Quantum Mechanics in Three-Dimensions: The Radial Equation

Hananish Joy G. Odarve and Majvell Kay G. Odarve

The wavefunction, or quantum state, is a complete description that can be given into a physical system. The Schrodinger equation can describe how the wavefunction changes as time propagates.

A particle state, for example, can be determined by solving the Schrodinger equation. Since the potential of a system, $V(r)$ , is a function of the distance from the origin, the shperical coordinates $(r, \;\theta, \;\phi)$ can be employed . The time-independent Schrodinger equation in spherical coordinates has the expression

$\frac{-\hbar^2}{2m}[{\frac{1}{r^2}}\frac{\partial}{\partial r}\;(r^2 \frac{\partial \Psi}{\partial r})$ $+\frac{1}{{r^2}\sin \theta}\;\frac{\partial}{\partial \theta}(\sin \theta \frac{\partial \Psi}{\partial \theta})$ $+\frac{1}{{r^2}\sin^2 \theta}(\frac{\partial^2 \Psi}{\partial \phi^2})] + V\Psi = E\Psi$ (1)

where $\Psi$ is a separable solution to Eqn.(1) given as

$\Psi (r,\theta,\phi) = R(r)Y(\theta,\phi)$ .

Performing the separation of variables leads to the determination of the angular and the radial equations given, respectively as

$\frac{1}{Y}\left[\frac{1}{\sin \theta}\frac{\partial}{\partial \theta}(\sin \theta \frac{\partial Y}{\partial \theta})+\frac{1}{\sin^2 \theta}\frac{\partial^2 Y}{\partial \phi^2}\right]= -l(l+1)$ (2)

and

$\frac{1}{R}\frac{d}{dr}(r^2\frac{dR}{dr})-\frac{2mr^2}{\hbar^2}\left(V(r)-E\right)=l(l+1)\;.$ (3)

The actual shape of a given potential, $V(r)$ , only affects the radial part of the wavefunction, $R(r)$ , which can be determined from Eqn. (3). This can be further be simplified by changing of variables where we let

$U(r)=rR(r),$

leading to the general expression of the radial equation,

$\frac{-\hbar^2}{2m}\;\frac{d^2U}{dr^2}+\left[V(r)+\frac{\hbar^2}{2m}\;\frac{l(l+1)}{r^2}\right]U = EU.$ (4)

From here, we can compute the ground state of a particle with $l=0$ . To demonstrate this, we can study a particle of mass m placed in a finite spherical well having a potential

$V(r) = \left\{<br /> \begin{array}{rl}<br /> -V_o, & if\;\;\;r\leq a\\<br /> 0, & if\;\;\;r > a<br /> \end{array} \right$ .

We also have to show that no bound states exists if

$V_o\;\;a^2\;\;<\;\;\frac{\pi^2\hbar^2}{8m}.$

Now, the Schrodinger equation yields bound states when $E < V_o$ . We first investigate the region at $r\leq a$ . Eqn. (4) with $l=0$ becomes,

$\frac{-\hbar^2}{2m}\;\frac{d^2U}{dr^2} - V_o U = EU$

$\frac{d^2U}{dr^2} = -\frac{2m}{\hbar^2}(E+V_o)U$

where we let $\epsilon = \sqrt{{\frac{2m}{\hbar^2}}(E+V_o)}$ . Now we have,

$\frac{d^2U}{dr^2}+\epsilon^2 U = 0$

The general solution for this expression is given as

$U_{in}(r)=A\sin(\epsilon r)+ B\cos(\epsilon r).$

Thus, the radial equation inside the finite spherical well can be expressed as

$R_{in}(r)=\frac{U_{in}(r)}{r}= \frac{A\sin(\epsilon r)}{r}+ \frac{B\cos(\epsilon r)}{r}.$ (5)

We impose the boundary condition that as $r \rightarrow 0$ , the potential has to have a finite value. However, the term $\frac{B\cos(\epsilon r)}{r}$ blows up so we set B=0. Thus,

$R_{in}(r)= \frac{A\sin(\epsilon r)}{r}.$ (6)

Next, we investigate the region where $r>a$ . In this region, $V(r)$ is zero so Eqn. (4) becomes

