# Free particle moving in the z direction

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.6.10.

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The radial function for a free particle can be either a spherical Bessel function ${j_{l}}$ or a spherical Neumann function ${n_{l}}$. If the solution space includes the origin, then only ${j_{l}}$ is acceptable since the ${n_{l}}$ functions diverge as ${r\rightarrow0}$.

In rectangular coordinates, a free particle wave function has the form

$\displaystyle \psi_{E}\left(x,y,z\right)=\frac{1}{\left(2\pi\hbar\right)^{3/2}}e^{i\mathbf{p}\cdot\mathbf{r}/\hbar} \ \ \ \ \ (1)$

where the energy ${E}$ is

$\displaystyle E=\frac{p^{2}}{2\mu}=\frac{\hbar^{2}k^{2}}{2\mu} \ \ \ \ \ (2)$

For a free particle travelling in the ${z}$ direction, this becomes

$\displaystyle \psi_{E}\left(r,\theta,\phi\right)=\frac{1}{\left(2\pi\hbar\right)^{3/2}}e^{ikr\cos\theta} \ \ \ \ \ (3)$

since ${z=r\cos\theta}$.

Since the solutions of the free-particle Schrödinger equation in spherical coordinations form a complete set, we must be able to express this wave function as a linear combination of these solutions, so that

$\displaystyle e^{ikr\cos\theta}=\sum_{l=0}^{\infty}\sum_{m=-l}^{l}C_{l}^{m}j_{l}\left(kr\right)Y_{l}^{m}\left(\theta,\phi\right) \ \ \ \ \ (4)$

where the ${C_{l}^{m}}$ are constants. Because we’re looking at motion in the ${z}$ direction, there is no angular momentum about the ${z}$ axis, which is reflected in the fact that ${\psi_{E}}$ does not depend on ${\phi}$. Thus ${L_{z}=m\hbar=0}$ and ${m=0}$. We therefore have

 $\displaystyle e^{ikr\cos\theta}$ $\displaystyle =$ $\displaystyle \sum_{l=0}^{\infty}C_{l}^{0}j_{l}\left(kr\right)Y_{l}^{0}\left(\theta,\phi\right)\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{l=0}^{\infty}\sqrt{\frac{2l+1}{4\pi}}C_{l}^{0}j_{l}\left(kr\right)P_{l}\left(\cos\theta\right)\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{l=0}^{\infty}C_{l}j_{l}\left(kr\right)P_{l}\left(\cos\theta\right) \ \ \ \ \ (7)$

where

$\displaystyle C_{l}\equiv\sqrt{\frac{2l+1}{4\pi}}C_{l}^{0} \ \ \ \ \ (8)$

The problem, of course, is to find these constants. We can do this using the identities given by Shankar in his problem 12.6.10, which are

 $\displaystyle \int_{-1}^{1}P_{l}\left(x\right)P_{l^{\prime}}\left(x\right)dx$ $\displaystyle =$ $\displaystyle \frac{2\delta_{ll^{\prime}}}{2l+1}\ \ \ \ \ (9)$ $\displaystyle P_{l}\left(x\right)$ $\displaystyle =$ $\displaystyle \frac{1}{2^{l}l!}\frac{d^{l}\left(x^{2}-1\right)^{l}}{dx^{l}}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\left(-1\right)^{l}}{2^{l}l!}\frac{d^{l}\left(1-x^{2}\right)^{l}}{dx^{l}}\ \ \ \ \ (11)$ $\displaystyle \int_{0}^{1}\left(1-x^{2}\right)^{m}dx$ $\displaystyle =$ $\displaystyle \frac{\left(2m\right)!!}{\left(2m+1\right)!!}\ \ \ \ \ (12)$ $\displaystyle \int_{-1}^{1}\left(1-x^{2}\right)^{m}dx$ $\displaystyle =$ $\displaystyle \frac{2\left(2m\right)!!}{\left(2m+1\right)!!} \ \ \ \ \ (13)$

The last line follows because ${\left(1-x^{2}\right)^{m}}$ is an even function and is therefore symmetric about ${x=0}$.

We can use the standard procedure for isolating ${C_{l}}$ by multiplying both sides by ${C_{a}}$ and using 9.

 $\displaystyle \int_{-1}^{1}P_{a}\left(x\right)e^{ikrx}dx$ $\displaystyle =$ $\displaystyle \sum_{l=0}^{\infty}C_{l}j_{l}\left(kr\right)\int_{-1}^{1}P_{a}\left(x\right)P_{l}\left(x\right)dx\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2}{2a+1}C_{a}j_{a}\left(kr\right) \ \ \ \ \ (15)$

This relation must be true for all values of ${r}$, so we can look at the limit of small (but not zero, since both sides are then zero) ${r}$. We have the asymptotic relation for the spherical Bessel functions

 $\displaystyle j_{l}$ $\displaystyle \underset{\rho\rightarrow0}{\longrightarrow}$ $\displaystyle \frac{\rho^{l}}{\left(2l+1\right)!!} \ \ \ \ \ (16)$

We thus have

$\displaystyle \int_{-1}^{1}P_{a}\left(x\right)e^{ikrx}dx=\underset{r\rightarrow0}{\longrightarrow}\frac{2}{2a+1}\frac{k^{a}r^{a}}{\left(2a+1\right)!!}C_{a} \ \ \ \ \ (17)$

We can then look at the integral on the LHS and hope that, when we expand the exponential, that the terms in ${\left(kr\right)^{n}}$ for ${n vanish. We can then match the coefficients of ${\left(kr\right)^{a}}$ on both sides to find ${C_{a}}$.

