Why is quantum entanglement surprising?Quantum Entanglement - What's the big deal?How do we know that entanglement allows measurement to instantly change the other particle's state?Understanding quantum entanglement.. help me validate this analogy!Can entanglement be explained as a consequence of conservation laws?Why do we think that quantum entanglement implies action at a distance?Controlling the outcome of a quantum measurement through translational entanglementWouldn't 3 or more particle entanglement allow passing classical information?What's the Cause of Quantum Entanglement?Quantum Entanglement - How To InterpretBasic Quantum EntanglementPrepared states and quantum entanglementDoes quantum entanglement really send a message?Quantum entanglement actually can affect particles across distance?Alignment of axes of measuring devices in a quantum entanglement experiment
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Why is quantum entanglement surprising?
Quantum Entanglement - What's the big deal?How do we know that entanglement allows measurement to instantly change the other particle's state?Understanding quantum entanglement.. help me validate this analogy!Can entanglement be explained as a consequence of conservation laws?Why do we think that quantum entanglement implies action at a distance?Controlling the outcome of a quantum measurement through translational entanglementWouldn't 3 or more particle entanglement allow passing classical information?What's the Cause of Quantum Entanglement?Quantum Entanglement - How To InterpretBasic Quantum EntanglementPrepared states and quantum entanglementDoes quantum entanglement really send a message?Quantum entanglement actually can affect particles across distance?Alignment of axes of measuring devices in a quantum entanglement experiment
$begingroup$
What about quantum entanglement is suprising, and what makes it a strictly quantum effect?.
Suppose we have an particle of spin 0 which decays into two other particles, each with either spin up or down. Supposedly the particles are "entangled". The angular momentum of the system is conserved, so measuring one particle's spin as up would mean if we measure the other particle's spin, it must be down, and vice versa. Why is this suprising, and what does this have to do with entanglement or information about the measurement performed passing from one particle to the other.
As far as I can see, it's just momentum conservation. And I don't see what's quantum about this either, as we could do the same thing with classical particles (say an object which breaks into two parts) and the momentum measurements would be anti-correlated.
quantum-mechanics quantum-entanglement
$endgroup$
|
show 7 more comments
$begingroup$
What about quantum entanglement is suprising, and what makes it a strictly quantum effect?.
Suppose we have an particle of spin 0 which decays into two other particles, each with either spin up or down. Supposedly the particles are "entangled". The angular momentum of the system is conserved, so measuring one particle's spin as up would mean if we measure the other particle's spin, it must be down, and vice versa. Why is this suprising, and what does this have to do with entanglement or information about the measurement performed passing from one particle to the other.
As far as I can see, it's just momentum conservation. And I don't see what's quantum about this either, as we could do the same thing with classical particles (say an object which breaks into two parts) and the momentum measurements would be anti-correlated.
quantum-mechanics quantum-entanglement
$endgroup$
$begingroup$
Someone might have a more eloquent answer, but essentially the weird features of entangled states are that they do not factor into (Particle A is in state 1)(Particle B is in state 2) which would be a satisfactory classical description. Instead entangled states look like (One of A/B is in state 1, then other is in 2). There are factored states as well, but they're much rarer and they would behave differently in terms of interference effects.
$endgroup$
– jacob1729
8 hours ago
3
$begingroup$
Possible duplicate of Can entanglement be explained as a consequence of conservation laws?
$endgroup$
– knzhou
8 hours ago
$begingroup$
I don't find the answer given there to be satisfactory.
$endgroup$
– Joshua Benabou
8 hours ago
$begingroup$
Though it might seem trivial and that entanglement is a consequence of conservation of momentum, but wait, there is more to it. Have you considered the non-locality of space which is inherent to quantumness of your entanglement experiment that you have described(above)? You aren't surprised by that aren't you?
$endgroup$
– EverydayFoolish
8 hours ago
$begingroup$
what do you mean by non locality. could you identify the problem with the momentum conservation explanation ?
$endgroup$
– Joshua Benabou
8 hours ago
|
show 7 more comments
$begingroup$
What about quantum entanglement is suprising, and what makes it a strictly quantum effect?.
Suppose we have an particle of spin 0 which decays into two other particles, each with either spin up or down. Supposedly the particles are "entangled". The angular momentum of the system is conserved, so measuring one particle's spin as up would mean if we measure the other particle's spin, it must be down, and vice versa. Why is this suprising, and what does this have to do with entanglement or information about the measurement performed passing from one particle to the other.
As far as I can see, it's just momentum conservation. And I don't see what's quantum about this either, as we could do the same thing with classical particles (say an object which breaks into two parts) and the momentum measurements would be anti-correlated.
quantum-mechanics quantum-entanglement
$endgroup$
What about quantum entanglement is suprising, and what makes it a strictly quantum effect?.
Suppose we have an particle of spin 0 which decays into two other particles, each with either spin up or down. Supposedly the particles are "entangled". The angular momentum of the system is conserved, so measuring one particle's spin as up would mean if we measure the other particle's spin, it must be down, and vice versa. Why is this suprising, and what does this have to do with entanglement or information about the measurement performed passing from one particle to the other.
As far as I can see, it's just momentum conservation. And I don't see what's quantum about this either, as we could do the same thing with classical particles (say an object which breaks into two parts) and the momentum measurements would be anti-correlated.
quantum-mechanics quantum-entanglement
quantum-mechanics quantum-entanglement
edited 8 hours ago
Qmechanic♦
109k122081281
109k122081281
asked 9 hours ago
Joshua BenabouJoshua Benabou
1,77342550
1,77342550
$begingroup$
Someone might have a more eloquent answer, but essentially the weird features of entangled states are that they do not factor into (Particle A is in state 1)(Particle B is in state 2) which would be a satisfactory classical description. Instead entangled states look like (One of A/B is in state 1, then other is in 2). There are factored states as well, but they're much rarer and they would behave differently in terms of interference effects.
$endgroup$
– jacob1729
8 hours ago
3
$begingroup$
Possible duplicate of Can entanglement be explained as a consequence of conservation laws?
$endgroup$
– knzhou
8 hours ago
$begingroup$
I don't find the answer given there to be satisfactory.
$endgroup$
– Joshua Benabou
8 hours ago
$begingroup$
Though it might seem trivial and that entanglement is a consequence of conservation of momentum, but wait, there is more to it. Have you considered the non-locality of space which is inherent to quantumness of your entanglement experiment that you have described(above)? You aren't surprised by that aren't you?
