Two of the commutation relations for the components of J are, [ J y , J z ] = i ℏ ℏ J x , [ J z , J x ] = i ℏ ℏ J y . They can be combined to obtain two equations, which are written together using ± ± signs in the following, J z ( J x ± ± i J y ) = ( J x ± ± i J y ) ( J z ± ± ℏ ℏ ) , where one of the equations uses the + signs and the other uses the − − signs. Applying both sides of the above to ψ ψ ( J 2 ′ J z ′ ) , J z ( J x ± ± i J y ) ψ ψ ( J 2 ′ J z ′ ) = ( J x ± ± i J y ) ( J z ± ± ℏ ℏ ) ψ ψ ( J 2 ′ J z ′ ) = ( J z ′ ± ± ℏ ℏ ) ( J x ± ± i J y ) ψ ψ ( J 2 ′ J z ′ ) . The above shows that ( J x ± ± i J y ) ψ ψ ( J 2 ′ J z ′ ) are two eigenfunctions of J z with respective eigenvalues J z ′ ± ± ℏ ℏ , unless one of the functions is zero, in which case it is not an eigenfunction. For the functions that are not zero, ψ ψ ( J 2 ′ J z ′ ± ± ℏ ℏ ) = ( J x ± ± i J y ) ψ ψ ( J 2 ′ J z ′ ) . Further eigenfunctions of J z and corresponding eigenvalues can be found by repeatedly applying J x ± ± i J y as long as the magnitude of the resulting eigenvalue is ≤ ≤ J 2 ′ . Since the eigenvalues of J z are bounded, let J z 0 be the lowest eigenvalue and J z 1 be the highest. Then ( J x − − i J y ) ψ ψ ( J 2 ′ J z 0 ) = 0 and ( J x + i J y ) ψ ψ ( J 2 ′ J z 1 ) = 0 , since there are no states where the eigenvalue of J z is < J z 0 or > J z 1 . By applying ( J x + i J y ) to the first equation, ( J x − − i J y ) to the second, using J x 2 + J y 2 = J 2 − − J z 2 , and using also J + J − − = J x 2 + J y 2 − − i [ J x , J y ] = J x 2 + J y 2 + J z , it can be shown that J 2 ′ − − ( J z 0 ) 2 + ℏ ℏ J z 0 = 0 and J 2 ′ − − ( J z 1 ) 2 − − ℏ ℏ J z 1 = 0. Subtracting the first equation from the second and rearranging, ( J z 1 + J z 0 ) ( J z 0 − − J z 1 − − ℏ ℏ ) = 0. Since J z 1 ≥ ≥ J z 0 , the second factor is negative. Then the first factor must be zero and thus J z 0 = − − J z 1 .