- #51

- 667

- 44

^{ikx}leading to

$$

P = \begin{bmatrix} 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \end{bmatrix}

$$

but then you state that for the symmetric potential well the basis vectors in momentum space are

cos(kx) and sin (kx) so why do we not have the parity operator matrix form shown above in this situation ? In both cases we have basis vectors in momentum space.