The property of hermitian is the sufficient condition for eigenvalue being real. Is there any non-hermitian operator on Hilbert Space with all real eigenvalues? If there exist, then can all eigenstates be orthogonal to each other? And these operators have any application in Quantum mechanics?
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$\begingroup$ Related: physics.stackexchange.com/q/16678/2451 , physics.stackexchange.com/q/39602/2451 , physics.stackexchange.com/q/105563/2451 , and links therein. $\endgroup$– Qmechanic ♦Commented Apr 22, 2014 at 6:20
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2 Answers
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It is indeed possible to have a non-hermitian operator with all real eigenvalues. However, in that case at least two of its eigenstates must be non-orthogonal: having real eigenvalues and orthogonal eigenstates is sufficient for hermiticity.
For an example of the former, try e.g. $$ \begin{pmatrix} 2 & - 1 \\ 0 & 1 \end{pmatrix}, $$ whose second eigenvector is $(1,1)$.
Note, however, that in a case like this (non-hermitian diagonalizable operator with all real eigenvalues) you can always find a second inner product with respect to which the operator is hermitian, essentially by declaring its eigenbasis to be orthogonal, by decree, on the new inner product.
answered Apr 22, 2014 at 6:04
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$\begingroup$ @user56751 Ask separately - probably on Mathematics. $\endgroup$ Commented Jun 16, 2020 at 7:37
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A Hermitian operator is diagonalizable and has real eigenvalues. If you have an operator with real eigenvalues and such that all its eigenstates are orthonormal and, moreover, form a complete basis, then this operator is Hermitian. This is because you can pick its eigenbasis as an orthonormal basis and in that basis it will be represented by a manifestly Hermitian matrix. If eigenstates do not form a complete basis or cannot be orthonormalized, it is not Hermitian.
If you have a diagonalizable operator with real eigenvalues, there is an inner product for which this operator is Hermitian. To construct it, go to the basis which diagonalizes the operator and declare the inner product to be given by the standard inner product for this basis (sum of products of corresponding components, the component expression you learn in kindergarten).
Finally, you can have an non-diagonalizable operator with real eigenvalues. Its Jordan normal form must then contain non-trivial Jordan cells. An example would be $$ \begin{pmatrix} 0&1\\ 0&0 \end{pmatrix}, $$ also known as the fermion creation operator $a^\dagger$ or as the momentum rasing operator $l_+$ in spin-1/2 representation.
