11.6: Motion in Central Field

To further study the motion of an electron in a central field, whose Hamiltonian is

\( \mbox{\boldmath\)$r = (x^{2}+y^{2}+z^{2})^{1/2},$ $\ref{1219}$

and

$[r,p_r] = {\rm i} \,\hbar,$ $\ref{1221}$

which implies that in the Schrödinger representation

${\bf J} = {\bf L} + \frac{\hbar}{2}\,$$.$ $\ref{1223}$

Furthermore, we know from general principles that the eigenvalues of $j\,(j+1)\,\hbar^2$ , where $j=\vert l+1/2\vert$ , where $($$\cdot {\bf L})\,($$\cdot{\bf L}) = L^{\,2} + {\rm i}\,$$\times ({\bf L}\times {\bf L}).$ $\ref{1224}$

However, because ${\bf L}$ is an angular momentum, its components satisfy the standard commutation relations

$($$\cdot {\bf L})\,($$\cdot{\bf L}) = L^{\,2} -\hbar\,$$\cdot{\bf L} = J^{\,2} - 2\,\hbar\,$$\cdot{\bf L} -\frac{\hbar^2}{4}\,\Sigma^{\,2}.$ $\ref{1226}$

However, $($$\cdot {\bf L} + \hbar)^2 = J^{\,2}+\frac{1}{4}\,\hbar^2.$ $\ref{1227}$

Further application of $\ref{1192}$ yields

\( \mbox{\boldmath\)\( \mbox{\boldmath\)$= {\bf L}\cdot {\bf p} + {\rm i}\,$$\cdot{\bf L}\times {\bf p}= {\rm i}\,$$\cdot{\bf L}\times{\bf p},$ $\ref{1228}$ \( \mbox{\boldmath\)\( \mbox{\boldmath\)$= {\bf p}\cdot {\bf L} + {\rm i}\,$$\cdot{\bf p}\times {\bf L}= {\rm i}\,$$\cdot{\bf p}\times{\bf L},$ $\ref{1229}$

However, it is easily demonstrated from the fundamental commutation relations between position and momentum operators that

$($$\cdot {\bf L})\,($$\cdot{\bf p}) + ($$\cdot {\bf p})\,($$\cdot{\bf L}) =-2\,\hbar\,$$\cdot{\bf p},$ $\ref{1231}$

which implies that

\( \mbox{\boldmath\)\( \mbox{\boldmath\)$\gamma^5\,$$\Sigma$ $=$ $\gamma^5$ commutes with ${\bf L}$ , and $\Sigma$ . Hence, we conclude that

\( \mbox{\boldmath\)\( \mbox{\boldmath\)$\beta$ commutes with ${\bf L}$ , but anti-commutes with the components of $[\zeta,$$\cdot {\bf p}] = 0,$ $\ref{1234}$

where

${\bf x}$ for ${\bf L}\times {\bf x} + {\bf x}\times{\bf L} = 2\,{\rm i}\,\hbar\,{\bf x},$ $\ref{1236}$

we find that

\( \mbox{\boldmath\)$r$ commutes with $\Sigma$ and ${\bf L}$ . Hence,

$\beta$ commutes with the components of ${\bf L}$ , and can easily be shown to commute with all components of $\Sigma$ . It follows that

$[\zeta, H] =0.$ $\ref{1240}$

In other words, an eigenstate of the Hamiltonian is a simultaneous eigenstate of $\zeta^{\,2} = [\beta\,($$\cdot{\bf L}+\hbar)]^{\,2} = ($$\cdot{\bf L}+\hbar)^2 = J^{\,2}+\frac{1}{4}\,\hbar^2,$

$\ref{1241}$

where use has been made of Equation $\ref{1227}$, as well as $\zeta^{\,2}$ are $\zeta$ can be written $k=\pm(j+1/2)$ is a non-zero integer.

Equation $\ref{1192}$ implies that

\( \mbox{\boldmath\)\( \mbox{\boldmath\)$= {\bf x}\cdot{\bf p} + {\rm i}\,$$\cdot{\bf x}\times {\bf p} = r\,p_r + {\rm i}\,$$\cdot{\bf L}$ $\epsilon$, where

\( \mbox{\boldmath\)$[\epsilon,r] = 0.$ $\ref{1244}$

Hence,

\( \mbox{\boldmath\)$\epsilon^2 = 1.$ $\ref{1246}$

We have already seen that $\alpha$ $\cdot{\bf x}$ and $r$ . Thus,

$($$\cdot{\bf x})\,({\bf x}\cdot{\bf p}) - ({\bf x}\cdot{\bf p})\,($$\cdot{\bf x}) =$ $\cdot\left[{\bf x}\,({\bf x}\cdot{\bf p})- ({\bf x}\cdot{\bf p})\,{\bf x}\right] = {\rm i}\,\hbar\,$$\cdot{\bf x},$ $\ref{1248}$

where use has been made of the fundamental commutation relations for position and momentum operators. However, $\Sigma$ $\gamma^5$, we get

$[\epsilon,p_r]= 0.$ $\ref{1250}$

Equation $\ref{1242}$ implies that

\( \mbox{\boldmath\)\( \mbox{\boldmath\)\( \mbox{\boldmath\)$H= - e\,\phi(r) + c\,\epsilon\,(p_r-{\rm i}\,\hbar/r) + {\rm i}\,c\,\epsilon\,\beta\,\zeta/r + \beta\,m_e\,c^2.$ $\ref{1253}$

Now, we wish to solve the energy eigenvalue problem

$E$ is the energy eigenvalue. However, we have already shown that an eigenstate of the Hamiltonian is a simultaneous eigenstate of the $k\,\hbar$ , where $\left[- e\,\phi(r) + c\,\epsilon\,(p_r-{\rm i}\,\hbar/r) + {\rm i}\,c\,\hbar\,k\,\epsilon\,\beta/r + \beta\,m_e\,c^2\right]\psi = E\,\psi,$ $\ref{1255}$

which only involves the radial coordinate $r$ . It is easily demonstrated that $\beta$ . Hence, given that $\epsilon^2=1$ , we can represent $\epsilon = \left(\begin{array}{rr}0&-{\rm i}\\ [0.5ex]{\rm i}&0\end{array}\right).$

$\ref{1256}$

Thus, writing $\psi = \left(\begin{array}{c} \psi_a(r)\\ [0.5ex]\psi_b(r)\end{array}\right),$ $\ref{1257}$

and making use of $\ref{1222}$, the energy eigenvalue problem for an electron in a central field reduces to the following two coupled radial differential equations:

$= 0,$ $\ref{1258}$ $= 0.$ $\ref{1259}$

Contributors

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

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