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11.6: Motion in Central Field

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

 

$\displaystyle H = - e\,\phi(r) + c\,$$\displaystyle \mbox{\boldmath$\alpha$}$$\displaystyle \cdot{\bf p} + \beta\,m_e\,c^2,$ (1218)

 

 

it is convenient to transform to polar coordinates. Let

 

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

 

 

and

 

$\displaystyle p_r = {\bf x}\cdot{\bf p}.$ (1220)

 

 

It is easily demonstrated that

 

$\displaystyle [r,p_r] = {\rm i} \,\hbar,$ (1221)

 

 

which implies that in the Schrödinger representation

 

$\displaystyle p_r = -{\rm i}\,\hbar\,\frac{\partial}{\partial r}.$ (1222)

 

 

Now, by symmetry, an energy eigenstate in a central field is a simultaneous eigenstate of the total angular momentum

 

$\displaystyle {\bf J} = {\bf L} + \frac{\hbar}{2}\,$$\displaystyle \mbox{\boldmath$\Sigma$}$$\displaystyle .$ (1223)

 

 

Furthermore, we know from general principles that the eigenvalues of $ J^{\,2}$ are $ j\,(j+1)\,\hbar^2$ , where $ j$ is a positive half-integer (since $ j=\vert l+1/2\vert$ , where $ l$ is the standard non-negative integer quantum number associated with orbital angular momentum.)

It follows from Equation (1192) that

 

$\displaystyle ($$\displaystyle \mbox{\boldmath$\Sigma$}$$\displaystyle \cdot {\bf L})\,($$\displaystyle \mbox{\boldmath$\Sigma$}$$\displaystyle \cdot{\bf L}) = L^{\,2} + {\rm i}\,$$\displaystyle \mbox{\boldmath$\Sigma$}$$\displaystyle \times ({\bf L}\times {\bf L}).$ (1224)

 

 

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

 

$\displaystyle {\bf L}\times {\bf L} = {\rm i}\,\hbar \,{\bf L}.$ (1225)

 

 

Thus, we obtain

 

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

 

 

However, $ \Sigma^{\,2}=3$ , so

 

$\displaystyle ($$\displaystyle \mbox{\boldmath$\Sigma$}$$\displaystyle \cdot {\bf L} + \hbar)^2 = J^{\,2}+\frac{1}{4}\,\hbar^2.$ (1227)

 

 

Further application of (1192) yields

 

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

 

 

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

 

$\displaystyle {\bf L}\times {\bf p} + {\bf p}\times{\bf L} = 2\,{\rm i}\,\hbar\,{\bf p}.$ (1230)

 

 

Thus,

 

$\displaystyle ($$\displaystyle \mbox{\boldmath$\Sigma$}$$\displaystyle \cdot {\bf L})\,($$\displaystyle \mbox{\boldmath$\Sigma$}$$\displaystyle \cdot{\bf p}) + ($$\displaystyle \mbox{\boldmath$\Sigma$}$$\displaystyle \cdot {\bf p})\,($$\displaystyle \mbox{\boldmath$\Sigma$}$$\displaystyle \cdot{\bf L}) =-2\,\hbar\,$$\displaystyle \mbox{\boldmath$\Sigma$}$$\displaystyle \cdot{\bf p},$ (1231)

 

 

which implies that

 

$\displaystyle \{$$\displaystyle \mbox{\boldmath$\Sigma$}$$\displaystyle \cdot {\bf L}+ \hbar,\,$   $\displaystyle \mbox{\boldmath$\Sigma$}$$\displaystyle \cdot {\bf p}\} = 0.$ (1232)

 

 

Now, $ \gamma^5\,$$ \Sigma$ $ =$   $ \alpha$ . Moreover, $ \gamma^5$ commutes with $ {\bf p}$ , $ {\bf L}$ , and $ \Sigma$ . Hence, we conclude that

 

