QM M97 1

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Ionizing the delta function. It is not difficult to show (and you can take as a given) that an attractive one-dimensional delta function potential V(x) = − λδ(x), with λ a positive, real constant, has one bound state with energy E=-m\lambda^{2}/2\hbar^{2}. The normalized wave function for this state is:

\psi(x)=\left\{ \begin{matrix} \sqrt{\alpha}e^{\alpha x},\quad\mathrm{for}\, x&lt;0\\ \sqrt{\alpha}e^{-\alpha x},\quad\mathrm{for}\, x\geq0\end{matrix}\right.

with α a real, positive constant and E=-\hbar^{2}\alpha^{2}/2m.

Consider placing electrons in this state. We can ionize the system by applying an oscillating electric field along the x direction \mathcal{E} a constant, the 2 present for later convenience, and \mathcal{E} is small enough that first-order perturbation theory can be applied. For ease of calculation, you may ignore the effect of the original potential on the ionized (positive energy) states and treat them as one-dimensional plane waves. For a given (fixed) value of ω, and an oscillating electric field that turns on at t=0, what is the ionization rate for the system after a time t that is large?

We can write the perturbing potential as:

V(t)=q\hat{\mathbf{E}}\cdot\mathbf{x}=2\mathcal{E}qx\sin\omega t

Using Fermi's golden rule for periodic perturbations:

\Gamma_{mk}=\frac{2\pi}{\hbar}\left(\rho\left(E_{m}-\hbar\omega\right)\left|\langle n|F|m\rangle\right|^{2}+\rho\left(E_{m}+\hbar\omega\right)\left|\langle n|F^{\dagger}|m\rangle\right|^{2}\right)

Where ρ is the density of final states, and the F's are defined:

V(t)=Fe^{i\omega t}+F^{\dagger}e^{-i\omega t}

So that:

F=-\mathcal{E}qxi;\qquad F^{\dagger}=\mathcal{E}qxi

Which boils down to calculating the matrix element \langle n|x|m\rangle . The free particle has wavefunction:

\psi_{n}(x)=\frac{1}{\sqrt{2\pi}}e^{ikx}

(where we have normalized for convenience later), so that:

\langle n|x|m\rangle=2\sqrt{\frac{\alpha}{2\pi}}\int_{0}^{\infty}e^{ikx}xe^{-\alpha x}

Integrating by parts:

\langle n|x|m\rangle=2\sqrt{\frac{\alpha}{2\pi}}\left(\left.e^{ikx}xe^{-\alpha x}\right|_{0}^{\infty}-\int_{0}^{\infty}\frac{1}{(ik-\alpha)}e^{(ik-\alpha)x}\right)
\langle n|x|m\rangle=2\sqrt{\frac{\alpha}{2\pi}}\frac{1}{(ik-\alpha)^{2}}

So that:

\left|\langle n|F|m\rangle\right|^{2}=\left|\langle n|F^{\dagger}|m\rangle\right|^{2}=\frac{\alpha}{2\pi}\frac{4\mathcal{E}^{2}e^{2}}{\left(k^{2}+\alpha^{2}\right)^{2}}

In our approximation, the density of final states is given by:

\rho(E)=\delta\left(E_{n}-E\right)

where E_{n}=\hbar^{2}k^{2}/2m. Accordingly:

\Gamma_{mk}=\frac{2\pi}{\hbar}\frac{\alpha}{2\pi}\frac{4\mathcal{E}^{2}e^{2}}{\left(k^{2}+\alpha^{2}\right)^{2}}\left(\delta\left(E_{n}-E_{m}+\hbar\omega\right)+\delta\left(E_{n}-E_{m}-\hbar\omega\right)\right)

since \hbar\omega>|E_{m}|, and En > 0 for a free particle, only the second delta function will be nonzero in our range of interest. All that is left is to integrate over the states k:

\Gamma_{m}=\frac{2\pi}{2\pi}\int dk\frac{4\mathcal{E}^{2}e^{2}\alpha}{\hbar\left(k^{2}+\alpha^{2}\right)^{2}}\delta\left(\frac{\hbar^{2}k^{2}}{2m}-E_{m}-\hbar\omega\right)
\Gamma_{m}=\frac{4\mathcal{E}^{2}e^{2}\alpha}{\hbar\left(k^{2}+\alpha^{2}\right)^{2}}\frac{2m}{\hbar^{2}k}

where k^{2}=\frac{2m\omega}{\hbar}+\alpha^{2}. Solving also for α using the equivelence of energies:

-\frac{m\lambda^{2}}{2\hbar^{2}}=-\frac{\hbar^{2}\alpha^{2}}{2m}\Rightarrow\alpha^{2}=\frac{m^{2}\lambda^{2}}{\hbar^{4}}

So that finally:

\Gamma_{m}=\frac{4\mathcal{E}^{2}e^{2}}{\hbar\left(\frac{2m\omega}{\hbar}+2\frac{m^{2}\lambda^{2}}{\hbar^{4}}\right)^{2}}\frac{m\lambda}{\hbar^{2}}\frac{2m}{\hbar^{2}\sqrt{\frac{2m\omega}{\hbar}+\frac{m^{2}\lambda^{2}}{\hbar^{4}}}}
\Gamma_{m}=\frac{8\mathcal{E}^{2}e^{2}m^{2}\lambda}{\hbar^{5}\left(\frac{2m\omega}{\hbar}+2\frac{m^{2}\lambda^{2}}{\hbar^{4}}\right)^{2}\sqrt{\frac{2m\omega}{\hbar}+\frac{m^{2}\lambda^{2}}{\hbar^{4}}}}

This page was recovered in October 2009 from the Plasmagicians page on Prelim_M97_QM1 dated 01:35, 20 December 2005.

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