Motion of a Polarized Electron Possessing a Vacuum Magnetic Moment in Crossed Electromagnetic Fields

V. G. Bagrov, V. A. Bordovitsyn

New exact solutions of the Dirac equation are found for an electron in orthogonal constant and homogeneous electric and magnetic fields. The vacuum magnetic moment of the electron and the polarization of the electron spin are considered.

§ 1. The Dirac Equation for an Electron with an Anomalous Magnetic Moment

The motion of an electron in an external electromagnetic field is described by the Dirac equation

\[ i\hbar \frac{\partial \Psi}{\partial t}=H\Psi, \tag{1} \]

\[ H=c\rho_1(\vec{\sigma}\vec{P})+\rho_3mc^2-e\varphi, \tag{2} \]

where
\(\sigma, \rho\) are the Dirac matrices,

\[ \vec{P}=-i\hbar\nabla+\frac{e}{c}\vec{A} \]

is the kinetic momentum,

\(m\) is the rest mass of the electron,

\(e\) is the positive elementary charge,

\(\vec{A}\) and \(\varphi\) are the potentials of the external electromagnetic field,

\(c\) is the speed of light.

As Pauli showed [1], the Hamiltonian (2) can be generalized in such a way that the presence of an anomalous magnetic moment of the electron can be taken into account. The generalized Hamiltonian in this case has the form:

\[ H=c\rho_1(\vec{\sigma}\vec{P})+\rho_3mc^2-e\varphi+c\hbar\mu\,[\rho_3(\vec{\sigma}\vec{H})+\rho_2(\vec{\sigma}\vec{E})]. \tag{3} \]

Here

\[ c\hbar\mu \]

is the anomalous (vacuum) part of the moment of the electron,

\[ \vec{H}=\operatorname{rot}\vec{A} \quad \text{and} \quad \vec{E}=-\frac{1}{c}\frac{\partial \vec{A}}{\partial t}-\nabla\varphi \]

are the external magnetic and electric fields. Obviously, the Hamiltonian (2) is obtained from (3) in the special case \(\mu=0\).

Let us consider the motion of an electron in constant and homogeneous orthogonal electric and magnetic fields, and we shall assume \(E \ll H\).

We choose the coordinate system so that the \(z\)-axis is directed along the magnetic field and the \(y\)-axis along the electric-field vector. In accordance with our condition \(E < H\), we set \(E = H \sin \eta\), where \(\eta\) is some angle \(\left(-\dfrac{\pi}{2} < \eta < \dfrac{\pi}{2}\right)\).

It is convenient to choose the potentials of the electromagnetic field in the following form:

\[ A_x=-yH,\qquad A_y=A_z=0,\qquad \varphi=-yH\sin\eta . \]

It is easy to see that in this case the Hamiltonian (2) and (3) does not depend on time, and one may seek a stationary solution of equation (1).

It may also be established that the operators \(P_3=-i\hbar\dfrac{\partial}{\partial z}\) and \(p_1=-i\hbar\dfrac{\partial}{\partial x}\) commute with the Hamiltonian (2) and (3) and with each other and, consequently, are integrals of motion. Thus, the stationary solution of equation (1) may be subjected to two additional equations

\[ P_3\Psi=\hbar\kappa_3\Psi,\qquad p_1\Psi=\hbar\kappa_1\Psi . \tag{4} \]

Taking equations (1) into account, the stationary solution (1) is written in the form

\[ \Psi=L^{-1}\exp\{-icKt+i\kappa_1x+i\kappa_3z\}\psi(t), \]

\[ t=q\left(y-\frac{\kappa_2}{\gamma}\right),\qquad \gamma=\frac{eH}{c\hbar}. \tag{5} \]

Here

\(q\) and \(\kappa_2\) are as yet unknown constants,
the quantity \(K\) is related to the electron energy by \(\varepsilon=c\hbar K\).

