What is the proof of entanglement?
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Really I know entanglement exists, but what I have never found is experimental evidence that measuring one entangled particle causes the state of the other entangled particle to change, rather than just being revealed.
e.g. using the spin up spin down example we know that one of the particles will be spin up and the other spin down, so when we measure one and find it is spin up we know the other is spin down.
Is there any situation after creation that the particle with spin up will change to spin down.
quantum-mechanics hilbert-space quantum-entanglement quantum-interpretations bells-inequality
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up vote
2
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Really I know entanglement exists, but what I have never found is experimental evidence that measuring one entangled particle causes the state of the other entangled particle to change, rather than just being revealed.
e.g. using the spin up spin down example we know that one of the particles will be spin up and the other spin down, so when we measure one and find it is spin up we know the other is spin down.
Is there any situation after creation that the particle with spin up will change to spin down.
quantum-mechanics hilbert-space quantum-entanglement quantum-interpretations bells-inequality
New contributor
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Really I know entanglement exists, but what I have never found is experimental evidence that measuring one entangled particle causes the state of the other entangled particle to change, rather than just being revealed.
e.g. using the spin up spin down example we know that one of the particles will be spin up and the other spin down, so when we measure one and find it is spin up we know the other is spin down.
Is there any situation after creation that the particle with spin up will change to spin down.
quantum-mechanics hilbert-space quantum-entanglement quantum-interpretations bells-inequality
New contributor
Really I know entanglement exists, but what I have never found is experimental evidence that measuring one entangled particle causes the state of the other entangled particle to change, rather than just being revealed.
e.g. using the spin up spin down example we know that one of the particles will be spin up and the other spin down, so when we measure one and find it is spin up we know the other is spin down.
Is there any situation after creation that the particle with spin up will change to spin down.
quantum-mechanics hilbert-space quantum-entanglement quantum-interpretations bells-inequality
quantum-mechanics hilbert-space quantum-entanglement quantum-interpretations bells-inequality
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Qmechanic♦
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Qandry
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2 Answers
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active
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up vote
9
down vote
The assumption that a measurable property exists whether or not we measure it is inconsistent with the experimental facts.
Here's a relatively simple example. Suppose we have four observables, $A,B,C,D$, each of which has two possible outcomes. (For example, these could be single-photon polarization observables.) For mathematical convenience, label the two possible outcomes $+1$ and $-1$, for each of the four observables. Suppose for a moment that the act of measurement
merely reveals properties that would exist
anyway even if they were not measured.
If this were true, then any given state
of the system would have definite values
$a,b,c,d$ of the observables $A,B,C,D$.
Each of the four values $a,b,c,d$
could be either $+1$ or $-1$,
so there would be $2^4=16$ different possible
combinations of these values.
Any given state would have one of these 16 possible combinations.
Now consider the two quantities $a+c$ and $c-a$.
The fact that $a$ and $c$ both have magnitude $1$
implies that one of these two quantities must be zero,
and then the other one must be either $+2$ or $-2$.
This, in turn, implies that the quantity
$$
(a+c)b+(c-a)d
$$
is either $+2$ or $-2$.
This is true
for every one of the 16 possible combinations
of values for $a,b,c,d$,
so if we prepare many states,
then the average value of this quantity must be somewhere
between $+2$ and $-2$.
In particular, the average cannot
be any larger than $+2$.
This gives the CHSH inequality
$$
langle{AB}rangle
+langle{CB}rangle
+langle{CD}rangle
-langle{AD}rangleleq 2,
$$
where $langle{AB}rangle$ denotes the average of the
product of the values of $a$ and $b$
over all of the prepared states.
In the real world, the CHSH inequality can be violated, and quantum theory correctly predicts the observed violations. The quantity $langle{AB}rangle
+langle{CB}rangle
+langle{CD}rangle
-langle{AD}rangle$ can be as large as $2sqrt{2}$. Here are a few papers describing experiments that verify this:
Kwiat et al, 1995. “New high-intensity source of polarization-entangled photon pairs,” http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.75.4337
Kwiat et al, 1998. “Ultra-bright source of polarization-entangled photons,” http://arxiv.org/abs/quant-ph/9810003
Weihs et al, 1998. “Violation of Bell’s inequality under strict Einstein locality conditions,” http://arxiv.org/abs/quant-ph/9810080
The fact that the CHSH inequality is violated in the real world implies that the premise from which it was derived cannot be correct. The CHSH inequality was derived above by assuming that the act of measurement merely reveals properties that would exist anyway even if they were not measured. The inequality is violated in the real world, so this assumption must be wrong in the real world. Measurement plays a more active role.
