8.1: Derivation of Radial Equation

Now, we have seen that the Cartesian components of the momentum, ${\bf p}$, can be represented as (see Section [s7.2]) $$p_i = -{\rm i}\,\hbar\,\frac{\partial}{\partial x_i}$$ 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 previous expressions, and the basic definitions of the spherical coordinates [see Equations ([e8.21])–([e8zz])], that the radial component of the momentum can be represented as

$$\begin{equation}p_{r} \equiv \frac{\mathbf{p} \cdot \mathbf{r}}{r}=-\mathrm{i} \hbar \frac{\partial}{\partial r}\end{equation}$$

Recall that the angular momentum vector, ${\bf L}$, is defined

$$\begin{equation}\mathbf{L}=\mathbf{r} \times \mathbf{p}\end{equation}$$

[See Equation ([e8.0]).] This expression can also be written in the following form: $$\label{e9.6} L_i = \epsilon_{ijk}\,x_j\,p_k.$$ 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: $$\begin{equation}\epsilon_{i j k}=\left\{\begin{array}{cl} 0 & \text { if } i, j, k \text { not all different } \\ 1 & \text { if } i, j, k \text { are cyclic permutation of } 1,2,3 \\ -1 & \text { if } i, j, k \text { are anti-cyclic permutation of } 1,2,3 \end{array}\right.\end{equation}$$ Thus, $\epsilon_{123}=\epsilon_{231}=1$, $\epsilon_{321}=\epsilon_{132}=-1$, and $\epsilon_{112}=\epsilon_{131}=0$, et cetera. Equation ([e9.6]) 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 Equation ([e9.7]), it is easily seen that Equations ([e9.5]) and ([e9.6]) are indeed equivalent.

