sec3 - III. Exactly Solvable Problems The systems for which...

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Unformatted text preview: III. Exactly Solvable Problems The systems for which exact QM solutions can be found are few in number and they are not particularly interesting in and of themselves; typical experimental systems are much more complicated than any exactly solvable Hamiltonian. However, we must understand these simple problems if we are to have any hope of attacking more complicated systems. For example, we will see later that one can develop very accurate approximate solutions by examining the difference between a given Hamiltonian and some exactly solvable Hamiltionian. The two problems we will deal with here are: the Harmonic Oscillator and Piecewise Constant Potentials. A. Operators and States in Real Space To solve even simple one-dimensional problems, we need to be able to describe experiments in real space in terms of Hilbert space operators. The ansatz is fairly straightforward. A generic (classical) observable can be associated with some function of the position and momentum: ) , ( q p A Note that we assume (for now) that there is only one particle, so the observable only depends on one position and one momentum. The associated quantum operator is obtained by replacing the classical variables ) , ( q p with the corresponding quantum operators ) , ( q p : ) , ( q p A A = The position and momentum operators satisfy the canonical commutation relation : [ ] & i p q = , This association has a deep connection with the role of Poisson brackets in classical mechanics. Unfortunately, this connection is completely lost on the typical chemist, who is unfamiliar with Poisson brackets to begin with. Now, the fact that p and q do not commute poses an immediate problem. What if we want to associate a quantum operator with a classical observable like pq q p A = ) , ( ? We have more than one choice: we could choose q p A = or p q A = . For this simple product form, the dilemma is easily resolved by requiring A to be Hermitian , in which case the only possible choice is the symmetric form: ( 29 p q q p A 2 1 + = You can verify for yourself that this operator is, indeed Hermitian. When ) , ( q p A contains more complicated products of p and q (e.g. ( 29 q p q p A- = 1 cos ) , ( 3 ) the solution is, well, more complicated. In fact, there is no a general way to associate arbitrary (non-linear) products of p and q with a unique quantum operator. Fortunately, we will not be interested in classical observables that involve non-linear products of p and q in this course. Thus this is not a practical obstacle for us; however, there is an on-going debate within the physics community about how these products should be treated....
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sec3 - III. Exactly Solvable Problems The systems for which...

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