Relativistic linear oscillator under the action of a constant external force. Transition amplitudes and the Green’s function

Abstract

We discuss an exactly solvable relativistic model of a nonrelativistic linear harmonic oscillator in the presence of a constant external force. We show that as in the nonrelativistic case, the relativistic linear oscillator in an external uniform field is unitarily equivalent to the oscillator without this field. Using two methods, we calculate transition amplitudes between energy states of the discrete spectrum of the relativistic linear oscillator under the action of a suddenly applied uniform field. We find Barut–Girardello coherent states and the Green’s function in the coordinate and momentum representations. We obtain the linear and bilinear generating functions for the Meixner–Pollaczek polynomials. We prove that the relativistic wave functions, the generators of the dynamical symmetry group, and the transition amplitudes have the correct nonrelativistic limit.

Appendix

We prove equality (4.7). Using the Baker–Hausdorff formula [40]

$$ e^{zK_+^0 - z^{*}K_-^0} = e^{e^{i\theta}tK_+^0} e^{u\Gamma_0^0} e^{e^{-i\theta}tK_-^0},$$

(A.1)

where $z = re^{i\theta}$, $t = \operatorname{tanh} r$, $u = - 2\ln\cosh r = \ln(1 - t^2)$, we represent operator $S^g$ (4.7) in the disentangled normal form

$$ S^g = e^{itK_+^0}e^{u\Gamma_0^0}e^{itK_-^0}e^{i\gamma\Gamma_0^0},$$

(A.2)

where, in our case, $z = i\beta/2$, $t = \operatorname{tanh}(\beta/2) = \rho/(1 + \delta)$, and $u = - 2\ln\cosh(\beta/2)$. Operator (A.1) can also be written in the disentangled antinormal form

$$ e^{zK_+^0 - z^{*}K_-^0} = e^{-e^{-i\theta}tK_-^0} e^{-u\Gamma_0^0} e^{e^{i\theta}tK_+^0}.$$

(A.3)

On the other hand, using operator representation (4.6) for the orthonormalized states of the linear harmonic oscillator in a uniform external field, we obtain

$$ S^g\psi_n^0 = \frac{1}{\sqrt{n!(2\nu)_n}}S^g (-K_+^0)^n \psi^0_0 = \frac{1}{\sqrt{n!(2\nu)_n}} (-K_+^g)^n S^g\psi^0_0.$$

(A.4)

Now it suffices to show that $S^g\psi_0^0 = \psi_0^g$. We have

$$ S^g\psi_0^0 = e^{(i\gamma + u)\nu}\sum_{n = 0}^{\infty} (-it)^n \sqrt{\frac{(2\nu)_n}{n!}} \psi^0_n.$$

(A.5)

Below, we calculate in the $x$-space. We substitute the explicit form

$$\psi_n^0(x)=\psi_n^g(x)|_{g = 0}$$

in (A.5) and use the generating function for the Meixner–Pollaczek polynomials [38],

$$\sum_{n = 0}^{\infty}z^{n}P_n^{\nu}(x;\varphi) = (1 - ze^{i\varphi})^{- \nu + ix}(1 - ze^{- i\varphi})^{- \nu - ix},\qquad |z| < 1,$$

with $\varphi = \pi/2$. As a result, we arrive at the expression

$$ S^g\psi_0^0 = \frac{2^{\nu}}{\sqrt{2\pi\lambda_\mathrm{C}\Gamma(2\nu)}} e^{i\gamma\nu}\omega_0^{ix/\lambda_\mathrm{C}} \Gamma\biggl(\nu + \frac{ix}{\lambda_\mathrm{C}}\biggr)\biggl(\frac{1 - t^2}{1 + t^2}\biggr)^{\nu} \biggl(\frac{1 +it}{1 -it}\biggr)^{ix/\lambda_\mathrm{C}}.$$

(A.6)

Substituting the value of the parameter $t$ from (A.2) in (A.6), we find after simple transformations that the right-hand side of (A.6) indeed coincides with the ground-state wave function $\psi_0^g$ in of the relativistic linear oscillator in a uniform external field, Eq. (3.6). This completes the proof of relation (4.7).