I'm having trouble understanding the recursion formula. Using $\xi \equiv \sqrt{m\omega/\hbar}x$ and $K = 2E/\hbar\omega$, the time-independent Schrödinger equation becomes

$$\frac{d^2\psi}{d\xi ^2} = (\xi ^2 - K ) \psi.$$

In the limit of large $\xi$ (and therefore large $x$), this differential equation becomes $$\frac{d^2\psi}{d\xi ^2} \approx \xi ^2 \psi.$$

Which has the approximate solution,

$$\psi(\xi) \approx Ae^{-\xi ^2 /2} + Be^{\xi ^2 /2}.$$ $B$ must be zero since $\psi$ must be normalizable (an exponential to a positive value blows up). My book then says that at large $\xi$, $$\psi = h(\xi)e^{-\xi ^2 /2}.$$

Question 1: why does $h$ have to be a function of $\xi$?

Plugging in this $\psi$ into the Schrodinger Eq, gives the Hermite differential equation. Then we use the power series method to generate $h(\xi)$ ($h = a_0 + a_1\xi + a_2\xi ^2 + ...$). The recursive formula that results is

$$a_{j+2} = \frac{2j + 1 - K}{(j+1)(j+2)}a_j.$$

At large $j$, $a_{j+2} \approx \frac{2}{j}a_j$, with the approximate solution $a_j \approx \frac{C}{(j/2)!}$ where $C$ is a constant.

Question 2: How was the approximate solution of $a_j$ found?

So $$h(\xi) \approx C\sum \frac{1}{(j/2)!}\xi ^j \approx C\sum \frac{1}{j!}\xi ^{2j}.$$

Question 3: How was that 2 moved into the exponent?

  • 2
    $\begingroup$ Hint for Q2: Apply the recursion formula until you reach $a_0$, then set $a_0 = C$. Hint for Q3: Simply redefine the summation variable as $j' := 2j$, and drop the $'$ (pay attention to the step size, though). $\endgroup$ – ACuriousMind Oct 5 '14 at 14:10
  • $\begingroup$ "My book then says that at large ξ,"... Are you certain about that? It should say something like given the large $\xi$ form, we guess a full solution of the form $h(\xi)e^{-\xi^2/2}$ $\endgroup$ – Alfred Centauri Oct 5 '14 at 14:49
  • 1
    $\begingroup$ @DWade64 I updated my answer significantly, to show you how the solution of the $a$'s comes about. $\endgroup$ – Danu Oct 5 '14 at 16:48

Answer to part 1: This is a common method of solving differential equations, employed by physicists to quickly extract a solution without having to make slow progress using more rigorous methods. The first step is to look for an asymptotic solution (i.e. the limit of large $\xi$ in this case), where the equation is easily solvable. Now, we know that this will not give us an exact solution, so we can't just proceed to claim $$\psi (\xi)=Ae^{-\xi^2/2}$$ However, since this solution must be exact in the limit, we can try to look for a solution of the form $$\psi(\xi)=\phi(\xi)e^{-\xi^2/2}$$ Which, we hope, may allow us to recast the original equation in a simpler or well-known form. Really, we've done nothing but use a change in variables, motivated by the asymptotic solution. Ideally, one would like to rigorously justify things like this (why exactly is it reasonable to believe that this makes things easier?) but, being a good physicist, we just push on and hope for the best ;-)

Answer to part 2: This is quite simple. To see that this is true, just try to take the limit $j\to \infty$. Since 1, 2 and $K$ are all constants, these eventually become negligible, so that we may write $$a_{j+2}=\frac{2j}{j^2}a_j$$ Which, after canceling a $j$, gives your formula. To solve this for $a_j$, follow ACuriousMind's advice (watch out: You cannot apply the recursion formula all the way to $a_0$, because $2/0$ is undefined) and realize that \begin{align*}a_3&=2a_1,\ a_4=a_2\\ a_5&=\frac{2}{3}a_3=\frac{2}{3}*2a_1,\hspace{1cm}a_6=\frac{1}{2}a_2\\ a_7&=\frac{2}{5}*\frac{2}{3}*\frac{2}{1}a_1\hspace{1cm}a_8=\frac{2}{6}*\frac{2}{4}*\frac{2}{2}a_2\end{align*}

