How to solve this Schrödinger equation?

Question

I am taking an intro level quantum mechanics class. Our textbook gives a problem like this:

The deuteron is a nucleus of "heavy hydrogen" consisting of one proton and one neutron. As a simple model for this nucleus, consider a single particle of mass $m$ moving in a fixed spherically -symmetric potential $V(r)$, defined by $V(r)=-V_{0}$ for $r<r_{0}$ and $V(r)=0$ for $r>r_{0}$. This is called a spherical square-well potential. Assume that the particle is in a bound state with $l=0$.

(a) Find the general solution $R(r)$ of the Schrodinger equation for $r<r_{0}$ and $r>r_{0}$. Use the fact that the wave equation must be finite at 0 and $\infty$ to simplify the solution as much as possible. (You do not need to normalize the solutions).

(b) The deuteron is only just bound: ie., $E$ is nearly equal to $0$. Take $m$ to be the proton mass, $m=1.67*10^{-27}kg$. and take $r_{0}$ to be a typical nuclear radius, $r_{0}=1*10^{-15}m$. Find the value of $V_{0}$ (the depth of the potential well) in MeV (1 MeV=$1.6*10^{-13}$J (Hint: The continuity conditions at $r_{0}$ must be used. The radial wave function $R(r)$ and its dervivative $R'(r)$ must both be continuous at $r_{0}$; this is equivalent to requiring $u(r)$ and $u'(r)$ must both be continuous at $r_{0}$, where $u(r)=rR(r)$. The resulting equations cannot be solved explicitly but can be used to derive the value of $V_{0}$.

I found I need to solve an equation of this form when $r<r_{0}$:

$$-\frac{\hbar^{2}}{2mr}\frac{\partial}{\partial {r}^{2}}(rR(r))-V_{0}R(r)=ER(r)$$

After expansion this becomes a second order ODE I am not able to solve. But I am not able to use the condition $r=0$ to obtain any good results. At $r=0$ we should have $-V_{0}R(r)=ER(r)$, which implies $E=-V_{0}$. (Is this allowed?). Then I can get

$$-\frac{\hbar^{2}}{2mr}\frac{\partial}{\partial {r}^{2}}(rR(r))=0$$

yet solving this implies $rR(r)=c_{0}r+c_{1}$, which I "feel" is impossible since it would imply $R(r)=c_{0}+\frac{c_{1}}{r}$, which has a singularity at $r=0$ unless $c_{1}=0$.

On the other hand at $r=\infty$ we should have the equation become:

$$-\frac{\hbar^{2}}{2mr}\frac{\partial}{\partial {r}^{2}}(rR(r))=ER(r)$$

Hence naming $rR(r)=S(r)$, we should have

$$\frac{\partial}{\partial {r}^{2}}S(r)=kS(r), k=-\frac{2mE}{{\hbar}^{2}}$$

and solve it we have $S(r)=e^{\sqrt{k}r}\Rightarrow R(r)=\frac{e^{\sqrt{k}r}}{r}$. This function is obviously non-constant at the origin and since $k$ is negative, it actually "blow up". I do not know how to reconcile this desperate situation. So I need some help. Hence I cannot continue to part $(b)$ as well. I feel there is something very wrong in my approach. I also want to ask what does it mean "the deuteron is only just bound", why this implies $E$ is nearly equal to $0$? Also, if I cannot solve the equation explicitly, how can I get $V_{0}$'s value? I am very confused.

Answer 1

At $r=0$ we should have $−V_0 R(r)=ER(r)$, which implies $E=−V_0$. (Is this allowed?)

Nope, not allowed. In any case, that ODE isn't as bad as it looks. Change variables from $R(r)$ to $u(r)$ [which was defined in the problem as $u(r) = r R(r)$], and it will wind up looking very familiar. You'll come up with a second-order ODE that has two linearly independent solutions,

$$u_1(r) = \cdots,\quad u_2(r) = \cdots$$

Then you can get the two independent solutions to the original Schrodinger equation as $R_i(r) = \frac{u_i(r)}{r}$.

The final physical solution for $R(r)$ needs to be well-defined at the origin. (In fact you realized this in the second part, but you don't need to worry about it there because $r > r_0$ doesn't include the origin; but you do need to worry about it here) So you need to pick a linear combination of $R_1(r)$ and $R_2(r)$ that is not infinite at $r = 0$. Hint: what needs to be the numerical value of $u(0)$?

The other part, with $r > r_0$, is extremely similar. Again, remember that there are two linearly independent solutions. You only found one of them. Also, the solution you found doesn't blow up if $k$ is negative - but are you sure that $k$ is negative? What do you know about the value of $E$? (Specifically, what does it mean for the particle to be in a bound state?)

You don't need to care about what the solution to the second part does at the origin, because the region you're solving the equation in doesn't include $r = 0$. But it does include $r \to \infty$, so you'll need to pick a linear combination of the solutions that stays finite in that limit.