$endgroup$
– EverydayFoolish
8 hours ago
$begingroup$
what do you mean by non locality. could you identify the problem with the momentum conservation explanation ?
$endgroup$
– Joshua Benabou
8 hours ago
|
show 7 more comments
$begingroup$
Someone might have a more eloquent answer, but essentially the weird features of entangled states are that they do not factor into (Particle A is in state 1)(Particle B is in state 2) which would be a satisfactory classical description. Instead entangled states look like (One of A/B is in state 1, then other is in 2). There are factored states as well, but they're much rarer and they would behave differently in terms of interference effects.
$endgroup$
– jacob1729
8 hours ago
3
$begingroup$
Possible duplicate of Can entanglement be explained as a consequence of conservation laws?
$endgroup$
– knzhou
8 hours ago
$begingroup$
I don't find the answer given there to be satisfactory.
$endgroup$
– Joshua Benabou
8 hours ago
$begingroup$
Though it might seem trivial and that entanglement is a consequence of conservation of momentum, but wait, there is more to it. Have you considered the non-locality of space which is inherent to quantumness of your entanglement experiment that you have described(above)? You aren't surprised by that aren't you?
$endgroup$
– EverydayFoolish
8 hours ago
$begingroup$
what do you mean by non locality. could you identify the problem with the momentum conservation explanation ?
$endgroup$
– Joshua Benabou
8 hours ago
$begingroup$
Someone might have a more eloquent answer, but essentially the weird features of entangled states are that they do not factor into (Particle A is in state 1)(Particle B is in state 2) which would be a satisfactory classical description. Instead entangled states look like (One of A/B is in state 1, then other is in 2). There are factored states as well, but they're much rarer and they would behave differently in terms of interference effects.
$endgroup$
– jacob1729
8 hours ago
$begingroup$
Someone might have a more eloquent answer, but essentially the weird features of entangled states are that they do not factor into (Particle A is in state 1)(Particle B is in state 2) which would be a satisfactory classical description. Instead entangled states look like (One of A/B is in state 1, then other is in 2). There are factored states as well, but they're much rarer and they would behave differently in terms of interference effects.
$endgroup$
– jacob1729
8 hours ago
3
3
$begingroup$
Possible duplicate of Can entanglement be explained as a consequence of conservation laws?
$endgroup$
– knzhou
8 hours ago
$begingroup$
Possible duplicate of Can entanglement be explained as a consequence of conservation laws?
$endgroup$
– knzhou
8 hours ago
$begingroup$
I don't find the answer given there to be satisfactory.
$endgroup$
– Joshua Benabou
8 hours ago
$begingroup$
I don't find the answer given there to be satisfactory.
$endgroup$
– Joshua Benabou
8 hours ago
$begingroup$
Though it might seem trivial and that entanglement is a consequence of conservation of momentum, but wait, there is more to it. Have you considered the non-locality of space which is inherent to quantumness of your entanglement experiment that you have described(above)? You aren't surprised by that aren't you?
$endgroup$
– EverydayFoolish
8 hours ago
$begingroup$
Though it might seem trivial and that entanglement is a consequence of conservation of momentum, but wait, there is more to it. Have you considered the non-locality of space which is inherent to quantumness of your entanglement experiment that you have described(above)? You aren't surprised by that aren't you?
$endgroup$
– EverydayFoolish
8 hours ago
$begingroup$
what do you mean by non locality. could you identify the problem with the momentum conservation explanation ?
$endgroup$
– Joshua Benabou
8 hours ago
$begingroup$
what do you mean by non locality. could you identify the problem with the momentum conservation explanation ?
$endgroup$
– Joshua Benabou
8 hours ago
|
show 7 more comments
3 Answers
3
active
oldest
votes
$begingroup$
You are correct that the observation you mention is not surprising, but you have not mentioned the observation that lies at the heart of entanglement. Entanglement is interesting and surprising because it is owing to entanglement that further experiments can be done, in addition to the one you mention, and it is these further experiments that exhibit the surprising features. The further experiments can be, for example, measurements of pairs of spin-half particles, but with measurements along various different directions (e.g. 0, 120, 250 degrees to the $z$ axis if they are moving along the $x$ axis), or measurements on pairs of spin-1 particles, or measurements on triples of spin-half particles. Various scenarios break the Bell inequalities, and this means the measurement outcomes are inconsistent with a description in which each particle carries its properties with it in a local way.
In this answer I am not going to repeat the Bell arguments; you can look them up if you like (e.g. try CHSH inequality). I will simply present a nice argument involving symmetry which you may find interesting.
Suppose I have a single spin in the state $| uparrow rangle$. Then if I rotate it through 180 degrees then it will go to the state $| downarrow rangle$. One can do this in the lab and measure the outcome and thus confirm that the state does change in exactly this way. So far so good.
Now prepare two spins $A$ and $B$ in the state
$$
|Erangle equiv frac1sqrt2( | uparrow_Auparrow_B rangle + | downarrow_A downarrow_B rangle )
$$
This is an entangled state. Rotate the first spin: you then get
$$
|Rrangle = frac1sqrt2( | downarrow_A uparrow_B rangle + | uparrow_A downarrow_B rangle ).
$$
This is a different state, indeed it is orthogonal to the first, so the rotation certainly changed the system and one can perform measurements to confirm that it did change. Still no surprise. But see what comes next.
Now suppose you want to return the system to its initial state. You have a choice. You could rotate spin $A$ back again, undoing the change. OR you could instead approach spin $B$ and rotate that one. Then you would get
$$
frac1sqrt2( | downarrow_A downarrow_B rangle + | uparrow_A uparrow_B rangle ) = | E rangle .
$$
Thus this rotation returns the system to its original state.
Now think very carefully about what just happened. Those two spins could be in different places, say one in Athens and one in Bermuda. But to take the system from state $R$ to state $E$ you can rotate either the Athens spin or the Bermuda spin. These two operations, taking place on either side of the Atlantic Ocean, carry the joint system between the same two places ($R$ and $E$) in its state space. Try to imagine a classical scenario where this would happen---you will not be able to. Notice especially the sequence where first an operation is applied in Athens---an operation which certainly changes the system state---and then an operation in applied in Bermuda, and the overall outcome is no net change to the joint system.
I hope you are beginning to see how amazing entanglement is.
Its amazingness will soon be put to practical use in quantum computing. It also has deep philosophical implications, because it shows that the natural world is not completely decomposable into separate bits and pieces.
