# orderedItems[].object.content
[13/13] This is a proof by Peres of the Kochen-Specker Theorem!

Think of each triad as a quantum measurement where 1 of 3 orthogonal possibilities must be observed.

Then we just exhibited a set of 16 measurements whose results can’t be pre-determined, independent of what we do.

# orderedItems[].object.contentMap.en
[13/13] This is a proof by Peres of the Kochen-Specker Theorem!

Think of each triad as a quantum measurement where 1 of 3 orthogonal possibilities must be observed.

Then we just exhibited a set of 16 measurements whose results can’t be pre-determined, independent of what we do.

# orderedItems[].object.content
[12/13] And the coup-de-grace: triad 16 gets a green on the right face, reddifying an orthogonal vector in triad 5 on the top face.

Every ray has a colour now, and up to symmetry, we had no choice in what we did.

But … triad 5 now has THREE red rays.

The goal is impossible!

# orderedItems[].object.contentMap.en
[12/13] And the coup-de-grace: triad 16 gets a green on the right face, reddifying an orthogonal vector in triad 5 on the top face.

Every ray has a colour now, and up to symmetry, we had no choice in what we did.

But … triad 5 now has THREE red rays.

The goal is impossible!

# orderedItems[].object.content
[11/13] Surprising absolutely no one, triad 3 gets a green on the top face, reddifying an orthogonal vector in triad 16 on the right face.

# orderedItems[].object.contentMap.en
[11/13] Surprising absolutely no one, triad 3 gets a green on the top face, reddifying an orthogonal vector in triad 16 on the right face.

# orderedItems[].object.content
[10/13] Now triad 15 gets a green on the top face, reddifying an orthogonal vector in triad 3 on the right face.

# orderedItems[].object.contentMap.en
[10/13] Now triad 15 gets a green on the top face, reddifying an orthogonal vector in triad 3 on the right face.

# orderedItems[].object.content
[9/13] And the cascade continues: triad 8 falls to the inevitable, with a new green on the left face, reddifying a ray orthogonal to it in triad 15, on the top face.

# orderedItems[].object.contentMap.en
[9/13] And the cascade continues: triad 8 falls to the inevitable, with a new green on the left face, reddifying a ray orthogonal to it in triad 15, on the top face.

# orderedItems[].object.content
[8/13] And now that means two reds in triad 12, so the third ray, also on the left face, must turn green.

This in turn reddifies a ray orthogonal to it in triad 8, on the top face.

# orderedItems[].object.contentMap.en
[8/13] And now that means two reds in triad 12, so the third ray, also on the left face, must turn green.

This in turn reddifies a ray orthogonal to it in triad 8, on the top face.

# orderedItems[].object.content
[7/13] Now … since we had two reds in triad 2, the remaining ray (on the top face) must be green. This also reddifies a ray orthogonal to it in triad 12, on the left face.

# orderedItems[].object.contentMap.en
[7/13] Now … since we had two reds in triad 2, the remaining ray (on the top face) must be green. This also reddifies a ray orthogonal to it in triad 12, on the left face.

# orderedItems[].object.content
[6/13] Putting all these reds and greens together, we’ve made a fair bit of progress!

All of this colouring has been forced on us by the goals we’ve set, up to some arbitrary choices where we either had to do one thing, or another that would look the same up to a symmetry.

# orderedItems[].object.contentMap.en
[6/13] Putting all these reds and greens together, we’ve made a fair bit of progress!

All of this colouring has been forced on us by the goals we’ve set, up to some arbitrary choices where we either had to do one thing, or another that would look the same up to a symmetry.

# orderedItems[].object.content
[5/13] And, you guessed it, we can assume the green ray here is green, because if it wasn’t then the opposite point on the top face would have to be green (one ray in triad 14 must be green) and we could take the mirror image.

# orderedItems[].object.contentMap.en
[5/13] And, you guessed it, we can assume the green ray here is green, because if it wasn’t then the opposite point on the top face would have to be green (one ray in triad 14 must be green) and we could take the mirror image.

# orderedItems[].object.content
[4/13] Again, we can assume the green ray here is green, because if it wasn’t then the opposite point on the top face would have to be green (one ray in triad 6 must be green) and we could take the mirror image.

# orderedItems[].object.contentMap.en
[4/13] Again, we can assume the green ray here is green, because if it wasn’t then the opposite point on the top face would have to be green (one ray in triad 6 must be green) and we could take the mirror image.

