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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


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.

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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.

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hi, so the energy at the bound state is not positive? I am very surprised. –  Kerry Mar 17 '11 at 5:21
If that's the case, you should probably go back to your references and review the definition of a bound state. –  David Z Mar 17 '11 at 6:35
@David Zaslavsky♦:hi, can you give me some hint how to find $V_{0}$? I found I am stuck again... –  Kerry Mar 18 '11 at 3:24
I think there is something very wrong in my computation as my result turned out to be 0. –  Kerry Mar 18 '11 at 3:59
@user2597: I can't commit to helping you with your homework problems through email. As I said, you could try math.stackexchange.com, if you can formulate your questions clearly. Another possibility is physicsforums.com. It seems that you're going to need an extended discussion to help you work through the problem, and the forum format of PF is better suited to extended discussion than the Q&A format of Stack Exchange. (I'm an occasional contributor there as well) –  David Z Mar 18 '11 at 7:08
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