Quantum Mechanics in Three-Dimensions: The Radial Equation | Quantum Science Philippines

## 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, [eq] V(r) [/eq],  is a function of the distance from the origin, the shperical coordinates [eq] (r, \;\theta, \;\phi) [/eq] can be employed . The time-independent Schrodinger equation in spherical coordinates has the expression

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

where [eq]\Psi[/eq] is a separable solution to Eqn.(1) given as

[eq]\Psi (r,\theta,\phi) = R(r)Y(\theta,\phi)[/eq].

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

[eq]\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)[/eq]                                  (2)

and

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

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

[eq]U(r)=rR(r),[/eq]

[eq]\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.[/eq]                              (4)

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

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

We also have to show that no bound states exists if

[eq]V_o\;\;a^2\;\;<\;\;\frac{\pi^2\hbar^2}{8m}.[/eq]

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

[eq]\frac{-\hbar^2}{2m}\;\frac{d^2U}{dr^2} – V_o U = EU[/eq]

[eq]\frac{d^2U}{dr^2} = -\frac{2m}{\hbar^2}(E+V_o)U[/eq]

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

[eq]\frac{d^2U}{dr^2}+\epsilon^2 U = 0[/eq]

The general solution for this expression is given as

[eq]U_{in}(r)=A\sin(\epsilon r)+ B\cos(\epsilon r).[/eq]

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

[eq]R_{in}(r)=\frac{U_{in}(r)}{r}= \frac{A\sin(\epsilon r)}{r}+ \frac{B\cos(\epsilon r)}{r}.[/eq]                                   (5)

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

[eq]R_{in}(r)= \frac{A\sin(\epsilon r)}{r}.[/eq]                                                                          (6)

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

[eq]\frac{-\hbar^2}{2m}\frac{d^2U}{dr^2} = EU[/eq]

[eq]\frac{d^2U}{dr^2} = -\frac{2m}{\hbar^2}EU.[/eq]

We then let [eq]\beta = \sqrt{-\frac{2mE}{\hbar^2}}[/eq] so we have

[eq]\frac{d^2U}{dr^2}-\beta^2 U = 0.[/eq]

The general solution for this expression is

[eq]U_{out}(r)=Ce^{\beta r}+ De^{-\beta r}.[/eq]                                  (7)

The radial equation outside the spherical well  is then expressed as

[eq]R_{out}(r)=\frac{U_{out}(r)}{r}= \frac{Ce^{\beta r}}{r}+ \frac{De^{-\beta r}}{r}.[/eq]

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

[eq]R_{out}(r)=\frac{De^{-\beta r}}{r}.[/eq]                                                                       (8)

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

i. [eq]R_{in}(r) = R_{out}(r)[/eq] and

ii. [eq]\frac{dR_{in}(r)}{dr}[/eq] = [eq]\frac{dR_{out}(r)}{dr}[/eq]

From condition (i):

[eq]\frac{A\sin(\epsilon a)}{a}=\frac{De^{-\beta a}}{a}[/eq]

[eq]A\sin(\epsilon a) = {De^{-\beta a}}.[/eq]                                                (9)

and from condition (ii):

[eq]\frac{dR_{in}(r)}{dr} = \frac{Aa\epsilon \cos(\epsilon a) – A \sin (\epsilon a)}{a^2}[/eq]

[eq]\frac{dR_{out}(r)}{dr} = \frac{-Da\beta e^{-\beta a} – De^{-\beta a}}{a^2}.[/eq]

equating [eq]\frac{dR_{in}(r)}{dr}[/eq] and [eq]\frac{dR_{out}(r)}{dr}[/eq],

[eq]Aa\epsilon \cos(\epsilon a) – A \sin (\epsilon a)=-Da\beta e^{-\beta a} – De^{-\beta a}[/eq]

[eq]A(a\epsilon \cos(\epsilon a) – \sin (\epsilon a))=-De^{-\beta a}(a\beta +1).[/eq]                                 (10)

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

[eq]\frac{a\epsilon \cos(\epsilon a) – \sin (\epsilon a)}{\sin\epsilon a} = -(a\beta+1)[/eq]

[eq]a\epsilon\frac{\cos{\epsilon a}}{\sin{\epsilon a}} – 1= -a\beta – 1[/eq]

[eq]a\epsilon \cot(\epsilon a) = -a\beta[/eq]

we let [eq]k_1 = \epsilon a[/eq] and [eq]k_2 = \beta a[/eq] so,

[eq]k_1 \cot(k_1) = -k_2.[/eq]                                                                                   (11)

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

[eq]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][/eq][eq] = \frac{a^2}{\hbar^2} 2m (E+V_o-E)[/eq]

[eq]R^2 = \frac{2ma^2}{\hbar^2} V_o[/eq]

From the results, the allowable states only occurs when [eq]k_1[/eq] and [eq]k_2[/eq] 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 [eq]R<\frac{\pi}{2}[/eq].

[eq]\frac{2ma^2}{\hbar^2}V_o\;\;<\;\;\frac{\pi^2}{4}[/eq]

[eq]V_o a^2\;\;<\;\;\frac{\hbar^2}{2m}\frac{\pi^2}{4}[/eq]

[eq]V_o a^2\;\;<\;\;\frac{\pi^2 \hbar^2}{8m}.[/eq]

### 2 Responses to “Quantum Mechanics in Three-Dimensions: The Radial Equation”

1. Angie Says:

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…

2. Norbert Chencinski Says:

If k1 and k2 are both positive,the first solution for k1 lies between Pi/2 and Pi.