- 306

- 1

**X**representing the position operator

*x*. Let's say you do this in the basis: 1, t, t^2, ..., t^n,... you'll find that the matrix

**X**is not hermitian even though

*x*is hermitian on P(oo). Then you decide to calculate the matrix again, but now you do it with the orthonormal basis generated from the above basis via the Gram-schmidt method (i.e. the properly normalized Legendre polynomials) Now you find that

**X**is hermitian. What gives? should the hermiticity of an operator be independent of its representation in a particular basis? (maybe not and this is super obvious, feel free in that case to slap me).

Kevin