B Quantum Entanglement information transmission idea to knock down....

[PeroK]

LM rocks!

Yes, that's it exactly! My poor usage of terminology confused everyone.

So my question is: why is LM not QM? What am I doing wrong in the LM? Now that I've cleared up my confused attempt to pose a thought experiment and been understood, what specifically is actually wrong in the LM trail of logic?

Your thought experiment reveals several misconceptions about QM. First, I suggest, we need to look at quantum spin compared to classical spin.

For a classical object we can find an axis of rotation. E.g. the Earth has an axis of rotation, and so would a spinning basketball. And, the spin of either object is not of fixed magnitude: it may be increased or decreased if a torque is applied.

The spin on an electron is fundamentally different in a number of ways (which leads to the picture of a classically spinning object not being a good mental model):

The total amount of spin is fixed. The measured value is always $\dfrac {\sqrt 3 \hbar}{2}$. This cannot be increased or decreased. Moreover, the spin component when measured about any axis is always fixed in magnitude and only varies in direction. It's always $\dfrac{\hbar}{2}$.

This is completely non-classical and means we can never find an axis of rotation for the electron. Whatever axis we choose, the measured spin component is always effectively a third of the total spin.

These two characeristics lead to the electron's spin being described by a spin state. A study of the nature of electron spin and of electron spin states is a good place to start with QM.

Finally, if we measure the spin about a given axis, the state of the electron after the measurement correpsonds to a state where further measurements about the same axis will yield the same result. This is called an eigenstate.

For example, if we measure the spin about the z-axis and get $+\dfrac \hbar 2$, then the state of the electron after the measurement is z-spin-up.

If we prepare an ensemble of electrons in the z-spin-up state and measure spin about the x-axis or y-axis, then in each case we get a 50-50 split of x-spin-up and x-spin-down or y-spin-up and y-spin-down respectively. And, if we measure spin about an axis that makes an angle $\theta$ with the z-axis, then we get a measurement of spin-up with probability $\cos^2 theta$ (and spin-down with probability $\sin^2 \theta$).

Now, let's describe an experiment that does obey QM:

Two particles are created with zero total spin. This means that their spins must be opposite (if measured about any axis). This could be explained by each particle having a definite but opposite spin about each axis. This would be a so-called hidden variables theory. There's a long story here about how tests of Bell's inequality have ruled out such local hidden variable theories.

Instead, QM says that the spin of neither particle is determined until one particle is measured. This is called an entangled state and, indeed, the system may only be analysed as a two-particle system; not as two separate particles each with its own state.

If both particles are measured about the same axis, then they are always found to have opposite spins. QM is silent on how nature achieves this. QM simply says that is the what nature serves up when the particles are measured.

Now, we prepare an ensemble of such entangled pairs and carry out some spin measurements about different axes.

Let's say that one particle is always measured about the z-axis. We get a 50-50 split of z-spin-up and z-spin-down. If the other particle is measured about the x-axis, then we get a 50-50 split of x-spin-up and x-spin down. If the results are compared, we find no correlation between the results. I.e. 25% of the time we will get z-up, x-up; 25% of the time z-up, x-down; 25% z-down, x-up; and 25% z-down, x-down.

Note that measuring one particle about the z-axis has no effect on the 50-50 distribution of measurements about the x-axis.

Now, suppose we measure one particle about an axis an angle $\theta$ from the z-axis, where $\cos^2 \theta = 0.9$, say. Let's call this axis $w$.

As always, we will get 50% w-spin-up and 50% w-spin-down.

Finally, suppose we do the w-axis measurement after the first particle has been measured about the z-axis. We still get 50-50, but thse results are correleted with the random results that of the z-measurement. I.e.:

45% of the time we get z-up and w-down
5% of the time we get z-up and w-up
45% of the time we get z-down and w-up
5% of the time we get z-down and w-down

If the experimenter could force a result of z-up, then they could influence the distribution of w-axis measurements. But, they can't force the electron to be z-spin-up and thereby signal this to the second experimenter. The first measurement is random 50-50 of up or down. We have no control over this.

In conclusion, we have known correlation between measurement results; but no control over either set of measurement results, hence no signalling capability.

 

[LozD]

Your thought experiment reveals several misconceptions about QM. First, I suggest, we need to look at quantum spin compared to classical spin.

For a classical object we can find an axis of rotation. E.g. the Earth has an axis of rotation, and so would a spinning basketball. And, the spin of either object is not of fixed magnitude: it may be increased or decreased if a torque is applied.