1
No, please.I offered a simple case and would appreciate a simple answer. In my case the spin is either up or down. Can you provide an experimental fact that shows that the properties do not exists when they have not been measured?
– Qandry
4 hours ago
We don't know that particle A has spin up. But we know that it has either spin up or spin down. Therefore when we measure it we know that particle B has the opposite spin. Particle A had a spin when it was created whatever it was it will always have that spin until interfered with in some way, unless there is a way that it may change. Nevertheless, if Particle A has spin up, Particle B has spin down. If we measure one, we know the other. There is no need for communication between the particles.
– Qandry
4 hours ago
2
The thing is that the CHSH inequality mentioned (and its violation) mean that your premise cannot be true: particle A did not "have a spin when it was created". It only had a spin after it was measured.
– Remellion
4 hours ago
2
@Qandry If we know that particle A has either spin up or spin down, even if we don't know which one, then its spin is not entangled with the spin of any other particle. (I'm assuming that we're talking about spin 1/2 particles.) In order for us to know that the spin of A is $in{text{up},text{down}}$, we must know that it has already been measured along that axis, in which case it is no longer entangled. Conversely, if the spin of particle A is entangled with the spin of particle B, then we don't know that the spin of particle A is $in{text{up},text{down}}$.
– Dan Yand
4 hours ago
2
Qandry: @DanYand has taken the trouble to write an excellent answer for you; you'd do well to digest it. The key point you seem to be missing is that if you've got a particle and if the outcome of at least one local measurement on that particle is predetermined, then by definition that particle is not entangled. You can insist that the outcome of some certain measurement is predetermined (the jargon for this is that the particle is in an eigenstate of that measurement) or you can insist that the particle is entangled, but you cannot insist simultaneously on both.
– WillO
2 hours ago
|
show 3 more comments
up vote
1
down vote
Just for a moment, if one (you) imagine that the entangled particles are determinated in their states from the beginning of their entanglement, would it change something in the outcome of the experiment? In the pair production of two photons from nonlinear optical crystals the result of the experiments shows a statistical dependency; the orientation of the electric field components of the photons is orthogonal to each over.
Producing such a pair of photons we are not able to control the direction of the polarisation, we only able to design the process so that the polarisations of two particles are orthogonal on each over, the orientation by itself is randomly distributed around 360°. The uncertainty is in the phenomenon, that using a polariser to find the orientation of the particles, we do not get a result in 50% of our measurements (with the best designed polarizer and for the right wavelength).
Let the polarizer be orientated to the vertical (0°). Getting a result, we will know that the photon has had an orientation from -45° to 45° and from 135° to 225°. Not getting a result, we are allowed to say, that the photon has an orientation in the opposite to the above mentioned angles. To count for the statistics these not measured photons we are able only after getting the information from the second measurement device, that a photon in their setup was measured. In reality simply the number of measured incidents is compared.
What is the proof of entanglement
Firstly a lot of experiments for a given process, in this case for photons from nonlinear optical crystals in a well definded setup. And later the knowledge, that in this process are produced entangled particles.
Is there any situation after creation that the particle with spin up will change to spin down.
This happens all the time due to “impurities” in the experimental setup. The produced particles colliding with other particles and are under the influence of external fields. The longer the distance and the time between production and measurement, the weaker the measurable coincidence. But the swap of a spin by itself is impossible.
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2 Answers
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2 Answers
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up vote
9
down vote
The assumption that a measurable property exists whether or not we measure it is inconsistent with the experimental facts.
Here's a relatively simple example. Suppose we have four observables, $A,B,C,D$, each of which has two possible outcomes. (For example, these could be single-photon polarization observables.) For mathematical convenience, label the two possible outcomes $+1$ and $-1$, for each of the four observables. Suppose for a moment that the act of measurement
merely reveals properties that would exist
anyway even if they were not measured.
If this were true, then any given state
of the system would have definite values
$a,b,c,d$ of the observables $A,B,C,D$.
Each of the four values $a,b,c,d$
could be either $+1$ or $-1$,
so there would be $2^4=16$ different possible
combinations of these values.
Any given state would have one of these 16 possible combinations.
Now consider the two quantities $a+c$ and $c-a$.
The fact that $a$ and $c$ both have magnitude $1$
implies that one of these two quantities must be zero,
and then the other one must be either $+2$ or $-2$.
