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Is there any reason why such an operator cannot describe an actual physical quantity? Orthogonal complement in quantum mechanics,orthonormal bases. The book is Solid State and Quantum Theory for Optoelectronics by Michael A. Parker, page number 595, and here is the [Google book link](books.google.co.in/… $\endgroup$ – user65318 Nov 25 '14 at 19:26 Advantage of operator algebra is that it does not rely upon particular basis, e.g. It is therefore convenient to reformulate quantum mechanics in framework that involves only operators, e.g. When it comes to quantum mechanics and wavefunctions, still the most practical system is assuming orthogonality for the corresponding system. In quantum mechanics, a quantum state is typically represented as an element of a complex Hilbert space, for example, the infinite-dimensional vector space of all possible wavefunctions (square integrable functions mapping each point of 3D space to a complex number) or some more abstract Hilbert space constructed more algebraically. The concept of orthogonality goes back to vectors, like these: Geometrically, two vectors are orthogonal when they are perpendicular, i.e. Quantum mechanics and Green’s functions, at rst glance, seem entirely unrelated, however within the last 50 years Green’s functions have proven themselves to be a useful tool for solving many avors of boundary value problems within the realm of quantum mechanics. Imagine that we would have an operator (finite-dimensional) acting on a spin system that has real eigenvalues, but its eigenvectors are not perpendicular to each other. I was wondering if it is important in Quantum Mechanics to deal with operators that have an orthonormal basis of eigenstates? Hˆ . when there is an angle of 90 degrees between them.

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