The spin on an electron is fundamentally different in a number of ways (which leads to the picture of a classically spinning object not being a good mental model):

The total amount of spin is fixed. The measured value is always $\dfrac {\sqrt 3 \hbar}{2}$. This cannot be increased or decreased. Moreover, the spin component when measured about any axis is always fixed in magnitude and only varies in direction. It's always $\dfrac{\hbar}{2}$.

This is completely non-classical and means we can never find an axis of rotation for the electron. Whatever axis we choose, the measured spin component is always effectively a third of the total spin.

These two characeristics lead to the electron's spin being described by a spin state. A study of the nature of electron spin and of electron spin states is a good place to start with QM.

Finally, if we measure the spin about a given axis, the state of the electron after the measurement correpsonds to a state where further measurements about the same axis will yield the same result. This is called an eigenstate.

For example, if we measure the spin about the z-axis and get $+\dfrac \hbar 2$, then the state of the electron after the measurement is z-spin-up.

If we prepare an ensemble of electrons in the z-spin-up state and measure spin about the x-axis or y-axis, then in each case we get a 50-50 split of x-spin-up and x-spin-down or y-spin-up and y-spin-down respectively. And, if we measure spin about an axis that makes an angle $\theta$ with the z-axis, then we get a measurement of spin-up with probability $\cos^2 theta$ (and spin-down with probability $\sin^2 \theta$).

Now, let's describe an experiment that does obey QM:

Two particles are created with zero total spin. This means that their spins must be opposite (if measured about any axis). This could be explained by each particle having a definite but opposite spin about each axis. This would be a so-called hidden variables theory. There's a long story here about how tests of Bell's inequality have ruled out such local hidden variable theories.

Instead, QM says that the spin of neither particle is determined until one particle is measured. This is called an entangled state and, indeed, the system may only be analysed as a two-particle system; not as two separate particles each with its own state.

If both particles are measured about the same axis, then they are always found to have opposite spins. QM is silent on how nature achieves this. QM simply says that is the what nature serves up when the particles are measured.

Now, we prepare an ensemble of such entangled pairs and carry out some spin measurements about different axes.

Let's say that one particle is always measured about the z-axis. We get a 50-50 split of z-spin-up and z-spin-down. If the other particle is measured about the x-axis, then we get a 50-50 split of x-spin-up and x-spin down. If the results are compared, we find no correlation between the results. I.e. 25% of the time we will get z-up, x-up; 25% of the time z-up, x-down; 25% z-down, x-up; and 25% z-down, x-down.

Note that measuring one particle about the z-axis has no effect on the 50-50 distribution of measurements about the x-axis.

Now, suppose we measure one particle about an axis an angle $\theta$ from the z-axis, where $\cos^2 \theta = 0.9$, say. Let's call this axis $w$.

As always, we will get 50% w-spin-up and 50% w-spin-down.

Finally, suppose we do the w-axis measurement after the first particle has been measured about the z-axis. We still get 50-50, but thse results are correleted with the random results that of the z-measurement. I.e.:

45% of the time we get z-up and w-down
5% of the time we get z-up and w-up
45% of the time we get z-down and w-up
5% of the time we get z-down and w-down

If the experimenter could force a result of z-up, then they could influence the distribution of w-axis measurements. But, they can't force the electron to be z-spin-up and thereby signal this to the second experimenter. The first measurement is random 50-50 of up or down. We have no control over this.

In conclusion, we have known correlation between measurement results; but no control over either set of measurement results, hence no signalling capability.

That's a very cool "deeper dive", so-to-speak, explanation of the quantum phenomenon as described in a the shallower manner in the Quanta Magazine article - thank you!

It also helps me frame my question yet again, hopefully slightly more accurately than even the last attempt. This is rather like a heat seeking missile that shoots off to the left of target, corrects, but overshoots to the right of target (but not quite as far as it did to the left), corrects, and so on. We might hit the target eventually or I might not be brainy enough in which case we won't!

So spin - and its x, y, z ... axes and attendant rules regarding amount of spin, magnitude of spin and such - is one, correct me if I'm wrong, "degree of freedom" of the particle, in this case an electron. A "degree of freedom" is what I was previously calling a property of the particle. Degrees of freedom, like spin, location, velocity, are the top level characteristics of the particle. Each comes with its own payload of rules and constraints, like the ones you described for spin.

I think I get entanglement between a degree of freedom, like spin, between two particles now.

So here's two independent (unentangled particles). The spin value on particle One is a, and the spin on particle Two is b. There is no correlation or "sharedness" of the two results. They are both local, independent. Knowing the value of one shines no light on the value of the other.

