Daily Archives: Mon, 3 July 2017

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<a} 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<a}} \ \ \ \ \ (18)

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<a}, 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)