User:Lamals/sandbox/Fluorescence microscopy via coherent control

The field of fluorescence microscopy has seen many advances in the ability to create high resolution images in a non invasive manner that are important to various fields such as biology. Historically in light based microscopes, the limit on resolution of a point source fluorophore is the diffraction limit. The desire of investigating samples that are smaller then the diffraction limit, e.g. on a nanometer scale, have lead to techniques that can overcome the diffraction limit. In the past few decades some of the most researched techniques are based on the general concept of reversible saturable optical fluorescence transition (RESOLFT). These types of techniques do not generally exploit coherent quantum optical control. However, there has been intense activity in the spatial localization of atomic population in Λ-type systems coherently interacting with two electromagnetic fields. These types of methods are generally based on coherent population trapping (CPT), and have been shown to be useful for microscopy. Another technique based off of stimulated Raman adiabatic passage (STIRAP) is used to increase the resolution of fluorescence imaging to nanometer resolutions.

Background

Lambda type atomic energy levels

Both CPT and STIRAP utilize a Λ configuration of a three-level system. They both rely on the coherent dark state that can occur in this type of system. We start with a three-state Λ-type system (pictured right), where ${\textstyle |1\rangle \leftrightarrow |3\rangle }$ and ${\textstyle |2\rangle \leftrightarrow |3\rangle }$ are dipole-allowed transitions and ${\textstyle |1\rangle \leftrightarrow |2\rangle }$ is forbidden. In the rotating wave approximation, the semi-classical Hamiltonian is given by

$$H=H_{A}+H_{AF}$$

where

$${\begin{aligned}&H_{A}=E_{1}|1\rangle \langle 1|+E_{2}|2\rangle \langle 2|+E_{3}|3\rangle \langle 3|\\[5pt]&H_{AF}={\frac {\hbar \Omega _{1}}{2}}(e^{-i\omega _{1}t}|3\rangle \langle 1|+e^{i\omega _{1}t}|1\rangle \langle 3|)+{\frac {\hbar \Omega _{2}}{2}}(e^{-i\omega _{2}t}|3\rangle \langle 2|+e^{i\omega _{2}t}|2\rangle \langle 3|)\end{aligned}}$$

where ${\textstyle \Omega _{1}}$ and ${\textstyle \Omega _{2}}$ are the Rabi frequencies, we can write our Hamiltonian in the rotating frame, taking ${\textstyle E_{1}=E_{2}=0}$

$$H_{RF}={\frac {\hbar \Omega _{1}}{2}}(|3\rangle \langle 1|+|1\rangle \langle 3|)+{\frac {\hbar \Omega _{2}}{2}}(|3\rangle \langle 2|+|2\rangle \langle 3|)$$

The atomic wave function can be written as:

$$|\psi (t)\rangle =c_{1}(t)e^{-i\omega _{1}t}|1\rangle +c_{2}(t)e^{-i\omega _{2}t}|2\rangle +c_{3}(t)e^{-i\omega _{3}t}|3\rangle .$$

Solving the Schrödinger equation $i\hbar |{\dot {\psi }}\rangle =H|\psi \rangle$, we obtain the solutions

$${\begin{aligned}&{\dot {c}}_{1}={\frac {i}{2}}\Omega _{1}c_{3}\\[5pt]&{\dot {c}}_{2}={\frac {i}{2}}\Omega _{2}c_{3}\\[5pt]&{\dot {c}}_{3}={\frac {i}{2}}(\Omega _{1}c_{1}+\Omega _{2}c_{2})\end{aligned}}$$