$\displaystyle \{$$\displaystyle \mbox{\boldmath$\Sigma$}$$\displaystyle \cdot {\bf L}+ \hbar,\,$   $\displaystyle \mbox{\boldmath$\alpha$}$$\displaystyle \cdot {\bf p}\} = 0.$ (1233)

 

 

Finally, since $ \beta$ commutes with $ {\bf p}$ and $ {\bf L}$ , but anti-commutes with the components of $ \alpha$ , we obtain

 

$\displaystyle [\zeta,$$\displaystyle \mbox{\boldmath$\alpha$}$$\displaystyle \cdot {\bf p}] = 0,$ (1234)

 

 

where

 

$\displaystyle \zeta = \beta\left(\mbox{\boldmath$\Sigma$}\cdot{\bf L} + \hbar\right).$ (1235)

 

 

If we repeat the above analysis, starting at Equation (1228), but substituting $ {\bf x}$ for $ {\bf p}$ , and making use of the easily demonstrated result

 

$\displaystyle {\bf L}\times {\bf x} + {\bf x}\times{\bf L} = 2\,{\rm i}\,\hbar\,{\bf x},$ (1236)

 

 

we find that

 

$\displaystyle [\zeta,$$\displaystyle \mbox{\boldmath$\alpha$}$$\displaystyle \cdot {\bf x}] = 0.$ (1237)

 

 

Now, $ r$ commutes with $ \beta$ , as well as the components of $ \Sigma$ and $ {\bf L}$ . Hence,

 

$\displaystyle [\zeta,r] = 0.$ (1238)

 

 

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

 

$\displaystyle [\zeta,\beta]=0.$ (1239)

 

 

Hence, Equations (1218), (1234), (1238), and (1239) imply that

 

$\displaystyle [\zeta, H] =0.$ (1240)

 

 

In other words, an eigenstate of the Hamiltonian is a simultaneous eigenstate of $ \zeta$ . Now,

 

$\displaystyle \zeta^{\,2} = [\beta\,($$\displaystyle \mbox{\boldmath$\Sigma$}$$\displaystyle \cdot{\bf L}+\hbar)]^{\,2} = ($$\displaystyle \mbox{\boldmath$\Sigma$}$$\displaystyle \cdot{\bf L}+\hbar)^2 = J^{\,2}+\frac{1}{4}\,\hbar^2,$ (1241)

 

 

where use has been made of Equation (1227), as well as $ \beta^{\,2}=1$ . It follows that the eigenvalues of $ \zeta^{\,2}$ are $ j\,(j+1)\,\hbar^2 + (1/4)\,\hbar^2 = (j+1/2)^2\,\hbar^2$ . Thus, the eigenvalues of $ \zeta$ can be written $ k\,\hbar$ , where $ k=\pm(j+1/2)$ is a non-zero integer.

Equation (1192) implies that

 

$\displaystyle ($$\displaystyle \mbox{\boldmath$\Sigma$}$$\displaystyle \cdot{\bf x})\,($$\displaystyle \mbox{\boldmath$\Sigma$}$$\displaystyle \cdot{\bf p})$ $\displaystyle = {\bf x}\cdot{\bf p} + {\rm i}\,$$\displaystyle \mbox{\boldmath$\Sigma$}$$\displaystyle \cdot{\bf x}\times {\bf p} = r\,p_r + {\rm i}\,$$\displaystyle \mbox{\boldmath$\Sigma$}$$\displaystyle \cdot{\bf L}$    
  $\displaystyle = r\,p_r + {\rm i}\,(\beta\,\zeta-\hbar),$ (1242)

 

 

where use has been made of (1220) and (1235).

