Peres construction: proof of the Kochen-Specker theorem

We measure the squared spin components  $J_x^2$, $J_y^2$ and $J_z^2$ along the three perpendicular directions $x$, $y$ and $z$ of a spin one particle. The operators  $J_x^2$, $J_y^2$ and $J_z^2$ are compatible and therefore we are dealing with comeasurable propositions or observables.

The fact that this is a spin one system means that we can only obtain the outcomes $0$ (color white) and $1$ (color black). The spin one system must also satisfy the constraint  $J_x^2+j_y^2+J_z^2=2$ and hence we are essentially attempting to assign the numbers $\{1,0,1\}$ to the tripod $(x,y,z)$.

 In other words, the comeasurement of  the squared spin components along the three perpendicular directions $x$, $y$ and $z$ of a spin one particle is equivalent to the problem of coloring one leg of the tripod white while the other two orthogonal legs are colored black. This coloring problem is in fact equivalent to the measurement of a single  "Ur"-operator ${U}$.

By rotating the tripod $(x,y,z)$ through the application of arbitrary $SO(3)$ rotations we obtain different tripods with different incompatible "Ur"-operators ${U}$.

The goal  then is to assign the two colors black and white to all lines through the origin in $\mathbb{R}^3$ in such a  way that:

1. for any three mutually orthogonal lines  one line will be colored white and the other two lines will be colored black



2. and for any two orthogonal lines one at least is colored black.

This coloring is required to be consistent, i.e. every line is colored either black or white and this is the assumption of non-contextuality (the measurement of the squared spin component along any axis does not depend on the order in which the measurements of the tripods to which this axis belong is done).


As it turns, we do no need to consider an infinite number of lines through the origin in $\mathbb{R}^3$ to show the impossibility of such a colouring since a contradiction already arises for a finite number of lines. The most efficient construction of the Kochen-Specker theorem in three dimensions (which is sufficient to establish a contradiction in higher dimensions) is due to Peres and it involves $33$ lines.


These $33$ directions (or $66$  if we include the opposite directions) can be given in a coordinate basis as follows. First, we have the  $9$ obvious directions given by:

1. The lines through the origin (where the spin one particle lies) which go  through the six points $(\pm 1,0,0)$, $(0,\pm 1,0)$, $(0,0,\pm 1)$.



2. The lines through the origin which go  through the twelve points $(\pm 1,\pm 1,0)$, $(\pm 1,0,\pm 1)$, $(0,\pm 1,\pm 1)$.

 These $18$ points define $9$ directions through the origin which intersect the surface of a cube ${\cal C}$ of size $2$ at the points on the middle of each of the $12$ edges plus the points on the center of each of the $6$ faces.

 This cube ${\cal C}$ has $13$ axes of symmetry given by 1) the three directions through the centers of opposite faces, 2) the four directions along the long diagonals which go through opposite vertices and 3) the six directions through the midpoints of opposite edges. By rotating ${\cal C}$ through $45$ degrees about one of its axes (the directions through the centers of opposite faces) we obtain the three cubes ${\cal C}_1$, ${\cal C}_2$ and ${\cal C}_3$ with a total number of axes of symmetry equal $4\times 13$. Some of these symmetry axes coincide and we are finally left with only $33$ independent symmetry axes which are precisely Peres' $33$ directions.

 An equivalent and much more visual representation of the extra directions of Peres is obtained by considering the largest circle on each face and then draw the largest squares inside these circles. The extra points of Peres are then the intersection points on these inner squares. See the figure. We get then the following extra directions:

1. The lines through the origin which go  through the $24$ points $(\pm 1,\pm \alpha,0)$, $(\pm \alpha,\pm 1,0)$,  $(0,\pm 1,\pm \alpha)$, $(0,\pm \alpha,\pm 1)$,  $(\pm 1,0,\pm \alpha)$, $(\pm \alpha,0,\pm 1)$ where $\alpha=1/\sqrt{2}$. These are the midpoints of the sides of the inner squares.



2. The lines through the origin which go  through the $24$ points $(\pm 1,\pm \alpha,\pm \alpha)$, $(\pm \alpha,\pm 1,\pm \alpha)$,  $(\pm \alpha,\pm \alpha,\pm 1)$. These are the vertices of the inner squares.

