Page 126 - 35Linear Algebra
P. 126

126                                                                                      Matrices


                            where

                                                 0 1             0 −i              1   0
                                          σ x =        ,  σ y =          ,  σ z =          .
                                                 1 0             i   0             0 −1
                            These three matrices are the famous Pauli matrices; they are used to describe electrons
                            in quantum theory, or qubits in quantum computation. Let

                                                              −2 + i 1 + i
                                                        v =                  .
                                                               3 − i 2 − i
                            Find the column vector of v in the basis B.
                               For this we must solve the equation


                                       −2 + i 1 + i      x  0 1       y  0 −i       z  1    0
                                                      = α          + α           + α            .
                                        3 − i 2 − i         1 0           i   0        0 −1
                            This gives four equations, i.e. a linear systems problem, for the α’s

                                                      x      y
                                                    α    − iα         =     1 + i
                                                   
                                                      α   + iα         =     3 − i
                                                      x      y
                                                                   α z  = −2 + i
                                                   
                                                   
                                                                   z
                                                                 −α    =     2 − i
                            with solution
                                                            y
                                                  x
                                                                          z
                                                α = 2 ,   α = 2 − 2i ,   α = −2 + i .
                            Thus
                                                                               
                                                                            2

                                                      −2 + i 1 + i
                                                v =                  =    2 − i    .
                                                        3 − i 2 − i
                                                                         −2 + i
                                                                                  B
                               To summarize, the column vector of a vector v in an ordered basis B =
                            (b 1 , b 2 , . . . , b n ),
                                                                 1  
                                                                 α
                                                                α 2 
                                                               .  ,
                                                                    
                                                              
                                                                  .
                                                               . 
                                                                 α n
                            is defined by solving the linear systems problem
                                                                                n
                                                                              X
                                                                                    i
                                                                        n
                                                    1
                                                           2
                                              v = α b 1 + α b 2 + · · · + α b n =  α b i .
                                                                               i=1
                                                      126
   121   122   123   124   125   126   127   128   129   130   131