This, in turn, implies that the quantity
$$
(a+c)b+(c-a)d
$$
is either $+2$ or $-2$.
This is true
for every one of the 16 possible combinations
of values for $a,b,c,d$,
so if we prepare many states,
then the average value of this quantity must be somewhere
between $+2$ and $-2$.
In particular, the average cannot
be any larger than $+2$.
This gives the CHSH inequality
$$
langle{AB}rangle
+langle{CB}rangle
+langle{CD}rangle
-langle{AD}rangleleq 2,
$$
where $langle{AB}rangle$ denotes the average of the
product of the values of $a$ and $b$
over all of the prepared states.
In the real world, the CHSH inequality can be violated, and quantum theory correctly predicts the observed violations. The quantity $langle{AB}rangle
+langle{CB}rangle
+langle{CD}rangle
-langle{AD}rangle$ can be as large as $2sqrt{2}$. Here are a few papers describing experiments that verify this:
Kwiat et al, 1995. “New high-intensity source of polarization-entangled photon pairs,” http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.75.4337
Kwiat et al, 1998. “Ultra-bright source of polarization-entangled photons,” http://arxiv.org/abs/quant-ph/9810003
Weihs et al, 1998. “Violation of Bell’s inequality under strict Einstein locality conditions,” http://arxiv.org/abs/quant-ph/9810080
The fact that the CHSH inequality is violated in the real world implies that the premise from which it was derived cannot be correct. The CHSH inequality was derived above by assuming that the act of measurement merely reveals properties that would exist anyway even if they were not measured. The inequality is violated in the real world, so this assumption must be wrong in the real world. Measurement plays a more active role.
1
No, please.I offered a simple case and would appreciate a simple answer. In my case the spin is either up or down. Can you provide an experimental fact that shows that the properties do not exists when they have not been measured?
– Qandry
4 hours ago
We don't know that particle A has spin up. But we know that it has either spin up or spin down. Therefore when we measure it we know that particle B has the opposite spin. Particle A had a spin when it was created whatever it was it will always have that spin until interfered with in some way, unless there is a way that it may change. Nevertheless, if Particle A has spin up, Particle B has spin down. If we measure one, we know the other. There is no need for communication between the particles.
– Qandry
4 hours ago
2
The thing is that the CHSH inequality mentioned (and its violation) mean that your premise cannot be true: particle A did not "have a spin when it was created". It only had a spin after it was measured.
– Remellion
4 hours ago
2
@Qandry If we know that particle A has either spin up or spin down, even if we don't know which one, then its spin is not entangled with the spin of any other particle. (I'm assuming that we're talking about spin 1/2 particles.) In order for us to know that the spin of A is $in{text{up},text{down}}$, we must know that it has already been measured along that axis, in which case it is no longer entangled. Conversely, if the spin of particle A is entangled with the spin of particle B, then we don't know that the spin of particle A is $in{text{up},text{down}}$.
– Dan Yand
4 hours ago
2
Qandry: @DanYand has taken the trouble to write an excellent answer for you; you'd do well to digest it. The key point you seem to be missing is that if you've got a particle and if the outcome of at least one local measurement on that particle is predetermined, then by definition that particle is not entangled. You can insist that the outcome of some certain measurement is predetermined (the jargon for this is that the particle is in an eigenstate of that measurement) or you can insist that the particle is entangled, but you cannot insist simultaneously on both.
– WillO
2 hours ago
|
show 3 more comments
up vote
9
down vote
The assumption that a measurable property exists whether or not we measure it is inconsistent with the experimental facts.
Here's a relatively simple example. Suppose we have four observables, $A,B,C,D$, each of which has two possible outcomes. (For example, these could be single-photon polarization observables.) For mathematical convenience, label the two possible outcomes $+1$ and $-1$, for each of the four observables. Suppose for a moment that the act of measurement
merely reveals properties that would exist
anyway even if they were not measured.
If this were true, then any given state
of the system would have definite values
$a,b,c,d$ of the observables $A,B,C,D$.
Each of the four values $a,b,c,d$
could be either $+1$ or $-1$,
so there would be $2^4=16$ different possible
combinations of these values.
Any given state would have one of these 16 possible combinations.
Now consider the two quantities $a+c$ and $c-a$.
The fact that $a$ and $c$ both have magnitude $1$
implies that one of these two quantities must be zero,
and then the other one must be either $+2$ or $-2$.
This, in turn, implies that the quantity
$$
(a+c)b+(c-a)d
$$
is either $+2$ or $-2$.
