toward Bob</sub> ( between the source and Bob). The two polarizer are in opposite direction and they only pass spin-up along their respective direction so that the spin-up( up in the +ve Z direction) from P<sub>toward Alice</sub> is accompanied by spin-down (-ve Z) from P<sub>toward Bob</sub>.
Then Alice and Bob are measuring their spins at their detectors/polarizers as in standard Bell`s experiment with different angles of their detectors and compare their results. Will they come to a quantum correlation?
Let`s repeat the experiment by rotation the polarizers-set at many angles from 0 to 2π ( but always keeping them at opposite direction) and compare results again. Will the result be in keeping with quantum correlation?

Clearly the BI will be violated if we only compare P_{toward Alice} with the detector/polarizer at Bob or P_{toward Bob} with the detector/polarizer at Alice but how about the correlation between Alice and Bob outcomes?

On one hand, the correlation between Alice and Bob would be classical in nature because the spin status has been already collapsed at the polarizer near the source and the results at Alice and Bob places will be independent as if there would be taken at two non-entanged spins at different places.
On the other hand, the orientation of spins reaching Alice and Bob will be always in opposite direction and repeating experiment at different angles of the polarizer-set at the source with no information about which angle is choozen might be similar to the fully entanglement status which should mandate results in keeping with quantum correlation.
So where did I go wrong?

 

[Nugatory]

So where did I go wrong?

Alice's measurements will depend on the angle between her detector and the filtering device on her side, but not on Bob's settings. Likewise, Bob's results will depend on the angle between his detector and the filtering device on his side, but not on Alice's settings.

(You said "polarizer" but you're talking about particle spins not polarizations, so a polarizer is the wrong device to achieve your purpose. You need something like a Stern-Gerlach device, which deflects particles with spin-up along its axis in one direction and spin-down in the opposite direction).

 

[Adel Makram]

Alice's measurements will depend on the angle between her detector and the filtering device on her side, but not on Bob's settings. Likewise, Bob's results will depend on the angle between his detector and the filtering device on his side, but not on Alice's settings.

(You said "polarizer" but you're talking about particle spins not polarizations, so a polarizer is the wrong device to achieve your purpose. You need something like a Stern-Gerlach device, which deflects particles with spin-up along its axis in one direction and spin-down in the opposite direction).

So the correlation between Alice and Bob measurements will be classical.
But considering that both sides know nothing about the angle of the filtering device near the source relative to z-axis which is changing at random for each pair. This would simulate the mixed state of fully entangled particles, I expect the correlation to follow quantum mechanics prediction.

 

[Nugatory]

This would simulate the mixed state of fully entangled particles

The probability of Alice and Bob getting the same result will not be $\sin^2\frac{\alpha-\beta}{2}$. Bob and Alice will find this out when they get together afterwards and compare the results of their measurements.

 

[DrChinese]

But considering that both sides know nothing about the angle of the filtering device near the source relative to z-axis which is changing at random for each pair. This would simulate the mixed state of fully entangled particles, I expect the correlation to follow quantum mechanics prediction.

It will not simulate an entangled state at all. The stats may match at selectively chosen angles, but that's it. One is product state stats, the other is entangled state stats.

 

[mpc755]

In order for there to be conservation of momentum the downconverted photon pair are created with opposite angular momentums.

Each of the pair can determine the position and momentum of the other based upon their own position and momentum.

Entanglement is each of the pair being able to determine the state of the other.

They are not physically or superluminally connected.

Their ability to determine each other's state is non-local.

 

[DrChinese]

They are not physically or superluminally connected.

Welcome to PhysicsForums, mpc755!

There are QM interpretations that include such a remote connection. The underlying physical nature of entanglement of remote particles is something of a mystery.

 

[mpc755]

Welcome to PhysicsForums, mpc755!

Thanks!

There are QM interpretations that include such a remote connection.

Those interpretations are incorrect.

The underlying physical nature of entanglement of remote particles is something of a mystery.

It's not a mystery. Entanglement is each of the pair being able to determine the other's state based upon their own position and momentum.

