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Some Notes on taking the limit to infinite systems

K.E. Schmidt
Department of Physics and Astronomy
Arizona State University
Tempe, AZ U.S.A.

Many of you are having trouble taking the limits of a finite system to infinity and converting the sums to the corresponding integrals especially when Bessel functions are involved.

Let's look at examples of taking the limit of two simple systems as their size goes to infinity and converting the corresponding sum of basis functions to an integral. We covered the simplest example in class, the usual coulomb Green's function for free space,

G($\displaystyle\vec{r}$,$\displaystyle\vec{r}$') = $\displaystyle{\textstyle\frac{1}{\vert\vec r - \vec r'\vert}}$, (1)

written as a Fourier transform.

In the usual case, we imagine a cubic box of side L $\rightarrow$ $\infty$ , with periodic boundary conditions. The Laplacian separates in cartesian coordinates, and we need to solve the three equations

$\displaystyle{\frac{d^2 f(x_i)}{dx_i^2}}$ + ki2fn(xi) = 0 . (2)

The normalized eigenfunctions and eigenvalues of $\nabla$'2 which satisfy our boundary conditions are

$\displaystyle\Psi_{mno}^{}$ = $\displaystyle{\textstyle\frac{1}{L^{3/2}}}$exp(i$\displaystyle\vec{q}_{mno}^{}$$\displaystyle\vec{r}$') (3)

where

$\displaystyle\vec{q}_{mno}^{}$ = $\displaystyle{\frac{2 \pi}{L}}$(m$\displaystyle\hat{x}$ + n$\displaystyle\hat{y}$ + o$\displaystyle\hat{z}$) , (4)

and m , n , o are integers from minus infinity to infinity.

The Green's function is

G($\displaystyle\vec{r}$,$\displaystyle\vec{r}$') = 4$\displaystyle\pi$$\displaystyle\sum_{mno}^{}$$\displaystyle{\frac{ \Psi_{mno}(\vec r) \Psi_{mno}^*(\vec r')}{q_{mno}^2}}$ (5)

which can be verified by direct substitution into the differential equation for the Green's function and the boundary conditions. Plugging in the explicit form for $\Psi_{mno}^{}$ ,

G($\displaystyle\vec{r}$,$\displaystyle\vec{r}$') = $\displaystyle{\frac{4 \pi}{L^3}}$$\displaystyle\sum_{mno}^{}$$\displaystyle{\frac{\exp(i \vec q_{mno} \cdot [\vec r - \vec r'])}{\vert\vec q_{mno}\vert^2}}$ (6)

In the limit of L $\rightarrow$ $\infty$ for any nonzero q , the integers m , n , o go to infinity, and the spacing between adjacent q values goes to zero. This means we can convert the sum to an integral, and for sums spaced by 1,

$\displaystyle\sum_{n}^{}$ $\displaystyle\rightarrow$ $\displaystyle\int$dn (7)

The Green's function becomes

G($\displaystyle\vec{r}$,$\displaystyle\vec{r}$') = $\displaystyle{\frac{4 \pi}{L^3}}$$\displaystyle\int$ dn dm do $\displaystyle{\frac{\exp(i \vec q_{mno} \cdot [\vec r - \vec r'])}{\vert\vec q_{mno}\vert^2}}$ . (8)

Upon writing
dn = dqx$\displaystyle{\frac{dn}{dq_x}}$ = dqx$\displaystyle{\frac{L}{2\pi}}$   
dm = dqy$\displaystyle{\frac{dn}{dq_y}}$ = dqy$\displaystyle{\frac{L}{2\pi}}$   
do = dqz$\displaystyle{\frac{dn}{dq_z}}$ = dqz$\displaystyle{\frac{L}{2\pi}}$ (9)
we have the usual replacement

dn dm do  = $\displaystyle{\frac{L^3}{(2 \pi)^3}}$d 3q . (10)