$endgroup$
2
$begingroup$
Why is this outcome surprising? If the outcome is determined by the state of the system - by the orientation of A and B relative to one another - then why is it surprising that rotations performed independently and yet resulting in a particular mutual state, produce a particular outcome?
$endgroup$
– Steve
4 hours ago
$begingroup$
I think I see @Steve s question. If I have a quarter heads up in Athens and another heads down in Bermuda the system is in a mismatched state. If I flip Athens over the system is now in a matched state. I can return to the mismatched state by flipping Athens or Bermuda. So, Andrew, how is this mundane experiment any different than what you just said about quantum entanglement?
$endgroup$
– candied_orange
29 mins ago
add a comment |
$begingroup$
In quantum mechanics, a particle doesn't really have a property like "spin up". It only has a probability of having a certain spin and the actual spin is only determined when you make a "measurement" or observation.
Let's say we park a space ship halfway between earth and moon and we shoot two entangled particles: one towards earth, one towards the moon. When they arrive we measure their spin on earth and on the moon at the same time. As expected, the spins are opposite.
The quantum mechanical interpretation says "spin only gets determined when measured". That would basically require the two particles on earth and on the moon to communicate with each other and decide on the spot, which spin to assume, so they make sure the spins come out opposite. This would require faster than light communication or, as Einstein put it "Spooky action at a distance" or violation of "locality"
This was called out by Einstein Podolsky and Rosen in the so-called ERP experiment or ERP paradox https://plato.stanford.edu/entries/qt-epr/ and a clear challenge the Copenhagen interpretation (Bohr, Heisenberg, etc.). At the time, no one could come up with setup or experiment which would result in an observable difference.
However, in the 1960s John Stuart Bell came up with Bell's theorem that predicted a measurable difference. https://en.wikipedia.org/wiki/Bell%27s_theorem. In the 1990s, Alain Aspect et al. https://en.wikipedia.org/wiki/Alain_Aspect managed to execute a meaningful experiment and it clearly showed the violation of Bell's theorem. It's been confirmed many times after this as well.
In other words, Einstein was wrong and Bohr was right: apparently entangled particles can communicate instantaneously and we have clear experimental evidence that this entanglement cannot be explained by a-priori knowledge or some hidden state variable.
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1
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two things: why did the copenhagen guys believe that "an observable only gets determined when measured". second, if there is instantaneous communication, how is this not violating relativity?
$endgroup$
– Joshua Benabou
7 hours ago
1
$begingroup$
third question: what's this hidden variable business? what kind of hidden variable could be a priori influencing the measurements?
$endgroup$
– Joshua Benabou
7 hours ago
$begingroup$
1) Since it's one of the few ways to explain quantum behavior. It's weird, but less weird that the alternatives. Read en.wikipedia.org/wiki/Copenhagen_interpretation .
$endgroup$
– Hilmar
5 hours ago
1
$begingroup$
3) Hidden variable is a theory to explain quantum entanglement while maintaining locality at the same time. It's basically your initial assumption "all can be explain with conservation and conventional physics.". Doesn't work though. See en.wikipedia.org/wiki/Hidden-variable_theory
$endgroup$
– Hilmar
5 hours ago
2
$begingroup$
@Hilmar Bell's theorem does not show that entangled particles can communicate instantaneously, if by 'communicate' you mean 'communicate information'. No-communication theorems make it clear that the failure of local realism does not imply instantaneous communication of information. I think your answer should be edited to choose a different word than 'communicate'.
$endgroup$
– gabe
5 hours ago
|
show 3 more comments
$begingroup$
Maybe a good demonstration of why entanglement is so puzzling is the Mermin-Peres magic square game. There are three players, two of whom (A and B, say) have entangled states.
A and B are allowed to communicate and arrange their strategy in advance, but they cannot communicate once the game is in progress.
There is a $3 times 3$ grid. You can ask A for any column of the grid (but just one), and you can ask B for any row of the grid (but just one). The rules are that A and B must assign 0s and 1s for their cells in the grid, they must agree on the cell where the row and column intersect, and the number of 1s in a column is always odd, but the number of 1s in a row is always even.
For example, if you ask for column 1 and row 2, they might return:
A B
1xx xxx
0xx 011
0xx xxx
If they have entangled states, A and B can always win.
"Simple," you say, "A and B have agreed on which cells have 0s and 1s in advance, and they just return those values."
But if that's their strategy, does the master grid have an even or an odd number of 1's?
$endgroup$
add a comment |
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3 Answers
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3 Answers
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$begingroup$
You are correct that the observation you mention is not surprising, but you have not mentioned the observation that lies at the heart of entanglement. Entanglement is interesting and surprising because it is owing to entanglement that further experiments can be done, in addition to the one you mention, and it is these further experiments that exhibit the surprising features. The further experiments can be, for example, measurements of pairs of spin-half particles, but with measurements along various different directions (e.g. 0, 120, 250 degrees to the $z$ axis if they are moving along the $x$ axis), or measurements on pairs of spin-1 particles, or measurements on triples of spin-half particles. Various scenarios break the Bell inequalities, and this means the measurement outcomes are inconsistent with a description in which each particle carries its properties with it in a local way.
In this answer I am not going to repeat the Bell arguments; you can look them up if you like (e.g. try CHSH inequality). I will simply present a nice argument involving symmetry which you may find interesting.
Suppose I have a single spin in the state $| uparrow rangle$. Then if I rotate it through 180 degrees then it will go to the state $| downarrow rangle$. One can do this in the lab and measure the outcome and thus confirm that the state does change in exactly this way. So far so good.
Now prepare two spins $A$ and $B$ in the state
$$
|Erangle equiv frac1sqrt2( | uparrow_Auparrow_B rangle + | downarrow_A downarrow_B rangle )
$$
This is an entangled state. Rotate the first spin: you then get
$$
|Rrangle = frac1sqrt2( | downarrow_A uparrow_B rangle + | uparrow_A downarrow_B rangle ).
$$
This is a different state, indeed it is orthogonal to the first, so the rotation certainly changed the system and one can perform measurements to confirm that it did change. Still no surprise. But see what comes next.
Now suppose you want to return the system to its initial state. You have a choice. You could rotate spin $A$ back again, undoing the change. OR you could instead approach spin $B$ and rotate that one. Then you would get
$$
frac1sqrt2( | downarrow_A downarrow_B rangle + | uparrow_A uparrow_B rangle ) = | E rangle .
$$
Thus this rotation returns the system to its original state.
