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# 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

$\label{e9.4} p_r \equiv \frac ParseError: invalid DekiScript (click for details) Callstack: at (Bookshelves/Quantum_Mechanics/Book:_Introductory_Quantum_Mechanics_(Fitzpatrick)/8:_Central_Potentials/8.1:_Derivation_of_Radial_Equation), /content/body/p/span/span, line 1, column 1  {r} = -\rm i \hbar\frac{\partial}{\partial r}.$

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

$\label{e9.5} {\bf L} = {\bf r}\times {\bf p}.$

[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: $\label{e9.7} \epsilon_{ijk} = \left\{ \begin{array}{rcl} 0 &\mbox{\hspace{1cm}}&\mbox{if i,j,k not all different}\\[0.5ex] 1 &&\mbox{if i,j,k are cyclic permutation of 1,2,3}\\[0.5ex] -1 &&\mbox{if i,j,k are anti-cyclic permutation of 1,2,3} \end{array}\right. .$ 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 $\delta_{ij} = \left\{ \begin{array}{rcl} 1 &\mbox{\hspace{1cm}}&\mbox{if i and j the same}\\[0.5ex] 0 &&\mbox{if i and j different}\\[0.5ex] \end{array}\right. .$ 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

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

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