Timeline for Harmonic Oscillator - Energy quantisation
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Feb 2, 2014 at 15:11 | comment | added | user35952 | @JánLalinský : Right, Thanks for reminding that part, it seems to be true for any such bound system that can become free beyond some distance. | |
| Feb 2, 2014 at 15:02 | comment | added | Ján Lalinský | Discrete indexing may occur only for some part of the spectrum. For hydrogen atom, spectrum is discrete only for negative energies. For positive energies, there are continuously-indexed positive eigenvalues associated with improper eigenfunctions - functions that satisfy Schroedinger equation but not the normalization condition. These are important too to obtain good generalized basis for expressing integrable functions. | |
| Feb 2, 2014 at 15:00 | comment | added | Ruslan | @user35952 In general quantization appears as a property of the operator you select. When you define an operator, you must define its domain and thus boundary conditions. Eigenvectors of that operator are the solutions of time-independent Schroedinger equation. If the spectrum of the operator has discrete part, then you have some quantized states. | |
| Feb 2, 2014 at 14:58 | comment | added | user35952 | So can we say that, quantisation of a system in genereal is always because of boundary condition. | |
| Feb 2, 2014 at 14:56 | comment | added | Ruslan | @user35952 yes. Without boundary conditions you have general solution for any $\lambda$, it has two free parameters. When you impose boundary conditions, one of these parameters is used to satisfy one boundary condition, and another one is normalization coefficient. To satisfy second boundary condition you have to restrict $\lambda$ to some spectrum of values. | |
| Feb 2, 2014 at 14:56 | history | edited | Ján Lalinský | CC BY-SA 3.0 |
added 9 characters in body
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| Feb 2, 2014 at 14:35 | comment | added | user35952 | So in this case, the boundary condition is restricting the solutions to integer values ? | |
| Feb 2, 2014 at 14:31 | comment | added | user35952 | I am sorry about that, I meant not a quantum mechanically acceptable wavefunction !! I have made some edits. Thanks !! | |
| Feb 2, 2014 at 14:30 | history | answered | Ján Lalinský | CC BY-SA 3.0 |