Now think very carefully about what just happened. Those two spins could be in different places, say one in Athens and one in Bermuda. But to take the system from state $R$ to state $E$ you can rotate either the Athens spin or the Bermuda spin. These two operations, taking place on either side of the Atlantic Ocean, carry the joint system between the same two places ($R$ and $E$) in its state space. Try to imagine a classical scenario where this would happen---you will not be able to. Notice especially the sequence where first an operation is applied in Athens---an operation which certainly changes the system state---and then an operation in applied in Bermuda, and the overall outcome is no net change to the joint system.
I hope you are beginning to see how amazing entanglement is.
Its amazingness will soon be put to practical use in quantum computing. It also has deep philosophical implications, because it shows that the natural world is not completely decomposable into separate bits and pieces.
$endgroup$
2
$begingroup$
Why is this outcome surprising? If the outcome is determined by the state of the system - by the orientation of A and B relative to one another - then why is it surprising that rotations performed independently and yet resulting in a particular mutual state, produce a particular outcome?
$endgroup$
– Steve
4 hours ago
$begingroup$
I think I see @Steve s question. If I have a quarter heads up in Athens and another heads down in Bermuda the system is in a mismatched state. If I flip Athens over the system is now in a matched state. I can return to the mismatched state by flipping Athens or Bermuda. So, Andrew, how is this mundane experiment any different than what you just said about quantum entanglement?
$endgroup$
– candied_orange
29 mins ago
add a comment |
$begingroup$
You are correct that the observation you mention is not surprising, but you have not mentioned the observation that lies at the heart of entanglement. Entanglement is interesting and surprising because it is owing to entanglement that further experiments can be done, in addition to the one you mention, and it is these further experiments that exhibit the surprising features. The further experiments can be, for example, measurements of pairs of spin-half particles, but with measurements along various different directions (e.g. 0, 120, 250 degrees to the $z$ axis if they are moving along the $x$ axis), or measurements on pairs of spin-1 particles, or measurements on triples of spin-half particles. Various scenarios break the Bell inequalities, and this means the measurement outcomes are inconsistent with a description in which each particle carries its properties with it in a local way.
In this answer I am not going to repeat the Bell arguments; you can look them up if you like (e.g. try CHSH inequality). I will simply present a nice argument involving symmetry which you may find interesting.
Suppose I have a single spin in the state $| uparrow rangle$. Then if I rotate it through 180 degrees then it will go to the state $| downarrow rangle$. One can do this in the lab and measure the outcome and thus confirm that the state does change in exactly this way. So far so good.
Now prepare two spins $A$ and $B$ in the state
$$
|Erangle equiv frac1sqrt2( | uparrow_Auparrow_B rangle + | downarrow_A downarrow_B rangle )
$$
This is an entangled state. Rotate the first spin: you then get
$$
|Rrangle = frac1sqrt2( | downarrow_A uparrow_B rangle + | uparrow_A downarrow_B rangle ).
$$
This is a different state, indeed it is orthogonal to the first, so the rotation certainly changed the system and one can perform measurements to confirm that it did change. Still no surprise. But see what comes next.
Now suppose you want to return the system to its initial state. You have a choice. You could rotate spin $A$ back again, undoing the change. OR you could instead approach spin $B$ and rotate that one. Then you would get
$$
frac1sqrt2( | downarrow_A downarrow_B rangle + | uparrow_A uparrow_B rangle ) = | E rangle .
$$
Thus this rotation returns the system to its original state.
Now think very carefully about what just happened. Those two spins could be in different places, say one in Athens and one in Bermuda. But to take the system from state $R$ to state $E$ you can rotate either the Athens spin or the Bermuda spin. These two operations, taking place on either side of the Atlantic Ocean, carry the joint system between the same two places ($R$ and $E$) in its state space. Try to imagine a classical scenario where this would happen---you will not be able to. Notice especially the sequence where first an operation is applied in Athens---an operation which certainly changes the system state---and then an operation in applied in Bermuda, and the overall outcome is no net change to the joint system.
I hope you are beginning to see how amazing entanglement is.
Its amazingness will soon be put to practical use in quantum computing. It also has deep philosophical implications, because it shows that the natural world is not completely decomposable into separate bits and pieces.
$endgroup$
2
$begingroup$
Why is this outcome surprising? If the outcome is determined by the state of the system - by the orientation of A and B relative to one another - then why is it surprising that rotations performed independently and yet resulting in a particular mutual state, produce a particular outcome?
$endgroup$
– Steve
4 hours ago
$begingroup$
I think I see @Steve s question. If I have a quarter heads up in Athens and another heads down in Bermuda the system is in a mismatched state. If I flip Athens over the system is now in a matched state. I can return to the mismatched state by flipping Athens or Bermuda. So, Andrew, how is this mundane experiment any different than what you just said about quantum entanglement?
$endgroup$
– candied_orange
29 mins ago
add a comment |
$begingroup$
You are correct that the observation you mention is not surprising, but you have not mentioned the observation that lies at the heart of entanglement. Entanglement is interesting and surprising because it is owing to entanglement that further experiments can be done, in addition to the one you mention, and it is these further experiments that exhibit the surprising features. The further experiments can be, for example, measurements of pairs of spin-half particles, but with measurements along various different directions (e.g. 0, 120, 250 degrees to the $z$ axis if they are moving along the $x$ axis), or measurements on pairs of spin-1 particles, or measurements on triples of spin-half particles. Various scenarios break the Bell inequalities, and this means the measurement outcomes are inconsistent with a description in which each particle carries its properties with it in a local way.
In this answer I am not going to repeat the Bell arguments; you can look them up if you like (e.g. try CHSH inequality). I will simply present a nice argument involving symmetry which you may find interesting.
Suppose I have a single spin in the state $| uparrow rangle$. Then if I rotate it through 180 degrees then it will go to the state $| downarrow rangle$. One can do this in the lab and measure the outcome and thus confirm that the state does change in exactly this way. So far so good.
Now prepare two spins $A$ and $B$ in the state
$$
|Erangle equiv frac1sqrt2( | uparrow_Auparrow_B rangle + | downarrow_A downarrow_B rangle )
$$
This is an entangled state. Rotate the first spin: you then get
$$
|Rrangle = frac1sqrt2( | downarrow_A uparrow_B rangle + | uparrow_A downarrow_B rangle ).
$$
This is a different state, indeed it is orthogonal to the first, so the rotation certainly changed the system and one can perform measurements to confirm that it did change. Still no surprise. But see what comes next.
