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I am solving a special case of the radial component of Schroedinger's equation numerically. The equation looks like this: $$ u''(r) = \frac{2\mu}{\hbar^2}(V(r) - E)u(r). $$ $V(r)$ is a potential and $E$ is the energy of the ground state. This is part of an assignment, and the instructions say somewhat cryptically that in order to rewrite $K = \frac{2\mu}{\hbar^2}$ in suitable units (MeV and fm), we need to make use of $\hbar c = 197.327\ \rm{MeV}\cdot\rm{fm}$.

$\hbar$ is Planck's reduced constant and $\mu = \frac{m_p m_n}{m_p + m_n}$ where $m_p$ is the mass of a proton given in MeV/$c^2$ and $m_n$ is the mass of a neutron given in the same unit.

My question is, how can the quantity $\hbar c$ be used? I can write down $\hbar$ in the correct units without knowing what it is multiplied by $c$. Is there any use of the product with $c$ at all?

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  • $\begingroup$ $\hbar c$ could be understood as the energy associated with a particular (in this case fm) wavelength since $E=\frac{\hbar c}{\lambda}$ $\endgroup$
    – Triatticus
    May 14, 2018 at 23:15

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I guess the point is that the reduced mass is given in units of $\mathrm{MeV/c^2}$. If we call $\mu^*$ the adimensional value of $\mu$ in those units, so that $\mu=\mu^* \mathrm{MeV/c^2}$, then $$K=\frac{2\mu}{\hbar^2}=\frac{2\mu^*}{(\hbar c)^2}\mathrm{MeV}$$ and you can then use the expression provided.

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  • $\begingroup$ Could you clarify what the numerical value of $\mu^*$ is? If I know what the numerical value of $\mu$ is, then how do I find the numerical value of $\mu^*$? Thank you. $\endgroup$
    – user195364
    May 17, 2018 at 14:44
  • $\begingroup$ $\mu^*$ is just the value of $\mu$ in $\mathrm{MeV}/c^2$ stripped of its units. So if for example $\mu = 6.789\mathrm{MeV}/c^2$ then $\mu^* = 6.789$ $\endgroup$
    – John Donne
    May 17, 2018 at 19:49

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