The spinor \(\psi(t)\) satisfies the following system of equations:

\[ \begin{aligned} &\left[K-\kappa_0-\mu H-\left(\kappa_2+\frac{\gamma t}{q}\right)\sin\eta\right]\psi_1 \\ &\qquad -\left(\kappa_1-\kappa_2-\mu H\sin\eta-\frac{\gamma t}{q}-q\frac{d}{dt}\right)\psi_4 -\kappa_3\psi_3=0,\\[6pt] &\left[K-\kappa_0+\mu H-\left(\kappa_2+\frac{\gamma t}{q}\right)\sin\eta\right]\psi_2 \\ &\qquad -\left(\kappa_1-\kappa_2+\mu H\sin\eta-\frac{\gamma t}{q}+q\frac{d}{dt}\right)\psi_3 +\kappa_3\psi_4=0,\\[6pt] &\left[K+\kappa_0+\mu H-\left(\kappa_2+\frac{\gamma t}{q}\right)\sin\eta\right]\psi_3 \\ &\qquad -\left(\kappa_1-\kappa_2+\mu H\sin\eta-\frac{\gamma t}{q}-q\frac{d}{dt}\right)\psi_2 -\kappa_3\psi_1=0,\\[6pt] &\left[K+\kappa_0-\mu H-\left(\kappa_2+\frac{\gamma t}{q}\right)\sin\eta\right]\psi_4 \\ &\qquad -\left(\kappa_1-\kappa_2-\mu H\sin\eta-\frac{\gamma t}{q}+q\frac{d}{dt}\right)\psi_1 +\kappa_3\psi_2=0, \end{aligned} \tag{6} \]

where

\[ \kappa_0=\frac{mc}{\hbar}. \]

The solution of the system (6) is essentially different for \(|\sin \eta|<1\) and for

\[ |\sin \eta|=1 \left( \eta=\pm \frac{\pi}{2} \right). \]

§ 2. Solution of the Dirac equation for unequal fields

Consider the case \(|\sin \eta|<1\). In this case the solution of equations (6) can in principle be obtained by a Lorentz transformation of the solution of the Dirac equation for a purely magnetic field \((\eta=0)\). Such a solution is known [2]. It is easy to verify that in the case \(|\sin \eta|<1\) the solution of the system (6) should be sought in the form:

\[ \psi_i=\alpha_i U_n(t)+\beta_i U_{n-1}(t),\quad i=1,2,3,4. \tag{7} \]

Here \(\alpha_i\) and \(\beta_i\) are certain constants.

The Hermite functions \(U_n(t)\) are related to the Hermite polynomials \(H_n(t)\) by the following relation:

\[ U_n(t)=\frac{1}{\sqrt{2^n n!}}\sqrt[4]{\frac{q^2}{\pi}}\, e^{-\frac{t^2}{2}} H_n(t) \]

and satisfy the functional equations

\[ U_n'=\sqrt{2n}\,U_{n-1}-tU_n,\quad U_{n-1}'=tU_{n-1}-\sqrt{2n}\,U_n. \tag{8} \]

Substituting the expressions (7) into the system (6) and using the relations (3), we set equal to zero the coefficients in the equations at \(U_n'\), \(U_{n-1}\), \(tU_n\), \(tU_{n-1}\). We obtain altogether sixteen homogeneous linear equations for the eight unknowns \(\alpha_i\) and \(\beta_i\). Solving these equations, we come to the conclusion that a solution is possible if

\[ q=\sqrt{\gamma\cos\eta},\quad \kappa_2=\frac{\kappa_1-K\sin\eta}{\cos^2\eta}, \]

whence it follows, by virtue of (5),

\[ t=\sqrt{\gamma\cos\eta}\left(y+\frac{K\sin\eta-\kappa_1}{\cos^2\eta}\right). \tag{9} \]

The quantities \(\alpha_i\) and \(\beta_i\) then have the form:

\[ \alpha_1=c_4\sin\frac{\eta}{2},\quad \alpha_2=c_2\cos\frac{\eta}{2},\quad \alpha_3=c_2\sin\frac{\eta}{2},\quad \alpha_4=c_4\cos\frac{\eta}{2}; \tag{10} \]

\[ \beta_1=c_1\cos\frac{\eta}{2},\quad \beta_2=c_3\sin\frac{\eta}{2},\quad \beta_3=c_3\cos\frac{\eta}{2},\quad \beta_4=c_1\sin\frac{\eta}{2}. \]

