2019
DOI: 10.1007/978-3-030-24748-5_7
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Killing Vector Fields and Quantisation of Natural Hamiltonians

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Cited by 4 publications
(4 citation statements)
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“…This is an interesting property since it shows that, on spaces of constant curvature, the angular momentum has a direct contribution to the total energy of the system, and, as mentioning in the previous section, it is also important in the quantization process [55,60,61].…”
Section: 1mentioning
confidence: 72%
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“…This is an interesting property since it shows that, on spaces of constant curvature, the angular momentum has a direct contribution to the total energy of the system, and, as mentioning in the previous section, it is also important in the quantization process [55,60,61].…”
Section: 1mentioning
confidence: 72%
“…We close this section mentioning that this property, that is, expressing the Hamiltonian as a function of the Noether momenta (instead of the canonical momenta) is important for the process of quantization of the system; see [55] for the free particle and [60,61] for general properties of the Killing vector fields and Noether momenta approach to quantization.…”
Section: Hamiltonian Formalismmentioning
confidence: 99%
“…Its wavefunctions of the stationary states vanish at both values of the position x = −a and x → +∞. Such behavior is achieved thanks to the introduction of the position-dependent effective mass instead constant one [25][26][27][28][29][30][31][32][33][34][35][36][37]. Wavefunctions of the stationary states of the model are expressed via the generalized Laguerre polynomials, but, its energy spectrum completely overlaps with the energy spectrum of the non-relativistic canonical quantum harmonic oscillator described above.…”
Section: Introductionmentioning
confidence: 99%
“…Its value being different than zero is due to the asymmetrical behavior of the semiconfinement potential (32) and the mass (33), but the zero value of xn from (67) will be easily recovered under the following limit: lim a→∞ xn = 0. At the same time, the mean value of the momentum does not differ from (67) and it is simply zero as follows:…”
mentioning
confidence: 99%