Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I currently started learning Quantum Optics using the book "Atom Physics" of Christopher Foot. Currently I am working through Chapter 7 and a question occured concerning the intuitively physical understanding:

Assume a two level atom with energy levels $E_{1}$ and $E_{2}$ (with $E_{1}$<$E_{2}$ is given). The eigenstates of the system are given by $|1\rangle$ and $|2\rangle$ respectively. Now assume that we perturbe the system by using a time dependent Electric field of the form \begin{equation} \vec{E} = \vec E_{0} \cos\left(kz - \omega t\right) \end{equation} and expand the solution of the Schroedinger Equation in terms of the unperturbed eigenstates \begin{equation} \Psi(\vec r,t) = c_{1}(t) \ e^{-i (E_{1}/\hbar) t} \ |1\rangle + c_{2}(t) \ e^{-i (E_{2}/\hbar) t} \ |2\rangle \end{equation} Let's assume that $c_{1}(0)=1$ and $c_{2}(0)=0$. Using the dipole approximation as well as the rotating wave approximation as discussed for example in www.physics.ox.ac.uk/Users/lucas/BIII/QuantumNotes.pdf‎, we obtain that \begin{equation} |c_{2}(t)| \sim t^{2} \ \frac{\sin^{2}(x)}{x^{2}} \end{equation} with $x=(\omega-\omega_{0})t/2$ and $\hbar \omega_{0}:=E_{2}-E_{1}$. Plotting this as a function of $\omega-\omega_{0}$ we obtain a "Fraunhofer-diffraction"-like graph.

My Question:

1) As the interaction time increases the width of the maxima is decreasing. What is from a physical / intuitive perspective the reason for this.

2) There are also certain minima occuring for certain frequencies of the light. What is the physical interpretation (intuitively) for this?

I am looking forward to your responses. Thanks in advance.

share|improve this question
add comment

1 Answer 1

The atom is interacting with rectangular impulse of length t, modulated with frequency of $\omega$. The more energy is pumped to the atom on its resonance frequency, the more probability of its excitation. Spectral density (i.e, the energy density) of the impulse is $$ S(\omega_0) = \left| \int_{-t/2}^{t/2} dt' e^{-i\omega_0 t} \cos\omega t \right|^2 \approx \frac{sin^2 {\frac{\omega-\omega_0}2t}}{(\omega-\omega_0)^2} $$ where I've used the RWA dropping the high-frequency part of the spectra.

So,

1) Some frequencies are totally absent in the rectangular pulse spectrum (i.e., those with cosine having maximum at pulse flank). Thus atom having such a resonant frequency wouldn't feel the impulse at all.

2) Increasing impulse length leads to its spectrum getting narrower. Making the impulse more monochromatic one lowers the excitation probability of detuned atom.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.