Now suppose you want to return the system to its initial state. You have a choice. You could rotate spin $A$ back again, undoing the change. OR you could instead approach spin $B$ and rotate that one. Then you would get
$$
frac1sqrt2( | downarrow_A downarrow_B rangle + | uparrow_A uparrow_B rangle ) = | E rangle .
$$
Thus this rotation returns the system to its original state.
Now think very carefully about what just happened. Those two spins could be in different places, say one in Athens and one in Bermuda. But to take the system from state $R$ to state $E$ you can rotate either the Athens spin or the Bermuda spin. These two operations, taking place on either side of the Atlantic Ocean, carry the joint system between the same two places ($R$ and $E$) in its state space. Try to imagine a classical scenario where this would happen---you will not be able to. Notice especially the sequence where first an operation is applied in Athens---an operation which certainly changes the system state---and then an operation in applied in Bermuda, and the overall outcome is no net change to the joint system.
I hope you are beginning to see how amazing entanglement is.
Its amazingness will soon be put to practical use in quantum computing. It also has deep philosophical implications, because it shows that the natural world is not completely decomposable into separate bits and pieces.
$endgroup$
You are correct that the observation you mention is not surprising, but you have not mentioned the observation that lies at the heart of entanglement. Entanglement is interesting and surprising because it is owing to entanglement that further experiments can be done, in addition to the one you mention, and it is these further experiments that exhibit the surprising features. The further experiments can be, for example, measurements of pairs of spin-half particles, but with measurements along various different directions (e.g. 0, 120, 250 degrees to the $z$ axis if they are moving along the $x$ axis), or measurements on pairs of spin-1 particles, or measurements on triples of spin-half particles. Various scenarios break the Bell inequalities, and this means the measurement outcomes are inconsistent with a description in which each particle carries its properties with it in a local way.
In this answer I am not going to repeat the Bell arguments; you can look them up if you like (e.g. try CHSH inequality). I will simply present a nice argument involving symmetry which you may find interesting.
Suppose I have a single spin in the state $| uparrow rangle$. Then if I rotate it through 180 degrees then it will go to the state $| downarrow rangle$. One can do this in the lab and measure the outcome and thus confirm that the state does change in exactly this way. So far so good.
Now prepare two spins $A$ and $B$ in the state
$$
|Erangle equiv frac1sqrt2( | uparrow_Auparrow_B rangle + | downarrow_A downarrow_B rangle )
$$
This is an entangled state. Rotate the first spin: you then get
$$
|Rrangle = frac1sqrt2( | downarrow_A uparrow_B rangle + | uparrow_A downarrow_B rangle ).
$$
This is a different state, indeed it is orthogonal to the first, so the rotation certainly changed the system and one can perform measurements to confirm that it did change. Still no surprise. But see what comes next.
Now suppose you want to return the system to its initial state. You have a choice. You could rotate spin $A$ back again, undoing the change. OR you could instead approach spin $B$ and rotate that one. Then you would get
$$
frac1sqrt2( | downarrow_A downarrow_B rangle + | uparrow_A uparrow_B rangle ) = | E rangle .
$$
Thus this rotation returns the system to its original state.
Now think very carefully about what just happened. Those two spins could be in different places, say one in Athens and one in Bermuda. But to take the system from state $R$ to state $E$ you can rotate either the Athens spin or the Bermuda spin. These two operations, taking place on either side of the Atlantic Ocean, carry the joint system between the same two places ($R$ and $E$) in its state space. Try to imagine a classical scenario where this would happen---you will not be able to. Notice especially the sequence where first an operation is applied in Athens---an operation which certainly changes the system state---and then an operation in applied in Bermuda, and the overall outcome is no net change to the joint system.
I hope you are beginning to see how amazing entanglement is.
Its amazingness will soon be put to practical use in quantum computing. It also has deep philosophical implications, because it shows that the natural world is not completely decomposable into separate bits and pieces.
answered 5 hours ago
Andrew SteaneAndrew Steane
6,0871738
6,0871738
2
$begingroup$
Why is this outcome surprising? If the outcome is determined by the state of the system - by the orientation of A and B relative to one another - then why is it surprising that rotations performed independently and yet resulting in a particular mutual state, produce a particular outcome?
$endgroup$
– Steve
4 hours ago
$begingroup$
I think I see @Steve s question. If I have a quarter heads up in Athens and another heads down in Bermuda the system is in a mismatched state. If I flip Athens over the system is now in a matched state. I can return to the mismatched state by flipping Athens or Bermuda. So, Andrew, how is this mundane experiment any different than what you just said about quantum entanglement?
$endgroup$
– candied_orange
29 mins ago
add a comment |
2
$begingroup$
Why is this outcome surprising? If the outcome is determined by the state of the system - by the orientation of A and B relative to one another - then why is it surprising that rotations performed independently and yet resulting in a particular mutual state, produce a particular outcome?
$endgroup$
– Steve
4 hours ago
$begingroup$
I think I see @Steve s question. If I have a quarter heads up in Athens and another heads down in Bermuda the system is in a mismatched state. If I flip Athens over the system is now in a matched state. I can return to the mismatched state by flipping Athens or Bermuda. So, Andrew, how is this mundane experiment any different than what you just said about quantum entanglement?
$endgroup$
– candied_orange
29 mins ago
2
2
$begingroup$
Why is this outcome surprising? If the outcome is determined by the state of the system - by the orientation of A and B relative to one another - then why is it surprising that rotations performed independently and yet resulting in a particular mutual state, produce a particular outcome?
$endgroup$
– Steve
4 hours ago
$begingroup$
Why is this outcome surprising? If the outcome is determined by the state of the system - by the orientation of A and B relative to one another - then why is it surprising that rotations performed independently and yet resulting in a particular mutual state, produce a particular outcome?
$endgroup$
– Steve
4 hours ago
$begingroup$
I think I see @Steve s question. If I have a quarter heads up in Athens and another heads down in Bermuda the system is in a mismatched state. If I flip Athens over the system is now in a matched state. I can return to the mismatched state by flipping Athens or Bermuda. So, Andrew, how is this mundane experiment any different than what you just said about quantum entanglement?
