Dynamics of Time Evolution of Quantum Oscillator Excitation by Electromagnetic Pulses

Abstract

The dynamics of time evolution of quantum oscillator excitation by electromagnetic pulses is investigated theoretically for an arbitrary field amplitude in a pulse. We consider a harmonic oscillator without damping and excitation between stationary states. The general formula for the excitation of quantum states as a function of time is derived in terms of instantaneous energy of an associated classical oscillator in the field of an electromagnetic pulse. The derived expression is used in detailed analysis of the time dependence of the quantum oscillator excitation probability beyond the range of perturbation theory for various pulse parameters including total excitation from the ground state, excitation from excited states, and excitation spectra.

Appendices

APPENDIX A

We assume that prior to excitation (t → –∞), an oscillator has been in the nth stationary state with wave function (without the time factor) [22]

$$\begin{gathered} \psi _{n}^{{(0)}}(x) = {{\left( {\frac{{M\omega }}{{\pi \hbar }}} \right)}^{{1/4}}}\frac{1}{{\sqrt {{{2}^{n}}n!} }} \\ \times \exp \left( { - \frac{{M\omega }}{{2\hbar }}{{x}^{2}}} \right){{H}_{n}}\left( {x\sqrt {\frac{{M\omega }}{\hbar }} } \right), \\ \end{gathered} $$

(A.1)

where H_n are the Hermitian polynomials. We introduce dimensionless variable

$$\tilde {x} = x\sqrt {\frac{{M\omega }}{\hbar }} $$

(A.2)

and dimensionless function

$$\tilde {\psi } = {{\left( {\frac{\hbar }{{M\omega }}} \right)}^{{1/4}}}\psi .$$

(A.3)

Under the action of an EMP, the initial function is transformed as $\tilde {\psi }_{n}^{{(0)}}$(x) → ${{\tilde {\psi }}_{n}}$(x, t). Then we obtain the following expression for the probability of transition n → m between the stationary states of a quantum oscillator (at instant t):

$${{W}_{{mn}}}(t) = {{\left| {\int\limits_{ - \infty }^\infty {d\tilde {x}{{{\tilde {\psi }}}_{n}}(\tilde {x},t)\tilde {\psi }_{m}^{{(0)}}(\tilde {x})} } \right|}^{2}},\quad m > n.$$

(A.4)

Here, we have introduced function (see [2])

$${{\tilde {\psi }}_{n}}(\tilde {x},t) = \exp (i\varphi t)\tilde {\psi }_{n}^{{(0)}}(\tilde {x} - \tilde {\eta }(t))\exp (i\dot {\tilde {\eta }}\tilde {x}),$$

(A.5)

which is the analytic solution to the time-dependent Schrödinger equation for a 1D harmonic oscillator under the action of a time-dependent external force. We define dimensionless function as

$$\tilde {\eta }(t) = \frac{{M\omega }}{\hbar }\eta (t),$$

(A.6)

where η(t) is the solution to the equation for induced oscillations of the classical oscillator under the action of the external force,

$$\ddot {\eta } + {{\omega }^{2}}\eta = \frac{{f(t)}}{M},$$

(A.7)

with initial conditions η(–∞) = $\dot {\eta }$(–∞) = 0, as well as corresponding derivatives:

$$\dot {\tilde {\eta }} = \frac{{d\tilde {\eta }}}{{\omega dt}} = \frac{{d\tilde {\eta }}}{{d\tilde {t}}},$$

(A.8)

$$\dot {\tilde {\eta }} = \sqrt {\frac{M}{{\hbar \omega }}} \dot {\eta }.$$

(A.9)

With account for the above argument, we obtain

$$\begin{gathered} {{W}_{{mn}}}(t) = \frac{1}{{\pi {{2}^{n}}n!{{2}^{m}}m!}}\left| {\int\limits_{ - \infty }^{ + \infty } {d\tilde {x}\exp \left( { - \frac{{{{{(\tilde {x} - \tilde {\eta })}}^{2}}}}{2}} \right)} } \right. \\ {{\left. { \times \,{{H}_{n}}(\tilde {x} - \tilde {\eta }(t))\exp (i\dot {\tilde {\eta }}\tilde {x}){{H}_{m}}(\tilde {x})\exp \left( { - \frac{{{{{\tilde {x}}}^{2}}}}{2}} \right)} \right|}^{2}}. \\ \end{gathered} $$

