Exact solution of the 1D Dirac equation for a pseudoscalar interaction potential with the inverse-square-root variation law

The possibility of bound states is determined by the boundary conditions at $x = 0$. Let the solution for $x < 0$ that vanishes at infinity be $\psi_{{\text{L}}} = (\psi_{{{\text{1L}}}} ,\psi_{{{\text{2L}}}} )$ and it be proportional to a constant, say $C_{3}$. Then the continuity condition of the wave function:

$$\left. {\psi_{{{\text{1L}}}} } \right|_{x \to – 0} = \left. {\psi_{{{\text{1R}}}} } \right|_{x \to + 0} ,$$

(22)

$$\left. {\psi_{{{\text{2L}}}} } \right|_{x \to – 0} = \left. {\psi_{{{\text{2R}}}} } \right|_{x \to + 0} ,$$

(23)

presents a set of two homogeneous linear equations with respect to the constants $C_{1}$ and $C_{3}$. For a non-trivial solution, the determinant of this system must be zero:

$$\left. {\psi_{{{\text{1L}}}} } \right|_{x \to – 0} \left. {\psi_{{{\text{2R}}}} } \right|_{x \to + 0} – \left. {\psi_{{{\text{1R}}}} } \right|_{x \to + 0} \left. {\psi_{{{\text{2L}}}} } \right|_{x \to – 0} = 0.$$

(24)

Since we consider the odd extension, when $W\left( { – x} \right) = – W\left( x \right)$, the Dirac equation is covariant under the parity transformation $x \to \,- x$, and the solution for the negative $x$-region can be written as $\psi_{{\text{L}}} (x) = C_{3} \left( {\psi_{{{\text{1R}}}} ( – x), – \psi_{{{\text{2R}}}} ( – x)} \right)$. As a result, Eq. (24) reduces to $\psi_{{1{\text{R}}}} \left( 0 \right)\psi_{{2{\text{R}}}} \left( 0 \right) = 0$, and we obtain two branches of bound states, generated by $\psi_{{1{\text{R}}}} \left( 0 \right) = 0$ or $\psi_{{2{\text{R}}}} \left( 0 \right) = 0$. Let us consider these cases separately.

Bound states with $\psi_{{1{\text{R}}}} \left( 0 \right) = 0$

The equation for the bound states’ energy spectrum is

$$F\left( E \right) = 2\left( {a – 1} \right)H_{a – 2} (y_{0} ) + \left( {b_{0} – 2y_{0} } \right)H_{a – 1} (y_{0} ) = 0,\quad y_{0} = \frac{{4W_{0} W_{1} }}{{c^{2} \hbar^{2} \delta^{3/2} }}.$$

(25)

The behavior of $F$ as a function of energy is shown in Fig. 3.

Figure 3

The behavior of function $F(E)$ of Eq. (25) in the negative (left panel) and positive (right panel) energy regions. The points show the position of $\varepsilon = \pm mc^{2}$. The roots are all located in the intervals $\left( { – \sqrt {m^{2} c^{4} + W_{0}^{2} } , – mc^{2} } \right)$ and $\left( {mc^{2} ,\sqrt {m^{2} c^{4} + W_{0}^{2} } } \right)$. $f = 2^{a/2} e^{{y_{0}^{2} /2}} \sqrt {(a + 1)!} ,\quad \left( {m,c,\hbar ,W_{1} ,W_{2} } \right) = \left( {1,1,1, – 1/2,1} \right)$.

It can be shown that, provided $W_{0} W_{1} < 0$, the spectrum equation has infinitely many roots, all located in the intervals

$$E \in \left( { – \sqrt {m^{2} c^{4} + W_{0}^{2} } , – mc^{2} } \right)\quad and\quad E \in \left( {mc^{2} ,\sqrt {m^{2} c^{4} + W_{0}^{2} } } \right).$$

(26)

To construct an approximation for the spectrum, it is convenient to transform Eq. (25), using the recurrence relations between contiguous Hermite functions⁴⁰, into the form

$$\frac{{2ay_{0} + b_{0} \left( {a + 1 – 2y_{0}^{2} } \right)}}{{b_{0} y_{0} – a}}H_{a + 1} (y_{0} ) + H_{a + 2} (y_{0} ) = 0.$$

(27)

The advantage of this form is that the argument $y_{0}$ of the involved Hermite functions here is such that it belongs to the left transient region $y_{0} \approx \sqrt {2\nu – 1}$, where $\nu = a + 1$ or $\nu = a + 2$. Then, using the Airy-function approximation of the Hermite function for this region⁴¹, we arrive at an approximation of this equation as

$$\sin \left( {\pi a – \frac{{A_{0} \left( {E_{0} – E} \right) + A_{1} \left( {E – mc^{2} } \right)}}{{E_{0} – mc^{2} }}} \right) = 0,$$

(28)

where

$$E_{0} = \sqrt {m^{2} c^{4} + W_{0}^{2} } ,$$

(29)

$$A_{0} = \frac{\pi }{2} – \tan^{ – 1} \left( {\frac{{Ai\left( {2^{1/3} 3^{1/6} \left( {\sqrt 2 u^{3/4} – \sqrt 3 } \right)} \right)}}{{Bi\left( {2^{1/3} 3^{1/6} \left( {\sqrt 2 u^{3/4} – \sqrt 3 } \right)} \right)}}} \right),$$