In software development terms, these are two "variables by value", each has its own storage where the value is contained. (In contrast to two "variables by reference", see next) ...

Now, here's two entangled particles, entangled by their spin degree of freedom ...

The spin value for particles One and Two is maintained in one shared place - in the top box in the diagram entitled "Particle one & two" - and the value is c. Since the value of c is shared between both entangled particles I have greyed out the local c value of spin in each particle as a visual cue that the value is not really there but rather in the shared location above.

In software development terms the spin in each particle is a "variable by reference". That means the spin variable in each particle doesn't hold the value of the spin in storage but rather a pointer to the shared memory location where the value resides, in this case in the Spin value storage location in the "Particle one & two" box. Hence also the arrow heads on the grey lines pointing from the local particles to the shared box: quite literally software pointers in the programming analogy.

So by the great big gaps in each of those boxes representing the particles you can maybe see where I am going with this, but didn't explain myself very well previously.

I'm going to post this and then come back ...

 

[PeroK]

I'm not sure how far the programming analogy can be pushed, but to implement quantum entanglement in a program:

You would wait until a "measure spin about axis $\hat n$" command was received. Randomly select $\pm \dfrac \hbar 2$ for the measured particle, return this value; set the state of that particle to $\hat n$ up; set the state of the second particle to $\hat n$ down; and, break the entanglement.

Where $\hat n$ is some axis in 3D space.

There are complications brought about by SR and the relatively of which measurement was first, but that's the gist of it.

 

[LozD]

That's a very cool "deeper dive", so-to-speak, explanation of the quantum phenomenon as described in a the shallower manner in the Quanta Magazine article - thank you!

It also helps me frame my question yet again, hopefully slightly more accurately than even the last attempt. This is rather like a heat seeking missile that shoots off to the left of target, corrects, but overshoots to the right of target (but not quite as far as it did to the left), corrects, and so on. We might hit the target eventually or I might not be brainy enough in which case we won't!

So spin - and its x, y, z ... axes and attendant rules regarding amount of spin, magnitude of spin and such - is one, correct me if I'm wrong, "degree of freedom" of the particle, in this case an electron. A "degree of freedom" is what I was previously calling a property of the particle. Degrees of freedom, like spin, location, velocity, are the top level characteristics of the particle. Each comes with its own payload of rules and constraints, like the ones you described for spin.

I think I get entanglement between a degree of freedom, like spin, between two particles now.

So here's two independent (unentangled particles). The spin value on particle One is a, and the spin on particle Two is b. There is no correlation or "sharedness" of the two results. They are both local, independent. Knowing the value of one shines no light on the value of the other.

In software development terms, these are two "variables by value", each has its own storage where the value is contained. (In contrast to two "variables by reference", see next) ...

View attachment 295631

Now, here's two entangled particles, entangled by their spin degree of freedom ...

View attachment 295633
The spin value for particles One and Two is maintained in one shared place - in the top box in the diagram entitled "Particle one & two" - and the value is c. Since the value of c is shared between both entangled particles I have greyed out the local c value of spin in each particle as a visual cue that the value is not really there but rather in the shared location above.

In software development terms the spin in each particle is a "variable by reference". That means the spin variable in each particle doesn't hold the value of the spin in storage but rather a pointer to the shared memory location where the value resides, in this case in the Spin value storage location in the "Particle one & two" box. Hence also the arrow heads on the grey lines pointing from the local particles to the shared box: quite literally software pointers in the programming analogy.

So by the great big gaps in each of those boxes representing the particles you can maybe see where I am going with this, but didn't explain myself very well previously.

I'm going to post this and then come back ...

Now, here I'm introducing another degree of freedom (my aka "property") of the particle: velocity. And that's entangled between the two particles too ...

Again I've shown the local value boxes greyed out to signify the value is not really there but rather in the shared place above, attendant arrows pointing to the shared place too. (Note I've put a slight purple tinge on this greyed-ness for Velocity to distinguish it from Spin, which I'll use later).

So, we now have two entangled particles, One and Two, with their Spin and Velocity degrees of freedom entangled.

Next, let's say the Velocity can be some value between 0 and 1. When I prepare the particles, I set the probability like this: |v> = 99% |11> + 1% |00> . (This is LozD using ket-speak for the first time so beware more confusion by me, feel free to critique me, and hat-tip @ Nugatory!) In plain English I'm saying "I'm preparing the velocity degree of freedom of the entangled particles to 99% probability it is 1, 1% it is 0.

