We consider the Hamiltonian to be Hermitian to make energy eigenvalue real. But, say for some infinitesimal region we can't measure the energy. Can the Hamiltonian be non-hermitian in that case?
No. For one, it relies on momentum and the momentum operator is hermitian. For another, a postulate of QM requires that it be.
You may consider a detailed article by C M Bender, "Making sense of non-Hermitian Hamiltonians" Rep. Prog. Phys. 70 947 (2007); DOI: 10.1088/0034-4885/70/6/R03 (arXiv:hep-th/0703096). There the condition of PT-symmetry is introduced instead of "traditional" Hermiticity requirement.
Dear Andrew Messing ,
If the Hamiltonian is Hermitian , the energy eigenvalue in that region must be real and hence measurable. But we can't measure the energy in this case. The only possible constraint that can make the Hamiltonian, Hermitian is the unitarity of the time evolution operator in my opinion. But even in that case, I am not sure what happens if Hamiltonian has the form H=A/dt.
Dear Debajyoti Sarkar:
1) I'm a little wary of speaking about the region in which the energy eigenvalue need be granted anything about the Hamiltonian of a quantum system (mostly because I'm a little or more than a little wary of speaking about the interpretation of QM formalism in general, but particularly where it concerns the relationship between any mathematical representation of a quantum system to any actual physical system).
2) If, by "measurable", you mean "observable" (as opposed to measure theoretic notions, which are so far from relevant here that I can't imagine you mean this), then again a postulateof QM is that to any observable there corresponds a hermitian operator.
3) By definition, all measurable/observable properties of a quantum system are hermitian operators, and no measurable/observable properties of any quantum system actually correspond to values the way that they do in classical physics (i.e., while in classical physics, the position and velocity of a system have a value that correspond directly to the position and velocity of that system, while in QM all observables are functions that act on an abstract system in Hilbert space). The need for operators to be hermitian is part of the structure of QM, and H isn't defined.
4) Assuming you meant by "dt" the derivative with respect to time, then you require the time evolution (linear) operator U^ which you plug into the Schrödinger equation as a substitute that still requires H^ (the Hamiltonian) to be hermiitian. H^ is not and cannot be defined as A/dt, particularly since d/dt is defined in such a way that the partial derivatives of A^ with respect to define vanish, leaving us with an observable defined in terms of the Hamiltonian which is not explicitly dependent upon time.
Non-Hermitian Hamoltanian system is possible. Think a Hamoltinian operator whose all eigenvalues are real and positive but it it is not Hermitian...
What is that infinitesimal region? A region of what? Of the physical space? The volume of a nucleus, or of an elementary particle? A volume of the phase-space?
thanks dear profesor Wiwat...
Please see paper of Ali Mostafazadeh on Psuedo hermitian Hamiltonian systems..
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