2 Nucleon Potential I am looking at a 2 Nucleon potential of the form
$$V(r)=V_0(r)[a+bI_1\cdot I_2]$$
Where a and b are constants.  $I_1,I_2$ are isospins.  $V_0(r)$ is of the square well form.  My goal is to find an equality for a and b, given that deuteron exists, and that diproton and dineutrons do not.  
My approach has been the following:
2 nucleons can either be in an isosinglet or an isotriplet.  The Isosinglet has I=0 and is the following:
$$|00\rangle=\frac{1}{\sqrt2}(pn-np),I_1\cdot I_2=-\frac34$$
The isotriplet has I=1 and 
$$|11\rangle=pp,I_1\cdot I_2=\frac12$$
$$|10\rangle=\frac{1}{\sqrt2}(pn+np),I_1\cdot I_2=\frac12$$
$$|1-1\rangle=nn,I_1\cdot I_2=\frac12$$
So I get the following equations based on V
$$\text{Isosinglet:  }\frac{V(r)}{V_0(r)}=a-\frac34 b$$
$$\text{Isotriplet: }\frac{V(r)}{V_0(r)}=a+\frac12 b$$
Now I am supposed to use the fact that $V_0(r)$ is of the square well form to create inequalities, but I am unsure of where the inequalities come from.
 A: You're almost there.  I'm assuming your square well $V_0(r)$ is nonzero and negative only on some interval $-r_0 < r < +r_0$ about the origin.  In that case, for positive $a,b$ you already have that the isosinglet well is deeper than the isotriplet well.
Remember that the finite square well has a finite number of bound states, each with energy $-|V_0| < E < 0$.  Find the width and depth of a well with a single bound state (for fun, with the correct binding energy, 2.2 MeV).  Next find the minimum $b$ so that a same-radius well, shallower by $\frac{3}{4}V_0b$, has zero bound states.  Tada: a bound isosinglet with no excited states, and an unbound isotriplet.
A: In my search for the Hagen Boson, I came across a very similar problem.  It would be my honor to assist a young mind like yours through such treacherous waters.
In my experience, it is beneficial to consider why the $I=0$ deuteron can exist, while the diproton and dineutron do not.  Since the only thing holding these particles together is a binding potential, for the particle to exist means that the particle is bound.  Similarly, if the particle does not exist, it must be unbound.  The inequalities come from considering these two conditions with the potential you are given.
And so, we have
$\text{Isosinglet}: V(r) + K = E < 0$
and
$\text{Isotriplet}: V(r) + K = E > 0$
With proper reasoning, you can remove the $K$ term from both inequalities.  Therefore, you get the desired relation $-1/4 < a/b < 3/4$.
