Quantum Mechanics Integration Problem

[CrusaderSean]

I'm doing Griffith Problem 5.11 and I'm stuck on how to do the integration. My prof. gave hints on how to simplify the integration with the following formula:

$$\frac{1}{|{\vec r}_{1}-{\vec r}_{2}|} = \sum_{\ell = 0}^{\infty}<br /> P_{\ell}( \cos \theta) \frac{r_{<}^{\ell}}{r_{>}^{\ell +1}}$$
where
$$\cos \theta = \frac{{\vec r}_1 \cdot {\vec r}_2}{r_1 r_2}$$

$$P_{\ell}(x)=\frac{1}{2^{\ell}\ell !}(\frac{d}{dx})^{\ell}(x^{2}-1)^{\ell}$$

$$r_<$$ is the smaller of $$|{\vec r}_1|, |{\vec r_2}|$$

I haven't had advanced electrodynamics yet so it's my first time seeing this type of expansion.

Since I don't really know if $$\vec r}_1$$ is bigger or smaller than $$\vec r}_2$$, do I need to do expansion for both cases and add them?...

How many terms do I keep in Legendre polynomial? I recall professor saying something about only one or two terms, but I don't see how you can truncate the series like that...

 

[nrqed]

I'm doing Griffith Problem 5.11 and I'm stuck on how to do the integration. My prof. gave hints on how to simplify the integration with the following formula:

$$\frac{1}{|{\vec r}_{1}-{\vec r}_{2}|} = \sum_{\ell = 0}^{\infty}<br /> P_{\ell}( \cos \theta) \frac{r_{<}^{\ell}}{r_{>}^{\ell +1}}$$
where
$$\cos \theta = \frac{{\vec r}_1 \cdot {\vec r}_2}{r_1 r_2}$$

$$P_{\ell}(x)=\frac{1}{2^{\ell}\ell !}(\frac{d}{dx})^{\ell}(x^{2}-1)^{\ell}$$

$$r_<$$ is the smaller of $$|{\vec r}_1|, |{\vec r_2}|$$

I haven't had advanced electrodynamics yet so it's my first time seeing this type of expansion.

Since I don't really know if $$\vec r}_1$$ is bigger or smaller than $$\vec r}_2$$, do I need to do expansion for both cases and add them?...

How many terms do I keep in Legendre polynomial? I recall professor saying something about only one or two terms, but I don't see how you can truncate the series like that...

Well, it *is* possible to do the integral without ever introducing the above expansion! Does he require you to use it?

In either case, you *have* to break the integral in wto cases acoording to whether r_2 is larger than r_1 or smaller. There is no way around this.

If you have to use the abpve formula then you have to write the rest of the theta dependence of the integrand in terms of some P_l(cos theta). You will surely get just a few terms, probably a linear combination of P_0 and P_1. Then you can use orthonormality of the P_l to do the integral over theta easily.

But if you have never used those P_l, it would be simpler to do the integral directly.

 

[CrusaderSean]

well I don't *have* to use the expansion to do the integral. I guess I'll the integral directly first and see if I can understand how this expansion works (suppose to make the integral easier).