Reports of the Academy of Sciences of the USSR
1957. Volume 114, No. 1

PHYSICS

G. A. ZAITSEV

SHIFT OF THE ENERGY LEVELS OF A PARTICLE WITH SPIN \(1/2\) IN A COULOMB FIELD

(Presented by Academician N. N. Bogoliubov, 15 XII 1956)

Let a particle with spin \(1/2\) and negative charge (for example, a \(\mu^-\)-meson or an electron) be in a field with potential \(\varphi = Ze/r\), \(A_k = 0\). We shall denote the charge of the particle by \(-e\) \((e>0)\). Using the basic equation for a particle with spin \(1/2\) in an electromagnetic field, considered in \((^1)\), we find the corresponding energy levels.

We shall assume that the particle is in a stationary state, so that

\[ \hbar i \frac{\partial}{\partial t}\xi = E\xi,\qquad \xi = [\exp(-iEt/\hbar)]\xi_0 . \tag{1} \]

Equation (7) from \((^1)\) gives:

\[ \hat H \xi_0 = 0, \]

\[ \hat H = \frac{1}{r^2}\left(\frac{\partial}{\partial r}r^2\frac{\partial}{\partial r}\right) + \frac{1}{\hbar^2 c^2} \left[\left(E+\frac{Ze^2}{r}\right)^2 - m_0^2 c^4\right] + \frac{\nabla_{\vartheta\varphi}^2 + iZ\alpha(1+\delta)\sigma_{(r)}}{r^2}, \tag{2} \]

where

\[ \alpha=\frac{e^2}{\hbar c},\qquad \sigma_{(r)}=\frac{\sum\limits_k x^k\sigma_k}{r},\qquad \nabla_{\vartheta\varphi}^2 = \frac{1}{\sin\vartheta}\frac{\partial}{\partial\vartheta} \left(\sin\vartheta\frac{\partial}{\partial\vartheta}\right) + \frac{1}{\sin^2\vartheta}\frac{\partial^2}{\partial\varphi^2}. \]

We shall be able to achieve separation of variables if we put

\[ \xi_0 = f(r) \begin{pmatrix} y_1(\vartheta,\varphi)\\ y_2(\vartheta,\varphi) \end{pmatrix}, \]

where

\[ -\left[\nabla_{\vartheta\varphi}^2+iZ\alpha(1+\delta)\sigma_{(r)}\right] \begin{pmatrix} y_1\\ y_2 \end{pmatrix} = \hat L y = \lambda y . \tag{3} \]

Then the energy levels will be determined from the equation

\[ \left\{ \frac{\partial^2}{\partial r^2} + \frac{2}{r}\frac{\partial}{\partial r} + \frac{1}{\hbar^2 c^2} \left[\left(E+\frac{Ze^2}{r}\right)^2-m_0^2c^4\right] - \frac{\lambda}{r^2} \right\}f(r)=0. \tag{4} \]

It is not difficult to verify that the operators

\[ \hat M_k = \hbar\left( -i\sum_{j,s=1}^{3}\varepsilon_{kjs}x^j\frac{\partial}{\partial x^s} + \frac{1}{2}\sigma_k \right) \tag{5} \]

commute with \(\hat H\) and \(\hat L\).

Moreover, the operators \(\hat L\), \(\hat M_k\), and \(\hat M^2=\hat M_1^2+\hat M_2^2+\hat M_3^2\) commute with one another. Therefore we may choose \(y_1\) and \(y_2\) so that \(y\), and consequently \(\xi_0\), will simultaneously be an eigenfunction of the operators \(\hat L\), \(\hat M_3\), and \(\hat M^2\), i.e.

\[ \hat M_3 y = \hbar m y, \tag{6} \]

\[ \hat M^2 y = \hbar^2 j(j+1)y, \tag{7} \]

and the state of the particle will be characterized by the quantum numbers \(\lambda, m, j\) and by the radial quantum number, which we shall introduce below.

From the commutation relations between \(\hat M_k\), as is known, it follows that for a given \(j\), \(m\) can take the values \(m=-j,-j+1,\ldots,\ldots,j-1,j\). It is easy to verify that from conditions (6) and (7) it follows that \(y\) has the form

\[ y=\binom{y_1}{y_2}= \tag{8} \]

\[ = \begin{pmatrix} \left(C_{j-\frac12}P_{j-\frac12,\,m-\frac12}+C_{j+\frac12}P_{j+\frac12,\,m-\frac12}\right)e^{i(m-\frac12)\varphi} \\[6pt] \left( -\sqrt{\frac{j-m}{j+m}}\,C_{j-\frac12}P_{j-\frac12,\,m+\frac12} +\sqrt{\frac{j+m+1}{j-m+1}}\,C_{j+\frac12}P_{j+\frac12,\,m+\frac12} \right)e^{i(m+\frac12)\varphi} \end{pmatrix}, \]

where \(C_{j-\frac12}, C_{j+\frac12}\) are arbitrary constants;

\[ P_{j-\frac12,\,m-\frac12} = \sqrt{ j\,\frac{(j-m)!}{(j+m-1)!} } (1-x^2)^{\frac{m-\frac12}{2}} \frac{1}{2^{\,j-\frac12}(j-\frac12)!} \times \]

\[ \times \frac{d^{\,j+m-1}}{dx^{\,j+m-1}} (x^2-1)^{j-\frac12}, \qquad x=\cos\vartheta . \tag{9} \]

Using the properties of the functions \(P\) (see, for example, (2), pp. 383–384), from (3) we find the ratio \(C_{j+\frac12}:C_{j-\frac12}\) and the third quantum number \(\lambda\):

\[ \lambda=(j+\tfrac12)^2\mp \sqrt{(j+\tfrac12)^2-\alpha^2(1+\delta)^2} \tag{10} \]

(here and in the subsequent formulas, for simplicity, we put \(Z=1\)).

