Motion in Central Field

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

$$H = - e\,\phi(r) + c\, \mbox{\boldmath$\alpha$} \cdot{\bf p} + \beta\,m_e\,c^{\,2},$$ (11.128)

it is convenient to transform to polar coordinates. Let

$$r = \left(x^{\,2}+y^{\,2}+z^{\,2}\right)^{1/2},$$ (11.129)

and

$$r\,p_r = {\bf x}\cdot{\bf p}.$$ (11.130)

It is easily demonstrated that

$$[r,p_r] = {\rm i} \,\hbar,$$ (11.131)

which implies that in the Schrödinger representation

$$p_r = -{\rm i}\,\hbar\,\frac{\partial}{\partial r}.$$ (11.132)

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

$${\bf J} = {\bf L} + \frac{\hbar}{2}\, \mbox{\boldmath$\Sigma$} .$$ (11.133)

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 (because $j=\vert l+1/2\vert$ , where $l$ is the standard non-negative integer quantum number associated with orbital angular momentum.) (See Chapter 6.)

It follows from Equation (11.101) that

$$( \mbox{\boldmath$\Sigma$} \cdot {\bf L})\,( \mbox{\boldmath$\Sigma$} \cdot{\bf L}) = L^{\,2} + {\rm i}\, \mbox{\boldmath$\Sigma$} \times ({\bf L}\times {\bf L}).$$ (11.134)

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

$${\bf L}\times {\bf L} = {\rm i}\,\hbar \,{\bf L}.$$ (11.135)

(See Section 4.1.) Thus, we obtain

$$( \mbox{\boldmath$\Sigma$} \cdot {\bf L})\,( \mbox{\boldmath$\Sigma$} \cdot{\bf L}) = L^{\,2} -\hbar\, \mbox{\boldmath$\Sigma$} \cdot{\bf L} = J^{\,2} - 2\,\hbar\, \mbox{\boldmath$\Sigma$} \cdot{\bf L} -\frac{\hbar^{\,2}}{4}\,\Sigma^{\,2}.$$ (11.136)

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

$$( \mbox{\boldmath$\Sigma$} \cdot {\bf L} + \hbar)^2 = J^{\,2}+\frac{1}{4}\,\hbar^{\,2}.$$ (11.137)

Further application of Equation (11.101) yields

$$( \mbox{\boldmath$\Sigma$} \cdot {\bf L})\,( \mbox{\boldmath$\Sigma$} \cdot{\bf p}) = {\bf L}\cdot {\bf p} + {\rm i}\, \mbox{\boldmath$\Sigma$} \cdot{\bf L}\times {\bf p}= {\rm i}\, \mbox{\boldmath$\Sigma$} \cdot{\bf L}\times{\bf p},$$ (11.138)

$$( \mbox{\boldmath$\Sigma$} \cdot {\bf p})\,( \mbox{\boldmath$\Sigma$} \cdot{\bf L}) = {\bf p}\cdot {\bf L} + {\rm i}\, \mbox{\boldmath$\Sigma$} \cdot{\bf p}\times {\bf L}= {\rm i}\, \mbox{\boldmath$\Sigma$} \cdot{\bf p}\times{\bf L},$$ (11.139)

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

$${\bf L}\times {\bf p} + {\bf p}\times{\bf L} = 2\,{\rm i}\,\hbar\,{\bf p}.$$ (11.140)

(See Section 2.2.) Thus,

$$( \mbox{\boldmath$\Sigma$} \cdot {\bf L})\,( \mbox{\boldmath$\Sigma$} \cdot{\bf p}) + ( \mbox{\boldmath$\Sigma$} \cdot {\bf p})\,( \mbox{\boldmath$\Sigma$} \cdot{\bf L}) =-2\,\hbar\, \mbox{\boldmath$\Sigma$} \cdot{\bf p},$$ (11.141)

which implies that

$$\{ \mbox{\boldmath$\Sigma$} \cdot {\bf L}+ \hbar,\, \mbox{\boldmath$\Sigma$} \cdot {\bf p}\} = 0.$$ (11.142)

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

$$\{ \mbox{\boldmath$\Sigma$} \cdot {\bf L}+ \hbar,\, \mbox{\boldmath$\alpha$} \cdot {\bf p}\} = 0.$$ (11.143)

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

$$[\zeta, \mbox{\boldmath$\alpha$} \cdot {\bf p}] = 0,$$ (11.144)

where

$$\zeta = \beta\left(\mbox{\boldmath$\Sigma$}\cdot{\bf L} + \hbar\right).$$ (11.145)

If we repeat the previous analysis, starting at Equation (11.138), but substituting ${\bf x}$ for ${\bf p}$ , and making use of the easily demonstrated result

$${\bf L}\times {\bf x} + {\bf x}\times{\bf L} = 2\,{\rm i}\,\hbar\,{\bf x},$$ (11.146)

we find that

$$[\zeta, \mbox{\boldmath$\alpha$} \cdot {\bf x}] = 0.$$ (11.147)

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

$$[\zeta,r] = 0.$$ (11.148)

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

$$[\zeta,\beta]=0.$$ (11.149)

Hence, Equations (11.128), (11.144), (11.148), and (11.149) imply that

$$[\zeta, H] =0.$$ (11.150)

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

$$\zeta^{\,2} = [\beta\,( \mbox{\boldmath$\Sigma$} \cdot{\bf L}+\hbar)]^{\,2} = ( \mbox{\boldmath$\Sigma$} \cdot{\bf L}+\hbar)^2 = J^{\,2}+\frac{1}{4}\,\hbar^{\,2},$$ (11.151)

where use has been made of Equation (11.137), 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 (11.101) implies that

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

$$= r\,p_r + {\rm i}\,(\beta\,\zeta-\hbar),$$ (11.152)

where use has been made of Equations (11.130) and (11.145).

