The Context to my Question: I have a hydrogen nucleus, a deuteron, a proton and neutron, and I am trying to show that a binding energy of $\beta=2.2 MeV$ would be valid. This is a finite potential well problem. $$ \begin{align} U_0=36MeV\qquad &a=2.6fm\\ \psi(0)=0;\qquad&\psi(x)=-\psi(-x)\\ \end{align}$$ Where $U_0$ is the potential wall of the well, $a$ is the width of the well, $\psi$ is the anti-symmetric wave equation. I think the relationship of binding energy to potential and total is $$E_0=U_0 -\beta$$ Given that $\beta=2.2MeV$ and $U_0=36MeV$, then $E_0 = 33.8MeV$. Although I assume that this deuteron would have quantized energy because it is in a finite well problem, and because the nucleus's of atoms are described with the Schrodinger equation. This energy quantization should be defined by $$2cot(ka)=\frac{k}{\alpha}-\frac{\alpha}{k}$$ Where $k^2=\frac{2m}{\hbar}E$ and $\alpha^2=\frac{2m}{\hbar}(U_0-E)$. This is where I become confused.

Priming: A particle in a a finite well is said have quantized energy. Although, it seems that any value for $U_0$ where $0 < U_0$ and any value for $\beta$ ( the binding energy ) where $0 < \beta < U_0$ will result in the stable bound state. Quantum means a discrete value, although I do not see any condition that makes $E$ discrete.

Question: What condition is making the energy discrete? How do I verify that $E_0 = 33.8 MeV$ is one of these quantized energy values?

  • 2
    $\begingroup$ hyperphysics.phy-astr.gsu.edu/hbase/quantum/pfbox.html, for a single particle in a box. Your problem appears to be a multiple particle problem though. Please clarify what's in the box and what is being bound there, it's far from clear (at least to me) :-) $\endgroup$ – Gert Oct 15 '16 at 23:55
  • $\begingroup$ Yes there are two particles, but they can be treated as one. The force bining them is the strong-nuclear force. Because the proton and neutron are of similar mass, I can use relative mass, $m=(m_p*m_n)/(m_p +m_n)$ to treat this problem as a single particle in a finite well. The mass is approximately $m\approx\frac{m_p}{2}\approx\frac{m_n}{2}$. $\endgroup$ – Tsangares Oct 16 '16 at 3:47
  • $\begingroup$ I meant reduced mass. en.wikipedia.org/wiki/Reduced_mass $\endgroup$ – Tsangares Oct 16 '16 at 3:57
  • $\begingroup$ I believe if you plot the left side and the right side of your equation simultaneously on the ordinate and $k$ (or $E/U_0$) on the abscissa, you will find there are discrete values of $k$ (or ...) which will satisfy the equation. $\endgroup$ – Bill N Oct 17 '16 at 17:27