 

[DrChinese]

It's not a mystery. Entanglement is each of the pair being able to determine the other's state based upon their own position and momentum.

I guess mystery is in the eye of the beholder.

So I would ask, according to your line of reasoning: If I measure Alice's p, is entangled partner Bob's q indefinite? Because Alice's is. If I measure Alice's q, is Bob's p indefinite? Because Alice's is. If so, does the nature of a measurement on Alice change Bob? This is a bit of a mystery to myself and most.

 

[mpc755]

I guess mystery is in the eye of the beholder.

So I would ask, according to your line of reasoning: If I measure Alice's p, is entangled partner Bob's q indefinite? Because Alice's is.

Alice's measurement is indefinite to you. It's not indefinite to Alice or Bob.

If I measure Alice's q, is Bob's p indefinite? Because Alice's is. If so, does the nature of a measurement on Alice change Bob?

The nature of measurement on Alice does not change Bob. They are propagating with opposite angular momentums.

This is a bit of a mystery to myself and most.

Yes, because no one realized you have to understand entanglement from the point of view of Alice and Bob, not an external observer performing the measurement.

 

[DrChinese]

Alice's measurement is indefinite to you. It's not indefinite to Alice or Bob. The nature of measurement on Alice does not change Bob. They are propagating with opposite angular momentums. Yes, because no one realized you have to understand entanglement from the point of view of Alice and Bob, not an external observer performing the measurement.

What you are saying sounds suspiciously like the argument of a local realist. Hopefully I am misunderstanding you, because such is prohibited by Bell's Theorem. You are familiar with this, correct?

I can predict Bob's spin at any angle across the spectrum by measuring Alice. If Alice and Bob are independent, as you imply, and there is no superluminal interaction, as you also imply, then Bob's spin at any angle across the spectrum must actually be predetermined. (Which is contradicted by Bell tests.)

 

[Nugatory]

Those interpretations are incorrect.

We have many threads on the validity of quantum mechanical interpretations already; we don't need another one. There is no experimental basis for a flat assertion that some interpretations are simply incorrect.

Continued discussion of how Bell's theorem precludes an explanation of the type that mpc755 is advancing is welcome (although it would be good to first review the background of Bell's theorem including the material at http://www.drchinese.com/Bells_Theorem.htm and many other threads here).

 

[Nugatory]

OK, thread is open to all again.

 

[mpc755]

What you are saying sounds suspiciously like the argument of a local realist. Hopefully I am misunderstanding you, because such is prohibited by Bell's Theorem. You are familiar with this, correct?

I can predict Bob's spin at any angle across the spectrum by measuring Alice. If Alice and Bob are independent, as you imply, and there is no superluminal interaction, as you also imply, then Bob's spin at any angle across the spectrum must actually be predetermined. (Which is contradicted by Bell tests.)

Bell's theorem precludes local hidden variable theories. I am discussing a non-local hidden variable theory.

If Alice and Bob are a downconverted photon pair then they are propagating with opposite angular momentums.

The 'non-local' aspect of entanglement is the ability of Alice and Bob to know what they themselves and each other will be detected as based upon their own position and angular momentum.

They can determine each other's spin. Their spins are deterministic to each other.

This does not contradict Bell's Theorem as their ability to determine each other's spin is non-local.

 

[Nugatory]

If Alice and Bob are a downconverted photon pair then they are propagating with opposite angular momentums.

To be consistent with the observed violation of Bell's inequality, you need an additional assumption - not only are the angular momenta opposite, but also the direction is not set (note that I said "set", not "discovered" or "known") until the first measurement is made.

 

[mpc755]

To be consistent with the observed violation of Bell's inequality, you need an additional assumption - not only are the angular momenta opposite, but also the direction is not set (note that I said "set", not "discovered" or "known") until the first measurement is made.

When Alice and Bob are created they both know the direction they are traveling. They are both able to determine the direction they are traveling from their point of origin. Due to conservation of momentum they are propagating with opposite angular momentums. Alice and Bob are able to determine each other's location and calculate what it will be in the future based upon their own angular momentum.