(30)

$$A_{1} = \frac{\pi }{2} – \tan^{ – 1} \left( {\frac{{Ai^{\prime } \left( u \right) – \sqrt u Ai\left( u \right)}}{{Bi^{\prime } \left( u \right) – \sqrt u Bi\left( u \right)}}} \right),$$

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$${\text{and}}\quad u = \frac{{W_{1}^{4/3} }}{{\left( { – W_{0} c\hbar } \right)^{2/3} }}.$$

(32)

It is shown that $A_{0,1} < 1$ and for the solution of Eq. (29), $a > 1$. Furthermore, for large $a$, the energy $E$ is close to $E_{0}$: $E \approx E_{0}$. With these observations, for large a, we arrive at the approximate equation

$$\pi a – A_{1} = \pi n,\,\,\,\,\,n = 1,2,3,….$$

(33)

The dependence of $a$ on $E$ (see Eq. (19)) shows that this is a sextic polynomial equation in $E$ that is readily reduced to a cubic one. The real roots of the equation are given as

$$E_{n} = \pm \sqrt {m^{2} c^{4} + \left( {1 – e_{n} } \right)W_{0}^{2} } ,$$

(34)

where, using Cardano’s formula for the cubic,

$$\begin{gathered} e_{n} = \frac{1}{3b} + \frac{{2^{1/3} \left( {6b – 1} \right)}}{{3b\left( {18b – 2 – 27b^{2} + 3\sqrt {b^{3} \left( {81b – 12} \right)} } \right)^{1/3} }} – \\ \frac{{2^{ – 1/3} }}{3b}\left( {18b – 2 – 27b^{2} + 3\sqrt {b^{3} \left( {81b – 12} \right)} } \right)^{1/3} \\ \end{gathered}$$

(35)

$${\text{and}}\quad b = \left( {\frac{{c\hbar W_{0} }}{{W_{1}^{2} }}} \right)^{2} \left( {n – 1 + \frac{{A_{1} }}{\pi }} \right)^{2} .$$

(36)

This is a fairly accurate approximation. It provides the spectrum with relative error of the order of $10^{ – 4}$ or less (see Table 1 for a comparison with the exact numerical result). The normalized wave function on the entire $x$-axis is shown in Fig. 4 ($n = 3$, positive energy branch). It can be observed that the wave function is anti-symmetric with respect to the origin, as expected, and that the derivative of $\psi_{2} (x)$ is discontinuous at the origin.

Table 1 Comparison of approximation (34) with the numerical solution of exact Eq. (27). $\left( {m,c,\hbar ,W_{1} ,W_{2} } \right) = \left( {1,1,1, – 1/2,1} \right)$.

Bound states with $\psi_{{{\text{2R}}}} \left( 0 \right) = 0$

This time, the exact spectrum equation is written as

$$g\,H_{a + 1} (y_{0} ) + H_{a + 2} (y_{0} ) = 0,$$

(37)

$$g = \frac{{4\left( {2W_{1}^{2} – c^{2} \hbar^{2} \delta } \right)\left( {m^{2} c^{4} + W_{0}^{2} – E^{2} } \right) – 8W_{0} W_{1}^{2} \left( {c\hbar \delta – W_{0} } \right)}}{{W_{1} c^{2} \hbar^{2} \delta^{3/2} \left( {c\hbar \delta – 2W_{0} } \right)}},$$

(38)

Acting now essentially in the same way as in the previous case, we arrive at the spectrum expressed by the same formulas (34)–(36), with the parameter $A_{1}$ given as

$$A_{1} = – \tan^{ – 1} \left( {\frac{{\sqrt u Ai\left( u \right) + Ai^{\prime } \left( u \right)}}{{\sqrt u Bi\left( u \right) + Bi^{\prime } \left( u \right)}}} \right),$$

(39)

where $u$ is given by Eq. (32). The obtained result again is a fairly good approximation as seen from Table 2. The normalized wave function on the entire $x$-axis is shown in Fig. 5 ($n = 3$, positive energy branch). It can be observed that this time the derivative of $\psi_{1} (x)$ is discontinuous at the origin.

Table 2 Comparison of approximation (34) with the numerical solution of exact Eq. (37). $\left( {m,c,\hbar ,W_{1} ,W_{2} } \right) = \left( {1,1,1, – 1/2,1} \right)$.

As Eq. (36) shows, the Maslov index is given as

$$\gamma = – 1 + \frac{{A_{1} }}{\pi },$$

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where the parameter $A_{1}$ is different for the spectrum branches with $\psi_{1} (0) = 0$ and $\psi_{2} (0) = 0$. An interesting observation is that $A_{1}$ is not a constant but depends on the potential parameters $W_{0}$ and $W_{1}$. Figure 6 shows this dependence. As we can see, the Maslov index for the energy spectrum branch with $\psi_{1} (0) = 0$ starts from $- 1/6$, while that for the branch with $\psi_{2} (0) = 0$ starts from $- 5/6$ as $u = 0$. We note that both indices tend to $- 1$ as $u \to \infty$.