We introduce Alice and Bob and give them each a particle; Alice has One and Bob has Two, and we separate them. Here they are with the prepared pair of particles ...

We want Bob to communicate a 1 or 0 in the following manner ...

At midnight, if Bob wants to send a 0 he must not measure anything on his particle. Not Spin, not Velocity, nada. On the other hand, if he wants to send a 1 then he will examine the Spin. This protocol is pre-arranged with Alice and Bob.

At one minute past midnight, Alice will measure Velocity. If Bob didn't measure anything at midnight, then Alice's measuring of Velocity will have the probability |v> = 99% |11> + 1% |00> . She can interpret this as a zero being been sent by Bob. On the other hand, if Bob did measure Spin he will have collapsed the entanglement, then Alice's measuring of Velocity will have the probability |v> = 50% |11> + 50% |00> . She can interpret this as a one being been sent by Bob. Hence we have information communication between the two observers at a distance, and this will be happening instantaneously (or however many thousands of times faster than the speed of light someone posted that disentanglement happens at). And no subluminal back channel is required to send the information for comparison.

Now it might be said (someone did on this thread) that the entanglement collapse will only happen for the Spin, not the Velocity. So the spin will remain |v> = 99% |11> + 1% |00> as it was originally prepared and not revert to |v> = 50% |11> + 50% |00> .

Well, this is where the "degree-of-freedom to degree-of-freedom" entanglement comes in. If in addition to the particles being entangled, the Spin and Velocity are also entangled, then Bob's entanglement collapse of Spin will cascade to Velocity. See this diagram ...

The entanglement is shown by greying with a butterscotch tinge the two degrees of freedom, Spin and Velocity. (Perhaps I should have shown the entanglement in the shared place in black to be consistent with the notion that this is where the real values are and not on the particles?). But the entanglement collapse path if Bob transmits a 1 is: Bob measures Particle Two's Spin, that collapses the entanglement in the notional shared values Particle One & Two (right hand side butterscotch arrow), which in turn collapses the entanglement between Spin and Velocity in the shared notional particle (curved lines on either side of the shared notional particle One & Two), which in turn collapses the Spin and Velocity in the Particle One (left hand side butterscotch arrow), which means that when Alice measures Particle One's Velocity at a minute past midnight it will be |v> = 50% |11> + 50% |00> - Bob sent a 1. And if Bob doesn't measure Spin on Particle Two at midnight none of that cascading collapse will happen, and Alice's Velocity probability will be |v> = 99% |11> + 1% |00> - Bob sent a 0.

Is that clear now?

 

[PeroK]

That feels more like Alice in Wonderland, following Bob down a rabbit hole!

 

[LozD]

That feels more like Alice in Wonderland, following Bob down a rabbit hole!

 

[LozD]

Now, here I'm introducing another degree of freedom (my aka "property") of the particle: velocity. And that's entangled between the two particles too ...

View attachment 295634
Again I've shown the local value boxes greyed out to signify the value is not really there but rather in the shared place above, attendant arrows pointing to the shared place too. (Note I've put a slight purple tinge on this greyed-ness for Velocity to distinguish it from Spin, which I'll use later).

So, we now have two entangled particles, One and Two, with their Spin and Velocity degrees of freedom entangled.

Next, let's say the Velocity can be some value between 0 and 1. When I prepare the particles, I set the probability like this: |v> = 99% |11> + 1% |00> . (This is LozD using ket-speak for the first time so beware more confusion by me, feel free to critique me, and hat-tip @ Nugatory!) In plain English I'm saying "I'm preparing the velocity degree of freedom of the entangled particles to 99% probability it is 1, 1% it is 0.

We introduce Alice and Bob and give them each a particle; Alice has One and Bob has Two, and we separate them. Here they are with the prepared pair of particles ...

View attachment 295636

We want Bob to communicate a 1 or 0 in the following manner ...

At midnight, if Bob wants to send a 0 he must not measure anything on his particle. Not Spin, not Velocity, nada. On the other hand, if he wants to send a 1 then he will examine the Spin. This protocol is pre-arranged with Alice and Bob.

At one minute past midnight, Alice will measure Velocity. If Bob didn't measure anything at midnight, then Alice's measuring of Velocity will have the probability |v> = 99% |11> + 1% |00> . She can interpret this as a zero being been sent by Bob. On the other hand, if Bob did measure Spin he will have collapsed the entanglement, then Alice's measuring of Velocity will have the probability |v> = 50% |11> + 50% |00> . She can interpret this as a one being been sent by Bob. Hence we have information communication between the two observers at a distance, and this will be happening instantaneously (or however many thousands of times faster than the speed of light someone posted that disentanglement happens at). And no subluminal back channel is required to send the information for comparison.