This is true
for every one of the 16 possible combinations
of values for $a,b,c,d$,
so if we prepare many states,
then the average value of this quantity must be somewhere
between $+2$ and $-2$.
In particular, the average cannot
be any larger than $+2$.
This gives the CHSH inequality
$$
langle{AB}rangle
+langle{CB}rangle
+langle{CD}rangle
-langle{AD}rangleleq 2,
$$
where $langle{AB}rangle$ denotes the average of the
product of the values of $a$ and $b$
over all of the prepared states.
In the real world, the CHSH inequality can be violated, and quantum theory correctly predicts the observed violations. The quantity $langle{AB}rangle
+langle{CB}rangle
+langle{CD}rangle
-langle{AD}rangle$ can be as large as $2sqrt{2}$. Here are a few papers describing experiments that verify this:
Kwiat et al, 1995. “New high-intensity source of polarization-entangled photon pairs,” http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.75.4337
Kwiat et al, 1998. “Ultra-bright source of polarization-entangled photons,” http://arxiv.org/abs/quant-ph/9810003
Weihs et al, 1998. “Violation of Bell’s inequality under strict Einstein locality conditions,” http://arxiv.org/abs/quant-ph/9810080
The fact that the CHSH inequality is violated in the real world implies that the premise from which it was derived cannot be correct. The CHSH inequality was derived above by assuming that the act of measurement merely reveals properties that would exist anyway even if they were not measured. The inequality is violated in the real world, so this assumption must be wrong in the real world. Measurement plays a more active role.
1
No, please.I offered a simple case and would appreciate a simple answer. In my case the spin is either up or down. Can you provide an experimental fact that shows that the properties do not exists when they have not been measured?
– Qandry
4 hours ago
We don't know that particle A has spin up. But we know that it has either spin up or spin down. Therefore when we measure it we know that particle B has the opposite spin. Particle A had a spin when it was created whatever it was it will always have that spin until interfered with in some way, unless there is a way that it may change. Nevertheless, if Particle A has spin up, Particle B has spin down. If we measure one, we know the other. There is no need for communication between the particles.
– Qandry
4 hours ago
2
The thing is that the CHSH inequality mentioned (and its violation) mean that your premise cannot be true: particle A did not "have a spin when it was created". It only had a spin after it was measured.
– Remellion
4 hours ago
2
@Qandry If we know that particle A has either spin up or spin down, even if we don't know which one, then its spin is not entangled with the spin of any other particle. (I'm assuming that we're talking about spin 1/2 particles.) In order for us to know that the spin of A is $in{text{up},text{down}}$, we must know that it has already been measured along that axis, in which case it is no longer entangled. Conversely, if the spin of particle A is entangled with the spin of particle B, then we don't know that the spin of particle A is $in{text{up},text{down}}$.
– Dan Yand
4 hours ago
2
Qandry: @DanYand has taken the trouble to write an excellent answer for you; you'd do well to digest it. The key point you seem to be missing is that if you've got a particle and if the outcome of at least one local measurement on that particle is predetermined, then by definition that particle is not entangled. You can insist that the outcome of some certain measurement is predetermined (the jargon for this is that the particle is in an eigenstate of that measurement) or you can insist that the particle is entangled, but you cannot insist simultaneously on both.
– WillO
2 hours ago
|
show 3 more comments
up vote
9
down vote
up vote
9
down vote
The assumption that a measurable property exists whether or not we measure it is inconsistent with the experimental facts.
Here's a relatively simple example. Suppose we have four observables, $A,B,C,D$, each of which has two possible outcomes. (For example, these could be single-photon polarization observables.) For mathematical convenience, label the two possible outcomes $+1$ and $-1$, for each of the four observables. Suppose for a moment that the act of measurement
merely reveals properties that would exist
anyway even if they were not measured.
If this were true, then any given state
of the system would have definite values
$a,b,c,d$ of the observables $A,B,C,D$.
Each of the four values $a,b,c,d$
could be either $+1$ or $-1$,
so there would be $2^4=16$ different possible
combinations of these values.
Any given state would have one of these 16 possible combinations.
Now consider the two quantities $a+c$ and $c-a$.
The fact that $a$ and $c$ both have magnitude $1$
implies that one of these two quantities must be zero,
and then the other one must be either $+2$ or $-2$.
This, in turn, implies that the quantity
$$
(a+c)b+(c-a)d
$$
is either $+2$ or $-2$.
