Given a Hilbert space and a Hermitian operator defined on it, will the operator exhibit Hermiticity in any basis used to span the space? My thought on this is that this must be the case, after all, if an operator is Hermitian then it represent an observable. Surely, whether or not an operator is an observable cannot depend on the basis you expand it with? However, I know that there are non-Hermitian operators that have real eigenvalues, so perhaps, it is the case that Hermiticity merely guarantees that an operator is an observable, so that it is possible that an operator might be non-Hermitian in a given basis and yet still represent an observable? Thank you for your time and consideration.
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asked Feb 29 at 0:03
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$\begingroup$ Yes, the matrix representation of a hermitian operator is a hermitian matrix in any orthonormal bases. For non-orthonormal bases this must not hold, I think. $\endgroup$– Tobias FünkeCommented Feb 29 at 6:59
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2 Answers
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A linear operator on a Hilbert space $A:\mathcal H\to\mathcal H$ is Hermitian if for all $v,w\in\mathcal H$, $$\langle Av, w\rangle = \langle v,Aw\rangle$$
This definition makes no reference to any particular basis, since both the inner product and the linear operators on the Hilbert space are defined independently of any basis. Therefore Hermiticity is a property of the operator and not what basis it is represented in.
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$\begingroup$ You want to be careful here. There are plenty of realizations of operators that are equivalent to hermitian ones, but mathematicians don’t bother to make operators explicitly hermitian (v.g. all the Harish-Chandra stuff). So while you are technically right, the question can also be understood at the level of equivalence using a non-unitary change of basis. See this paper for instance. $\endgroup$ Commented Feb 29 at 4:15
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An operator is not Hermitian "in a basis". An operator $A$ on an inner product space is defined to be Hermitian iff it equals its own adjoint, $A = A^\dagger$. This condition has nothing to do with any basis. (This definition turns out to be equivalent to the one in Er Jio's answer.)
You may have seen the adjoint "defined" to be the conjugate transpose of a matrix, and a Hermitian (i.e. self-adjoint) operator "defined" to be a matrix that equals its own conjugate transpose. But these definitions are incorrect. The fundamental definition of the adjoint has nothing to with the conjugate transpose, and it applies to any operator on any inner product space - even if the inner product space is infinite-dimensional and so cannot be represented by a matrix.