We see that the quantum number \(\lambda\), for given \(j\) and \(m\), can take only two values, corresponding to the two possible choices of sign in (10). If the quantity \(\alpha^2\) is neglected and in (10) one sets \(\alpha^2=0\), then for this limiting case, instead of the quantum number \(\lambda\), we may introduce another quantum number \(l\), defined according to

\[ \lambda=l(l+1),\qquad l=j\mp \tfrac12 . \tag{11} \]

In order to preserve correspondence with the terminology adopted in spectroscopy, one may characterize the state of a particle with spin \(1/2\) by the number \(l\) (in addition to \(j\) and \(m\)) also in the case when we do not neglect the quantity \(\alpha^2\). To this end we put \(l=j-\tfrac12\) or \(l=j+\tfrac12\), depending on the sign in (10). Then the quantum number \(\lambda\) may be expressed through \(j\) and \(l\) by the formula

\[ \lambda=(j+\tfrac12)^2+2(l-j) \sqrt{(j+\tfrac12)^2-\alpha^2(1+\delta)^2}. \tag{12} \]

Now let us consider the equation for the radial function (4), writing it in the form:

\[ \left( \frac{\partial^2}{\partial r^2} +\frac{2}{r}\frac{\partial}{\partial r} -A+\frac{2B}{r}-\frac{C}{r^2} \right)f(r)=0, \tag{13} \]

where

\[ A=\frac{m_0^2c^2}{\hbar^2} \left[ 1-\left(1+\frac{\mathcal E}{m_0c^2}\right)^2 \right], \qquad \mathcal E=E-m_0c^2, \qquad B=\frac{m_0e^2}{\hbar^2} \left(1+\frac{\mathcal E}{m_0c^2}\right), \]

\[ C=\lambda-\alpha^2 =(j+\tfrac12)^2\mp \sqrt{(j+\tfrac12)^2-\alpha^2(1+\delta)^2} -\alpha^2 =l'(l'+1),\qquad l'>0 . \tag{14} \]

The solution of this equation is well known (see, for example, (3), p. 438). In particular, in order to obtain discrete energy levels \(\mathcal E\), it is necessary to introduce the radial quantum number \(k\) by the formula

\[ B/\sqrt{A}=l'+k+1. \tag{15} \]

\(k\) can take the values \(k=0,1,2,\ldots\). The energy levels are determined from the formula

\[ 1+\frac{\mathcal{E}}{m_0c^2} = \left[ 1+\frac{\alpha^2}{(l'+k+1)^2} \right]^{-1/2} = \left[ 1+\frac{\alpha^2}{\left(k+\frac12+\sqrt{\frac14+\lambda-\alpha^2}\right)^2} \right]^{-1/2}. \tag{16} \]

Expanding the right-hand side of (16) in powers of \(\alpha^2\) and retaining terms of order \(\alpha^4\), we shall have

\[ \frac{\mathcal{E}}{h} = -\frac{R}{n^2} \left[ 1+\frac{\alpha^2}{n^2} \left( \frac{n}{j+\frac12}-\frac34 \right) + \frac{\alpha^2\left(\delta+\frac12\delta^2\right)} {n\left(j+\frac12\right)\left(j+\frac12\mp\frac12\right)} \right], \tag{17} \]

where

\[ R=\frac{m_0e^4}{4\pi\hbar^3}=\frac{\alpha^2m_0c^2}{2h},\qquad n=j+1\mp\frac12+k=l+k+1. \]

Thus, the deviation of the magnetic moment of the particle from the corresponding magneton, characterized by the quantity \(\delta\), leads to a shift of the energy levels in comparison with the case when the Dirac equation is used without taking this deviation into account. In particular, the distance between the levels \(2^2S_{1/2}\) and \(2^2P_{1/2}\), according to (17), is now given by the following formula:

\[ h\left[\nu\left(2^2S_{1/2}\right)-\nu\left(2^2P_{1/2}\right)\right] = -\frac{h}{3}R\alpha^2\left(\delta+\frac12\delta^2\right). \tag{18} \]

Received
28 X 1954

REFERENCES

\(^{1}\) G. A. Zaitsev, DAN, 113, No. 6 (1957).
\(^{2}\) H. Bethe, Quantum Mechanics of the Simplest Systems, 1935.
\(^{3}\) A. A. Sokolov, D. D. Ivanenko, Quantum Field Theory, 1952.