 

[Nugatory]

When Alice and Bob are created they both know the direction they are traveling. They are both able to determine the direction they are traveling from their point of origin. Due to conservation of momentum they are propagating with opposite angular momentums. Alice and Bob are able to determine each other's location and calculate what it will be in the future based upon their own angular momentum.

They can. However those calculations don't match what happens when we actually do the experiment.

 

[Heinera]

mpc755: You seem to call the particles "Alice" and "Bob". In the Bell-terminology, Alice and Bob are hypothetical observers, and the particles are not usually elevated to a status where they are named.

 

[mpc755]

They can. However those calculations don't match what happens when we actually do the experiment.

They will if they use the calculations of quantum mechanics and not classical physics.

 

[mpc755]

You seem to call the particles "Alice" and "Bob". In the Bell-terminology, Alice and Bob are hypothetical observers, and the particles are not usually elevated to a status where they are named.

Okay, Photon1 and Photon2 then. Thanks.

Photon1 and Photon2 know their angular momentum from the time of their creation. They can also determine their pair's position and momentum based upon their own position and momentum because they know they are each propagating with opposite angular momentums.

Their position and momentum are not hidden from one another.

Their states are exposed to one another.

Entanglement is more correctly described as an 'exposed variable theory'.

 

[Nugatory]

They will if they use the calculations of quantum mechanics and not classical physics.

But the calculations of quantum mechanics include the assumption that the direction of spin is set by the first measurement. The quantum mechanical calculation of the probability distribution of Bob's results depends not just on what he knows about Alice's particle and his particle, but also on Alice's choice of measurement direction (made long after the photon pair is created).

 

[Adel Makram]

It will not simulate an entangled state at all. The stats may match at selectively chosen angles, but that's it. One is product state stats, the other is entangled state stats.

So if A is the product state of two particles U₁ and U₂, then A=U₁⊗U₂.

U₁=α_1uu+β_1dd
U₂=α_2uu+β_2dd, where u and d are basis vectors.

A=U₁⊗U₂=(α_1uu+β_1dd)(α_2uu+β_2dd)
= α_1uα_2uuu+α_1uβ_2dud+β_1dα_2udu+β_1dβ_2ddd

If we prepare particles with opposite momenta to each other by putting filtering devices at each side of the source with opposite direction then,

Only the coefficients with states ud, du should be only non-zero which should reduce the state into an entangled state with proper normalization ( multiplying by 1/√2). A=1/√2(ud+du)

 

[mpc755]

But the calculations of quantum mechanics include the assumption that the direction of spin is set by the first measurement. The quantum mechanical calculation of the probability distribution of Bob's results depends not just on what he knows about Alice's particle and his particle, but also on Alice's choice of measurement direction (made long after the photon pair is created).

Even if Photon1 does not know what Photon2's measurement direction will be Photon1 can still determine what the spin will be detected as ahead of time. Photon1 can calculate if Photon2 is detected with such-and-such measurement direction then it will have this spin. It's still deterministic.

 

[Adel Makram]

To be consistent with the observed violation of Bell's inequality, you need an additional assumption - not only are the angular momenta opposite, but also the direction is not set (note that I said "set", not "discovered" or "known") until the first measurement is made.

What is the difference between "set" and "known"?
In general what is the real difference between entangled particles and two prepared particles with opposite momenta and unknown direction relative to +z-axis?
1) The 2 entangled particles have always opposite angular momenta so do prepared particles.
2) The state of each of entangled particle is in superposition of spin-up and down with 50-50 chance, so do prepared particles (because Alice for example, will have an expected value of spin measured along her direction =1/2π∫cos²(θ/2)dθ (the interval 0 to 2π)=1/2 provided all angles of prepared particles are included by rotating the filtering device at the source at random).

 

[DrChinese]

If we prepare particles with opposite momenta to each other by putting filtering devices at each side of the source with opposite direction then,

Only the coefficients with states ud, du should be only non-zero which should reduce the state into an entangled state with proper normalization ( multiplying by 1/√2). A=1/√2(ud+du)

You will have something that is in a superposition of ud + du, true enough. But only at special angles. Those being related to whatever angle you filtered at. Any other angle will be some combination of uu + ud + dd +du. It's a Product State, after all.

