Analytical Results for Classical Axially Symmetric States of Relativistic Hydrogenic Ions in Collinear Electric and Magnetic Fields of Arbitrary Strengths

Abstract

We consider classical axially symmetric states of relativistic hydrogenic ions under collinear electric and magnetic fields of arbitrary strengths. For such Rydberg states, we study the role of relativistic effects. Our analytical results demonstrate that the primary outcome of the allowance for the relativism is an increase in the critical value of the electric field at which the ionization occurs.

1. Introduction

Axially symmetric highly excited states of hydrogen-like systems correspond to the quantum numbers |m| = n −1 >> 1, with m and n being magnetic and principal quantum numbers, respectively. Highly excited states of this symmetry are also called circular Rydberg states (CRS), and they are the subject of lots of theoretical and experimental works, e.g., papers [1,2,3,4,5,6,7,8] and the references therein.

In paper [9], one of us focused on the analytical classical description of CRS in collinear electric (F) and magnetic (B) fields. Exact analytical expressions were obtained in paper [9] for the energy, E, of the atomic electron under the collinear electric and magnetic fields of arbitrary strengths.

In the present paper, we consider the same system but with the allowance for relativistic effects. This research may contribute to studies of ion beam focusing in magnetic fusion devices, e.g., paper [10] and the references therein. The results of our study may have some applications to pulsar astronomy dealing with relativistic effects in super high magnetic fields, e.g., papers [11,12] and the references therein.

2. Results

We study axially symmetric highly excited states of a hydrogen-like system (an atom or an ion) with the nucleus of charge Z, which is at the origin, acted upon by the collinear electric and magnetic fields F || B accordingly. The Oz-axis is in the direction of the angular momentum, L, of the electron. The electric field, F, is assumed to be parallel to L, so that F_z > 0 in this coordinate system. Thus, the magnetic field, B, can be either parallel (B_z > 0) or antiparallel (B_z < 0) to L. The electron is in the circular state, and its orbit, whose radius is ρ, is perpendicular to the z-axis, with the center of the orbit having the coordinate z. The classical relativistic Hamiltonian for the electron in this configuration is

$$H=\sqrt{{\left(cP-eA\right)}^2+m^2{c}^4}-m{c}^2-\frac{Z{e}^2}{r}+Fz$$

(1)

where P is the canonical momentum, A is the vector potential, m and e are the electron mass and charge, respectively, c is the speed of light, Z is the nuclear charge, F is the magnitude of the electric field, z is the axial coordinate, and r is the distance of the electron from the nucleus. We use the atomic units m = e = ħ = 1. We take the vector potential to be A = 1/2 r × B, use the approximation P << mc and the fact that, for unperturbed circular orbits, P² = L²/(2ρ²), and write the Hamiltonian in cylindrical coordinates, where r = (z² + ρ²)^1/2:

$$\begin{array}{c}H=\left(1-\frac{\mathsf{\Omega }L}{c^2}\right)\frac{L^2}{2{\mathsf{\rho }}^2}-\frac{{\left(E_0+\frac{Z}{\sqrt{z^2+{\mathsf{\rho }}^2}}\right)}^2}{2c^2}+\Omega L+\\ +\frac{{\mathsf{\Omega }}^2{\mathsf{\rho }}^2}{2}-\frac{{\mathsf{\Omega }}^2{\mathsf{\rho }}^2\left(E_0+\frac{Z}{\sqrt{z^2+{\mathsf{\rho }}^2}}\right)}{2c^2}-\frac{{\mathsf{\Omega }}^2{L}^2}{2c^2}-\frac{{\mathsf{\Omega }}^{3}L{\mathsf{\rho }}^2}{2c^2}-\frac{{\mathsf{\Omega }}^4{\mathsf{\rho }}^4}{8c^2}-\frac{Z}{\sqrt{z^2+{\mathsf{\rho }}^2}}+Fz\end{array}$$

(2)

where Ω = B/(2c), ρ is the radial cylindrical coordinate, L is the magnitude of the angular momentum, and E₀ is the energy of the non-relativistic unperturbed electron. Using the scaled quantities

$$w=\frac{Z}{L^2}z,v=\frac{Z}{L^2}\mathsf{\rho },p=v^2,\mathsf{\omega }=\frac{L^3}{Z^2}\mathsf{\Omega },f=\frac{L^4}{Z^3}F,{\mathsf{\epsilon }}_0=\frac{L^2}{Z^2}{E}_0,\mathsf{\gamma }={\left(\frac{Z}{cL}\right)}^2,h=\frac{L^2}{Z^2}H$$