$endgroup$
– candied_orange
29 mins ago
$begingroup$
I think I see @Steve s question. If I have a quarter heads up in Athens and another heads down in Bermuda the system is in a mismatched state. If I flip Athens over the system is now in a matched state. I can return to the mismatched state by flipping Athens or Bermuda. So, Andrew, how is this mundane experiment any different than what you just said about quantum entanglement?
$endgroup$
– candied_orange
29 mins ago
add a comment |
$begingroup$
In quantum mechanics, a particle doesn't really have a property like "spin up". It only has a probability of having a certain spin and the actual spin is only determined when you make a "measurement" or observation.
Let's say we park a space ship halfway between earth and moon and we shoot two entangled particles: one towards earth, one towards the moon. When they arrive we measure their spin on earth and on the moon at the same time. As expected, the spins are opposite.
The quantum mechanical interpretation says "spin only gets determined when measured". That would basically require the two particles on earth and on the moon to communicate with each other and decide on the spot, which spin to assume, so they make sure the spins come out opposite. This would require faster than light communication or, as Einstein put it "Spooky action at a distance" or violation of "locality"
This was called out by Einstein Podolsky and Rosen in the so-called ERP experiment or ERP paradox https://plato.stanford.edu/entries/qt-epr/ and a clear challenge the Copenhagen interpretation (Bohr, Heisenberg, etc.). At the time, no one could come up with setup or experiment which would result in an observable difference.
However, in the 1960s John Stuart Bell came up with Bell's theorem that predicted a measurable difference. https://en.wikipedia.org/wiki/Bell%27s_theorem. In the 1990s, Alain Aspect et al. https://en.wikipedia.org/wiki/Alain_Aspect managed to execute a meaningful experiment and it clearly showed the violation of Bell's theorem. It's been confirmed many times after this as well.
In other words, Einstein was wrong and Bohr was right: apparently entangled particles can communicate instantaneously and we have clear experimental evidence that this entanglement cannot be explained by a-priori knowledge or some hidden state variable.
$endgroup$
1
$begingroup$
two things: why did the copenhagen guys believe that "an observable only gets determined when measured". second, if there is instantaneous communication, how is this not violating relativity?
$endgroup$
– Joshua Benabou
7 hours ago
1
$begingroup$
third question: what's this hidden variable business? what kind of hidden variable could be a priori influencing the measurements?
$endgroup$
– Joshua Benabou
7 hours ago
$begingroup$
1) Since it's one of the few ways to explain quantum behavior. It's weird, but less weird that the alternatives. Read en.wikipedia.org/wiki/Copenhagen_interpretation .
$endgroup$
– Hilmar
5 hours ago
1
$begingroup$
3) Hidden variable is a theory to explain quantum entanglement while maintaining locality at the same time. It's basically your initial assumption "all can be explain with conservation and conventional physics.". Doesn't work though. See en.wikipedia.org/wiki/Hidden-variable_theory
$endgroup$
– Hilmar
5 hours ago
2
$begingroup$
@Hilmar Bell's theorem does not show that entangled particles can communicate instantaneously, if by 'communicate' you mean 'communicate information'. No-communication theorems make it clear that the failure of local realism does not imply instantaneous communication of information. I think your answer should be edited to choose a different word than 'communicate'.
$endgroup$
– gabe
5 hours ago
|
show 3 more comments
$begingroup$
In quantum mechanics, a particle doesn't really have a property like "spin up". It only has a probability of having a certain spin and the actual spin is only determined when you make a "measurement" or observation.
Let's say we park a space ship halfway between earth and moon and we shoot two entangled particles: one towards earth, one towards the moon. When they arrive we measure their spin on earth and on the moon at the same time. As expected, the spins are opposite.
The quantum mechanical interpretation says "spin only gets determined when measured". That would basically require the two particles on earth and on the moon to communicate with each other and decide on the spot, which spin to assume, so they make sure the spins come out opposite. This would require faster than light communication or, as Einstein put it "Spooky action at a distance" or violation of "locality"
This was called out by Einstein Podolsky and Rosen in the so-called ERP experiment or ERP paradox https://plato.stanford.edu/entries/qt-epr/ and a clear challenge the Copenhagen interpretation (Bohr, Heisenberg, etc.). At the time, no one could come up with setup or experiment which would result in an observable difference.
However, in the 1960s John Stuart Bell came up with Bell's theorem that predicted a measurable difference. https://en.wikipedia.org/wiki/Bell%27s_theorem. In the 1990s, Alain Aspect et al. https://en.wikipedia.org/wiki/Alain_Aspect managed to execute a meaningful experiment and it clearly showed the violation of Bell's theorem. It's been confirmed many times after this as well.
In other words, Einstein was wrong and Bohr was right: apparently entangled particles can communicate instantaneously and we have clear experimental evidence that this entanglement cannot be explained by a-priori knowledge or some hidden state variable.
$endgroup$
1
$begingroup$
two things: why did the copenhagen guys believe that "an observable only gets determined when measured". second, if there is instantaneous communication, how is this not violating relativity?
$endgroup$
– Joshua Benabou
7 hours ago
1
$begingroup$
third question: what's this hidden variable business? what kind of hidden variable could be a priori influencing the measurements?
$endgroup$
– Joshua Benabou
7 hours ago
$begingroup$
1) Since it's one of the few ways to explain quantum behavior. It's weird, but less weird that the alternatives. Read en.wikipedia.org/wiki/Copenhagen_interpretation .
$endgroup$
– Hilmar
5 hours ago
1
$begingroup$
3) Hidden variable is a theory to explain quantum entanglement while maintaining locality at the same time. It's basically your initial assumption "all can be explain with conservation and conventional physics.". Doesn't work though. See en.wikipedia.org/wiki/Hidden-variable_theory
$endgroup$
– Hilmar
5 hours ago
2
$begingroup$
@Hilmar Bell's theorem does not show that entangled particles can communicate instantaneously, if by 'communicate' you mean 'communicate information'. No-communication theorems make it clear that the failure of local realism does not imply instantaneous communication of information. I think your answer should be edited to choose a different word than 'communicate'.
$endgroup$
– gabe
5 hours ago
|
show 3 more comments
$begingroup$
In quantum mechanics, a particle doesn't really have a property like "spin up". It only has a probability of having a certain spin and the actual spin is only determined when you make a "measurement" or observation.
Let's say we park a space ship halfway between earth and moon and we shoot two entangled particles: one towards earth, one towards the moon. When they arrive we measure their spin on earth and on the moon at the same time. As expected, the spins are opposite.
