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Jun 11, 2021 at 9:30 comment added Gandalf73 Boundary conditions are unique for a specific problem, you can't change. There are no general B.C. for SE but there are the wavefunction like the wavefunction has to continuous everywhere. My advice is solve a simple problem using both the solutions and apply appropriate B.C. and check if its true. Look how the wave is propagating.
Jun 10, 2021 at 11:44 comment added J. Murray @DipankarMitra Yes.
Jun 10, 2021 at 11:43 comment added Dipankar Mitra @J.Murray oh..I have understood. But can C and D in the second solution take complex values as A and B?
Jun 10, 2021 at 11:39 comment added J. Murray @DipankarMitra Yes. Wavefunctions are generally complex valued, which shouldn’t be surprising because the two complex exponentials are also solutions.
Jun 10, 2021 at 11:35 comment added Dipankar Mitra @J.Murray I am not sure A and B real-valued or not. Can A and B take complex values?
Jun 10, 2021 at 11:13 comment added J. Murray @DipankarMitra why should A and B be real-valued?
Jun 10, 2021 at 7:33 comment added Dipankar Mitra As you said both are the solutions to the TISE and P1(x) and P2(x) will be same after applying the boundary conditions How it can Be? Say for example, I have choosen two arbitrary values for C and D as C = 2 and D = 3. Now no real values of A and B can equalize P1(x) and P2(x). so how can P1(x) and P2(x) be same?
Jun 10, 2021 at 7:26 comment added Dipankar Mitra Then what are the boundary conditions of this equation? Does boundary conditions depends on systems? Or there is any general boundary condition for Schoedinger's equation?
Jun 9, 2021 at 15:26 vote accept Dipankar Mitra
Jun 10, 2021 at 8:26
Jun 9, 2021 at 15:25 vote accept Dipankar Mitra
Jun 9, 2021 at 15:25
Jun 9, 2021 at 15:25 vote accept Dipankar Mitra
Jun 9, 2021 at 15:25
Jun 9, 2021 at 7:28 history answered Gandalf73 CC BY-SA 4.0