This is true
for every one of the 16 possible combinations
of values for $a,b,c,d$,
so if we prepare many states,
then the average value of this quantity must be somewhere
between $+2$ and $-2$.
In particular, the average cannot
be any larger than $+2$.
This gives the CHSH inequality
$$
langle{AB}rangle
+langle{CB}rangle
+langle{CD}rangle
-langle{AD}rangleleq 2,
$$
where $langle{AB}rangle$ denotes the average of the
product of the values of $a$ and $b$
over all of the prepared states.
In the real world, the CHSH inequality can be violated, and quantum theory correctly predicts the observed violations. The quantity $langle{AB}rangle
+langle{CB}rangle
+langle{CD}rangle
-langle{AD}rangle$ can be as large as $2sqrt{2}$. Here are a few papers describing experiments that verify this:
Kwiat et al, 1995. “New high-intensity source of polarization-entangled photon pairs,” http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.75.4337
Kwiat et al, 1998. “Ultra-bright source of polarization-entangled photons,” http://arxiv.org/abs/quant-ph/9810003
Weihs et al, 1998. “Violation of Bell’s inequality under strict Einstein locality conditions,” http://arxiv.org/abs/quant-ph/9810080
The fact that the CHSH inequality is violated in the real world implies that the premise from which it was derived cannot be correct. The CHSH inequality was derived above by assuming that the act of measurement merely reveals properties that would exist anyway even if they were not measured. The inequality is violated in the real world, so this assumption must be wrong in the real world. Measurement plays a more active role.
The assumption that a measurable property exists whether or not we measure it is inconsistent with the experimental facts.
Here's a relatively simple example. Suppose we have four observables, $A,B,C,D$, each of which has two possible outcomes. (For example, these could be single-photon polarization observables.) For mathematical convenience, label the two possible outcomes $+1$ and $-1$, for each of the four observables. Suppose for a moment that the act of measurement
merely reveals properties that would exist
anyway even if they were not measured.
If this were true, then any given state
of the system would have definite values
$a,b,c,d$ of the observables $A,B,C,D$.
Each of the four values $a,b,c,d$
could be either $+1$ or $-1$,
so there would be $2^4=16$ different possible
combinations of these values.
Any given state would have one of these 16 possible combinations.
Now consider the two quantities $a+c$ and $c-a$.
The fact that $a$ and $c$ both have magnitude $1$
implies that one of these two quantities must be zero,
and then the other one must be either $+2$ or $-2$.
This, in turn, implies that the quantity
$$
(a+c)b+(c-a)d
$$
is either $+2$ or $-2$.
This is true
for every one of the 16 possible combinations
of values for $a,b,c,d$,
so if we prepare many states,
then the average value of this quantity must be somewhere
between $+2$ and $-2$.
In particular, the average cannot
be any larger than $+2$.
This gives the CHSH inequality
$$
langle{AB}rangle
+langle{CB}rangle
+langle{CD}rangle
-langle{AD}rangleleq 2,
$$
where $langle{AB}rangle$ denotes the average of the
product of the values of $a$ and $b$
over all of the prepared states.
In the real world, the CHSH inequality can be violated, and quantum theory correctly predicts the observed violations. The quantity $langle{AB}rangle
+langle{CB}rangle
+langle{CD}rangle
-langle{AD}rangle$ can be as large as $2sqrt{2}$. Here are a few papers describing experiments that verify this:
Kwiat et al, 1995. “New high-intensity source of polarization-entangled photon pairs,” http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.75.4337
Kwiat et al, 1998. “Ultra-bright source of polarization-entangled photons,” http://arxiv.org/abs/quant-ph/9810003
Weihs et al, 1998. “Violation of Bell’s inequality under strict Einstein locality conditions,” http://arxiv.org/abs/quant-ph/9810080
The fact that the CHSH inequality is violated in the real world implies that the premise from which it was derived cannot be correct. The CHSH inequality was derived above by assuming that the act of measurement merely reveals properties that would exist anyway even if they were not measured. The inequality is violated in the real world, so this assumption must be wrong in the real world. Measurement plays a more active role.
answered 4 hours ago
Dan Yand
4,9291422
4,9291422
1
No, please.I offered a simple case and would appreciate a simple answer. In my case the spin is either up or down. Can you provide an experimental fact that shows that the properties do not exists when they have not been measured?