# orderedItems[].object.content
[3/13] We can assume the green ray here is green, because if it wasn’t then the opposite point on the top face would have to be green (one ray in triad 10 must be green) and we could just take the mirror image.

# orderedItems[].object.contentMap.en
[3/13] We can assume the green ray here is green, because if it wasn’t then the opposite point on the top face would have to be green (one ray in triad 10 must be green) and we could just take the mirror image.

# orderedItems[].object.content
[2/13] Can we colour every ray green or red, so that:

• Every triad has exactly one green ray, and
• All rays orthogonal to any green ray are red?

We can assume the z-axis ray is green, because if it wasn’t the x- or y-axis would have to be, and we could redefine that as z.

# orderedItems[].object.contentMap.en
[2/13] Can we colour every ray green or red, so that:

• Every triad has exactly one green ray, and
• All rays orthogonal to any green ray are red?

We can assume the z-axis ray is green, because if it wasn’t the x- or y-axis would have to be, and we could redefine that as z.

# orderedItems[].object.content
[1/13] The black dots here mark points on a cube that intersect a set of 33 rays (two intersections are visible for some rays).

The cube has vertices (±√2,±√2,±√2), the grey lines are at 0,±1.

There are 16 orthogonal triads of rays; beside each ray is a list of its triads.

# orderedItems[].object.contentMap.en
[1/13] The black dots here mark points on a cube that intersect a set of 33 rays (two intersections are visible for some rays).

The cube has vertices (±√2,±√2,±√2), the grey lines are at 0,±1.

There are 16 orthogonal triads of rays; beside each ray is a list of its triads.

# orderedItems[].object.content
Quantum measurements are necessarily “contextual”.

However, there is a twist! The fact that quantum measurements behave this way means that if we give entangled quantum particles to two players and challenge them to tell us numbers like this — for one row and column each, at a time — they can use Mermin’s grid to agree on the overlapping square 100% of the time.

More: https://www. gregegan.net/SCIENCE/MPMS/MPMS.html

https://www. scientificamerican.com/article/researchers-use-quantum-telepathy-to-win-an-impossible-game/

# orderedItems[].object.contentMap.en
Quantum measurements are necessarily “contextual”.

However, there is a twist! The fact that quantum measurements behave this way means that if we give entangled quantum particles to two players and challenge them to tell us numbers like this — for one row and column each, at a time — they can use Mermin’s grid to agree on the overlapping square 100% of the time.

More: https://www. gregegan.net/SCIENCE/MPMS/MPMS.html

https://www. scientificamerican.com/article/researchers-use-quantum-telepathy-to-win-an-impossible-game/

# orderedItems[].object.content
What this means, though, is that given a pair of spins, you can’t say that the result you get for one of these measurements “would have been the same” if you’d done it as part of a different set of measurements: as part of the row it belongs to, rather than as part of its column. Because if you try to write a fixed 1 or –1 in each of the 9 squares, telling you the values for those measurements, it is impossible to make the products of all rows equal 1, but the products of all columns –1.

# orderedItems[].object.contentMap.en
What this means, though, is that given a pair of spins, you can’t say that the result you get for one of these measurements “would have been the same” if you’d done it as part of a different set of measurements: as part of the row it belongs to, rather than as part of its column. Because if you try to write a fixed 1 or –1 in each of the 9 squares, telling you the values for those measurements, it is impossible to make the products of all rows equal 1, but the products of all columns –1.

# orderedItems[].object.content
David Mermin came up with these matrices in 1990, in the form of measurements on a pair of quantum spins, as shown below.

Every measurement in this grid will give 1 or –1, and if you perform the three measurements in one row, or one column, at the same time, the product of the three results you get will always be 1 for a row, and –1 for a column. This follows from our original statements about the grid of matrices.

# orderedItems[].object.contentMap.en
David Mermin came up with these matrices in 1990, in the form of measurements on a pair of quantum spins, as shown below.

Every measurement in this grid will give 1 or –1, and if you perform the three measurements in one row, or one column, at the same time, the product of the three results you get will always be 1 for a row, and –1 for a column. This follows from our original statements about the grid of matrices.

# orderedItems[].object.content
If you take the three matrices in one row or one column of this 3×3 grid, they will all commute with each other, and the result of multiplying all three together will be the identity matrix for rows, but minus the identity matrix for columns.

# orderedItems[].object.contentMap.en
If you take the three matrices in one row or one column of this 3×3 grid, they will all commute with each other, and the result of multiplying all three together will be the identity matrix for rows, but minus the identity matrix for columns.