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Why is it that the SHO/SHM solution $$u(x)=H(x)\,e^{-x^2/2},\quad \Psi(x,t)=u(x)\,e^{-iEt/\hbar}$$ with $$H_n (x)=\sum_k^na_kx^k \tag{Hermite Polynomials}$$ is valid even when $x$ (displacement from equilibrium) becomes very large, in which case $H_n(x) \propto x^n$?

Isn't the Taylor expansion of the potential $$V(x_0+\delta x)\approx V(x_0)+\left.\frac{1}{2}\frac{d^2V}{dx^2}\right|_{x_0}(\delta x)^2$$ only valid for when $\delta x$ is small?

Cosmas Zachos
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Chern-Simons
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1 Answers1

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There are no approximations made in this solution.

The potential is exactly a quadratic. So, in the simple quantum harmonic oscillator system, you don't need to do the expansion of the potential you mention at the end of your question; it simply is exactly a quadratic.

There can be some confusion because people often say things like "this dip in X funny-looking potential is quadratic for small deviations from the middle of the dip," but the QHO is in an exactly quadratic potential, so it looks quadratic arbitrarily far from the middle.

flevinBombastus
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  • But.. how is discussion of the SHO going to address the question which is about departures from harmonicity? – Cosmas Zachos Nov 24 '21 at 15:50
  • I did not interpret the question that way. OP explicitly says "in the SHO system." – flevinBombastus Nov 24 '21 at 15:57
  • No, he is not restricting his problem to the SHO: why bother asking anything there? He is asking why he is allowed to apply SHO solutions extending to large distances to the small excursions dictated by his TISE proximation. – Cosmas Zachos Nov 24 '21 at 16:01
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    I guess we'll have to hear from OP, but I did not interpret the question that way. The answer seems trivial to "experts", but these types of confusions (specifically, not recognizing when one needs to make the approximation given in the question and what its effects are) are common for students. That is why someone might "bother asking anything there." – flevinBombastus Nov 24 '21 at 16:05