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Next: Fine Structure of Hydrogen Up: Relativistic Electron Theory Previous: Electron Spin

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},$ (11.128)

it is convenient to transform to polar coordinates. Let

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

and

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

It is easily demonstrated that

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

which implies that in the Schrödinger representation

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

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

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

$\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}).$ (11.134)

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

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

(See Section 4.1.) 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}.$ (11.136)

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

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

Further application of Equation (11.101) 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},$ (11.138)
$\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},$ (11.139)

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

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

(See Section 2.2.) 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},$ (11.141)

which implies that

$\displaystyle \{$$\displaystyle \mbox{\boldmath$\Sigma$}$$\displaystyle \cdot {\bf L}+ \hbar,\,$   $\displaystyle \mbox{\boldmath$\Sigma$}$$\displaystyle \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

$\displaystyle \{$$\displaystyle \mbox{\boldmath$\Sigma$}$$\displaystyle \cdot {\bf L}+ \hbar,\,$   $\displaystyle \mbox{\boldmath$\alpha$}$$\displaystyle \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

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

where

$\displaystyle \zeta = \beta\left(\mbox{\boldmath$\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

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

we find that

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

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

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

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

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

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

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},$ (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

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

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

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

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

Moreover, it is evident that

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

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}\,x^{\,i}=r^{\,2},$ (11.155)

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

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

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

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

$\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}.$ (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

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

Equation (11.131) then yields

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

Equation (11.152) 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).$ (11.161)

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

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

Hence, the Hamiltonian (11.128) becomes

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

Now, we wish to solve the energy eigenvalue problem

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

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

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

Thus, writing $ \psi$ in the spinor form

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

$\displaystyle \hbar\,c\left(\frac{d}{d r} + \frac{k+1}{r}\right)\psi_b + (E-m_e\,c^{\,2}+e\,\phi)\,\psi_a$ $\displaystyle =0,$ (11.168)
$\displaystyle \hbar\,c\left(\frac{d}{d r}-\frac{k-1}{r}\right)\psi_a - (E+m_e\,c^{\,2}+e\,\phi)\,\psi_b$ $\displaystyle = 0.$ (11.169)


next up previous
Next: Fine Structure of Hydrogen Up: Relativistic Electron Theory Previous: Electron Spin
Richard Fitzpatrick 2016-01-22