Abstract

One of the possible versions of quantum mechanics, known as Kuryshkin-Wodkiewicz quantum mechanics, is considered. In this version, the quantum distribution function is positive, but, as a retribution for this, the von Neumann quantization rule is replaced by a more complicated rule, in which an observed value AA is associated with a pseudodifferential operator O^(A){\hat{O}(A)}. This version is an example of a dissipative quantum system and, therefore, it was expected that the eigenvalues of the Hamiltonian should have imaginary parts. However, the discrete spectrum of the Hamiltonian of a hydrogen-like atom in this theory turned out to be real-valued. In this paper, we propose the following explanation for this paradox. It is traditionally assumed that in some state ψ{\psi} the quantity AA is equal to λ{\lambda} if ψ{\psi} is an eigenfunction of the operator O^(A){\hat{O}(A)}. In this case, the variance O^((A-λ)2)ψ{\hat{O}((A-\lambda)2)\psi} is zero in the standard version of quantum mechanics, but nonzero in Kuryshkins mechanics. Therefore, it is possible to consider such a range of values and states corresponding to them for which the variance O^((A-λ)2){\hat{O}((A-\lambda)2)} is zero. The spectrum of the quadratic pencil O^(A2)-2O^(A)λ+λ2E^{\hat{O}(A2)-2\hat{O}(A)\lambda + \lambda 2 \hat{E}} is studied by the methods of perturbation theory under the assumption of small variance D^(A)=O^(A2)-O^(A)2{\hat{D}(A) = \hat{O}(A2) - \hat{O}(A) 2} of the observable AA. It is shown that in the neighborhood of the real eigenvalue λ{\lambda} of the operator O^(A){\hat{O}(A)}, there are two eigenvalues of the operator pencil, which differ in the first order of perturbation theory by ±i⟨D^⟩{\pm i \sqrt{\langle \hat{D} \rangle}}.

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