We can see that this will work because the Legendre polynomials ${P_{l}}$ are a complete set of functions, and the polynomial ${P_{l}}$ has degree ${l}$. This means that any polynomial of degree ${a-1}$ can be written as a linear combination of the ${P_{l}}$, where ${l=0,\ldots,a-1}$. Because of 9, this means that

$\displaystyle \int_{-1}^{1}x^{l}P_{a}\left(x\right)dx=0\;\;\mbox{if \ensuremath{l

Therefore, when we expand ${e^{ikrx}}$ in a power series, we have

 $\displaystyle \int_{-1}^{1}P_{a}\left(x\right)e^{ikrx}dx$ $\displaystyle =$ $\displaystyle \int_{-1}^{1}P_{a}\left(x\right)\left(1+ikrx+\frac{\left(ikrx\right)^{2}}{2!}+\ldots\right)dx\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int_{-1}^{1}P_{a}\left(x\right)\left(\frac{\left(ikrx\right)^{a}}{a!}+\ldots\right)dx \ \ \ \ \ (20)$

In the limit of small ${r}$, higher order terms in the sum on the RHS can be ignored, so we get

 $\displaystyle \frac{\left(ikr\right)^{a}}{a!}\int_{-1}^{1}x^{a}P_{a}\left(x\right)dx$ $\displaystyle =$ $\displaystyle \frac{2}{2a+1}\frac{k^{a}r^{a}}{\left(2a+1\right)!!}C_{a}\ \ \ \ \ (21)$ $\displaystyle C_{a}$ $\displaystyle =$ $\displaystyle \frac{i^{a}\left(2a+1\right)\left(2a+1\right)!!}{2a!}\int_{-1}^{1}x^{a}P_{a}\left(x\right)dx \ \ \ \ \ (22)$

Now consider the integral in the last line. Using 11 we have

$\displaystyle \int_{-1}^{1}x^{a}P_{a}\left(x\right)dx=\frac{\left(-1\right)^{a}}{2^{a}a!}\int_{-1}^{1}x^{a}\frac{d^{a}\left(1-x^{2}\right)^{a}}{dx^{a}}dx \ \ \ \ \ (23)$

We can integrate by parts repeatedly until the derivative in the integrand disappears. Note that the ${n}$th derivative of ${\left(1-x^{2}\right)^{a}}$ will always contain a factor of ${\left(1-x^{2}\right)}$ to some power for any ${n, and thus is zero at both limits of integration. Since the integrated term in the integration by parts always contains such a derivative, all integrated terms are zero at both limits. We therefore integrate ${\frac{d^{a}\left(1-x^{2}\right)^{a}}{dx^{a}}}$ (${a}$ times) and differentiate ${x^{a}}$ (${a}$ times) and keep only the residual integral after each iteration. The differentiation of ${x^{a}}$ (${a}$ times) introduces a factor of ${a!}$. Since the sign of the residual integral alternates as we perform each integration by parts, the final result is

 $\displaystyle \int_{-1}^{1}x^{a}P_{a}\left(x\right)dx$ $\displaystyle =$ $\displaystyle \frac{\left(-1\right)^{2a}}{2^{a}a!}a!\int_{-1}^{1}\left(1-x^{2}\right)^{a}dx\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2^{a}}\frac{2\left(2a\right)!!}{\left(2a+1\right)!!} \ \ \ \ \ (25)$

where we used 13 in the last line. The double factorial in the numerator can be written as

 $\displaystyle \left(2a\right)!!$ $\displaystyle =$ $\displaystyle \left(2a\right)\left(2a-2\right)\ldots\left(4\right)\left(2\right)\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2^{a}a\left(a-1\right)\ldots\left(2\right)\left(1\right)\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2^{a}a! \ \ \ \ \ (28)$

We therefore have

 $\displaystyle \int_{-1}^{1}x^{a}P_{a}\left(x\right)dx$ $\displaystyle =$ $\displaystyle \frac{1}{2^{a}}\frac{2\times2^{a}a!}{\left(2a+1\right)!!}\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2a!}{\left(2a+1\right)!!} \ \ \ \ \ (30)$

Plugging this back into 22 we have

$\displaystyle C_{a}=i^{a}\left(2a+1\right) \ \ \ \ \ (31)$

The wave function for a free particle moving in the ${z}$ direction is therefore

$\displaystyle \psi_{E}\left(r,\theta,\phi\right)=\frac{1}{\left(2\pi\hbar\right)^{3/2}}\sum_{l=0}^{\infty}i^{a}\left(2a+1\right)j_{l}\left(kr\right)P_{l}\left(\cos\theta\right) \ \ \ \ \ (32)$