The constants \(c_i\) satisfy the following system of equations:

\[ (d-\kappa_0-\mu H\cos\eta)c_1+\sqrt{2\gamma n\cos\eta}\,c_4-\kappa_3 c_3=0, \]

\[ (d-\kappa_0+\mu H\cos\eta)c_2+\sqrt{2\gamma n\cos\eta}\,c_3+\kappa_3 c_4=0, \]

\[ (d+\kappa_0+\mu H\cos\eta)c_3+\sqrt{2\gamma n\cos\eta}\,c_2-\kappa_3 c_1=0, \]

\[ (d+\kappa_0-\mu H\cos\eta)c_4+\sqrt{2\gamma n\cos\eta}\,c_1+\kappa_3 c_2=0, \]

\[ d=\frac{K-\kappa_1\sin\eta}{\cos\eta}. \tag{11} \]

For the function (5) to be normalized to unity, the relation

\[ \sum_{i=1}^{4} |c_i|^2 = 1. \tag{12} \]

must be satisfied.

The condition for solvability of equations (11) is that the determinant of the system vanish. Hence we obtain the electron energy levels

\[ K = \kappa_1 \sin \eta + \cos \eta \sqrt{\kappa_3^2 + \left(\sqrt{\kappa_0^2 + 2\gamma h \cos \eta} + \zeta_\mu H \cos \eta\right)^2}. \]

Here \(\zeta = \pm 1\). We shall subsequently relate the quantity \(\zeta\) to the orientation of the electron spin. Thus, taking account of the anomalous magnetic moment removes the spin degeneracy.

If in (11) one sets \(\mu = 0\) (i.e., does not take the anomalous magnetic moment of the electron into account), then, together with the determinant of the system, all third-order determinants vanish, and equations (11), even under condition (12), have no unique solution. This is due to the fact that, along with equations (4), the wave function may be subjected to an additional equation (describing the orientation of the electron spin). In the case \(\mu \ne 0\), the system (11), under condition (12), has a unique solution; however, as we shall see below, in this case as well the wave function may be subjected to an additional spin equation.

§ 3. Solution of the Dirac Equation for Fields Equal in Magnitude

Equations (6) also admit an exact solution in the case \(E = H\) \(\left(\eta = \dfrac{\pi}{2}\right)\). This case cannot be obtained by a Lorentz transformation from a purely magnetic field. It turns out that, in contrast to the case \(E < H\), the electron energy spectrum is continuous. The solution of equations (6) in this case should be sought in the form

\[ \psi_i = Q \left[\alpha_i \Phi(t) + \beta_i \Phi'(t)\right], \tag{13} \]

where \(\alpha_i\) and \(\beta_i\) are constant numbers,

\(Q\) is a certain normalization factor,

and \(\Phi(t)\) and \(\Phi'(t)\) are the Airy function and its derivative [3],

\[ \Phi(t) = \frac{1}{2\sqrt{\pi}} \int_{-\infty}^{\infty} e^{\,i\left(\frac{x^3}{3}+tx\right)}\,dx. \tag{14} \]

The Airy function satisfies the differential equation

\[ \Phi'' - t\Phi = 0. \tag{15} \]

Substituting expressions (13) into (6) under the condition \(\eta = \dfrac{\pi}{2}\) and setting equal to zero the coefficients of \(\Phi\), \(\Phi'\), \(t\Phi\), and \(t\Phi'\), we again obtain sixteen equations for the eight quantities \(\alpha_i\) and \(\beta_i\), which have a solution provided

\[ q = \sqrt[3]{2\gamma (K - \kappa_1)}, \]

\[ K = \kappa_2 + \zeta_\mu H + \sqrt{\kappa_0^2 + \kappa_3^2 + (\kappa_2 + \zeta_\mu H - \kappa_1)^2}. \]

In this case, from (5) it follows that

\[ t = \sqrt[3]{2\gamma (K - \kappa_1)} \left(y - \frac{\kappa_2}{\gamma}\right). \]

Here \(\zeta=\pm 1\) is connected, as will become clear below, with the orientation of the electron spin (therefore, for \(\mu\ne 0\), the spin degeneracy is also removed here). The quasi-momentum \(\kappa_2\) may be regarded here as a new quantum number. The sign before the root in the expression for the energy \(K\) has been chosen from the condition that the functions (13) must decrease in the positive direction of the \(y\)-axis (the electron moves against the electric field!). As is known, for large positive \(t\) the Airy function and its derivative decrease rapidly [3].