On the other hand, entangled states will maintain the ud + du superposition at any angle.

So no, they are not at all alike and I have no idea why you think they would be. This goes blatantly against everything ever written (at least that can be referenced here).

 

[Adel Makram]

You will have something that is in a superposition of ud + du, true enough. But only at special angles. Those being related to whatever angle you filtered at. Any other angle will be some combination of uu + ud + dd +du.

So does entangled state!
If u, d represent basis vector of spin-up and spin-down, respectively, then entangled state has 4 components of various combinations of spin up and down of two particles at Alice and Bob places.

 

[Adel Makram]

You will have something that is in a superposition of ud + du, true enough. But only at special angles.

Also it s true for the frame of reference at this angle. For other angles, rotational transformation matrix is applied and yield similar coefficients to entangled state with sin²(θ₂-θ₁/2) and cos²(θ₂-θ₁/2) , where θ₁ and θ₂ are angles of Alice and Bob detectors, respectively. If θ₁=θ₂, the equation will reduced to the usual entangled state.​

 

[Adel Makram]

It will not simulate an entangled state at all. The stats may match at selectively chosen angles, but that's it. One is product state stats, the other is entangled state stats.

I got it at last.

Two particles with opposite spins after being filtered out will not simulate the entangled state.

Consider particle-1 state, P₁= 1/√2(U₊+U_-)
and particle-2 state, P₂= 1/√2(V₊+V_-)
where U₊, U_-, V₊, V_- are states of spin up and down for particle-1 and up and down for particle-2, respectively.
The state of two particles in opposite momenta after exisiting a filtering device is a product state,
lP₁>⊗lP₂>=1/√2[U₊V_-+U_-V₊]
All four possible states of two particles can be also represented onto the space of spin-up and spin-down at Alice and Bob devices.
U₊=cos(θ₁/2) u₁ + sin(θ₁/2) v₁
U_-=sin(θ₁/2) u₁ - cos(θ₁/2) v₁
V₊=cos(θ₂/2) u₂ + sin(θ₂/2) v₂
V_-=sin(θ₂/2) u₂ - cos(θ₂/2) v₂

lP₁>⊗lP₂>=1/√2[(cos(θ₁/2) u₁ + sin(θ₁/2) v₁)(sin(θ₂/2) u₂ - cos(θ₂/2) v₂) + (sin(θ₁/2) u₁ - cos(θ₁/2) v₁)(cos(θ₂/2) u₂ + sin(θ₂/2) v₂)]

after multiplying and collecting like terms,
lP₁>⊗lP₂>=1/√2{ [(cos(θ₁/2)(sin(θ₂/2)+(sin(θ₁/2)(cos(θ₂/2)] u₁u₂ + [-(cos(θ₁/2)(cos(θ₂/2)+(sin(θ₁/2)(sin(θ₂/2)] u₁v₂ +[(sin(θ₁/2)(sin(θ₂/2)-(cos(θ₁/2)(cos(θ₂/2)] v₁u₂ - [(sin(θ₁/2)(cos(θ₂/2)+(cos(θ₁/2)(sin(θ₂/2)] v₁v₂}

Bob should set his angle θ₂=π in order to get his spin up.
In doing so, only coefficients of u₁u₂ and v₁v₂ are non-zero. This explains why at those angles only the state may simulate the entangled state. For all other angles, it is not. And even though, the probability of both Alice and Bob spins-up in this product state is not equal to its counterpart in the entangled state.

So the probability of getting spin-up for Alice (θ₁=0) and for Bob (θ₂=π),
P_up,up= (1/√2[(cos(θ₁/2)(sin(θ₂/2)+(sin(θ₁/2)(cos(θ₂/2))² = 1/2 at (θ₁=0) and for Bob (θ₂=π)
And the probability of getting spin-down for Alice and Bob;
P_down,down= (-1/√2[(sin(θ₁/2)(cos(θ₂/2)+(cos(θ₁/2)(sin(θ₂/2))² =1/2 at (θ₁=0) and for Bob (θ₂=π)