Using the initial condition

$$|\psi (0)\rangle =c_{1}(0)|1\rangle +c_{2}(0)|2\rangle +c_{3}(0)|3\rangle$$

we can solve these equations to obtain

$${\begin{aligned}&c_{1}(t)=c_{1}(0)\left[{\frac {\Omega _{2}^{2}}{\Omega ^{2}}}+{\frac {\Omega _{1}^{2}}{\Omega ^{2}}}\cos \left({\frac {\Omega t}{2}}\right)\right]+c_{2}(0)\left[-{\frac {\Omega _{1}\Omega _{2}}{\Omega ^{2}}}+{\frac {\Omega _{1}\Omega _{2}}{\Omega ^{2}}}\cos \left({\frac {\Omega t}{2}}\right)\right]-ic_{3}(0){\frac {\Omega _{1}}{\Omega }}\sin \left({\frac {\Omega t}{2}}\right)\\[5pt]&c_{2}(t)=c_{1}(0)\left[-{\frac {\Omega _{1}\Omega _{2}}{\Omega ^{2}}}+{\frac {\Omega _{1}\Omega _{2}}{\Omega ^{2}}}\cos \left({\frac {\Omega t}{2}}\right)\right]+c_{2}(0)\left[{\frac {\Omega _{1}^{2}}{\Omega ^{2}}}+{\frac {\Omega _{2}^{2}}{\Omega ^{2}}}\cos \left({\frac {\Omega t}{2}}\right)\right]-ic_{3}(0){\frac {\Omega _{2}}{\Omega }}\sin \left({\frac {\Omega t}{2}}\right)\\[5pt]&c_{3}(t)=-ic_{1}(0){\frac {\Omega _{1}}{\Omega }}\sin \left({\frac {\Omega t}{2}}\right)-ic_{2}(0){\frac {\Omega _{2}}{\Omega }}\sin \left({\frac {\Omega t}{2}}\right)+c_{3}(0)\cos \left({\frac {\Omega t}{2}}\right)\end{aligned}}$$

with ${\textstyle \Omega ={\sqrt {\Omega _{1}^{2}+\Omega _{2}^{2}}}}$. We observe that we can choose the initial conditions

$$c_{1}(0)={\frac {\Omega _{2}}{\Omega }},\qquad c_{2}(0)=-{\frac {\Omega _{1}}{\Omega }},\qquad c_{3}(0)=0$$

one of the time-independent solutions to these equations with no probability of the system being in state $|3\rangle$

$$|\psi \rangle ={\frac {\Omega _{2}|1\rangle -\Omega _{1}|2\rangle }{\Omega }}$$

Atoms in this state will stay in this state indefinitely because the laser fields cannot excite the electrons into state $|3\rangle$

$${\begin{aligned}&\langle 3|H_{RF}|\psi \rangle ={\frac {\hbar }{2\Omega }}(\Omega _{1}\Omega _{2}-\Omega _{2}\Omega _{1})=0\\[5pt]&|\psi \rangle =|D\rangle \end{aligned}}$$

This is a dark state, because it can not absorb or emit any photons from the applied fields. Spontaneous emission from ${\textstyle |3\rangle }$ can result in an atom being in this dark state or another coherent state, known as a bright state. Therefore, in a collection of atoms that decay into the dark state, it will inevitably result in the system being "trapped" coherently in that state. This happens once the system reaches a steady state.

Process

CPT

This dark state, $|D\rangle$, can be used to improve resolution past the diffraction limit in microscopy If you can remove most of the population from ${\textstyle |1\rangle }$ while leaving a small spatial spot still in ${\textstyle |1\rangle }$ to do fluorescence microscopy on. This can be done if you use laser fields with spatial varying profiles like the ones below.

$${\begin{aligned}&\Omega _{1}(\nu ,t)=\Omega _{1peak}\left[F(\nu ,\delta )+F(\nu ,-\delta )\right]e^{-{\frac {(t-t_{1})^{2}}{\sigma ^{2}}}}\\[5pt]&\Omega _{2}(\nu ,t)=\Omega _{2peak}F(\nu ,0)e^{-{\frac {(t-t_{2})^{2}}{\sigma ^{2}}}}\end{aligned}}$$

where

$$F(\nu ,r)\equiv {\frac {2J_{1}(\nu +r)}{\nu +r}},\quad \nu ={\frac {2\pi x{\text{NA}}}{\lambda }},\quad \delta =1.22\pi$$

and ${\textstyle \sigma }$ is the temporal width of the pulses, ${\textstyle J_{1}(\nu )}$ is the first order Bessel function.This will cause the dynamics to become position dependent, and population will only go to the dark state in positions around the node of the standing wave field(${\textstyle \Omega _{2}}$). At the node of ${\textstyle \Omega _{2}}$, we will have almost 0 intensity and the population in this spot will not get trapped in the dark state, and can be smaller than the classical diffraction limit. We can see this if we monitor the population through fluorescence of level ${\textstyle |1\rangle }$:

$$\rho _{11}={\frac {1}{1+\left[\ \Omega _{2}(x)/\Omega _{1}\right]^{2}}}={\frac {1}{1+R\sin ^{2}{kx}}}$$

where ${\textstyle R=\left|\Omega _{2}(x)\right|^{2}/\left|\Omega _{1}\right|^{2}}$, the intensity ratio between the two laser pulses. From this we can derive a modified Abbe's diffraction limit

$${\begin{aligned}&D_{Classical}={\frac {\lambda }{2{\text{NA}}}}\\[5pt]&D_{CPT}={\frac {\lambda }{2{\text{NA}}}}{\frac {2\delta }{\pi }}\left({\sqrt {1+2{\sqrt {R}}}}\right)^{-1/2}\end{aligned}}$$

