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There are few choice here. One length scale is the Bohr radius $a_0$. By using this scale, the corresponding frequency scale and energy scale can be defined as $a_0 = \sqrt{\hbar/(2m \omega_0)}$ and $E_0 = \hbar \omega_0$ respectively. Substitute these back, you will get the dimensionless equation. Suppose you are trying to find the ground state, you can ...


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You can substitute: $$u(r) = f(\lambda r)$$ in the differential equation and choose $\lambda$ so that the left hand side becomes dimensionless up to some overall factor (you can then absorb what's left into E and f). You can also reason as follows. You know that $\lambda= \frac{\hbar}{m c}$ has the dimensions of a length, it is the Compton wavelength of ...


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"Presumably, the usual method for speeds v≪c is to consider a Newtonian n-body approach and simply integrate the equations of motion with the gravitational inverse square law, using a numerical scheme such as Runge-Kutta or, something more sophisticated, like a symplectic integrator?" It might work for non-relativistic but for more than two bodies it is ...


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If we vary conditions (temperature/pressure), such that an FCC-BCC phase transition occurred, how would we know the BCC lattice formed? In this paper Zhang & Chen used Möbius pair-potentials to model NaCl phase transitions in an MD simulation, so I am not sure how relevant it is to your question, but they use several indicators of the FCC-BCC ...


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With forward Euler, you're simply out of luck. Because of the way the scheme is constructed, you always make an error in momentum in the same direction, which then compounds, leading to exponential behavior. You need a symplectic integrator, the simplest being leapfrog, or any other Verlet integrator. These still won't conserve momentum on a timestep to ...


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TL;DR: Yes. Although in reality you don't have unlimited floating point precision, and this will almost always break time-reversibility. I should point out that not all integrators are time-reversible. For example, predictor-corrector schemes, and most schemes that deal with constraints. The Verlet method, however, is time-reversible, even for large ...


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Okay, I don't know if this can still help you, as your question is really old at this point. I am currently doing a thesis on Monte Carlo Methods, so I might help you. To me it seems that you used the formula for a normalized function, while yours isn't. So, I would start looking at your $\pi$ in the $\tan$ term. Just calculating on the fly would give ...


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I assume you're talking about the numerical instabilities that arise from having an infinite potential at $r=0$. Here are three common solutions: Use a soft-core potential that behaves like $1/r$ except very close to $r=0$ where it levels off to a finite value. For example, $1/\sqrt{\epsilon+r^2}$ instead of $1/r$ is common. Add hard sphere collision ...


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The Schrödinger equation is $$ \left[\frac{1}{2m}(-\Delta)+V(r)\right]\psi=E\psi $$ If we let $\psi=R(r)Y_\ell^m(\theta,\phi)$, we find $$ \left[\frac{-1}{2m}\frac{\mathrm d}{\mathrm dr}r^2\frac{\mathrm d}{\mathrm dr}+\frac{\ell(\ell+1)}{2mr^2}+V(r)\right]R(r)=E R(r) $$ Finally, let $R(r)=u(r)/r$, so that $$ -\frac{1}{2m} u''(r)+\hat V(r)u(r)=Eu(r) $$ ...



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