(A.10)

Further, we must evaluate integral

$$\begin{gathered} {{I}_{1}} = \int\limits_{ - \infty }^\infty {du\exp \left[ { - {{u}^{2}} + u\tilde {\eta } - \frac{{{{{\tilde {\eta }}}^{2}}}}{2} + i\dot {\tilde {\eta }}u} \right]} \\ \times \,{{H}_{m}}(u){{H}_{n}}(u - \tilde {\eta }). \\ \end{gathered} $$

(A.11)

We introduce variable

$$y = \frac{i}{2}\dot {\tilde {\eta }} + \frac{1}{2}\tilde {\eta },$$

(A.12)

which gives

$${{I}_{1}} = \exp \left( {\frac{i}{2}\dot {\tilde {\eta }}\tilde {\eta } - \frac{{{{{\tilde {\eta }}}^{2}}}}{4} - \frac{{{{{\dot {\tilde {\eta }}}}^{2}}}}{4}} \right){{I}_{2}}(\tilde {\eta },y),$$

(A.13)

where

$$\begin{gathered} {{I}_{2}}(\tilde {\eta },y) \equiv \int\limits_{ - \infty }^\infty {du{{e}^{{ - {{{(u - y)}}^{2}}}}}{{H}_{m}}(u){{H}_{n}}(u - \tilde {\eta })} \\ \mathop = \limits^{(u = x + y)} \int\limits_{ - \infty }^\infty {dx{{e}^{{ - {{x}^{2}}}}}{{H}_{m}}(x + y){{H}_{n}}(x + y - \tilde {\eta }).} \\ \end{gathered} $$

(A.14)

Let us use tabulated integral (7.378) from [23],

$$\begin{gathered} \int\limits_{ - \infty }^\infty {{{e}^{{ - {{x}^{2}}}}}{{H}_{m}}(x + w){{H}_{n}}(x + z)dx} \\ = {{2}^{m}}\sqrt \pi n!{{z}^{{m - n}}}L_{n}^{{m - n}}( - 2wz),\quad m > n, \\ \end{gathered} $$

(A.15)

where $L_{n}^{{m - n}}$ is the generalized Laguerre polynomial.

For reducing I₂($\tilde {\eta }$, y) to tabulated integral (A.15), we must perform the substitution

$$\begin{gathered} w = y, \\ z = y - \tilde {\eta }. \\ \end{gathered} $$

(A.16)

This gives

$${{I}_{2}}(\tilde {\eta },y) = {{2}^{m}}\sqrt \pi n!{{(y - \tilde {\eta })}^{{m - n}}}L_{n}^{{m - n}}( - 2y(y - \tilde {\eta })).$$

(A.17)

Introducing notation arg = –2y(y – $\tilde {\eta }$) and using relations (A.16), we obtain

$$\begin{gathered} \arg = - 2\left( {\frac{i}{2}\dot {\tilde {\eta }} + \frac{1}{2}\tilde {\eta }} \right)\left( {\frac{i}{2}\dot {\tilde {\eta }} + \frac{1}{2}\tilde {\eta } - \tilde {\eta }} \right) \\ = - 2\left( {\frac{i}{2}\dot {\tilde {\eta }} + \frac{1}{2}\tilde {\eta }} \right)\left( {\frac{i}{2}\dot {\tilde {\eta }} - \frac{1}{2}\tilde {\eta }} \right) = \frac{{{{{\dot {\tilde {\eta }}}}^{2}} + {{{\tilde {\eta }}}^{2}}}}{2}, \\ \end{gathered} $$

(A.18)

or, in dimensional variables,

$$\arg = \frac{{M{{{\dot {\eta }}}^{2}} + {{\omega }^{2}}M{{\eta }^{2}}}}{{2\hbar \omega }} \equiv \frac{{\varepsilon (t)}}{{\hbar \omega }} = \frac{{A(t)}}{{\hbar \omega }} = \nu (t),$$

