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Next: Infinite Spherical Potential Well Up: Central Potentials Previous: Introduction

Derivation of Radial Equation

Now, we have seen that the Cartesian components of the momentum, ${\bf p}$, can be represented as (see Sect. 7.2)
p_i = -{\rm i} \hbar \frac{\partial}{\partial x_i}
\end{displaymath} (624)

for $i=1,2,3$, where $x_1\equiv x$, $x_2\equiv y$, $x_3\equiv z$, and ${\bf r}\equiv (x_1, x_2, x_3)$. Likewise, it is easily demonstrated, from the above expressions, and the basic definitions of the spherical polar coordinates [see Eqs. (545)-(550)], that the radial component of the momentum can be represented as
p_r \equiv \frac{{\bf p}\cdot{\bf r}}{r} = -{\rm i} \hbar \frac{\partial}{\partial r}.
\end{displaymath} (625)

Recall that the angular momentum vector, ${\bf L}$, is defined [see Eq. (526)]

{\bf L} = {\bf r}\times {\bf p}.
\end{displaymath} (626)

This expression can also be written in the following form:
L_i = \epsilon_{ijk} x_j p_k.
\end{displaymath} (627)

Here, the $\epsilon_{ijk}$ (where $i,j,k$ all run from 1 to 3) are elements of the so-called totally anti-symmetric tensor. The values of the various elements of this tensor are determined via a simple rule:
\epsilon_{ijk} = \left\{
0 &\mbox{\hspace...
... are anti-cyclic permutation of $1,2,3$}
\end{array}\right. .
\end{displaymath} (628)

Thus, $\epsilon_{123}=\epsilon_{231}=1$, $\epsilon_{321}=\epsilon_{132}=-1$, and $\epsilon_{112}=\epsilon_{131}=0$, etc. Equation (627) also makes use of the Einstein summation convention, according to which repeated indices are summed (from 1 to 3). For instance, $a_i b_i\equiv a_1 b_1+a_2 b_2+a_3 b_3$. Making use of this convention, as well as Eq. (628), it is easily seen that Eqs. (626) and (627) are indeed equivalent.

Let us calculate the value of $L^2$ using Eq. (627). According to our new notation, $L^2$ is the same as $L_i L_i$. Thus, we obtain

L^2 = \epsilon_{ijk} x_j p_k \epsilon_{ilm} x_l p_m =
\epsilon_{ijk} \epsilon_{ilm} x_j p_k x_l p_m.
\end{displaymath} (629)

Note that we are able to shift the position of $\epsilon_{ilm}$ because its elements are just numbers, and, therefore, commute with all of the $x_i$ and the $p_i$. Now, it is easily demonstrated that
\epsilon_{ijk} \epsilon_{ilm}\equiv \delta_{jl} \delta_{km}-\delta_{jm} \delta_{kl}.
\end{displaymath} (630)

Here $\delta_{ij}$ is the usual Kronecker delta, whose elements are determined according to the rule
\delta_{ij} = \left\{
1 &\mbox{\hspace{1c...{if $i$ and $j$ different}\ [0.5ex]
\end{array}\right. .
\end{displaymath} (631)

It follows from Eqs. (629) and (630) that
L^2 = x_i p_j x_i p_j - x_i p_j x_j p_i.
\end{displaymath} (632)

Here, we have made use of the fairly self-evident result that $\delta_{ij} a_i b_j
\equiv a_i b_i$. We have also been careful to preserve the order of the various terms on the right-hand side of the above expression, since the $x_i$ and the $p_i$ do not necessarily commute with one another.

We now need to rearrange the order of the terms on the right-hand side of Eq. (632). We can achieve this by making use of the fundamental commutation relation for the $x_i$ and the $p_i$ [see Eq. (483)]:

\begin{displaymath}[x_i,p_j]= {\rm i} \hbar \delta_{ij}.
\end{displaymath} (633)

$\displaystyle L^2$ $\textstyle =$ $\displaystyle x_i\left(x_i p_j - [x_i,p_j]\right) p_j
- x_i p_j \left(p_i x_j+[x_j,p_i]\right)$  
  $\textstyle =$ $\displaystyle x_i x_i p_j p_j - {\rm i} \hbar \delta_{ij} x_i p_j
-x_i p_j p_i x_j - {\rm i} \hbar \delta_{ij} x_i p_j$  
  $\textstyle =$ $\displaystyle x_i x_i p_j p_j -x_i p_i p_j x_j - 2 {\rm i} \hbar x_i p_i.$ (634)

Here, we have made use of the fact that $p_j p_i=p_i p_j$, since the $p_i$ commute with one another [see Eq. (482)]. Next,
L^2 = x_i x_i p_j p_j - x_i p_i\left(x_j p_j - [x_j,p_j]\right) - 2 {\rm i} \hbar x_i p_i.
\end{displaymath} (635)