The solution of the equations for \(\alpha_i\) and \(\beta_i\) under these conditions has the form:

\[ \begin{aligned} \alpha_1&=A(K-\kappa_1+\kappa_0)+B\kappa_3,\\ \alpha_2&=A\kappa_3-B(K-\kappa_1+\kappa_0),\\ \alpha_3&=A\kappa_3+B(K-\kappa_1-\kappa_0),\\ \alpha_4&=-A(K-\kappa_1+\kappa_0)+B\kappa_3,\\ \beta_1&=\beta_4=\sqrt[3]{2\gamma(K-\kappa_1)}\,A,\\ \beta_2&=\beta_3=\sqrt[3]{2\gamma(K-\kappa_1)}\,B. \end{aligned} \tag{16} \]

The quantities \(A\) and \(B\) may be subjected to the condition

\[ AA^{+}+BB^{+}=1, \tag{17} \]

while the normalization will be taken into account by the factor \(Q\) in formula (13). If \(\mu=0\), then the coefficients \(A\) and \(B\), even under condition (17), are not determined uniquely from the Dirac equations and require an additional spin equation for their determination. If \(\mu\ne 0\), then, under condition (17), the following values are obtained for \(A\) and \(B\):

\[ A=\frac{1+\zeta}{2},\qquad B=\frac{1-\zeta}{2}. \tag{18} \]

Let us carry out the normalization of the wave functions (13). In states of the continuous spectrum the normalization may be performed either to \(\delta(K-K')\), or to \(\delta(\kappa_2-\kappa_2')\). The normalization of the functions (13) may be found in the following way.

We form the expression

\[ I(\chi)=\int e^{\chi y}\sum_{i=1}^{4}\psi_i^{\prime +}(t)\psi_i(t)\,dy, \]

where the prime on \(\psi_i\) denotes another state \((K')\), and \(\chi\) is some positive constant. We then compute this integral and require that, as \(\chi\to 0\), the function \(I(\chi)\to \delta(K-K')\). The expression \(I(\chi)\) can be calculated by using the following integral

\[ J(\mu,\nu;\mu',\nu';\chi)=\int_{-\infty}^{\infty}e^{\chi y}\Phi(\mu y-\nu)\Phi(\mu' y-\nu')\,dy, \]

which can be evaluated by means of an artificial device. We replace the function \(\Phi(\mu' y-\nu')\) by the integral representation (14) and change the order of integration. Then using the known integral

\[ \int_{-\infty}^{\infty} e^{\alpha x}\Phi(x)\,dx=\sqrt{\pi}\,e^{\frac{\alpha^3}{3}}, \]

one can finally obtain the integral \(J(\mu,\nu;\mu',\nu';\chi)\) in the following form:

\[ J(\mu,\nu;\mu',\nu';\chi) = \frac{\sqrt{\pi}}{\left|\mu^3-\mu'^3\right|^{1/3}}\,e^S \Phi(M), \]

\[ S=\chi\,\frac{\mu^2\nu-\mu'^2\nu'}{\mu^3-\mu'^3} + \frac{\chi^3}{3}\, \frac{\mu^3+\mu'^3}{(\mu^3-\mu'^3)^3}, \]

\[ M= \frac{(\mu'\nu-\mu\nu')(\mu^3-\mu'^3)+\mu\mu'\chi^2} {(\mu^3-\mu'^3)^{4/3}}. \]

After this, the calculation of \(I(\chi)\) presents no substantial difficulties. Knowing the integral \(I(\chi)\), one can obtain the normalization factor \(Q\) (and at the same time prove the orthogonality of the functions (13)).

\[ Q=\left[2\pi(K-\kappa_1)\sqrt[3]{2\gamma(K-\kappa_1)}\right]^{-1/2} \quad \text{for normalization to } \delta(K-K'), \]

\[ Q=\left[2\pi(K-\kappa_2)\sqrt[3]{2\gamma(K-\kappa_1)}\right]^{-1/2} \quad \text{for normalization to } \delta(\kappa_2-\kappa_2). \]

