Comparison between quantum entanglement and a classical version

[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)