(3)

we write the scaled Hamiltonian, h:

$$h=\frac{1}{2p}-\frac{1}{\sqrt{w^2+p}}+\mathsf{\omega }+\frac{p{\mathsf{\omega }}^2}{2}+fw-\frac{\mathsf{\gamma }}{2}\left({\left(\frac{1}{\sqrt{w^2+p}}+{\mathsf{\epsilon }}_0\right)}^2+\left(\frac{1}{\sqrt{w^2+p}}+{\mathsf{\epsilon }}_0\right)p{\mathsf{\omega }}^2+p{\mathsf{\omega }}^3+\frac{p^2{\mathsf{\omega }}^4}{4}+\frac{\mathsf{\omega }}{p}+{\mathsf{\omega }}^2\right)$$

(4)

The unperturbed non-relativistic orbit has the eccentricity

$${\mathsf{\epsilon }}_e=\sqrt{1+\frac{2E_0{L}^2}{Z^2}}=\sqrt{1+2{\mathsf{\epsilon }}_0}$$

(5)

so the circular orbit, whose eccentricity is zero, corresponds to the scaled unperturbed energy ε₀ = −1/2, which we substitute into (4):

$$h=\frac{1}{2p}-\frac{1}{\sqrt{w^2+p}}+\mathsf{\omega }+\frac{p{\mathsf{\omega }}^2}{2}+fw-\frac{\mathsf{\gamma }}{2}\left({\left(\frac{1}{\sqrt{w^2+p}}-\frac{1}{2}\right)}^2+\left(\frac{1}{\sqrt{w^2+p}}-\frac{1}{2}\right)p{\mathsf{\omega }}^2+p{\mathsf{\omega }}^3+\frac{p^2{\mathsf{\omega }}^4}{4}+\frac{\mathsf{\omega }}{p}+{\mathsf{\omega }}^2\right)$$

(6)

For the non-relativistic case, corresponding to γ = 0, (6) coincides with Equation (3) in paper [9].

To find the equilibrium points in the scaled (w, p) coordinate space, the derivatives of (6) with respect to both coordinates should vanish, which gives us the following two equations:

$$\frac{\partial h}{\partial w}=\frac{w}{{\left(w^2+p\right)}^{3/2}}\left(1+\frac{{\left(w^2+p\right)}^{3/2}}{w}f+\mathsf{\gamma }\left(\frac{1}{\sqrt{w^2+p}}+\frac{p{\mathsf{\omega }}^2-1}{2}\right)\right)=0$$

(7)

$$\frac{\partial h}{\partial p}=\frac{1}{2}\left(\frac{1}{{\left(w^2+p\right)}^{3/2}}-\frac{1}{p^2}+{\mathsf{\omega }}^2-\mathsf{\gamma }\left(\left(\frac{1}{\sqrt{w^2+p}}+\frac{p{\mathsf{\omega }}^2-1}{2}\right)\left({\mathsf{\omega }}^2-\frac{1}{{\left(w^2+p\right)}^{3/2}}\right)+\mathsf{\omega }\left({\mathsf{\omega }}^2-\frac{1}{p^2}\right)\right)\right)=0$$

(8)

We solve (7) for f:

$$f\left(w,p,\mathsf{\omega },\mathsf{\gamma }\right)=-\frac{w}{{\left(w^2+p\right)}^{3/2}}\left(1+\mathsf{\gamma }\left(\frac{1}{\sqrt{w^2+p}}+\frac{p{\mathsf{\omega }}^2-1}{2}\right)\right)$$

(9)

Then, if we define

$$s=\frac{1}{\sqrt{w^2+p}}$$

(10)

and substitute it into (8), we obtain an equation that is a 4th-degree polynomial with respect to s:

$$\mathsf{\gamma }{s}^4+\left(1+\frac{\mathsf{\gamma }}{2}\left(p{\mathsf{\omega }}^2-1\right)\right)s^3-{\mathsf{\gamma }\mathsf{\omega }}^{2}s+\left(1-\mathsf{\gamma }\mathsf{\omega }\right)\left({\mathsf{\omega }}^2-\frac{1}{p^2}\right)-\frac{\mathsf{\gamma }}{2}{\mathsf{\omega }}^2\left(p{\mathsf{\omega }}^2-1\right)=0$$

(11)

By solving this equation for s and selecting the relevant roots, we obtain the solution s(p, ω, γ), and we express the solution for w from (10):

$$w\left(p,\mathsf{\omega },\mathsf{\gamma }\right)=-\sqrt{\frac{1}{s^2\left(p,\mathsf{\omega },\mathsf{\gamma }\right)}-p}$$

(12)

