Lecture9Notes-Part2 - Chem 120A 02/08/06 Spring 2006...

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Chem 120A The Schrodinger Equation 02/08/06 Spring 2006 Lecture 9 READING: QCS Chapter 3; FL3 Chapter 8-1 to 8-6; Last time we ended by introducing the time-dependent Schrodinger Equation: i ¯ h t ± ± Ψ ( t ) ² = ˆ H ± ± Ψ ( t ) ² We will assume that the Hamiltonian operatator, ˆ H , itself is not a function of time. In order to solve the time-dependent Schrodinger equation, we must apply an initial condition. Thus given an intitial state at t=0, ± ± Ψ ( 0 ) ² , then the state at any other time, t, is totally determined. Probability is still there- just that, there is no fuzziness as to what the soulution is. We will use the energy representation method to solve the time-dep Schr Eqn. This means, we will pick a basis set, ± ± n ² with n = 1. ..N, which makes up an energy eigenspace. This will give us a form of the time-dep Schr Eqn that is particularly easy to solve (a big part of solving problems in quantum mechanics is choos- ing the right basis). The reason the energy representation simplifies things because when the Hamiltonian operator acts on a state, its eigenvalue is the energy of that state (which is why the Hamiltonian operator is so important). ˆ H ± ± n ² = E n ± ± n ² Expanding a state, ± ± Ψ ² in terms of such basis states ± ± Ψ ( t ) ² = n C n ( t ) ± ± n ² Assuming orthonormality: ³ m ± ± n ² = δ mn Then C m ( t ) = ³ m ± ± Ψ ( t ) ² is the probability amplitude that ± ± Ψ ² will be found on axis ± ± m ² in the eigenspace.
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This note was uploaded on 09/29/2009 for the course CHEM 120A taught by Professor Whaley during the Spring '07 term at University of California, Berkeley.

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Lecture9Notes-Part2 - Chem 120A 02/08/06 Spring 2006...

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