$\frac{-\hbar^2}{2m}\frac{d^2U}{dr^2} = EU$

$\frac{d^2U}{dr^2} = -\frac{2m}{\hbar^2}EU.$

We then let $\beta = \sqrt{-\frac{2mE}{\hbar^2}}$ so we have

$\frac{d^2U}{dr^2}-\beta^2 U = 0.$

The general solution for this expression is

$U_{out}(r)=Ce^{\beta r}+ De^{-\beta r}.$ (7)

The radial equation outside the spherical well is then expressed as

$R_{out}(r)=\frac{U_{out}(r)}{r}= \frac{Ce^{\beta r}}{r}+ \frac{De^{-\beta r}}{r}.$

From here, we impose another boundary condition that as $r\rightarrow\infty$ the potential must be finite. However, the term $e^{\beta r} \rightarrow\infty$ , so we set $C=0$ . Thus,

$R_{out}(r)=\frac{De^{-\beta r}}{r}.$ (8)

The continuity of $R$ and $\frac{dR}{dr}$ at the interface region, $r=a$ , requires that,

i. $R_{in}(r) = R_{out}(r)$ and

ii. $\frac{dR_{in}(r)}{dr}$ = $\frac{dR_{out}(r)}{dr}$

From condition (i):

$\frac{A\sin(\epsilon a)}{a}=\frac{De^{-\beta a}}{a}$

$A\sin(\epsilon a) = {De^{-\beta a}}.$ (9)

and from condition (ii):

$\frac{dR_{in}(r)}{dr} = \frac{Aa\epsilon \cos(\epsilon a) - A \sin (\epsilon a)}{a^2}$

$\frac{dR_{out}(r)}{dr} = \frac{-Da\beta e^{-\beta a} - De^{-\beta a}}{a^2}.$

equating $\frac{dR_{in}(r)}{dr}$ and $\frac{dR_{out}(r)}{dr}$ ,

$Aa\epsilon \cos(\epsilon a) - A \sin (\epsilon a)=-Da\beta e^{-\beta a} - De^{-\beta a}$

$A(a\epsilon \cos(\epsilon a) - \sin (\epsilon a))=-De^{-\beta a}(a\beta +1).$ (10)

We divide Eqn. (10) by (9),

$\frac{a\epsilon \cos(\epsilon a) - \sin (\epsilon a)}{\sin\epsilon a} = -(a\beta+1)$

$a\epsilon\frac{\cos{\epsilon a}}{\sin{\epsilon a}} - 1= -a\beta - 1$

$a\epsilon \cot(\epsilon a) = -a\beta$

we let $k_1 = \epsilon a$ and $k_2 = \beta a$ so,

$k_1 \cot(k_1) = -k_2.$ (11)

From our representation that $\epsilon = \frac{k_1}{a}$ and $\beta=\frac{k_2}{a}$ , it tells us that the value of $k_1$ and $k_2$ should be positive.

$k_1^2 + k_2^2 =\epsilon^2 a^2 + \beta^2 a^2 = a^2 \left[\frac{2m}{\hbar^2} (E+V_o) + (\frac{-2mE}{\hbar^2})\right]$ $= \frac{a^2}{\hbar^2} 2m (E+V_o-E)$

$R^2 = \frac{2ma^2}{\hbar^2} V_o$

From the results, the allowable states only occurs when $k_1$ and $k_2$ are positive. That argumnet would only be possible if the constants are located in the first quadrant. The solutions thus can only be found at $R<\frac{\pi}{2}$ .

$\frac{2ma^2}{\hbar^2}V_o\;\;<\;\;\frac{\pi^2}{4}$

$V_o a^2\;\;<\;\;\frac{\hbar^2}{2m}\frac{\pi^2}{4}$

$V_o a^2\;\;<\;\;\frac{\pi^2 \hbar^2}{8m}.$

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One Response to “Quantum Mechanics in Three-Dimensions: The Radial Equation”

Angie Says:
February 10th, 2010 at 9:04 pm
Clean, nice presentation of equations! I would like to see more of some hidden, convincing arguments like why k1 and k2 should be chosen positive etc…