- #1

tylerscott

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Here is a grab of it:

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- Thread starter tylerscott
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- #1

tylerscott

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Here is a grab of it:

- #2

Chopin

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Oftentimes, the best way to solve these sorts of equations is just by the time-honored technique of taking an informed guess (a procedure which you justify by giving it the official-sounding name of "ansatz" ), and then checking whether it works. In this case, it's fairly straightforward to check that

[tex]\frac{d^2}{d\xi^2} (A e^{-\xi^2/2} + B e^{\xi^2/2}) = A(\xi^2-1)e^{-\xi^2/2} + B(\xi^2+1)e^{\xi^2/2}[/tex]

Which is an approximate solution to the equation as long as [itex]\xi[/itex] is large.

P.S. I wouldn't worry too much about this approximate solution for anything past a general characterization of the solutions. You're almost certainly going to turn around in the very next section and redo the whole problem using the ladder operator formalism, which allows you to convert the problem into a much simpler first-order equation that you can solve exactly.

[tex]\frac{d^2}{d\xi^2} (A e^{-\xi^2/2} + B e^{\xi^2/2}) = A(\xi^2-1)e^{-\xi^2/2} + B(\xi^2+1)e^{\xi^2/2}[/tex]

Which is an approximate solution to the equation as long as [itex]\xi[/itex] is large.

P.S. I wouldn't worry too much about this approximate solution for anything past a general characterization of the solutions. You're almost certainly going to turn around in the very next section and redo the whole problem using the ladder operator formalism, which allows you to convert the problem into a much simpler first-order equation that you can solve exactly.

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- #3

jtbell

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- #4

tylerscott

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beforedescribing the analytic method that is being discussed here. Unless he's switched them around for the second edition; I have only the first edition at hand.

Yeah, you're correct. The ladder operator formalism makes more sense to me than the analytic method.

- #5

Chopin

- 368

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In any case, since you're only calculating an approximation here, it's difficult to provide any hard and fast rules about how to come up with solutions to the equation. Intuitively, though, you know that the harmonic oscillator is a potential well, so you know the solution is going to be some sort of peaked waveform. The simplest peaked waveform which still tends towards 0 at infinity (so that it's normalizable) is the Gaussian, so something with that form is a logical place to start. Then you just plug it in, and see if it works.

- #6

tylerscott

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Ok, that makes sense. Thanks

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