Sign up ×

In his discussion of the radial wave function of hydrogen Griffiths (Introduction to Quantum Mechanics, 2nd ed, p.146) gives the general solution of$$\frac{d^{2}u}{d\rho^{2}}=\frac{l\left(l+1\right)}{\rho^{2}}u$$ as$$u\left(\rho\right)=C\rho^{l+1}+D\rho^{-l}.$$ Why is this? There's not even a $u$ on the right-hand side! says this general solution can be verified by direct substitution, ie $$\frac{d^{2}u}{d\rho^{2}}=Cl\left(l+1\right)\rho^{l-1}+D\left(-l\right)\left(-l-1\right)\rho^{-l-2}=\frac{l\left(l+1\right)}{\rho^{2}}u.$$

I've fiddled around with this all afternoon (again there's no $u$ in the second expression, which I find ultra-confusing) but can't see how these expressions are equal.

share|cite|improve this question

migrated from Jan 16 '13 at 18:06


Personally, I'm... not sure I understand your point :-/ What's exactly the matter? I see that this ODE in the radial part of the wavefunction is clearly satisfied by direct substitution. Maybe you lost a little bit lost in the algebra? ó_ò –  user17581 Jan 16 '13 at 17:36
Obviously, what's obvious to you is not obvious to me. –  Peter4075 Jan 16 '13 at 19:09

5 Answers 5

up vote 1 down vote accepted

You just need to use substitution:


from the third term which is equal to the second expression. The second set of terms (in parenthesis) is the u-term that you are looking for. Is this what you wanted?

share|cite|improve this answer
Yes, I just couldn't see it. Now it seems embarrassingly obvious. –  Peter4075 Jan 16 '13 at 19:59

It's really very simple! In your first equation, $u = u(\rho)$ is some function of $\rho$ you want to determine. So of course a solution will not 'depend on $u$' but on $\rho$.

For the second part of your question, let's consider only the first term, i.e. $u(\rho) = C \rho^{l+1}.$ If you plug that into the left-hand side of the differential equation, you find $$\frac{d^2 u(\rho)}{d \rho^2} = \frac{d}{d \rho} \frac{d u(\rho)}{d \rho} = \frac{d}{d \rho} \left[C (l+1) \rho^l\right] = C (l+1) l \rho^{l-1}.$$ But this is almost $u(\rho)$ itself, only multiplied by $l(l+1)$ and divided by $\rho^2.$ So we could continue $$ \ldots = \frac{l(l+1)}{\rho^2} u(\rho).$$ This proves that for any $C$, $u(\rho) = C \rho^{l+1}$ solves your differential equation. You can do the same for the second term and you can also check that the sum of both terms works.

share|cite|improve this answer
Thanks. I couldn't see the wood for the trees. –  Peter4075 Jan 16 '13 at 20:00

Go on to simplify this expression (-l)(-l-1)=l(l+1). Factor this out, factor out to powers of rho and you're done.

share|cite|improve this answer

Another way to look at it is: suppose $u(\rho)$ is $e^{something}$. The differential equation tells us that the second derivative is $\frac{cu}{\rho^2}$ with $c=l(l+1)$. You can see that "something" should be a logarithm, because it's derivative is $1/\rho$, so suppose something equals $a\cdot \ln \rho$ . Just try to differentiate: $\frac{du}{d\rho}=\frac{a\cdot \exp(a\ln \rho)}{\rho}$. Differentiate this again: $\frac{d^2u}{d\rho^2}=-\frac{a\cdot \exp(a\ln \rho)}{\rho^2}+\frac{a^2\cdot \exp(a\ln \rho)}{\rho^2}= \exp(a\ln \rho)\left(\frac{a(a-1)}{\rho^2}\right)$. But $a(a-1)$ must equal $l(l+1)$. So $-l=a$. Also $l+1=a$ is a solution. So $u=A\exp(-l\ln\rho)+B \exp((l+1)\ln\rho)$. This is equal to: $u=A\rho^{-l}+B\rho^{l+1}$ because $e^{\ln x}=x$.

share|cite|improve this answer

This differential equation can be transformed into an ordinary differential equation with constant coefficients by means of the substitution $x = \mathrm{ln} \rho$. The reason for this is that the equation is homogeneous in $\rho$, i.e., invariant under the transformation $\rho \rightarrow \lambda \rho$ ($\lambda = \mathrm{const.}$).

share|cite|improve this answer
Afraid I don't understand that. But I got there eventually. –  Peter4075 Jan 16 '13 at 20:02
By the way, although someone has tagged this as Homework, it isn't for me. In my sixtieth decade, I'm simply trying to struggle through Griffiths' book. –  Peter4075 Jan 16 '13 at 20:24

Your Answer