Now, according to (633),
\begin{displaymath}[x_j,p_j]\equiv [x_1,p_1]+[x_2,p_2]+[x_3,p_3] = 3 {\rm i} \hbar.
\end{displaymath} (636)

Hence, we obtain
L^2 = x_i x_i p_j p_j - x_i p_i x_j p_j + {\rm i} \hbar x_i p_i.
\end{displaymath} (637)

When expressed in more conventional vector notation, the above expression becomes
L^2 = r^2 p^2 - ({\bf r}\cdot{\bf p})^2 + {\rm i} \hbar {\bf r}\cdot{\bf p}.
\end{displaymath} (638)

Note that if we had attempted to derive the above expression directly from Eq. (626), using standard vector identities, then we would have missed the final term on the right-hand side. This term originates from the lack of commutation between the $x_i$ and $p_i$ operators in quantum mechanics. Of course, standard vector analysis assumes that all terms commute with one another.

Equation (638) can be rearranged to give

p^2 = r^{-2}\left[({\bf r}\cdot{\bf p})^2- {\rm i} \hbar {\bf r}\cdot{\bf p}+L^2\right].
\end{displaymath} (639)

{\bf r}\cdot{\bf p} = r p_r = -{\rm i} \hbar r \frac{\partial}{\partial r},
\end{displaymath} (640)

where use has been made of Eq. (625). Hence, we obtain
p^2 = -\hbar^2\left[\frac{1}{r}\frac{\partial}{\partial r}\l...
...}\frac{\partial}{\partial r}- \frac{L^2}{\hbar^2 r^2}\right].
\end{displaymath} (641)

Finally, the above equation can be combined with Eq. (623) to give the following expression for the Hamiltonian:
H = -\frac{\hbar^2}{2 m}\left(\frac{\partial^2}{\partial r^...
...{\partial}{\partial r}- \frac{L^2}{\hbar^2 r^2}\right)
\end{displaymath} (642)

Let us now consider whether the above Hamiltonian commutes with the angular momentum operators $L_z$ and $L^2$. Recall, from Sect. 8.3, that $L_z$ and $L^2$ are represented as differential operators which depend solely on the angular spherical polar coordinates, $\theta $ and $\phi$, and do not contain the radial polar coordinate, $r$. Thus, any function of $r$, or any differential operator involving $r$ (but not $\theta $ and $\phi$), will automatically commute with $L^2$ and $L_z$. Moreover, $L^2$ commutes both with itself, and with $L_z$ (see Sect. 8.2). It is, therefore, clear that the above Hamiltonian commutes with both $L_z$ and $L^2$.

Now, according to Sect. 4.10, if two operators commute with one another then they possess simultaneous eigenstates. We thus conclude that for a particle moving in a central potential the eigenstates of the Hamiltonian are simultaneous eigenstates of $L_z$ and $L^2$. Now, we have already found the simultaneous eigenstates of $L_z$ and $L^2$--they are the spherical harmonics, $Y_{l,m}(\theta,\phi)$, discussed in Sect. 8.7. It follows that the spherical harmonics are also eigenstates of the Hamiltonian. This observation leads us to try the following separable form for the stationary wavefunction:

\psi(r,\theta,\phi) = R(r) Y_{l,m}(\theta,\phi).
\end{displaymath} (643)

It immediately follows, from (556) and (557), and the fact that $L_z$ and $L^2$ both obviously commute with $R(r)$, that
$\displaystyle L_z \psi$ $\textstyle =$ $\displaystyle m \hbar \psi,$ (644)
$\displaystyle L^2 \psi$ $\textstyle =$ $\displaystyle l (l+1) \hbar^2 \psi.$ (645)

Recall that the quantum numbers $m$ and $l$ are restricted to take certain integer values, as explained in Sect. 8.6.

Finally, making use of Eqs. (622), (642), and (645), we obtain the following differential equation which determines the radial variation of the stationary wavefunction:

-\frac{\hbar^2}{2 m}\left(\frac{d^2}{d r^2}
+ \frac{2}{r}\f...
...- \frac{l (l+1)}{r^2}\right)R_{n,l}
+V R_{n,l} = E R_{n,l}.
\end{displaymath} (646)

Here, we have labeled the function $R(r)$ by two quantum numbers, $n$ and $l$. The second quantum number, $l$, is, of course, related to the eigenvalue of $L^2$. [Note that the azimuthal quantum number, $m$, does not appear in the above equation, and, therefore, does not influence either the function $R(r)$ or the energy, $E$.] As we shall see, the first quantum number, $n$, is determined by the constraint that the radial wavefunction be square-integrable.

next up previous
Next: Infinite Spherical Potential Well Up: Central Potentials Previous: Introduction
Richard Fitzpatrick 2010-07-20