We can see that as long as week keep an intensity ratio greater then 300, we will get better resolution then the classical diffraction limit.

STIRAP

Counterintuitive Pulse sequence

In contrast to CPT, STIRAP can be used in microscopy to achieve resolutions better then the diffraction limit, and not rely on the system reaching steady state. When used for microscopy is called subwavelength localizaion via adiabatic passage (SLAP). This scheme does not require the system to reach steady state, and does not trap any population into a dark state indefinitely. Rather, it transfers population from ${\textstyle |1\rangle \rightarrow |2\rangle }$, by adiabatically following a dark state, and using what ever population is left in ${\textstyle |1\rangle }$ to perform fluorescence microscopy on. This is done by sending in the laser pulses, ${\textstyle \Omega _{1}}$ & ${\textstyle \Omega _{2}}$, in a counterintuitive temporal sequence (pictured to the right).

The counterintuitive pulse sequence allows you to start in a dark state, then follow it.

1. Start with both lasers off, ${\textstyle \Omega _{1}=\Omega _{2}=0}$, and all population in ${\textstyle |1\rangle }$
2. Turn laser 2 on, ${\textstyle \Omega _{1}=0}$
   • ${\textstyle |D\rangle ={\frac {\Omega _{2}|1\rangle -\Omega _{1}|2\rangle }{\Omega }}=|1\rangle }$
3. Slowly turn off laser 2 while slowly turning on laser 1 (adiabatic condition)
   • ${\textstyle \Omega _{1}=\Omega _{2}}$
   • ${\textstyle |D\rangle ={\frac {\Omega _{2}|1\rangle -\Omega _{1}|2\rangle }{\Omega }}={\frac {|1\rangle -|2\rangle }{\sqrt {2}}}}$
4. Completely turn laser 2, ${\textstyle \Omega _{2}=0}$
   • ${\textstyle |D\rangle ={\frac {\Omega _{2}|1\rangle -\Omega _{1}|2\rangle }{\Omega }}=|2\rangle }$
5. Turn off laser 1, ${\textstyle \Omega _{1}=\Omega _{2}=0}$, All the population is now in ${\textstyle |2\rangle }$

We must also fulfill a global adiabatic condition to ensure no coupling between the different energy eigenstates, ${\textstyle \Omega (x)T\geq A}$ (where A is a dimensionless constant that for optimal Gaussian profiles and delay times takes values around 10, and ${\textstyle T=t_{1}-t_{2}}$ is the temporal delay between the pulses). Using the same spatial profiles for the fields, ${\textstyle \Omega _{1}(\nu ,t)}$ & ${\textstyle \Omega _{2}(\nu ,t)}$, and pulsing the laser fields according to the counterintuitive temporal sequence, population is transfered from ${\textstyle |1\rangle \rightarrow |2\rangle }$, except at the position of the node in ${\textstyle \Omega _{1}(\nu ,t)}$, and we therefore obtain a narrow peak of population remaining in ${\textstyle |1\rangle }$. We can then pulse the excitation beam, E, to excite the population from ${\textstyle |1\rangle \rightarrow |3\rangle }$, and then allow fluorescence. This allows us to get another modified Abbe's diffraction limit for SLAP

$$D_{SLAP}={\frac {\lambda }{2{\text{NA}}}}{\frac {\delta }{\pi }}\left({\sqrt {\frac {4R}{k^{-2}-1}}}+1\right)^{-1/2}$$

Where ${\textstyle k\equiv \Omega _{2peak}T/A}$, and ${\textstyle 0<k<1}$. We can see that, depending on the adiabaticity of the process, the 3rd term can be less than 1, and tends to 0 in the limit ${\textstyle R\rightarrow \infty }$.