# Hydrogen atom in quantum mechanics

I have problems following the calculations in Griffiths' Introduction to Quantum Mechanics (Chapter 4.2.1):

If you apply the Schrödinger equation to the Coulomb potential you get the following differential equation: $$\tag{4.56} \frac{d^2u}{d\rho^2} ~=~ [1-\frac{\rho_0}{\rho} + \frac{\ell(\ell+1)}{\rho^2}]u.$$

A few steps later my books substitutes $u(\rho)$ without explaining why this is allowed:

The next step is to peel of the asymptotic behavior, introducing a new function $v(\rho)$:

$$\tag{4.60} u(\rho) ~=~ \rho^{\ell+1}e^{-\rho}v(\rho).$$

How is this function still related to the Hydrogen atom? To me it looks like something total random.

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Aside: Guess-and-check is a legitimate (if often inefficient) method of finding solutions to many classes of problems, including differential equations. Of course, it helps to have a uniqueness theorem for the answers or you never know when you are done that way. –  dmckee Jan 25 '13 at 17:48

As Lubos points out, you are allowed to make any substitutions you want as long as the new function is equivalent to the old one. In this case, substituting $u(\rho)=\rho^l e^{-\rho}v(\rho)$ is allowed because you can recover $u$ if you know $v$.

The question is, though, why would you choose exactly such a form? If you're just given it outright, it does look pretty random. However, there's method behind it.

Essentially, you've got an equation that's got very particular behaviour at both of its endpoints. Take, for example the $\rho\ll1$ limit: then the equation reads $$\frac{d^2u}{d\rho^2}\approx\frac{\ell(\ell+1)}{\rho^2}u,$$ whose solutions are $u=\rho^l$ and $u=\rho^{-(l+1)}$ (which we reject as unphysical). Of course, this says nothing exact about the full solution but it does inform us about its asymptotic behaviour as $\rho\rightarrow0$.

Similarly, in the $\rho\gg l$ limit you have $\frac{d^2u}{d\rho^2}\approx u$, so $u=e^{-\rho}$ (as $u=e^\rho$ is unphysical). Again, this only gives information about the asymptotics about $\rho\rightarrow\infty$.

So what have our limit games gained us? Formally, we know nothing about the exact solution, but we do know that it has to behave something like $u=\rho^le^{-\rho}\times\mathrm{some\, function}$. Plus, this new function will be better behaved than $u$ at the endpoints - constant at the origin and up to polynomial at infinity.

Thus, we make an informed guess and substitute $u=\rho^le^{-\rho}v(\rho)$, transforming the equation to the new variables, and hoping as we do so that the result will be a simpler equation. As it happens, it is both simpler and better behaved, with better-behaved singularities, and its solutions are now simple enough (the complexity being absorbed into the prefactors just discussed) that they are in fact polynomials.

In general, the exact form of the substitution will depend on the exact problem. One usually tries to do this kind of asymptotic analysis to try and distil away, slowly, the problem's complexity.

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As Lubos mentioned (in a now apparently deleted answer) the substitution $$u(\rho) = \rho^{\ell+1} \mathrm{e}^{-\rho} v(\rho)$$ is always a valid mathematical substitution you are allowed to make in any equation. The question is why is this a good substitution in this instance. It is justified by asymptotic analysis which, unfortunately, is only taught in bits and pieces in undergraduate physics. In particular, see "Method of Dominant Balance" for the general procedure behind these sorts of substitutions. The basic idea is to look at limiting cases. Often you find that the limiting cases simplify the equation by removing unecessary detail, allowing you to easily deduce things about the behaviour about the system even away from the limits.

In this example we have one crucial thing we know about wavefunctions that we can use to deduce something useful: the wavefunction must be normalisable. Now look at the D.E. in the $\rho\rightarrow\infty$ limit. We know that $u(\rho)$ must vanish in this limit. The $\rho^{-1}$ and $\rho^{-2}$ terms on the right hand side disappear in the limit and the equation becomes

$$\frac{\mathrm{d}^2 u}{\mathrm{d}\rho^2} = u$$

which has two solutions $u = \exp(\pm \rho)$. If we take the plus sign $u$ blows up as $\rho\rightarrow\infty$, so we must take the minus sign. This means that $u(\rho)$ can be written as $f(\rho) \exp(-\rho)$ where $f(\rho)$ is some function which doesn't grow exponentially fast at large $\rho$. From a similar argument in the $\rho\rightarrow 0$ limit you can deduce the $\rho^{\ell+1}$ behaviour. There are two solutions to the second order D.E. at small $\rho$ but only one of them is normalisable.

The substitution $u(\rho) = \rho^{\ell+1} \mathrm{e}^{-\rho} v(\rho)$ extracts the dominant behaviour in the limits. By comparison to $u$, $v$ has a mild behaviour that isn't particularly interesting unless you need exact answers. Another interesting thing to note is where the dominant terms came from. In the $\rho\rightarrow\infty$ limit the dominant term comes from the $E \psi$ term of the Schrodinger equation, and the $\rho\rightarrow 0$ limit comes from the angular momentum part $L^2 / 2mr$. Neither piece depends on the potential! Only the function $v$ depends on the potential. The simple requirement of normalisability is enough to impose the asymptotic form of the wavefunction.

This same technique works in any situation where there is a dominant behaviour coming from some simple required behaviour in a limit (e.g. normalisability restricting the form of $u$ as $\rho\rightarrow\infty$), and a subdominant behaviour having to do with the details of the situation ($v$ depends on Coulomb potential, whereas the asymptotic behaviour of $u$ is universal to all central potentials).

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You beat me by a bit. Sorry! I didn't see your answer. –  Emilio Pisanty Jan 25 '13 at 16:45
Why the downvote? –  Michael Brown Aug 12 '13 at 1:04
Again, why the (additional) downvotes? If there is a problem with my answer I would like to fix it. –  Michael Brown Sep 9 '13 at 2:16
Beats me, man. Potential downvoters: this answer is perfectly fine. It is equivalent to, at least as good is, and probably better than mine. –  Emilio Pisanty Sep 9 '13 at 9:07
Yes this answer is very good. I don't know why it was downvoted without comments. –  Prathyush Sep 9 '13 at 9:51