Let us calculate the value of $L^2$ using Equation ([e9.6]). According to our new notation, $L^2$ is the same as $L_i\,L_i$. Thus, we obtain $$\label{e9.8} L^2 = \epsilon_{ijk}\,x_j\,p_k\,\epsilon_{ilm}\,x_l\,p_m = \epsilon_{ijk}\,\epsilon_{ilm}\,x_j\,p_k\,x_l\,p_m.$$ 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 $$\label{e9.9} \epsilon_{ijk}\,\epsilon_{ilm}\equiv \delta_{jl}\,\delta_{km}-\delta_{jm}\,\delta_{kl}.$$ Here $\delta_{ij}$ is the usual Kronecker delta, whose elements are determined according to the rule $$\begin{equation}\delta_{i j}=\left\{\begin{array}{ll} 1 & \text { if } i \text { and } j \text { the same } \\ 0 & \text { if } i \text { and } j \text { different } \end{array}\right.\end{equation}$$ It follows from Equations ([e9.8]) and ([e9.9]) that $$\label{e9.11} L^2 = x_i\,p_j\,x_i\,p_j - x_i\,p_j\,x_j\,p_i.$$ 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 previous expression, because 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 Equation ([e9.11]). We can achieve this goal by making use of the fundamental commutation relation for the $x_i$ and the $p_i$: $$\label{e9.12} [x_i,p_j] = {\rm i}\,\hbar\,\delta_{ij}.$$ [See Equation ([commxp]).] Thus, $$\begin{aligned} L^2 &= x_i\left(x_i\,p_j - [x_i,p_j]\right) p_j - x_i\,p_j\,\left(p_i\,x_j+[x_j,p_i]\right)\nonumber\\[0.5ex] &=x_i\,x_i\,p_j\,p_j - {\rm i}\,\hbar\,\delta_{ij}\,x_i\,p_j -x_i\,p_j\,p_i\,x_j - {\rm i}\,\hbar\,\delta_{ij}\,x_i\,p_j\nonumber\\[0.5ex] &=x_i\,x_i\,p_j\,p_j -x_i\,p_i\,p_j\,x_j - 2\,{\rm i}\,\hbar\,x_i\,p_i.\end{aligned}$$ Here, we have made use of the fact that $p_j\,p_i=p_i\,p_j$, because the $p_i$ commute with one another. [See Equation ([commpp]).] Next, $$L^2 = x_i\,x_i\,p_j\,p_j - x_i\,p_i\left(x_j\,p_j - [x_j,p_j]\right) - 2\,{\rm i}\,\hbar\,x_i\,p_i.$$ Now, according to Equation ([e9.12]), $$[x_j,p_j]\equiv [x_1,p_1]+[x_2,p_2]+[x_3,p_3] = 3\,{\rm i}\,\hbar.$$ Hence, we obtain $$L^2 = x_i\,x_i\,p_j\,p_j - x_i\,p_i\,x_j\,p_j + {\rm i}\,\hbar\,x_i\,p_i.$$ When expressed in more conventional vector notation, the previous expression becomes $$\label{e9.17} L^2 = r^{\,2}\,p^{\,2} - ({\bf r}\cdot{\bf p})^2 + {\rm i}\,\hbar\,{\bf r}\cdot{\bf p}.$$ Note that if we had attempted to derive the previous expression directly from Equation ([e9.5]), 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 ([e9.17]) can be rearranged to give $$p^{\,2} = r^{\,-2}\left[({\bf r}\cdot{\bf p})^2- {\rm i}\,\hbar\,{\bf r}\cdot{\bf p}+L^2\right].$$ Now, $${\bf r}\cdot{\bf p} = r\,p_r = -{\rm i}\,\hbar\,r\,\frac{\partial}{\partial r},$$ where use has been made of Equation ([e9.4]). Hence, we obtain $$p^{\,2} = -\hbar^{\,2}\left[\frac{1}{r}\frac{\partial}{\partial r}\left(r\,\frac{\partial}{\partial r}\right) + \frac{1}{r}\frac{\partial}{\partial r}- \frac{L^2}{\hbar^{\,2}\,r^{\,2}}\right].$$ Finally, the previous equation can be combined with Equation ([e9.2]) to give the following expression for the Hamiltonian: $$\label{e9.21} H = -\frac{\hbar^{\,2}}{2\,m}\left(\frac{\partial^{\,2}}{\partial r^{\,2}} + \frac{2}{r}\frac{\partial}{\partial r}- \frac{L^2}{\hbar^{\,2}\,r^{\,2}}\right) +V(r).$$

Let us now consider whether the previous Hamiltonian commutes with the angular momentum operators $L_z$ and $L^2$. Recall, from Section [s8.3], that $L_z$ and $L^2$ are represented as differential operators that depend solely on the angular spherical coordinates, $\theta$ and $\phi$, and do not contain the radial 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 Section [s8.2].) It is, therefore, clear that the previous Hamiltonian commutes with both $L_z$ and $L^2$.

According to Section [smeas], 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 Section [sharm]. 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: $$\label{e9.22} \psi(r,\theta,\phi) = R(r)\,Y_{l,m}(\theta,\phi).$$ It immediately follows, from Equation ([e8.29]) and ([e8.30]), and the fact that $L_z$ and $L^2$ both obviously commute with $R(r)$, that $$\begin{aligned} L_z\,\psi &= m\,\hbar\,\psi,\\[0.5ex] L^2\,\psi&= l\,(l+1)\,\hbar^{\,2}\,\psi.\label{e9.24}\end{aligned}$$ Recall that the quantum numbers $m$ and $l$ are restricted to take certain integer values, as explained in Section [slsq].

Finally, making use of Equations ([e9.1]), ([e9.21]), and ([e9.24]), we obtain the following differential equation which determines the radial variation of the stationary wavefunction: $$-\frac{\hbar^{\,2}}{2\,m}\left[\frac{d^{\,2}}{d r^{\,2}} + \frac{2}{r}\frac{d}{d r}- \frac{l\,(l+1)}{r^{\,2}}\right]R_{n,l} +V\,R_{n,l} = E\,R_{n,l}.$$ 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 previous 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.

Contributors and Attributions

• Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)

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