# In the context of atom interferometry, how are $\pi$ and $\frac{\pi}{2}$ pulse generated?

First of all, I should probably let you know that my knowledge of Quantum Optics is terribly shallow.

I am studying atom interferometry and I am having trouble understanding what is exactly behind the $\frac{\pi}{2}$ and $\pi$ pulses that commonly appear on diagrams of interferometer setups, as shown below (let's assume that the atoms used are a simple two-level system, the ground being called $\left|\mathrm{red}\right\rangle$ and the excited state $\left| \mathrm{blue} \right\rangle$).

To be more precise, while I understand (after reading this post How do Rabi oscillations act on the relative phase of state kets) that applying a $r\pi$ pulse means that we apply the following transformation

$$U(r) = e^{-ir\pi X/2} = \begin{pmatrix} \cos(r\pi/2) & -i\sin(r\pi/2) \\ -i\sin(r\pi/2) & \phantom{-i}\cos(r\pi/2)\end{pmatrix}$$

to a statistical ensemble of $\left|\mathrm{red}\right\rangle$ and $\left|\mathrm{blue}\right\rangle$, I do not understand how this is actually done.

Basically, I am searching for a very technical explanation stating how are the lasers operated. Just saying

We apply a $\frac{\pi}{2}$ pulse to the atomic wave packet, and thus half of the atoms, which were all initially in the $\left| \mathrm{red} \right\rangle$ state, are excited to the $\left|\mathrm{blue} \right\rangle$ state and the two resulting beams are now spatially separated because of the momentum gained when absorbing the photons

doesn't speak to me because I don't understand how you apply a $\frac{\pi}{2}$ pulse, technically speaking.

In general, you apply $r\pi$ pulses by turning on an interaction hamiltonian $H_I$, which couples both states, for long enough (or with a strong enough coupling) to achieve the desired effect.

This builds on the standard representation for the propagator operator under that hamiltonian for a time difference $\Delta t$, which reads $U(\Delta t)=\exp(-i H_I\Delta t/\hbar)$, and which is obviously parallel to the expression $U=e^{-ir\pi X/2}$ you quote. In particular, we use pulses of radiation (lasers or microwaves or whatever is convenient) because they have the hamiltonian $$\hat H_I=\mathbf E\cdot e\hat{\mathbf r},$$ in the dipole approximation and in the length gauge (and, also, applying the rotating-wave approximation), with $\mathbf E$ the electric field strength and $e\hat{\mathbf r}$ the atomic electric dipole operator. Typically, both eigenstates $\left|\mathrm{red}\right\rangle$ and $\left|\mathrm{blue}\right\rangle$ will have zero intrinsic dipole moment, which means that only the off-diagonal elements of $H_I$ are nonzero, and it is therefore of the form $$\hat H_I=F\hat X=F\begin{pmatrix}0&1\\1&0\end{pmatrix}.$$

(Note that I'm setting the off-diagonal elements to be real, which I can do (once) by shifting the phase of the basis states. This is what fixes the axis of the $r\pi$ pulse rotation as the $X$ axis, and it can be changed by changing the phase of the laser (which introduces opposite complex phases on the off-diagonal elements, and moves the rotation axis along the $xy$ plane of the Bloch sphere).)

Given the hamiltonian above, you just simply let it run for a set time $\Delta t$, and it gives you a unitary operator $$e^{-iF\Delta t X/\hbar} = e^{-ir\pi X/2},$$ which represents an $r\pi$ pulse with rotation amount $$r\pi = \frac{2}{\hbar} F\,\Delta t.$$ To tune this rotation amount, you can change $\Delta t$ (run the pulse for longer) or $F$ (use a stronger pulse), and depending on experimental conditions you might choose either of them.

• I hope this clears it up but let me know if you need further clarifications and, if so, in what direction. Aug 17 '16 at 21:18
• Just an addendum: realistically the pulse strength will be a time dependent function $F(t)$, not least because the laser and associated electronics will have a finite response time. In this case, one has a time-dependent Hamiltonian $H(t)$ generating a unitary evolution operator $T\exp\left[\frac{1}{i\hbar}\int_0^{\Delta t}\mathrm{d} t \, H(t)\right]$, so that ultimately $r\pi = \frac{2}{\hbar}\int_0^{\Delta t}\mathrm{d} t\, F(t)$. Possibly more information than the OP wanted to know... Aug 17 '16 at 21:22

In the context of atom interferometry, the tool usually (always?) used for these manipulations is simulated Raman transitions*. These are two photon transitions, which are very versatile: one tunes the difference between the two laser frequencies to be resonant to the energy difference between two states to couple them, and the momentum difference between a photon in each beam determines the momentum kick given to the atoms.

So, in your diagram, what actually happens at the $\pi/2$ or $\pi$ pulses is that two lasers are turned on, for the time that corresponds to a partial or full transition as Emilio has described. The lasers will have some nonzero angle between them, which causes a momentum kick, and are tuned to the frequency different between the ground and excited state. As a result, the states $|g,k\rangle$ and $|e, k+\delta k\rangle$ are coupled, where the first quantum number is the internal state of the atom and the second one is its momentum.

Some further reading: a important PRL about the technique, and many papers on Stimulated Raman adiabatic passage (STIRAP), which is a slight variation on the same idea.

*in the case in which the internal state does not change, essentially the same process is called "Bragg scattering."

• No, they're not always stimulated Raman transitions. That applies for small energy separations (i.e. fine and hyperfine structure) while still allowing for the large relative stability of the optical domain. However, it's perfectly possible to do it directly via microwaves (which is useful in its own right), or to just use an optical transition. Aug 24 '16 at 17:06
• Hi @EmilioPisanty - To be clear, by "these manipulations" I mean in the context of atom interferometry specifically. The general scheme I have in mind, as practiced e.g. in the Kasevich group and shown in the OP's diagram, requires the transition to give a significant momentum impulse to the atoms, which is very very small for microwaves. This isn't a problem for optical transitions, but it seems to me that the short lifetime would generally keep those from being feasible for this application. Aug 25 '16 at 23:31
• ...All that said, maybe there is some other kind of atom interferometry that I am forgetting or am ignorant of- I would be interested in some examples! Aug 25 '16 at 23:31
• The specific scheme referred to by the OP probably does involve Raman transitions, but the concept of $r\pi$ pulses is very general - you can use them for an enormous host of applications where the momentum transfer is not a factor. On the direct microwave route, you can go from the original caesium clock to, say, this scheme. On the optical domain, you can try the initial Cirac-Zoller realization, which used a metastable state with a long lifetime, but even with a microsecond transition you have time to do stuff. Aug 25 '16 at 23:45
• Yeah, I guess my point was that "usually" is context-dependent, and usually (heh) best teamed up with an indication of what the context is. And "always" is a very strong word for such casual, off-handed use. Aug 26 '16 at 1:37