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

$$r\,\epsilon = \mbox{\boldmath$\alpha$} \cdot{\bf x}.$$ (11.153)

Moreover, it is evident that

$$[\epsilon,r] = 0.$$ (11.154)

Hence,

$$r^{\,2}\,\epsilon^{\,2} = ( \mbox{\boldmath$\alpha$} \cdot{\bf x})^2 = \frac{1}{2}\sum_{i,j=1,3}\{\alpha_i,\alpha_j\}\,x^{\,i}\,x^{\,j} = \sum_{i=1,3}x^{\,i}\,x^{\,i}=r^{\,2},$$ (11.155)

where use has been made of Equation (11.24). It follows that

$$\epsilon^{\,2} = 1.$$ (11.156)

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

$$[\zeta,\epsilon] = 0.$$ (11.157)

Because $\Sigma$ commutes with ${\bf x}$ and ${\bf p}$ , and ${\bf x}\cdot{\bf p} =-{\rm i}\,\hbar\,r\,\partial/\partial r$ , as well as $r\,\partial{\bf x}/\partial r = {\bf x}$ , we obtain

$$( \mbox{\boldmath$\Sigma$} \cdot{\bf x})\,({\bf x}\cdot{\bf p}) - ({\bf x}\cdot{\bf p})\,( \mbox{\boldmath$\Sigma$} \cdot{\bf x}) = \mbox{\boldmath$\Sigma$} \cdot\left[{\bf x}\,({\bf x}\cdot{\bf p})- ({\bf x}\cdot{\bf p})\,{\bf x}\right] = {\rm i}\,\hbar\, \mbox{\boldmath$\Sigma$} \cdot{\bf x}.$$ (11.158)

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

$$r^{\,2}\,\epsilon\,p_r - r\,p_r\,r\,\epsilon = {\rm i}\,\hbar\,r\,\epsilon.$$ (11.159)

Equation (11.131) then yields

$$[\epsilon,p_r]= 0.$$ (11.160)

Equation (11.152) implies that

$$( \mbox{\boldmath$\alpha$} \cdot{\bf x})\,( \mbox{\boldmath$\alpha$} \cdot{\bf p}) = r\,p_r+{\rm i}\,(\beta\,\zeta-\hbar).$$ (11.161)

Making use of Equations (11.148), (11.153), (11.154), and (11.156), we get

$$\mbox{\boldmath$\alpha$} \cdot {\bf p} = \epsilon\,(p_r-{\rm i}\,\hbar/r) + {\rm i}\,\epsilon\,\beta\,\zeta/r.$$ (11.162)

Hence, the Hamiltonian (11.128) becomes

$$H= - e\,\phi(r) + c\,\epsilon\,(p_r-{\rm i}\,\hbar/r) + {\rm i}\,c\,\epsilon\,\beta\,\zeta/r + \beta\,m_e\,c^{\,2}.$$ (11.163)

Now, we wish to solve the energy eigenvalue problem

$$H\,\psi = E\,\psi,$$ (11.164)

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

$$\left[- e\,\phi(r) + c\,\epsilon\,(p_r-{\rm i}\,\hbar/r) + {\rm i}\,c\,\hbar\,k\,\epsilon\,\beta/r + \beta\,m_e\,c^{\,2}\right]\psi = E\,\psi,$$ (11.165)

which only involves the radial coordinate, $r$ . It is easily demonstrated that $\epsilon$ anti-commutes with $\beta$ . Hence, given that $\beta$ takes the form (11.32), and that $\epsilon^{\,2}=1$ , we can represent $\epsilon$ as the matrix

$$\epsilon = \left(\begin{array}{rr}0,&-{\rm i}\\ [0.5ex]{\rm i},&0\end{array}\right).$$ (11.166)

Thus, writing $\psi$ in the spinor form

$$\psi = \left(\begin{array}{c} \psi_a(r)\\ [0.5ex]\psi_b(r)\end{array}\right),$$ (11.167)

and making use of Equation (11.132), the energy eigenvalue problem for an electron in a central field reduces to the following two coupled radial differential equations:

$$\hbar\,c\left(\frac{d}{d r} + \frac{k+1}{r}\right)\psi_b + (E-m_e\,c^{\,2}+e\,\phi)\,\psi_a =0,$$ (11.168)

$$\hbar\,c\left(\frac{d}{d r}-\frac{k-1}{r}\right)\psi_a - (E+m_e\,c^{\,2}+e\,\phi)\,\psi_b = 0.$$ (11.169)