(A.19)

$$\nu (t) = \frac{{A(t)}}{{\hbar \omega }}.$$

(A.20)

Here, A(t) is the work done on the classical oscillator η(t) under the action of an EMP. Considering that

$${\text{|}}y - \tilde {\eta }{{{\text{|}}}^{{\text{2}}}} = \frac{{{{{\dot {\tilde {\eta }}}}^{2}} + {{{\tilde {\eta }}}^{2}}}}{4} = \frac{{M({{{\dot {\eta }}}^{2}} + {{\omega }^{2}}{{\eta }^{2}})}}{{4\hbar \omega }} = \frac{\nu }{2},$$

(A.21)

we obtain

$$\begin{gathered} {\text{|}}{{I}_{2}}{{{\text{|}}}^{2}} = {{2}^{{2m}}}\pi {{(n!)}^{2}}\frac{{{{\nu }^{{m - n}}}}}{{{{2}^{{m - n}}}}}{\text{|}}L_{n}^{{m - n}}(\nu ){{{\text{|}}}^{2}} \\ = {{2}^{{n + m}}}\pi {{(n!)}^{2}}{{\nu }^{{m - n}}}{\text{|}}L_{n}^{{m - n}}(\nu ){{{\text{|}}}^{2}}. \\ \end{gathered} $$

(A.22)

Since

$${\text{|}}{{I}_{1}}{{{\text{|}}}^{2}} = \exp ( - \nu ){\text{|}}{{I}_{2}}{{{\text{|}}}^{2}},$$

(A.23)

we get

$${{W}_{{mn}}}(t) = \frac{1}{{\pi {{2}^{n}}n!{{2}^{m}}m!}}{\text{|}}{{I}_{1}}{{{\text{|}}}^{2}}.$$

(A.24)

Finally, we obtain the QO excitation probability on transition n → m, m > n:

$$\begin{gathered} {{W}_{{mn}}}(t) = {{{\tilde {W}}}_{{mn}}}[\nu (t)] \\ = \frac{{n!}}{{m!}}\nu {{(t)}^{{m - n}}}\exp ( - \nu (t)){\text{|}}L_{n}^{{m - n}}(\nu (t)){{{\text{|}}}^{{\text{2}}}}. \\ \end{gathered} $$

(A.25)

For deriving formula (11), we use the following representation of the generalized Laguerre polynomial [23]:

$$L_{n}^{\alpha }(x) = \frac{1}{{n!}}{{e}^{x}}{{x}^{{ - \alpha }}}\frac{{{{d}^{n}}}}{{d{{x}^{n}}}}({{e}^{{ - x}}}{{x}^{{n + \alpha }}}).$$

(A.26)

Substituting this relation into (A.25), we get

$${{\tilde {W}}_{{mn}}}(\nu ) = \frac{1}{{n!m!}}{{e}^{\nu }}{{\nu }^{{n - m}}}{{\left| {\frac{{{{d}^{n}}}}{{d{{\nu }^{n}}}}({{e}^{{ - \nu }}}{{\nu }^{m}})} \right|}^{2}}.$$

(A.27)

Considering that

$$\begin{gathered} {{\nu }^{{n - m}}}{{\left| {\frac{{{{d}^{n}}}}{{d{{\nu }^{n}}}}({{e}^{{ - \nu }}}{{\nu }^{m}})} \right|}^{2}} \\ = {{\nu }^{{m + n}}}{{e}^{{ - 2\nu }}}{{\left| {\sum\limits_{k = 0}^{\min (n,m)} {\frac{{n!m!{{{( - 1)}}^{k}}{{\nu }^{{ - k}}}}}{{k!(m - k)!(n - k)!}}} } \right|}^{2}}, \\ \end{gathered} $$

(A.28)

we obtain

$$\begin{gathered} {{{\tilde {W}}}_{{mn}}}(\nu ) = n!m!{{\nu }^{{m + n}}}{{e}^{{ - \nu }}} \\ \times {{\left| {\sum\limits_{k = 0}^{\min (n,m)} {\frac{{{{{( - 1)}}^{k}}{{\nu }^{{ - k}}}}}{{k!(m - k)!(n - k)!}}} } \right|}^{2}}. \\ \end{gathered} $$

(A.29)

This relation implies, in particular, that

$${{\tilde {W}}_{{mn}}}(\nu ) = {{\tilde {W}}_{{nm}}}(\nu ).$$

(A.30)

Thus, the time dependence of the probability of transition n → m, m > n, coincides with that for reverse transition m → n.