– Qandry
4 hours ago
We don't know that particle A has spin up. But we know that it has either spin up or spin down. Therefore when we measure it we know that particle B has the opposite spin. Particle A had a spin when it was created whatever it was it will always have that spin until interfered with in some way, unless there is a way that it may change. Nevertheless, if Particle A has spin up, Particle B has spin down. If we measure one, we know the other. There is no need for communication between the particles.
– Qandry
4 hours ago
2
The thing is that the CHSH inequality mentioned (and its violation) mean that your premise cannot be true: particle A did not "have a spin when it was created". It only had a spin after it was measured.
– Remellion
4 hours ago
2
@Qandry If we know that particle A has either spin up or spin down, even if we don't know which one, then its spin is not entangled with the spin of any other particle. (I'm assuming that we're talking about spin 1/2 particles.) In order for us to know that the spin of A is $in{text{up},text{down}}$, we must know that it has already been measured along that axis, in which case it is no longer entangled. Conversely, if the spin of particle A is entangled with the spin of particle B, then we don't know that the spin of particle A is $in{text{up},text{down}}$.
– Dan Yand
4 hours ago
2
Qandry: @DanYand has taken the trouble to write an excellent answer for you; you'd do well to digest it. The key point you seem to be missing is that if you've got a particle and if the outcome of at least one local measurement on that particle is predetermined, then by definition that particle is not entangled. You can insist that the outcome of some certain measurement is predetermined (the jargon for this is that the particle is in an eigenstate of that measurement) or you can insist that the particle is entangled, but you cannot insist simultaneously on both.
– WillO
2 hours ago
|
show 3 more comments
1
No, please.I offered a simple case and would appreciate a simple answer. In my case the spin is either up or down. Can you provide an experimental fact that shows that the properties do not exists when they have not been measured?
– Qandry
4 hours ago
We don't know that particle A has spin up. But we know that it has either spin up or spin down. Therefore when we measure it we know that particle B has the opposite spin. Particle A had a spin when it was created whatever it was it will always have that spin until interfered with in some way, unless there is a way that it may change. Nevertheless, if Particle A has spin up, Particle B has spin down. If we measure one, we know the other. There is no need for communication between the particles.
– Qandry
4 hours ago
2
The thing is that the CHSH inequality mentioned (and its violation) mean that your premise cannot be true: particle A did not "have a spin when it was created". It only had a spin after it was measured.
– Remellion
4 hours ago
2
@Qandry If we know that particle A has either spin up or spin down, even if we don't know which one, then its spin is not entangled with the spin of any other particle. (I'm assuming that we're talking about spin 1/2 particles.) In order for us to know that the spin of A is $in{text{up},text{down}}$, we must know that it has already been measured along that axis, in which case it is no longer entangled. Conversely, if the spin of particle A is entangled with the spin of particle B, then we don't know that the spin of particle A is $in{text{up},text{down}}$.
– Dan Yand
4 hours ago
2
Qandry: @DanYand has taken the trouble to write an excellent answer for you; you'd do well to digest it. The key point you seem to be missing is that if you've got a particle and if the outcome of at least one local measurement on that particle is predetermined, then by definition that particle is not entangled. You can insist that the outcome of some certain measurement is predetermined (the jargon for this is that the particle is in an eigenstate of that measurement) or you can insist that the particle is entangled, but you cannot insist simultaneously on both.
– WillO
2 hours ago
1
1
No, please.I offered a simple case and would appreciate a simple answer. In my case the spin is either up or down. Can you provide an experimental fact that shows that the properties do not exists when they have not been measured?
– Qandry
4 hours ago
No, please.I offered a simple case and would appreciate a simple answer. In my case the spin is either up or down. Can you provide an experimental fact that shows that the properties do not exists when they have not been measured?
– Qandry
4 hours ago
We don't know that particle A has spin up. But we know that it has either spin up or spin down. Therefore when we measure it we know that particle B has the opposite spin. Particle A had a spin when it was created whatever it was it will always have that spin until interfered with in some way, unless there is a way that it may change. Nevertheless, if Particle A has spin up, Particle B has spin down. If we measure one, we know the other. There is no need for communication between the particles.
– Qandry
4 hours ago
We don't know that particle A has spin up. But we know that it has either spin up or spin down. Therefore when we measure it we know that particle B has the opposite spin. Particle A had a spin when it was created whatever it was it will always have that spin until interfered with in some way, unless there is a way that it may change. Nevertheless, if Particle A has spin up, Particle B has spin down. If we measure one, we know the other. There is no need for communication between the particles.