(w is negative for F_z > 0). Then, we substitute (12) into the equation for f given in (9) and obtain the expression for the scaled electric field depending on the squared scaled orbit radius, p, for the given values of ω and γ:

$$f\left(p,\mathsf{\omega },\mathsf{\gamma }\right)=f\left(w\left(p,\mathsf{\omega },\mathsf{\gamma }\right),p,\mathsf{\omega },\mathsf{\gamma }\right)$$

(13)

Then, we substitute the expression for w from (12) and the expression for f from (13) into (6) and obtain the energy of the electron, depending on the squared scaled orbit radius, p, for the given values of ω and γ:

$$\mathsf{\epsilon }\left(p,\mathsf{\omega },\mathsf{\gamma }\right)=h\left(w\left(p,\mathsf{\omega },\mathsf{\gamma }\right),p,f\left(p,\mathsf{\omega },\mathsf{\gamma }\right),\mathsf{\omega },\mathsf{\gamma }\right)$$

(14)

Equations (13) and (14) represent a parametric dependence of the scaled energy, ε, on the scaled electric field, f, with the parameter, p, for the given values of ω and γ.

In the absence of the electric field, w = 0 from (9). Substituting w = 0 into (8), we obtain an implicit dependence, p(ω), for the given γ in the absence of the electric field. Figure 1 presents this dependence for γ = 0.1, corresponding to the nuclear charge Z = 43 and the angular momentum L = 1.

It is seen that the relativistic effect compresses the orbit in the case of positive ω and, starting from some negative value of ω, expands the orbit, as the absolute value of ω increases from that point. The dependence of the effect on whether the magnetic field is parallel or antiparallel to the orbital momentum is caused by the term ΩL in the Hamiltonian from Equation (2), which is the term whose sign depends on the direction of the magnetic field.

By equating the values of (8) at γ = 0 and a non-zero γ (both for the case of f = 0, i.e., w = 0), this point is numerically found to be at ω_c = −0.6753 and p_c = 0.6509 (we remind the readers that these, as well as the other quantities introduced in Equation (3), are dimensionless). It is also seen that the magnetic field shifts the maximum of the curve in Figure 1 to some negative value of ω, corresponding to the antiparallel configuration of vectors B and L.

Solving (8) numerically at w = 0 for p, we find the minimum value of p corresponding to given ω and γ values, and we take the maximum value to be 1/|ω|, as in the non-relativistic case. We plot the parametric dependence, ε(f), of the scaled energy of the electron on the scaled electric field, with the parameter, p, varying in the range determined by the above-mentioned limits, for various values of the magnetic field in the case of Z = 14 and L = 1 (γ = 0.01). We compare the relativistic plots with the non-relativistic plots in Figure 2.

It is seen that the allowance for the relativism decreases the energy for ω > ω_c and increases the energy for ω < ω_c. Again, the dependence of the effect on whether the magnetic field is parallel or antiparallel to the orbital momentum is caused by the term ΩL in the Hamiltonian from Equation (2).

It is important to clarify the following. In each of the plots in Figure 2, there are two branches of the energy. The lower branch corresponds to the stable motion (bound state), while the upper branch corresponds to the unstable motion leading to the ionization. The value of the electric field where the two branches meet represents the ionization threshold. Thus, in Figure 2 it is also seen that the relativistic effects increase the critical electric field required for ionization.

Figure 3 shows the dependence of the electric field at the ionization threshold on the scaled magnetic field, ω. It is seen that the relativistic effect increases the critical electric field, f_c, corresponding to the ionization threshold. Physically, this is because, with the allowance for the relativism, the mass of the bound electron increases so that it becomes more difficult to push it away from the nucleus.

Figure 4 shows the dependence of the energy at the ionization threshold on the scaled magnetic field, ω. It is seen that the relativism decreases the energy at the ionization threshold for the scaled magnetic field above the critical value and increases it for the scaled magnetic field below the critical value. The dependence of the effect on whether the magnetic field is parallel or antiparallel to the orbital momentum is caused by the term ΩL in the Hamiltonian from Equation (2).

3. Conclusions

We studied the role of relativistic effects for the axially symmetric states of hydrogenic ions under collinear electric and magnetic fields of arbitrary strengths. Here are the main conclusions:

We demonstrated analytically that the primary outcome of the allowance for the relativism is an increase in the critical value of the electric field at which the ionization occurs. In other words, relativistic effects work as the stabilizing factor. Physically, this is because, with the allowance for the relativism, the mass of the bound electron increases so that it becomes more difficult to push it away from the nucleus.

We also demonstrated analytically the existence of the critical value of the magnetic field. The sign of the effect of the relativism on the orbit radius of the electron and on the electron energy (including its energy at the classical ionization threshold) depends on whether the magnetic field is above or below the critical value.