The Green's function is

 
G($\displaystyle\vec{r}$,$\displaystyle\vec{r}$') = $\displaystyle{\frac{4 \pi}{(2 \pi)^3}}$$\displaystyle\int$d 3q$\displaystyle{\frac{\exp(i \vec q \cdot [\vec r - \vec r'])}{\vert\vec q\vert^2}}$ (11)

The integration is straightforward. Taking $\vec{r}$ - $\vec{r}$' as the z axis for the q coordinate system, the angular integrations are easy and the result is

G($\displaystyle\vec{r}$,$\displaystyle\vec{r}$') = $\displaystyle{\textstyle\frac{2}{\pi}}$$\displaystyle\int_{0}^{\infty}$dq$\displaystyle{\frac{\sin( q \vert\vec r- \vec r'\vert)}{q \vert\vec r - \vec r'\vert}}$ . (12)

Changing variables from q to q|$\vec{r}$ - $\vec{r}$'| , gives

G($\displaystyle\vec{r}$,$\displaystyle\vec{r}$') = $\displaystyle{\textstyle\frac{2}{\pi \vert\vec r - \vec r'\vert}}$$\displaystyle\int_{0}^{\infty}$du$\displaystyle{\frac{\sin(u)}{u}}$ , (13)

which shows the Green's function is proportional to 1/|$\vec{r}$ - $\vec{r}$'| .

The last integral

I = $\displaystyle\int_{0}^{\infty}$du$\displaystyle{\frac{\sin(u)}{u}}$ (14)

is easily done using contour integration. Since it is symmetric in u , we can take half the integral from - $\infty$ to $\infty$ . Since the integrand is analytic at the origin, we can distort the contour to go around the origin with an infinitesimally small semicircle in the upper half plane. Replacing the sin by (exp(iu) - exp (- iu))/(2i) , we note that we can close the contour with a large semicircle at infinity in the upper half plane for the first term and in the lower half plane for the second term giving no contribution on these parts. The second integral encloses the pole at the origin and the result of the contour integration is

I = $\displaystyle{\frac{\pi}{2}}$ (15)

and the Green's function is

G($\displaystyle\vec{r}$,$\displaystyle\vec{r}$') = $\displaystyle{\textstyle\frac{1}{\vert\vec r - \vec r'\vert}}$ (16)

as expected.

Now that we did it the easy way, let's use a harder method that may help in taking other limits. Let's again calculate the free space Green's function, but let's take the limit using Dirichlet boundary conditions on a sphere of radius R that we will let go to infinity. Separating variables as in class, the eigenfunctions of $\nabla^{2}_{}$ are

$\displaystyle\Psi_{mno}^{}$($\displaystyle\vec{r}$) = $\displaystyle\sqrt{\frac{2}{R^3 j^2_{l+1}(q_{lo}R)}}$jl(qlor)Ylm($\displaystyle\theta$,$\displaystyle\phi$) (17)

where qloR is the o th zero of jl , and the normalization integral evaluated by substituting into the corresponding result for Jm Bessel functions derived in Jackson and the homework.

The Green's function is

G($\displaystyle\vec{r}$,$\displaystyle\vec{r}$') = 4$\displaystyle\pi$$\displaystyle\sum_{lmo}^{}$$\displaystyle{\textstyle\frac{2}{q_{lo}^2 R^3 j^2_{l+1}(q_{lo}R)}}$jl(qlor)jl(qlor')Ylm($\displaystyle\theta$,$\displaystyle\phi$)Y *lm($\displaystyle\theta$',$\displaystyle\phi$') , (18)

where the l sum runs from 0 to $\infty$ , the o sum from 1 to $\infty$ , and the m sum from - l to l .

In the limit of R $\rightarrow$ $\infty$ for any nonzero q value, the integer o must go to infinity, and the spacing between adjacent q values goes to zero. This means the the o sum can be converted to an integral. Just as in the cartesian case, we need to calculate do/dql to change to an integration over ql . Since o goes to $\infty$ for any finite ql value, the Bessel function zeroes can be evaluated for large argument, since the small arguments contribute an amount of measure zero as R $\rightarrow$ $\infty$ .

