Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

I'm trying to understand the calculation of spherical Bessel functions in chapter four of Griffiths' Introduction to Quantum Mechanics (2nd ed, p142). He gives $$j_{2}\left(x\right)=\left(-x\right)^{2}\left(\frac{1}{x}\frac{d}{dx}\right)^{2}\frac{\sin x}{x}=x^{2}\left(\frac{1}{x}\frac{d}{dx}\right)\frac{x\cos x-\sin x}{x^{3}}$$

$$=\frac{3\sin x-3x\cos x-x^{2}\sin x}{x^{3}}.$$

I can't see how he arrives at this answer. I think my problem is the $\left(\frac{1}{x}\frac{d}{dx}\right)^{2}$ bit (the general term for $j_{l}\left(x\right)$ is $\left(\frac{1}{x}\frac{d}{dx}\right)^{l}$ ). I'm assuming this means $1/x^{2}$ multiplied by the second derivative of $\frac{\sin x}{x}$ but I make that $$\left(\frac{1}{x^{2}}\right)\left(-\frac{\sin x}{x}+\frac{2\sin x}{x^{3}}-\frac{2\cos x}{x^{2}}\right).$$

Any idea what I'm doing wrong?

share|improve this question

1 Answer 1

up vote 6 down vote accepted

No. The term $\left(\frac1x \frac d{dx}\right)^2$ should be understood to mean the operator $$\left(\frac1x \frac d{dx}\right)^2=\frac1x \frac d{dx}\frac1x \frac d{dx}.$$ Otherwise one would simply say $\frac1{x^2} \frac {d^2}{dx^2}$. Check the derivation of the formula you use for the spherical Bessel functions to get a feel for why this must be the case.

share|improve this answer
Thanks. So, in this case, you'd simply take the derivative, then multiply by 1/x, then take the derivative again and then multiply by 1/x? Afraid I'm not too familiar with operators. –  Peter4075 Jan 15 '13 at 17:28
Yes Peter, thats what you do. $\frac{1}{x}\frac{d}{dx} \frac{1}{x} \frac{d}{dx} \phi= \frac{1}{x}\frac{d}{dx} \frac{1}{x} \phi^\prime $. Mind that you need to apply of the product rule for derivations when you use the second $\frac{d}{dx}$ term –  elcojon Jan 15 '13 at 17:58

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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