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Unformatted text preview: Schrödinger equation, so is . This is useful in defining a probability, since we would like (3.17) Given we can thus use this freedom to "normalise" the wave function! (If the integral over is finite, i.e., if is ``normalisable''.) As an example suppose that we have a Hamiltonian that has the function as eigen function. This function is not normalised since (3.18) The normalised form of this function is (3.19) We need to know a bit more about the structure of the solution of the Schrödinger equation  boundary conditions and such. Here I shall postulate the boundary conditions, without any derivation. 1. is a continuous function, and is single valued. 2. must be finite, so that (3.20) 3. 4. is a probability density. 5. is continuous except where has an infinite discontinuity....
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This note was uploaded on 11/09/2011 for the course PHY PHY2053 taught by Professor Davidjudd during the Fall '10 term at Broward College.
 Fall '10
 DavidJudd
 Physics, Energy

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