Now it might be said (someone did on this thread) that the entanglement collapse will only happen for the Spin, not the Velocity. So the spin will remain |v> = 99% |11> + 1% |00> as it was originally prepared and not revert to |v> = 50% |11> + 50% |00> .

Well, this is where the "degree-of-freedom to degree-of-freedom" entanglement comes in. If in addition to the particles being entangled, the Spin and Velocity are also entangled, then Bob's entanglement collapse of Spin will cascade to Velocity. See this diagram ...

View attachment 295640
The entanglement is shown by greying with a butterscotch tinge the two degrees of freedom, Spin and Velocity. (Perhaps I should have shown the entanglement in the shared place in black to be consistent with the notion that this is where the real values are and not on the particles?). But the entanglement collapse path if Bob transmits a 1 is: Bob measures Particle Two's Spin, that collapses the entanglement in the notional shared values Particle One & Two (right hand side butterscotch arrow), which in turn collapses the entanglement between Spin and Velocity in the shared notional particle (curved lines on either side of the shared notional particle One & Two), which in turn collapses the Spin and Velocity in the Particle One (left hand side butterscotch arrow), which means that when Alice measures Particle One's Velocity at a minute past midnight it will be |v> = 50% |11> + 50% |00> - Bob sent a 1. And if Bob doesn't measure Spin on Particle Two at midnight none of that cascading collapse will happen, and Alice's Velocity probability will be |v> = 99% |11> + 1% |00> - Bob sent a 0.

Is that clear now?

Damn, there's a confusing typo.

Where I said ...

Now it might be said (someone did on this thread) that the entanglement collapse will only happen for the Spin, not the Velocity. So the spin will remain |v> = 99% |11> + 1% |00> as it was originally prepared and not revert to |v> = 50% |11> + 50% |00> .

I maent ...

Now it might be said (someone did on this thread) that the entanglement collapse will only happen for the Spin, not the Velocity. So the velocity will remain |v> = 99% |11> + 1% |00> as it was originally prepared and not revert to |v> = 50% |11> + 50% |00> .

 

[PeroK]

Honestly, I think you would be better focusing on the basics of QM for now. You can't really learn QM very well by trying to do FTL messaging by entanglement.

 

[DrChinese]

... as it was originally prepared ...

Any quantum property that is entangled cannot also be in a "prepared" state. To the extent you know what the value of an observable is, it is NOT entangled. Any measurement on a particle's observable will "collapse" the entanglement for that particle (and that observable). It will also affect the conjugate observables (which will then take on undefined values that will read as being random if subsequently measured).

It is possible to measure spin without affecting velocity (momentum), because spin and velocity are not conjugate.

 

[LozD]

Honestly, I think you would be better focusing on the basics of QM for now. You can't really learn QM very well by trying to do FTL messaging by entanglement.

I might well do, but it's a big ask. Keeping up with my day job tech and dev is tough enough.

Thanks for your input PeroK, and all others who chipped in.

Did y'all read about the tardigrade? How does an entire critter become entangled with a qubit? Bet the quantum maths and physicals around that gets pretty hectic?! ...

https://www.livescience.com/tardigrade-quantum-entangled-experiment

I think I'll quit while I'm ahead - OK, while I'm behind but not as far as I will be if I carry on!

Cheers!

 

[Nugatory]

Is that clear now?

Yes, but it's also clear that it is based on a misunderstanding of how QM works.

To get past this point you will have to develop the mathematical background needed to follow the presentation in https://en.wikipedia.org/wiki/No-communication_theorem (which unfortunately is one of the less good wikipedia articles - I'm not aware of any better readily accessible proof).

Without digging into the math, the theorem says that the quantum mechanical calculation will always show that no matter how cleverly we set up our experiment, the probabilities on one side of an entanglement are not affected by anything done or not done at the other side. Thus, your idea can only work if the quantum mechanical calculation is wrong.

This thread has drifted to the wrong side of the forum rules about personal speculation and challenges to mainstream physics so it is closed. As with all thread closures it can be reopened if someone has something more to add - suggestions for better explanations of the no-communication theorem would be particularly welcome.

 

[Nugatory]

Did y'all read about the tardigrade? How does an entire critter become entangled with a qubit? Bet the quantum maths and physicals around that gets pretty hectic?! ...

OK, one last pre-closure post: We already have at least one long thread about the tardigrade: https://www.physicsforums.com/threa...lar-organism-to-be-quantum-entangled.1010369/