The quantum mechanical interpretation says "spin only gets determined when measured". That would basically require the two particles on earth and on the moon to communicate with each other and decide on the spot, which spin to assume, so they make sure the spins come out opposite. This would require faster than light communication or, as Einstein put it "Spooky action at a distance" or violation of "locality"
This was called out by Einstein Podolsky and Rosen in the so-called ERP experiment or ERP paradox https://plato.stanford.edu/entries/qt-epr/ and a clear challenge the Copenhagen interpretation (Bohr, Heisenberg, etc.). At the time, no one could come up with setup or experiment which would result in an observable difference.
However, in the 1960s John Stuart Bell came up with Bell's theorem that predicted a measurable difference. https://en.wikipedia.org/wiki/Bell%27s_theorem. In the 1990s, Alain Aspect et al. https://en.wikipedia.org/wiki/Alain_Aspect managed to execute a meaningful experiment and it clearly showed the violation of Bell's theorem. It's been confirmed many times after this as well.
In other words, Einstein was wrong and Bohr was right: apparently entangled particles can communicate instantaneously and we have clear experimental evidence that this entanglement cannot be explained by a-priori knowledge or some hidden state variable.
$endgroup$
In quantum mechanics, a particle doesn't really have a property like "spin up". It only has a probability of having a certain spin and the actual spin is only determined when you make a "measurement" or observation.
Let's say we park a space ship halfway between earth and moon and we shoot two entangled particles: one towards earth, one towards the moon. When they arrive we measure their spin on earth and on the moon at the same time. As expected, the spins are opposite.
The quantum mechanical interpretation says "spin only gets determined when measured". That would basically require the two particles on earth and on the moon to communicate with each other and decide on the spot, which spin to assume, so they make sure the spins come out opposite. This would require faster than light communication or, as Einstein put it "Spooky action at a distance" or violation of "locality"
This was called out by Einstein Podolsky and Rosen in the so-called ERP experiment or ERP paradox https://plato.stanford.edu/entries/qt-epr/ and a clear challenge the Copenhagen interpretation (Bohr, Heisenberg, etc.). At the time, no one could come up with setup or experiment which would result in an observable difference.
However, in the 1960s John Stuart Bell came up with Bell's theorem that predicted a measurable difference. https://en.wikipedia.org/wiki/Bell%27s_theorem. In the 1990s, Alain Aspect et al. https://en.wikipedia.org/wiki/Alain_Aspect managed to execute a meaningful experiment and it clearly showed the violation of Bell's theorem. It's been confirmed many times after this as well.
In other words, Einstein was wrong and Bohr was right: apparently entangled particles can communicate instantaneously and we have clear experimental evidence that this entanglement cannot be explained by a-priori knowledge or some hidden state variable.
edited 5 hours ago
Cosmas Zachos
18.7k241140
18.7k241140
answered 7 hours ago
HilmarHilmar
80945
80945
1
$begingroup$
two things: why did the copenhagen guys believe that "an observable only gets determined when measured". second, if there is instantaneous communication, how is this not violating relativity?
$endgroup$
– Joshua Benabou
7 hours ago
1
$begingroup$
third question: what's this hidden variable business? what kind of hidden variable could be a priori influencing the measurements?
$endgroup$
– Joshua Benabou
7 hours ago
$begingroup$
1) Since it's one of the few ways to explain quantum behavior. It's weird, but less weird that the alternatives. Read en.wikipedia.org/wiki/Copenhagen_interpretation .
$endgroup$
– Hilmar
5 hours ago
1
$begingroup$
3) Hidden variable is a theory to explain quantum entanglement while maintaining locality at the same time. It's basically your initial assumption "all can be explain with conservation and conventional physics.". Doesn't work though. See en.wikipedia.org/wiki/Hidden-variable_theory
$endgroup$
– Hilmar
5 hours ago
2
$begingroup$
@Hilmar Bell's theorem does not show that entangled particles can communicate instantaneously, if by 'communicate' you mean 'communicate information'. No-communication theorems make it clear that the failure of local realism does not imply instantaneous communication of information. I think your answer should be edited to choose a different word than 'communicate'.
$endgroup$
– gabe
5 hours ago
|
show 3 more comments
1
$begingroup$
two things: why did the copenhagen guys believe that "an observable only gets determined when measured". second, if there is instantaneous communication, how is this not violating relativity?
$endgroup$
– Joshua Benabou
7 hours ago
1
$begingroup$
third question: what's this hidden variable business? what kind of hidden variable could be a priori influencing the measurements?
$endgroup$
– Joshua Benabou
7 hours ago
$begingroup$
1) Since it's one of the few ways to explain quantum behavior. It's weird, but less weird that the alternatives. Read en.wikipedia.org/wiki/Copenhagen_interpretation .
$endgroup$
– Hilmar
5 hours ago
1
$begingroup$
3) Hidden variable is a theory to explain quantum entanglement while maintaining locality at the same time. It's basically your initial assumption "all can be explain with conservation and conventional physics.". Doesn't work though. See en.wikipedia.org/wiki/Hidden-variable_theory
$endgroup$
– Hilmar
5 hours ago
2
$begingroup$
@Hilmar Bell's theorem does not show that entangled particles can communicate instantaneously, if by 'communicate' you mean 'communicate information'. No-communication theorems make it clear that the failure of local realism does not imply instantaneous communication of information. I think your answer should be edited to choose a different word than 'communicate'.
$endgroup$
– gabe
5 hours ago
1
1
$begingroup$
two things: why did the copenhagen guys believe that "an observable only gets determined when measured". second, if there is instantaneous communication, how is this not violating relativity?
$endgroup$
– Joshua Benabou
7 hours ago
$begingroup$
two things: why did the copenhagen guys believe that "an observable only gets determined when measured". second, if there is instantaneous communication, how is this not violating relativity?
$endgroup$
– Joshua Benabou
7 hours ago
1
1
$begingroup$
third question: what's this hidden variable business? what kind of hidden variable could be a priori influencing the measurements?
$endgroup$
– Joshua Benabou
7 hours ago
$begingroup$
third question: what's this hidden variable business? what kind of hidden variable could be a priori influencing the measurements?
$endgroup$
– Joshua Benabou
7 hours ago
$begingroup$
1) Since it's one of the few ways to explain quantum behavior. It's weird, but less weird that the alternatives. Read en.wikipedia.org/wiki/Copenhagen_interpretation .