APPENDIX B

The EMP energy absorbed by a charged classical oscillator by instant t is given by

$${{\varepsilon }_{{{\text{clas}}}}}(t) = q\int\limits_{ - \infty }^t {\dot {x}(t')E(t')dt'.} $$

(B.1)

The solution to the equation for a harmonic oscillator has form (see, for example, [24])

$$x(t) = \frac{q}{M}\int\limits_{ - \infty }^{ + \infty } {\frac{{E(\omega ')\exp ( - i\omega 't)}}{{\omega _{0}^{2} - \omega {\kern 1pt} {{'}^{2}} - 2i\gamma \omega '}}} \frac{{d\omega '}}{{2\pi }},$$

(B.2)

where γ is the relaxation constant. Formula (B.1) leads to expression

$$\dot {x}(t) = - i\frac{q}{M}\int\limits_{ - \infty }^{ + \infty } {\frac{{\omega 'E(\omega ')\exp ( - i\omega 't)}}{{\omega _{0}^{2} - \omega {\kern 1pt} {{'}^{2}} - 2i\gamma \omega '}}} \frac{{d\omega '}}{{2\pi }}.$$

(B.3)

Substituting this expression into (B.1), we get

$$\begin{gathered} {{\varepsilon }_{{{\text{clas}}}}}(t) = \frac{{{{q}^{2}}}}{M}\int\limits_{ - \infty }^t {dt'E(t')} \\ \times \int\limits_{ - \infty }^{ + \infty } {\frac{{d\omega '}}{{2\pi }}E(\omega ')\frac{{i\omega '\exp ( - i\omega 't)}}{{\omega {\kern 1pt} {{'}^{2}} - \omega _{0}^{2} + 2i\gamma \omega '}}} . \\ \end{gathered} $$

(B.4)

The frequency integral on the right-hand side of this equation can be evaluated using the residue theorem. In the limit γ → 0, this integral has form

$$\begin{gathered} \int\limits_{ - \infty }^{ + \infty } {\frac{{d\omega '}}{{2\pi }}\frac{{i\omega '{\text{exp}}( - i\omega '(t'\, - \,t''))}}{{\omega {\kern 1pt} {{'}^{2}} - \omega _{0}^{2} - 2i\gamma \omega '}}} \\ = \,\theta (t'\, - \,t''){\text{cos}}[{{\omega }_{0}}(t'\, - \,t'')], \\ \end{gathered} $$

(B.5)

where θ(τ) is the Heaviside theta function. Substituting the right-hand side of expression (B.5) into (B.4), we obtain

$$\begin{gathered} {{\varepsilon }_{{{\text{clas}}}}}(t) = \frac{{{{q}^{2}}}}{M}\int\limits_{ - \infty }^t {dt'E(t')\int\limits_{ - \infty }^{t'} {dt''E(t'')\cos [{{\omega }_{0}}(t'\, - t'')]} } \\ = \frac{{{{q}^{2}}}}{{2M}}\int\limits_{ - \infty }^t {dt'E(t')\int\limits_{ - \infty }^t {dt''E(t'')\cos [{{\omega }_{0}}(t'\, - t'')]} } \\ = \frac{{{{q}^{2}}}}{{2M}}{{\left| {\int\limits_{ - \infty }^t {dt'E(t')\exp (i{{\omega }_{0}}t')} } \right|}^{2}}. \\ \end{gathered} $$

(B.6)

In the transition to the second equality in (B.6), we have used the fact that the integrand is symmetric relative to the transposition of integration variables (t' ↔ t'').