Now we attempt to color these $33$ lines as discussed above. As it turns this coloring problem admits no solution which is the essence of the Peres proof of the Kochen-Specker theorem. Since this coloring is impossible for these $33$ directions it must be impossible for the infinite number of lines going through the origin and since it is impossible in a three dimensional real Hilbert space it must be impossible in a Higher dimensional complex space which is the substance of the Kochen-Specker theorem. The proof consists in producing a contradiction as follows.

1. By rotational symmetry we can start from any point on the cube ${\cal C}$. We start with the direction which goes through the point $X=(1,0,0)$ and color it with white, i.e. the lines through the points $Y=(0,1,0)$ and $Z=(0,0,1)$ must be colored black. Then immediately all orthogonal directions must be colored black. These are given by the $16$ lines
   \begin{eqnarray}
     (0,\pm 1,0)~,~(0,0,\pm 1)~,~(0,\pm 1,\pm 1)~,~(0,\pm\alpha,\pm 1)~,~(0,\pm 1,\pm\alpha).
   \end{eqnarray}
   



2. From the six points $(1,\pm\alpha,\pm\alpha)$ and $(0,1,\pm 1)$ we construct the two tripods
     \begin{eqnarray}
       &&(0,1,1)~,~(1,\alpha,-\alpha)~,~(1,-\alpha,\alpha)\nonumber\\
       &&(0,1,-1)~,~(1,\alpha,\alpha)~,~(1,-\alpha,-\alpha).
     \end{eqnarray}
   Obviously, the two points $(0,1,\pm 1)$ are black.  Thus one of the two points $(1,\alpha,-\alpha)$, $(1,-\alpha,\alpha)$ must be black and the other one must be white and also one of the two points $(1,\alpha,\alpha)$, $(1,-\alpha,-\alpha)$ must be black and the other one must be white. The possible pairs of white points are therefore given by the following four possibilities
     \begin{eqnarray}
       &&(1,\alpha,-\alpha)~,~(1,\alpha,\alpha)\nonumber\\
       &&(1,\alpha,-\alpha)~,~(1,-\alpha,-\alpha)\nonumber\\
       &&(1,-\alpha,\alpha)~,~(1,\alpha,\alpha)\nonumber\\
       &&(1,-\alpha,\alpha)~,~(1,-\alpha,-\alpha).
     \end{eqnarray}
     By rotating the cube ${\cal C}$ around the direction going through the point $(1,0,0)$ we can easily see that all these four possibilities are actually equivalent. The two resulting white directions can then be taken to be given by the lines going through the two points
     \begin{eqnarray}
       &&C=(1,-\alpha,\alpha)~,~C^{\prime}=(1,\alpha,\alpha).
     \end{eqnarray}
    



3. In this last part of the argument we will employ the following $10$ lines:
    
    
   • The line through the point $D=(\alpha,1,0)$ is orthogonal to the line through the point $C$ and since $C$ is white we must color $D$ black.
   
   
   • The line through the point $D^{\prime}=(-\alpha,1,0)$ is orthogonal to the line through the point $C^{\prime}$ and since $C^{\prime}$ is white we must color $D^{\prime}$ black.
   
   
   • The lines through the points $Z$, $D$ and $E=(1,-\alpha,0)$ form a triple and since $Z$ and $D$ are colored black we must color $E$ white.
   • The lines through the points $Z$, $D^{\prime}$ and $E^{\prime}=(1,\alpha,0)$ form a triple and since $Z$ and $D^{\prime}$ are colored black we must color $E^{\prime}$ white.
   
   
   • The lines through the points $E$, $F=(\alpha,1,-\alpha)$, $G=(\alpha,1,\alpha)$ form a triple and since $E$ is white we must color both $F$ and $G$ black.
           
   • The lines through the points $E^{\prime}$, $F^{\prime}=(-\alpha,1,\alpha)$, $G^{\prime}=(-\alpha,1,-\alpha)$ form a triple and since $E^{\prime}$ is white we must color both $F^{\prime}$ and $G^{\prime}$ black.
           
   • The lines through the points $F$, $F^{\prime}$ and $U=(1,0,1)$ form a triple and since $F$ and $F^{\prime}$ are black we must color $U$ white.
           
   • The lines through the points $G$, $G^{\prime}$ and $V=(1,0,-1)$ form a triple and since $G$ and $G^{\prime}$ are black we must color $V$ white.
             
   • Finally, the lines through the points $U$, $V$ are mutually orthogonal but both points  $U$ and $V$ are colored white which is impossible.