– Qandry
4 hours ago
2
2
The thing is that the CHSH inequality mentioned (and its violation) mean that your premise cannot be true: particle A did not "have a spin when it was created". It only had a spin after it was measured.
– Remellion
4 hours ago
The thing is that the CHSH inequality mentioned (and its violation) mean that your premise cannot be true: particle A did not "have a spin when it was created". It only had a spin after it was measured.
– Remellion
4 hours ago
2
2
@Qandry If we know that particle A has either spin up or spin down, even if we don't know which one, then its spin is not entangled with the spin of any other particle. (I'm assuming that we're talking about spin 1/2 particles.) In order for us to know that the spin of A is $in{text{up},text{down}}$, we must know that it has already been measured along that axis, in which case it is no longer entangled. Conversely, if the spin of particle A is entangled with the spin of particle B, then we don't know that the spin of particle A is $in{text{up},text{down}}$.
– Dan Yand
4 hours ago
@Qandry If we know that particle A has either spin up or spin down, even if we don't know which one, then its spin is not entangled with the spin of any other particle. (I'm assuming that we're talking about spin 1/2 particles.) In order for us to know that the spin of A is $in{text{up},text{down}}$, we must know that it has already been measured along that axis, in which case it is no longer entangled. Conversely, if the spin of particle A is entangled with the spin of particle B, then we don't know that the spin of particle A is $in{text{up},text{down}}$.
– Dan Yand
4 hours ago
2
2
Qandry: @DanYand has taken the trouble to write an excellent answer for you; you'd do well to digest it. The key point you seem to be missing is that if you've got a particle and if the outcome of at least one local measurement on that particle is predetermined, then by definition that particle is not entangled. You can insist that the outcome of some certain measurement is predetermined (the jargon for this is that the particle is in an eigenstate of that measurement) or you can insist that the particle is entangled, but you cannot insist simultaneously on both.
– WillO
2 hours ago
Qandry: @DanYand has taken the trouble to write an excellent answer for you; you'd do well to digest it. The key point you seem to be missing is that if you've got a particle and if the outcome of at least one local measurement on that particle is predetermined, then by definition that particle is not entangled. You can insist that the outcome of some certain measurement is predetermined (the jargon for this is that the particle is in an eigenstate of that measurement) or you can insist that the particle is entangled, but you cannot insist simultaneously on both.
– WillO
2 hours ago
|
show 3 more comments
up vote
1
down vote
Just for a moment, if one (you) imagine that the entangled particles are determinated in their states from the beginning of their entanglement, would it change something in the outcome of the experiment? In the pair production of two photons from nonlinear optical crystals the result of the experiments shows a statistical dependency; the orientation of the electric field components of the photons is orthogonal to each over.
Producing such a pair of photons we are not able to control the direction of the polarisation, we only able to design the process so that the polarisations of two particles are orthogonal on each over, the orientation by itself is randomly distributed around 360°. The uncertainty is in the phenomenon, that using a polariser to find the orientation of the particles, we do not get a result in 50% of our measurements (with the best designed polarizer and for the right wavelength).
Let the polarizer be orientated to the vertical (0°). Getting a result, we will know that the photon has had an orientation from -45° to 45° and from 135° to 225°. Not getting a result, we are allowed to say, that the photon has an orientation in the opposite to the above mentioned angles. To count for the statistics these not measured photons we are able only after getting the information from the second measurement device, that a photon in their setup was measured. In reality simply the number of measured incidents is compared.
What is the proof of entanglement
Firstly a lot of experiments for a given process, in this case for photons from nonlinear optical crystals in a well definded setup. And later the knowledge, that in this process are produced entangled particles.
Is there any situation after creation that the particle with spin up will change to spin down.
This happens all the time due to “impurities” in the experimental setup. The produced particles colliding with other particles and are under the influence of external fields. The longer the distance and the time between production and measurement, the weaker the measurable coincidence. But the swap of a spin by itself is impossible.
add a comment |
up vote
1
down vote
Just for a moment, if one (you) imagine that the entangled particles are determinated in their states from the beginning of their entanglement, would it change something in the outcome of the experiment? In the pair production of two photons from nonlinear optical crystals the result of the experiments shows a statistical dependency; the orientation of the electric field components of the photons is orthogonal to each over.
Producing such a pair of photons we are not able to control the direction of the polarisation, we only able to design the process so that the polarisations of two particles are orthogonal on each over, the orientation by itself is randomly distributed around 360°. The uncertainty is in the phenomenon, that using a polariser to find the orientation of the particles, we do not get a result in 50% of our measurements (with the best designed polarizer and for the right wavelength).