The assymptotic expansion of jl(x) is

jl(x) $\displaystyle\rightarrow$ $\displaystyle{\textstyle\frac{1}{x}}$sin(x - l$\displaystyle\pi$/2) , (19)

so that the zeroes are

qlo = $\displaystyle{\frac{\pi}{2R}}$(l + 2o) (20)

and dql/do = $\pi$/R . Plugging into the Green's function expression, using the assymptotic expansion of jl + 1 ,

G($\displaystyle\vec{r}$,$\displaystyle\vec{r}$') = 8$\displaystyle\pi^{2}_{}$$\displaystyle\sum_{lm}^{}$$\displaystyle\int_{0}^{\infty}$dqljl(qlr)jl(qlr')Ylm($\displaystyle\theta$,$\displaystyle\phi$)Y *lm($\displaystyle\theta$',$\displaystyle\phi$') . (21)

Notice that the ql integral is a dummy variable, so we can just write q instead of ql , The result is

 
G($\displaystyle\vec{r}$,$\displaystyle\vec{r}$') = 8$\displaystyle\sum_{lm}^{}$$\displaystyle\int_{0}^{\infty}$dqjl(qr)jl(qr')Ylm($\displaystyle\theta$,$\displaystyle\phi$)Y *lm($\displaystyle\theta$',$\displaystyle\phi$') . (22)

Rather than perform the integration in Eq. 22, we can verify that this is the correct expression by comparing with our previous result. Jackson Eq. 10.43 gives the expansion of a plane wave in terms of spherical Bessel functions,

exp(i$\displaystyle\vec{k}$ $\displaystyle\cdot$ $\displaystyle\vec{r}$) = 4$\displaystyle\pi$$\displaystyle\sum_{lm}^{}$i ljl(kr)Ylm($\displaystyle\theta$,$\displaystyle\phi$)Ylm*($\displaystyle\theta_{k}^{}$,$\displaystyle\phi_{k}^{}$) (23)

where $\theta_{k}^{}$ and $\phi_{k}^{}$ are the spherical angles for the $\vec{k}$ vector. Note however, that his method is essentially the same as mine above in that he first calculates the Green's function for Helmholz equation.

We can therefore write

exp(i$\displaystyle\vec{q}$ $\displaystyle\cdot$ [$\displaystyle\vec{r}$ - $\displaystyle\vec{r}$']) = (4$\displaystyle\pi$)2$\displaystyle\sum_{lm}^{}$i ljl(qr)Ylm($\displaystyle\theta$,$\displaystyle\phi$)Ylm*($\displaystyle\theta_{q}^{}$,$\displaystyle\phi_{q}^{}$)$\displaystyle\sum_{l'm'}^{}$(- i)l'jl'(qr')Y *l'm'($\displaystyle\theta$',$\displaystyle\phi$')Yl'm'($\displaystyle\theta_{q}^{}$,$\displaystyle\phi_{q}^{}$) (24)

and substituting into Eq. 11 and integrating over the angular q variables gives Eq. 22.

It is also possible to do the integrals, but we know the result that must occur, and that means the integrals will be somewhat ugly. For example for l = 0 , the integral can be done readily with contour integration similarly to the contour integral above to give

$\displaystyle\int_{0}^{\infty}$dq$\displaystyle{\frac{\sin(qr) \sin(q r')}{ q^2 r r'}}$ = $\displaystyle{\frac{\pi}{4 r r'}}$$\displaystyle\left [ r + r' - \vert r -r'\vert \right]$   
  = $\displaystyle{\frac{\pi }{2 r_\gt}}$ (25)
and we recover the standard l = 0 term in the spherical harmonic expansion of 1/|$\vec{r}$ - $\vec{r}$'| .

The other integrals could be done similarly, but this is a hard way of getting the standard result.


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