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I'm trying to do an exercise in which you solve the Schrödinger equation for the hydrogen atom. Through the exercise, I've already shown that the wavefunction is:

$$ \psi_{n\ell m}(r,\theta,\varphi) = R_{n\ell}(r)Y^m_\ell (\theta,\varphi)$$

and that $Y^m_\ell (\theta,\varphi)$ are the spherical harmonics. Then, when solving for the radial part, the exercise tells me we need to study the asymptotic behaviour of $R(r)$ for large and small $r$. I had no problem showing that $R(r) \overset{\underset{+\infty}{}}{\sim} e^{-kr}$ but then the exercise gets completely bananas for the asymptotic part for small $r$.

It tells me to introduce $u(r) \doteqdot rR(r)$, and then I managed to show the Schrödinger equation becomes:

$$-\frac{\hbar^2}{2m}{\mathrm{d}^2 u(r) \over \mathrm{d}r^2} - \frac{e^2}{4\pi\epsilon_0r}u(r)+\frac{\hbar^2}{2m}\frac{\ell(\ell+1)}{r^2}u(r) = E_n u(r)$$

Ok. Then the exercise tells me to suppose $u(r) \overset{\underset{0}{}}{\sim} r^\lambda$ and to prove $\lambda > -\tfrac{1}{2}$. No problem, this follows from the fact that $R(r)$ must be a $L^2$ function. Then, as long as $\ell \neq 0$, we can solve the above equation for $r\to 0$, showing that $\lambda = \ell + 1$. The next part of the exercise says:

Apply Stokes's theorem on a sphere to show that

$$\nabla^2 \left ( \frac{1}{r} \right ) = -4\pi\delta({\mathbf r}) $$

then use this result to prove that $\lambda = \ell + 1$ even when $\ell = 0$.

And I get completely stumped. I can prove the above formula, but I have no idea what this Laplacian has to do with anything.

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Hmm could you clarify what the $\mapsto$ is supposed to mean in the quoted identity? –  joshphysics Feb 16 '13 at 20:11
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It's supposed to mean "the function that maps $\mathbf r$ to $\frac{1}{r}$" but I thought the exercise was being too pedantic with the notation so I edited the original post to remove it and make it clearer. –  Sandra Feb 16 '13 at 20:25
    
Ok cool thanks for the clarification. –  joshphysics Feb 16 '13 at 20:33
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2 Answers

up vote 3 down vote accepted

First, it is not true that $R(r)$ has to be $L^2$. Because the integration measure is $dV = r^2\cdot dr\cdot d\Omega$, and we integrate $|R|^2\cdot |Y|^2$ with this measure, it is $R(r)r=u(r)$ and not $R(r)$ itself that must be $L^2$.

Now, this is not just a correction of an unrelated minor mistake in your comments; it actually answers your main question. Why? Because the $L^2$ integrability would actually allow $u(r)$ to go like $r^0$ i.e. constant near $r\to 0$: the constant would clearly be square-integrable. However, if $u(r)$ were a constant, $R(r)$ would scale as $1/r$.

The whole wave function would go like $1/r$ near $r\to 0$; it would have this factor. Note that assuming $l=0$ which is the only case for which this singular $1/r$ behavior is imaginable, the angular part $Y$ goes like a constant, too. It is a completely legitimate candidate wave function, $\psi(r)\sim 1/r$.

However, the wave function doesn't solve the Schrödinger's equation exactly because the Laplacian term in the equation produces a new term proportional to the delta-function (that's why the identity, justified by Stokes' theorem, is mentioned) which isn't canceled, and the equation therefore fails to hold. This wave function of the sort $1/r$ should better be checked separately because its Laplacian is "almost zero". However, when we check it carefully, we confirm that the Ansatz $R(r)\sim r^l$ is completely general, whether $l$ is zero or positive, and the $1/r$ wave function that obeys the required normalization conditions and that could be an "exceptional extra solution" actually isn't a solution.

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Oh, thanks, I managed to solve the rest of the exercise after reading your comment! –  Sandra Feb 20 '13 at 18:05
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This note is just a restatement of Lubos' answer (a tyro's attempt to understand his answer) and comes courtesy of Dirac:

To investigate the appropriate boundary conditions at $r=0$, consider the (different) situation of a free particle with $l=0$ and $E=0$. In this case Schrodinger's equation reduces to Laplace's equation $\nabla^2 \Psi=0$, and the reduced one-dimension equation becomes $d^2u(r)/dr^2=0$.

Now, while $R=1/r$ (or $u=rR=1$) obviously satisfies the 1-D equation, it does not satisfy the 3-D Laplace equation because, as you showed, a delta-function appears at the origin, violating Laplace's equation there.

The point is that not every solution of the simple 1-D equation satisfies the 3-D Laplace's equation, and likewise, since the Laplacian is a term in the Schrodinger equation, not every solution of the "1-D Schrodinger equation" will satisfy the 3-D Schrodinger's equation. In particular, as $r \rightarrow 0$, $R$ must not tend to $\infty$ as fast as $1/r$, or, equivalently, $rR=u \rightarrow 0$, in order to avoid the delta-function.

Not quite done, since we haven't showed $u$ goes like $r$ near $r=0$, just that it goes to 0. Assume a solution with the proper large-$r$ behavior ($u= \sum c_sr^s \exp{-kr}$), where consecutive values of $s$ differ by unity, although these values may not be integers. From the above, the minimum $s$ must be positive. Sub'ing and plugging, one finds that the minimum of $s$ must be either $l+1$ or $-l$. The lower value is ruled out, even when $l=0$, since it is not positive, so u goes like $r^{l+1}$ in all cases, including $l=0$.

I'll reiterate that I've just restated Lubos' answer (which I now understand!).

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Thank you, Art, +1. –  Luboš Motl Feb 18 '13 at 7:26
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