$endgroup$
– Hilmar
5 hours ago
$begingroup$
1) Since it's one of the few ways to explain quantum behavior. It's weird, but less weird that the alternatives. Read en.wikipedia.org/wiki/Copenhagen_interpretation .
$endgroup$
– Hilmar
5 hours ago
1
1
$begingroup$
3) Hidden variable is a theory to explain quantum entanglement while maintaining locality at the same time. It's basically your initial assumption "all can be explain with conservation and conventional physics.". Doesn't work though. See en.wikipedia.org/wiki/Hidden-variable_theory
$endgroup$
– Hilmar
5 hours ago
$begingroup$
3) Hidden variable is a theory to explain quantum entanglement while maintaining locality at the same time. It's basically your initial assumption "all can be explain with conservation and conventional physics.". Doesn't work though. See en.wikipedia.org/wiki/Hidden-variable_theory
$endgroup$
– Hilmar
5 hours ago
2
2
$begingroup$
@Hilmar Bell's theorem does not show that entangled particles can communicate instantaneously, if by 'communicate' you mean 'communicate information'. No-communication theorems make it clear that the failure of local realism does not imply instantaneous communication of information. I think your answer should be edited to choose a different word than 'communicate'.
$endgroup$
– gabe
5 hours ago
$begingroup$
@Hilmar Bell's theorem does not show that entangled particles can communicate instantaneously, if by 'communicate' you mean 'communicate information'. No-communication theorems make it clear that the failure of local realism does not imply instantaneous communication of information. I think your answer should be edited to choose a different word than 'communicate'.
$endgroup$
– gabe
5 hours ago
|
show 3 more comments
$begingroup$
Maybe a good demonstration of why entanglement is so puzzling is the Mermin-Peres magic square game. There are three players, two of whom (A and B, say) have entangled states.
A and B are allowed to communicate and arrange their strategy in advance, but they cannot communicate once the game is in progress.
There is a $3 times 3$ grid. You can ask A for any column of the grid (but just one), and you can ask B for any row of the grid (but just one). The rules are that A and B must assign 0s and 1s for their cells in the grid, they must agree on the cell where the row and column intersect, and the number of 1s in a column is always odd, but the number of 1s in a row is always even.
For example, if you ask for column 1 and row 2, they might return:
A B
1xx xxx
0xx 011
0xx xxx
If they have entangled states, A and B can always win.
"Simple," you say, "A and B have agreed on which cells have 0s and 1s in advance, and they just return those values."
But if that's their strategy, does the master grid have an even or an odd number of 1's?
$endgroup$
add a comment |
$begingroup$
Maybe a good demonstration of why entanglement is so puzzling is the Mermin-Peres magic square game. There are three players, two of whom (A and B, say) have entangled states.
A and B are allowed to communicate and arrange their strategy in advance, but they cannot communicate once the game is in progress.
There is a $3 times 3$ grid. You can ask A for any column of the grid (but just one), and you can ask B for any row of the grid (but just one). The rules are that A and B must assign 0s and 1s for their cells in the grid, they must agree on the cell where the row and column intersect, and the number of 1s in a column is always odd, but the number of 1s in a row is always even.
For example, if you ask for column 1 and row 2, they might return:
A B
1xx xxx
0xx 011
0xx xxx
If they have entangled states, A and B can always win.
"Simple," you say, "A and B have agreed on which cells have 0s and 1s in advance, and they just return those values."
But if that's their strategy, does the master grid have an even or an odd number of 1's?
$endgroup$
add a comment |
$begingroup$
Maybe a good demonstration of why entanglement is so puzzling is the Mermin-Peres magic square game. There are three players, two of whom (A and B, say) have entangled states.
A and B are allowed to communicate and arrange their strategy in advance, but they cannot communicate once the game is in progress.
There is a $3 times 3$ grid. You can ask A for any column of the grid (but just one), and you can ask B for any row of the grid (but just one). The rules are that A and B must assign 0s and 1s for their cells in the grid, they must agree on the cell where the row and column intersect, and the number of 1s in a column is always odd, but the number of 1s in a row is always even.
For example, if you ask for column 1 and row 2, they might return:
A B
1xx xxx
0xx 011
0xx xxx
If they have entangled states, A and B can always win.
"Simple," you say, "A and B have agreed on which cells have 0s and 1s in advance, and they just return those values."
But if that's their strategy, does the master grid have an even or an odd number of 1's?
$endgroup$
Maybe a good demonstration of why entanglement is so puzzling is the Mermin-Peres magic square game. There are three players, two of whom (A and B, say) have entangled states.
A and B are allowed to communicate and arrange their strategy in advance, but they cannot communicate once the game is in progress.
There is a $3 times 3$ grid. You can ask A for any column of the grid (but just one), and you can ask B for any row of the grid (but just one). The rules are that A and B must assign 0s and 1s for their cells in the grid, they must agree on the cell where the row and column intersect, and the number of 1s in a column is always odd, but the number of 1s in a row is always even.
For example, if you ask for column 1 and row 2, they might return:
A B
1xx xxx
0xx 011
0xx xxx
If they have entangled states, A and B can always win.
"Simple," you say, "A and B have agreed on which cells have 0s and 1s in advance, and they just return those values."
But if that's their strategy, does the master grid have an even or an odd number of 1's?
answered 5 hours ago
Peter Shor Peter Shor
8,8302356
8,8302356
add a comment |
add a comment |
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Required, but never shown
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Someone might have a more eloquent answer, but essentially the weird features of entangled states are that they do not factor into (Particle A is in state 1)(Particle B is in state 2) which would be a satisfactory classical description. Instead entangled states look like (One of A/B is in state 1, then other is in 2). There are factored states as well, but they're much rarer and they would behave differently in terms of interference effects.
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– jacob1729
8 hours ago
3
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Possible duplicate of Can entanglement be explained as a consequence of conservation laws?
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– knzhou
8 hours ago
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I don't find the answer given there to be satisfactory.
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– Joshua Benabou
8 hours ago
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Though it might seem trivial and that entanglement is a consequence of conservation of momentum, but wait, there is more to it. Have you considered the non-locality of space which is inherent to quantumness of your entanglement experiment that you have described(above)? You aren't surprised by that aren't you?
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– EverydayFoolish
8 hours ago
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what do you mean by non locality. could you identify the problem with the momentum conservation explanation ?
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– Joshua Benabou
8 hours ago