Let us note that the solution of the Schrödinger equation for crossed fields (see [4], p. 146), for any relation between \(E\) and \(H\), leads to a discrete spectrum. Such a situation indicates that the case \(E=H,\ E\perp H\) has no satisfactory nonrelativistic approximation. This is all the more surprising since the case of a purely electric field has a reasonable nonrelativistic approximation (see [3], motion in a homogeneous field), whereas the Dirac equation for a purely electric field has no correct solutions at all. Thus, the new exact correct solution of the Dirac equation obtained here is of undoubted physical interest.

§ 4. Polarization of the Electron Spin

As is known, the electron possesses four degrees of freedom. Therefore, along with equations (1) and (4), the wave function may satisfy one more additional equation describing the orientation of the electron spin. There are two types of covariant spin operators that are integrals of motion in our case.

It is easy to establish that the operator

\[ T=(\boldsymbol{\sigma}\mathbf{P})\sin\varphi+T_3\cos\varphi-T_1\sin\eta\sin\varphi, \qquad \mathbf{T}=mc\,\rho_3\boldsymbol{\sigma}+\rho_1\mathbf{P}, \]

which is the contraction of the four-dimensional spin vector-operator [5] with a certain constant 4-vector (\(\varphi\) is an arbitrary constant phase), commutes with the Hamiltonian (2) for \(\mu=0\), and also with the operators (4), and, consequently, is an additional spin integral of motion. In exactly the same way one can find that the operator

\[ L=M_2\sin\varphi\sin\eta+M_3\cos\varphi+\varepsilon_2\cos\varphi\sin\eta-\varepsilon_3\sin\varphi, \]

\[ \mathbf{M}=mc\,\boldsymbol{\sigma}+\rho_2(\boldsymbol{\sigma}\mathbf{P}) +\hbar\mu\{\mathbf{H}-(\boldsymbol{\sigma}\mathbf{E})\}, \qquad \boldsymbol{\varepsilon}=-\rho_3(\boldsymbol{\sigma}\mathbf{P}), \]

which is the contraction of the four-dimensional spin tensor of rank two [5] with a certain constant tensor (\(\varphi\) is again an arbitrary phase), commutes with the Hamiltonian (2) for \(\mu=0\) and with the operators (4), i.e., it is also a spin integral of motion. Thus, for \(\mu=0\) there are two spin operators. However, they do not commute with one another, and therefore, to describe spin states it is necessary to choose only one of them.

For \(\mu \ne 0\), the operator \(T\) ceases to be an integral of motion. The operator \(L\) remains an integral of motion only for \(\varrho=0\). Thus, the presence of an anomalous magnetic moment sharply restricts the choice of possible spin operators.

Let us first consider the case of unequal fields \((|\sin\eta|<1)\). Subjecting the wave function to the equation \((\mu=0)\),

\[ T\Psi=\zeta\hbar\lambda\Psi \quad (\zeta=\pm 1), \tag{19} \]

for the coefficients \(c_i\) we obtain, in addition to (11), the following system of equations:

\[ \begin{aligned} (\kappa_0\cos\varphi-\zeta\lambda)c_4 &+\bigl[(d-\kappa_0)\cos\eta\sin\varphi+\kappa_3\cos\varphi\bigr]c_2=0,\\ \bigl[(d+\kappa_0)\cos\eta\sin\varphi+\kappa_3\cos\varphi\bigr]c_4 &-(\kappa_0\cos\varphi+\zeta\lambda)c_2=0,\\ (\kappa_0\cos\varphi-\zeta\lambda)c_1 &+\bigl[(d+\kappa_0)\cos\eta\sin\varphi+\kappa_3\cos\varphi\bigr]c_3=0,\\ \bigl[(d-\kappa_0)\cos\eta\sin\varphi+\kappa_3\cos\varphi\bigr]c_1 &-(\kappa_0\cos\varphi+\zeta\lambda)c_3=0. \end{aligned} \tag{20} \]

