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Lecture9

# Lecture9 - CHEM 356 Lecture 9 Fall 2009 1 Exact Solutions...

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Unformatted text preview: CHEM 356, Lecture 9, Fall 2009 1 Exact Solutions of the Schr¨ odinger Equation. Four Steps in the solution procedure for time–independent Schr¨ odinger equation, namely, (a) Write as a function of r [or of ?,?,? ], and form ℋ ; (b) Solve the eigenvalue equation ℋ = , to obtain the general solution; (c) Adjust the solution for so that all boundary conditions are satisfied; (d) Carry out the normalization of , if possible. The Free Particle in 1D. (a) Hamiltonian: ( ? ) = 0 , ˆ = ˆ 2 2 = − ℏ 2 2 2 ? 2 , so that ℋ = ˆ + ˆ = − ℏ 2 2 2 ? 2 . (b) Eigenvalue equation: ℋ = ⇒ 2 ? 2 = − ( √ 2 ℏ ) 2 ≡ − 2 , Solution: Let us now solve the Schr¨ odinger equation for this case: the differential equation (DE) that we have obtained is 2 ? 2 + 2 = 0 . CHEM 356, Lecture 9, Fall 2009 2 To solve this DE, let us assume that ( ? ) can be written in the form ( ? ) = e , so that it becomes 2 ? 2 + 2 = 0 ⇒ ( 2 + 2 )e = 0 , from which we see that 2 + 2 = 0 (this equation is referred to in the theory of differential equations as the characteristic equation ). The solution to the char- acteristic equation is = ± . We obtain one particular solution e for each value of obtained from the characteristic equation. The general solution to our DE will then be given as a linear combination of the particular solutions, which we shall write as ( ? ) = ? e + ? e − . To proceed further, we need to examine the boundary conditions connected with the particular problem. (c) For the free particle, there are no boundary conditions, as the particle can travel in either direction to infinity: that is, once the particle has been set in motion in either the + ? or − ? direction, it will continue forever in that direction. There are therefore no restrictions on or , so that the required solutions are simply: ( ? ) = ?...
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Lecture9 - CHEM 356 Lecture 9 Fall 2009 1 Exact Solutions...

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