It is helpful to define the new operator $ \epsilon$ , where

 

$\displaystyle r\,\epsilon =$   $\displaystyle \mbox{\boldmath$\alpha$}$$\displaystyle \cdot{\bf x}.$ (1243)

 

 

Moreover, it is evident that

 

$\displaystyle [\epsilon,r] = 0.$ (1244)

 

 

Hence,

 

$\displaystyle r^2\,\epsilon^2 = ($$\displaystyle \mbox{\boldmath$\alpha$}$$\displaystyle \cdot{\bf x})^2 = \frac{1}{2}\sum_{i,j=1,3}\{\alpha_i,\alpha_j\}\,x^i\,x^j = \sum_{i=1,3}(x^i)^{\,2}=r^2,$ (1245)

 

 

where use has been made of (1117). It follows that

 

$\displaystyle \epsilon^2 = 1.$ (1246)

 

 

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

 

$\displaystyle [\zeta,\epsilon] = 0.$ (1247)

 

 

Equation (1192) gives

 

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

 

 

where use has been made of the fundamental commutation relations for position and momentum operators. However, $ {\bf x}\cdot{\bf p}=r\,p_r$ and $ \Sigma$ $ \cdot{\bf x} = \gamma^5\,r\,\epsilon$ , so, multiplying through by $ \gamma^5$ , we get

 

$\displaystyle r^2\,\epsilon\,p_r - r\,p_r\,r\,\epsilon = {\rm i}\,\hbar\,r\,\epsilon.$ (1249)

 

 

Equation (1221) then yields

 

$\displaystyle [\epsilon,p_r]= 0.$ (1250)

 

 

Equation (1242) implies that

 

$\displaystyle ($$\displaystyle \mbox{\boldmath$\alpha$}$$\displaystyle \cdot{\bf x})\,($$\displaystyle \mbox{\boldmath$\alpha$}$$\displaystyle \cdot{\bf p}) = r\,p_r+{\rm i}\,(\beta\,\zeta-\hbar).$ (1251)

 

 

Making use of Equations (1238), (1243), (1244), and (1246), we get

 

$\displaystyle \mbox{\boldmath$\alpha$}$$\displaystyle \cdot {\bf p} = \epsilon\,(p_r-{\rm i}\,\hbar/r) + {\rm i}\,\epsilon\,\beta\,\zeta/r.$ (1252)

 

 

Hence, the Hamiltonian (1218) becomes

 

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

 

 

Now, we wish to solve the energy eigenvalue problem

 

$\displaystyle H\,\psi = E\,\psi,$ (1254)

 

 

where $ E$ is the energy eigenvalue. However, we have already shown that an eigenstate of the Hamiltonian is a simultaneous eigenstate of the $ \zeta$ operator belonging to the eigenvalue $ k\,\hbar$ , where $ k$ is a non-zero integer. Hence, the eigenvalue problem reduces to

 

$\displaystyle \left[- e\,\phi(r) + c\,\epsilon\,(p_r-{\rm i}\,\hbar/r) + {\rm i}\,c\,\hbar\,k\,\epsilon\,\beta/r + \beta\,m_e\,c^2\right]\psi = E\,\psi,$ (1255)

 

 

which only involves the radial coordinate $ r$ . It is easily demonstrated that $ \epsilon$ anti-commutes with $ \beta$ . Hence, given that $ \beta$ takes the form (1125), and that $ \epsilon^2=1$ , we can represent $ \epsilon$ as the matrix

 

$\displaystyle \epsilon = \left(\begin{array}{rr}0&-{\rm i}\\ [0.5ex]{\rm i}&0\end{array}\right).$ (1256)

 

 

Thus, writing $ \psi$ in the spinor form

 

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

 

 

and making use of (1222), the energy eigenvalue problem for an electron in a central field reduces to the following two coupled radial differential equations:

 

$\displaystyle \hbar\,c\left(\frac{d}{d r} + \frac{k+1}{r}\right)\psi_b + (E-m_e\,c^2+e\,\phi)\,\psi_a$ $\displaystyle = 0,$ (1258)
$\displaystyle \hbar\,c\left(\frac{d}{d r}-\frac{k-1}{r}\right)\psi_a - (E+m_e\,c^2+e\,\phi)\,\psi_b$ $\displaystyle = 0.$ (1259)

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