From the solvability condition for system (20) we obtain

\[ \lambda=\sqrt{\kappa_0^{\,2}\bigl(\cos^2\varphi+\cos^2\eta\,\sin^2\varphi\bigr) +\bigl[(K-\kappa_1\sin\eta)+\kappa_3\cos\eta\bigr]^2}. \]

Solving (20) and (11) jointly, and taking (12) into account, we obtain the following form of the coefficients \(c_i\):

\[ c_1=A\sqrt{\frac{1}{2}\left(1+\frac{\kappa_0}{d}\right)}, \]

\[ c_2=B\sqrt{\frac{1}{2}\left(1+\frac{\kappa_0}{d}\right)}, \]

\[ c_3=\frac{-\sqrt{2\gamma n\cos\eta}\,B+\kappa_3 A} {\sqrt{2d(\kappa_0+d)}}, \]

\[ c_4=-\frac{\sqrt{2\gamma n\cos\eta}\,A+\kappa_3 B} {\sqrt{2d(\kappa_0+d)}}, \]

\[ A=\delta\sqrt{\frac{1}{2}(1+\zeta q)}, \]

\[ B=-\zeta\sqrt{\frac{1}{2}(1-\zeta q)}, \]

\[ q=\frac{(d+\kappa_0)(\kappa_3\cos\eta\sin\varphi+\kappa_0\cos\varphi)+\kappa_3^{\,2}\cos\varphi} {\lambda(d+\kappa_0)}, \]

\[ \delta=\frac{(d+\kappa_0)\cos\eta\sin\varphi+\kappa_3\cos\varphi} {\left|(d+\kappa_0)\cos\eta\sin\varphi+\kappa_3\cos\varphi\right|}. \]

Subjecting the wave function to the equation \((\mu=0)\)

\[ L\Psi=\zeta\hbar\lambda\Psi \quad (\zeta=\pm 1), \tag{22} \]

for the coefficients \(c_i\), in addition to (11), we obtain

\[ (d\cos\varphi\cos\eta-\zeta\lambda+i\kappa_3\cos\eta\sin\varphi)c_4+ \]

\[ +\bigl[i(d-\kappa_0\cos\eta\sin\varphi+\kappa_3\cos\eta\cos\varphi\bigr]c_2=0, \]

\[ \bigl[i(d+\kappa_0)\cos\eta\sin\varphi+\kappa_3\cos\eta\cos\varphi\bigr]c_4+ \]

\[ \begin{gathered} + (d\cos\eta\cos\varphi+\zeta\lambda+i\kappa_{3}\cos\eta\sin\varphi)c_{2}=0,\\ (d\cos\eta\cos\varphi-\zeta\lambda-i\kappa_{3}\cos\eta\sin\varphi)c_{1}+\\ + [i(d+\kappa_{0})\cos\eta\sin\varphi-\kappa_{3}\cos\eta\cos\varphi]c_{3}=0,\\ [i(d-\kappa_{0})\cos\eta\sin\varphi-\kappa_{3}\cos\eta\cos\varphi]c_{1}+\\ + (d\cos\eta\cos\varphi+\zeta\lambda-i\kappa_{3}\cos\eta\sin\varphi)c_{3}=0. \end{gathered} \tag{23} \]

From the condition that the determinant of the system (23) is equal to zero we have

\[ \lambda=\sqrt{(K-\kappa_{1}\sin\eta)^{2}-\cos^{2}\eta\left(\kappa_{3}^{2}+\kappa_{0}^{2}\sin^{2}\varphi\right)}. \]

The simultaneous solution of (23) and (11) leads to the result:

