Let's say you're wondering around P(oo) (which I'll use to represent the space of polymomials of any degree on the interval [-1,1]) and you decide to calculate the matrix(adsbygoogle = window.adsbygoogle || []).push({}); Xrepresenting the position operatorx. Let's say you do this in the basis: 1, t, t^2, ..., t^n,... you'll find that the matrixXis not hermitian even thoughxis 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 thatXis 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

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# Hermitian and not Hermitian

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