Let the polarizer be orientated to the vertical (0°). Getting a result, we will know that the photon has had an orientation from -45° to 45° and from 135° to 225°. Not getting a result, we are allowed to say, that the photon has an orientation in the opposite to the above mentioned angles. To count for the statistics these not measured photons we are able only after getting the information from the second measurement device, that a photon in their setup was measured. In reality simply the number of measured incidents is compared.
What is the proof of entanglement
Firstly a lot of experiments for a given process, in this case for photons from nonlinear optical crystals in a well definded setup. And later the knowledge, that in this process are produced entangled particles.
Is there any situation after creation that the particle with spin up will change to spin down.
This happens all the time due to “impurities” in the experimental setup. The produced particles colliding with other particles and are under the influence of external fields. The longer the distance and the time between production and measurement, the weaker the measurable coincidence. But the swap of a spin by itself is impossible.
add a comment |
up vote
1
down vote
up vote
1
down vote
Just for a moment, if one (you) imagine that the entangled particles are determinated in their states from the beginning of their entanglement, would it change something in the outcome of the experiment? In the pair production of two photons from nonlinear optical crystals the result of the experiments shows a statistical dependency; the orientation of the electric field components of the photons is orthogonal to each over.
Producing such a pair of photons we are not able to control the direction of the polarisation, we only able to design the process so that the polarisations of two particles are orthogonal on each over, the orientation by itself is randomly distributed around 360°. The uncertainty is in the phenomenon, that using a polariser to find the orientation of the particles, we do not get a result in 50% of our measurements (with the best designed polarizer and for the right wavelength).
Let the polarizer be orientated to the vertical (0°). Getting a result, we will know that the photon has had an orientation from -45° to 45° and from 135° to 225°. Not getting a result, we are allowed to say, that the photon has an orientation in the opposite to the above mentioned angles. To count for the statistics these not measured photons we are able only after getting the information from the second measurement device, that a photon in their setup was measured. In reality simply the number of measured incidents is compared.
What is the proof of entanglement
Firstly a lot of experiments for a given process, in this case for photons from nonlinear optical crystals in a well definded setup. And later the knowledge, that in this process are produced entangled particles.
Is there any situation after creation that the particle with spin up will change to spin down.
This happens all the time due to “impurities” in the experimental setup. The produced particles colliding with other particles and are under the influence of external fields. The longer the distance and the time between production and measurement, the weaker the measurable coincidence. But the swap of a spin by itself is impossible.
Just for a moment, if one (you) imagine that the entangled particles are determinated in their states from the beginning of their entanglement, would it change something in the outcome of the experiment? In the pair production of two photons from nonlinear optical crystals the result of the experiments shows a statistical dependency; the orientation of the electric field components of the photons is orthogonal to each over.
Producing such a pair of photons we are not able to control the direction of the polarisation, we only able to design the process so that the polarisations of two particles are orthogonal on each over, the orientation by itself is randomly distributed around 360°. The uncertainty is in the phenomenon, that using a polariser to find the orientation of the particles, we do not get a result in 50% of our measurements (with the best designed polarizer and for the right wavelength).
Let the polarizer be orientated to the vertical (0°). Getting a result, we will know that the photon has had an orientation from -45° to 45° and from 135° to 225°. Not getting a result, we are allowed to say, that the photon has an orientation in the opposite to the above mentioned angles. To count for the statistics these not measured photons we are able only after getting the information from the second measurement device, that a photon in their setup was measured. In reality simply the number of measured incidents is compared.
What is the proof of entanglement
Firstly a lot of experiments for a given process, in this case for photons from nonlinear optical crystals in a well definded setup. And later the knowledge, that in this process are produced entangled particles.
Is there any situation after creation that the particle with spin up will change to spin down.
This happens all the time due to “impurities” in the experimental setup. The produced particles colliding with other particles and are under the influence of external fields. The longer the distance and the time between production and measurement, the weaker the measurable coincidence. But the swap of a spin by itself is impossible.
edited 3 hours ago
answered 3 hours ago
HolgerFiedler
4,04331133
4,04331133
add a comment |
add a comment |
Qandry is a new contributor. Be nice, and check out our Code of Conduct.
Qandry is a new contributor. Be nice, and check out our Code of Conduct.
Qandry is a new contributor. Be nice, and check out our Code of Conduct.
Qandry is a new contributor. Be nice, and check out our Code of Conduct.
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