If Det |HH^†  | = 1,H  is a unitary matrix & H^† = H^-1

+ i

+i

HH^† =H^†H = H²

[σ_i]  *  [σ_j]  +  [σ_j]  *  [σ_i]      ={σ_i,σ_j}= 2δ_ijI  and [σ_i,σ_j]= 2iε_ijkσ_k

Pauli eigen vectors are

ψ_x⁺=(1/√2 ) 1

                     1

ψ_x^-=(1/√2 ) 1

                  -1

ψ_y⁺=(1/√2 ) 1

                    i

ψ_y^-=(1/√2 ) 1

                  -i

ψ_z⁺ =         1

                  0

ψ_z^- =          0

                  1

and

ψ_x⁺=iσ₂ ψ_x^-  

ψ_y⁺=σ₃ ψ_y^-

ψ_z⁺=σ₁ ψ_z^-

ψ_x^-=-iσ₂ ψ_x⁺  

ψ_y^-=  σ₃ ψ_y⁺

ψ_z^-=  σ₁ ψ_z⁺

ψ_y⁺=iσ₁ ψ_x⁺

Hence there can be transformation from ψ_x  to ψ_y    state and vice versa but not to ψ_z    state.

   A =C₀σ₀  +C₁σ₁ + C₂σ₂  + C₃σ₃      where   C₀, C₁ ,C₂ ,C₃  are the linear co-efficient and are represented by real numbers. A 3-D Pauli vector is represented as

  σ= σ₁x‾ + σ₂y‾ +σ₃ z‾ , x‾ , y‾ ,z‾ being basis vectors in x,y,z axis respectively. If P is a point in 3-D Eucledean space with co-ordinates (x,y,z)

 then vector P = xx‾+yy‾+zz‾ where x‾ , y‾ ,z‾ being basis vectors in x,y,z axis respectively. Now P.σ =xσ₁ + yσ₂+zσ₃ ;which is a matrix representation of vector P & the determinant of this matrix is the negative norm square of the vector . Thus we have mapping from vector basis to the Pauli matrix basis for representing a 3-D vector . Now P.σ = z        x-iy

                             x+iy      -z  and determinant = -x² -y²-z²

where x,y,z are co-ordinates of point P. This matrix is a Hermitian matrix with zero trace.

If we take σ_1'=0  -1   σ_2'=0  i   σ_3'=-1 0 

                           -1  0       -i  0         0 1

then 

P.σ = -z        -x+iy

       -x-iy        z        and determinant = -x² -y²-z²

Hence it is all the same whether we take (σ₁,σ₂ ,σ₃ ) or (σ_1',σ_2' ,σ_3' ) as the set of Pauli matrices each having determinant -1. .

If we want make the determinant x² +y²+z²  , the matrix will be

P.σ =   z          x+iy

       -(x-iy)        z 

and Pauli matrices will be

rσ₁= 0   1   rσ₂= 0  i    rσ₃= 1  0    each having determinant +1.

        -1  0           i   0            0   1

Alternatively,

P.σ =   -z          -(x+iy)

       (x-iy)         -z 

and Pauli matrices will be

rσ_1'= 0   -1   rσ_2'= 0  -i    rσ_3'= -1  0    each having determinant +1.

        1    0           -i    0            0   -1

Hence it is all the same whether we take (rσ₁,rσ₂ ,rσ₃ ) or (rσ_1',rσ_2' ,rσ_3' ) as the set of Pauli matrices each having determinant 1.

Transformation matrix  for Lorentz boost is given by

L = cosθ    i sinθ  =  cosθ*I + iσ₁sinθ  where σ₁= 0   1  and I = 1  0

      i sinθ    cosθ                                                     1   0               0  1

* We know that for spin 1/2 particles like electrons, these must be rotated by an angle 4π radian in order to return to their original configuration, since these rotations are not in 2-D but in 3-D space ( dimensionality matters ) which means there is a 2 to 1 correspondence between su(2) and so(3). Hence for rotation θ  for so(3), su(2) rotation will be θ/2. Lie algebra su(2) is isomorphic to lie algebra so(3) which corresponds to Lie Group SO(3).For spin 1/2 particles, the spin operator is given by J=(h/2)σ which is a fundamental representation of SU(2).

*The real linear span of {I, iσ₁, iσ₂, iσ₃} is isomorphic to the real algebra of quaternions ℍ. The isomorphism from ℍ to this set is given by the following map (notice the reversed signs for the Pauli matrices):

  1 -> I , i ->-iσ₁   ,j ->-iσ₂ , k ->-iσ₃

Alternatively, the isomorphism can be achieved by a map using the Pauli matrices in reversed order,

1 -> I , i ->iσ₃   ,j ->iσ₂ , k ->iσ₁

So transformation matrix Q for rotation of spin 1/2 particles about x-axis by an angle θ may be written in terms of Pauli Matrices as

cos θ/2    i sin θ/2  =  cosθ/2*I + iσ₁sinθ/2  where σ₁= 0   1  and I = 1  0

i sin θ/2    cos θ/2                                                           1   0               0  1

* If n is a unit vector (in any orientation) , then n=xx‾ +yy‾ +zz‾  and   x² +y²+z²=1, and

  n.σ = -z          -(x+iy)  and (n.σ)² =x² +y²+z²          0         =   1      0  = I or (n.σ)^2p =I where p is any integer and  (n.σ)^2p+1 = n.σ

           (x-iy)         -z                              0              x² +y²+z²      0      1

   &   are eigen states of S_z operator and these spinors form a basis for 2-D spinor space. For σ_x  , the eigen values are =  and λ=+ 1 and -1.

For  λ=+ 1 , the eigenvector equation is which gives a=b. To normalize the eigen state,    so that the state is 

For λ=- 1 , the eigenvector equation is  which gives a=-b. And the normalized eigen state is .   Similarly for σ_y , we get  2 normalized eigen states as for  λ= 1 and -1 respectively. This can be written in terms of σ_z  eigen states  as  . Since the coefficients of the two σ_z  eigen states are equal in magnitude, if σ_z  is measured for a particle in σ_y   eigen state, it is equally likely to be spin up or down.