\[ \begin{gathered} c_{1}=\frac{\zeta}{2}\sqrt{\frac{1}{2}\left(1+\zeta\frac{\kappa_{0}}{\lambda_{0}}\cos\varphi\right)} \left( \sqrt{1+\frac{\kappa_{3}}{d}}\,e^{-i\frac{\varphi-\zeta\Phi}{2}} +\zeta\sqrt{1-\frac{\kappa_{3}}{d}}\,e^{i\frac{\varphi-\zeta\Phi}{2}} \right),\\ c_{2}=\frac{1}{2}\sqrt{\frac{1}{2}\left(1-\zeta\frac{\kappa_{0}}{\lambda_{0}}\cos\varphi\right)} \left( \sqrt{1+\frac{\kappa_{3}}{d}}\,e^{i\frac{\varphi+\zeta\Phi}{2}} -\zeta\sqrt{1-\frac{\kappa_{3}}{d}}\,e^{-i\frac{\varphi+\zeta\Phi}{2}} \right),\\ c_{3}=\frac{\zeta}{2}\sqrt{\frac{1}{2}\left(1+\zeta\frac{\kappa_{0}}{\lambda_{0}}\cos\varphi\right)} \left( \sqrt{1+\frac{\kappa_{3}}{d}}\,e^{-i\frac{\varphi-\zeta\Phi}{2}} -\zeta\sqrt{1-\frac{\kappa_{3}}{d}}\,e^{i\frac{\varphi-\zeta\Phi}{2}} \right),\\ c_{4}=-\frac{1}{2}\sqrt{\frac{1}{2}\left(1-\zeta\frac{\kappa_{0}}{\lambda_{0}}\cos\varphi\right)} \left( \sqrt{1+\frac{\kappa_{3}}{d}}\,e^{i\frac{\varphi+\zeta\Phi}{2}} +\zeta\sqrt{1-\frac{\kappa_{3}}{d}}\,e^{-i\frac{\varphi+\zeta\Phi}{2}} \right), \end{gathered} \tag{24} \]

where it is denoted that

\[ \lambda_{0}=\sqrt{\kappa_{0}^{2}\cos^{2}\varphi+2\gamma n\cos\eta}, \]

and the angle \(\Phi\) is determined by the relation:

\[ \sin\Phi=\frac{\kappa_{0}\sin\varphi}{\sqrt{\kappa_{0}^{2}+2\gamma n\cos\eta}}. \]

When the anomalous magnetic moment of the electron is taken into account, formulas (24) retain their form if one sets \(\varphi=\Phi=0\) in them.

Carrying out analogous calculations for the case of fields equal in magnitude, we obtain for the operator \(T\) (for \(\mu=0\)) the previous value \(\lambda\) (19), (20) under the condition \(\eta=\frac{\pi}{2}\) and the following coefficients \(A\) and \(B\):

\[ A=\zeta\sqrt{\frac{1}{2}\left(1+\zeta\frac{\kappa_{0}}{\lambda}\cos\varphi\right)}, \]

\[ B=\lambda\sqrt{\frac{1}{2}\left(1-\zeta\frac{\kappa_{0}}{\lambda}\cos\varphi\right)}, \]

\[ \lambda=\frac{(K-\kappa_{1})\sin\varphi+\kappa_{3}\cos\varphi} {\left|(K-\kappa_{1})\sin\varphi+\kappa_{3}\cos\varphi\right|}. \]

For the operator \(L\) (22) in the case \(\mu=0\) we have

\[ A=i\sqrt{\frac{1+\zeta\cos\varphi}{2}},\qquad B=\zeta\sqrt{\frac{1-\zeta\cos\varphi}{2}}. \]

For $\mu \ne 0$, $A$ and $B$ are determined by formulas (18). Thus, the problem of finding solutions of the Dirac equation for a polarized electron moving in crossed fields has been completely solved.

We see that taking the orientation of the electron spin into account makes it possible to determine the spin coefficients in the wave function uniquely. The presence of the electron’s anomalous magnetic moment sharply restricts the choice of possible spin operators and leads to the removal of spin degeneracy.

REFERENCES

1. W. Pauli. Rev. of Mod. Phys. 13, 203, 1941.

2. A. I. Akhiezer, V. B. Berestetskii. Quantum Electrodynamics. Fizmatgiz, 1959.

3. L. D. Landau, E. M. Lifshitz. Quantum Mechanics. Fizmatgiz, 1963.

4. I. I. Gol’dman, V. D. Krivchenkov. Collection of Problems in Quantum Mechanics. GITTL, 1967.

5. I. M. Ternov, V. G. Bagrov, V. A. Bordovitsyn. Izv. vyssh. uch. zav., “Fizika,” No. 4, 41, 1967.