$$ a_{2j+1}=\frac{2^j}{(2j+1)(2j-1)\dots1}a_1\hspace{1cm}a_{2j}=\frac{2^{j-1}}{2j(2j-2)\dots 2}a_2 $$ Now, for even terms $a_{2j}$, this is quite easily solved:$$a_{2j}=\frac{2^{j-1}}{2^jj!}a_2=\frac{a_2}{2}\frac{1}{j!}\implies a_k=\frac{C}{(k/2)!},\hspace{1cm}k=2j$$ Here, I used the fact that, for even $k=2n$, the double factorial $k!!$ can be easily expressed in terms of $n!$: $$k!!=2n(2n-2)*\dots *2=2(n)*2(n-1)*\dots*2(1)=2^n n!$$

For the odd terms, everything gets a bit messier. We first use the odd-numbered ($k=2n+1$) expression for $k!!$ in terms of $n!$: $$k!!=(2n+1)(2n-1)*\dots * 1=\frac{(2n+1)(2n)(2n-1)*\dots*1}{(2n)(2n-2)*\dots*2}=\frac{(2n+1)!}{2^n n!} $$ Plugging in this result, we have: $$a_{2j+1}=\frac{2^{2j}j!}{(2j+1)(2j)!}a_1 $$ Regrettably, it seems that this expression is not easy to evaluate (in terms of elementary functions). You'll need to use the gamma function instead, and most intuition is lost... Regardless, I hope that my treatment was sufficient to give you some intuition.


The previous answer does a great job of explaining the derivation up until the even and odd recursion formulae. The iterated coefficients are from 3 to 8 are correct, but if you check, the recursive formulae do not match the iterated examples. The correct recursion formulae are:



$$a_{2j+1}=\frac{2^{2j-1} (j-1)! a_1}{ (2j-1)!}$$

To derive the approximation for $$a_j$$, you can use either formula, but the even formula is easier. In case you would like to derive the other as well, the general process is the same. Remembering that the approximation is in the regime of very large $$j$$, you can invoke Stirling's approximation for very large factorials. This is:

$$n!\approx \sqrt[]{2\pi}\, n^{n+\frac{1}{2}}e^{-n}$$

So, for the even coefficients:

$$ a_{2j}\approx\\ \\ \frac{a_2}{\sqrt[]{2\pi}\, {(j-1)}^{(j-\frac{1}{2})} e^{(-j+1)}}\approx $$

$$\frac{a_2}{{\sqrt[]{2\pi}\,e^1{j}^{j}e^{-j}}} \approx$$


$$\frac{a_2}{{e\,(j)!}} = \frac{C}{{(j)!}}$$

Note that some of the fractions in the exponents were dropped or added since, in the approximation for very large numbers, these numbers become negligible; it has no effect on the approximation. Lastly, this approximation was carried out for the coefficient $$a_{2j}$$. To find it for any arbitrary coefficient, you can perform a change of variables:

$$n\equiv 2j$$


$$a_n = a_{2j} \approx \frac{C}{{(j)!}} = \frac{C}{{(\frac{n}{2})!}}$$

One could derive the same result for the odd formula, but it is a little bit of a lengthy derivation; it has the same procedure, but one extra step. Use Stirling's approximation (it will lead to the same form as the even approximation) and then perform a change of variables. Once the variables have been changed, it will require one more Stirling approximation.

As far as your third question:

$$h(\xi) = \sum\, a_n \xi^{n} \approx C\, \sum\, \frac{\xi^{n}}{{(\frac{n}{2})!}}$$

Performing a change of variables:

$$m\equiv \frac{n}{2}\,\,\, \Rightarrow\,\,\, 2m=n$$


$$h(\xi) = \sum\, a_n \xi^{n} \approx C\, \sum\, \frac{\xi^{2m}}{{(m)!}} $$

$$=\,C\, \sum\, \frac{({\xi^{2}})^{m}}{{(m)!}}\, =\, C\,e^{\xi^2}$$

Hope this helped!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.