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Forums
Physics
Quantum Physics
Orthonormal Basis of Wavefunctions in Hilbert Space
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[QUOTE="fresh_42, post: 6062873, member: 572553"] The position vector is irrelevant as far as it concerns the basis. It is merely a variable of the functions involved. However, depending on how the inner product is defined, it plays a role for that, e.g. ##\langle u_i,u_j \rangle = \int_M u_i(\vec{r})u_j^\dagger(\vec{r}) \,d\vec{r}##. But as soon as you have a orthonormal basis, the variable has nothing to do with it. As to whether such a basis can be chosen canonically, well, you can apply the Gram-Schmidt algorithm in case of a separable space. [/QUOTE]
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Forums
Physics
Quantum Physics
Orthonormal Basis of Wavefunctions in Hilbert Space
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