Light forces from steady-state solutions

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  Introduction to Atomic Physics

  Quantum light: states and dynamics

  Optical Bloch equations

  Photon-atom interactions

  Frontiers in Atomic Physics

  Ion traps and quantum information

  Quantum gases

Today, we'll discuss light forces; this is from API p.370-378. We'll also need the amount of light absorbed from an atom; this is from API p.369.

So far, we have discussed internal states of atoms, and their changes under interactions with light. This formalism turns out to also include everything we need to also include external degrees of freedom, such as the motion of the atom. Using the Heisenberg equations of motion, starting from the dipole interaction

$H_{I}=-{\vec {d}}\cdot {\vec {E}}_{\perp }\,,$

we find that if the electic field ${\vec {E}}_{\perp }$ has a spatial dependence, then the spatial derivative of $H_{I}$ gives us forces which result from the interaction.

This looks like a simple classical expression, but it is actually much more. Recall that $d$ is the atomic dipole operator, and $E$ is the operator of the quantized electromagnetic field, related to $a+a^{\dagger }$.

Transform of coherent states into c-numbers

A good place to begin to understand light forces is to assume that the electromagnetic field is in the coherent state $|\alpha {\rangle }$.

One nice approach to this problem is to utilize an exact unitary transform (see Exercise 17 of API), which transforms from an atom interacting with various Fock states of the electromagnetic field, to an equivalent Hamiltonian, with the following properties. In this equivalent picture, $|\alpha {\rangle }$ is transformed to be the vacuum state $|0{\rangle }$; in a sense, it transforms the laser field away, leaving the atom interacting only with the vacuum. And the electric field ${\vec {E}}_{\perp }$ becomes a classical c-number, plus a term representing just vacuum fluctuations, ${\vec {\mathcal {E}}}_{0}\cos \omega _{L}t+{\vec {E}}_{\perp }$. This transformation allows us to treat one part of the electromagnetic field classically, but keeps fluctuations due to quantum physics, so that the model is still an exact quantum-mechanical treatment.

Optical Bloch Equations -- Review

Let $\sigma$ be the reduced density matrix for the atom. The optical bloch equation is:

${\dot {\sigma }}={\frac {1}{i\hbar }}\left[{H_{0}+H_{I},\sigma }\right]-\Gamma (\sigma -\sigma _{e}q)\,,$

where $\Gamma$ is a matrix representing the dissipation or loss. Recall that we previously used

$\rho ={\frac {I+{\vec {r}}\cdot \sigma }{2}}\,.$

Note that API uses $\sigma$ for the reduced density matrix, whereas we've used it for the Pauli matrices; be careful and don't be confused by this. The specific elements of the Bloch vector are

${\begin{array}{rcl}r_{x}&=&\rho _{ge}+\rho _{eg}\\r_{y}&=&(\rho _{ge}-\rho _{eg})/i\\r_{z}&=&\rho _{ee}-\rho _{gg}\,.\end{array}}$

Often in the literature, and in API, these components are written as $u$, $v$, and $w$, where $u=r_{x}/2=\langle S_{x}{\rangle }$, $v=r_{y}/2=\langle S_{y}{\rangle }$, and $w=r_{z}/2=\langle S_{z}{\rangle }$. Note the extra factor of $1/2$.

Explicitly, using API's notation of $\sigma$ for the reduced density matrix (note that this atomic density matrix is in the rotating wave approximation, as well as in the rotating frame of the laser),

${\begin{array}{rcl}u&=&{\frac {1}{2}}(\sigma _{ge}+\sigma _{eg})\\v&=&{\frac {1}{2}}(\sigma _{ge}-\sigma _{eg})/i\\w&=&{\frac {1}{2}}(\sigma _{ee}-\sigma _{gg})\,.\end{array}}$

In terms of these components, the optical Bloch equations are

${\begin{aligned}{\dot {u}}&=\delta _{L}v-{\frac {\Gamma }{2}}u\\{\dot {v}}&=-\delta _{L}u-\Omega _{1}w-{\frac {\Gamma }{2}}v\\{\dot {w}}&=\Omega _{1}v-\Gamma w-{\frac {\Gamma }{2}}\end{aligned}}$

where we will use $\delta _{L}=\omega _{L}-\omega _{0}$ as the detuning of the laser from the atom, $\Omega _{1}$ as the Rabi frequency such that $\hbar \Omega _{1}=-{\vec {d}}_{ab}{\vec {E}}_{0}$, and the electric field is ${\vec {E}}={\vec {E}}_{0}\cos \omega _{L}t$. These may be compared and contrasted with the notation previously used.

What can we get from the optical bloch equations? One set of solutions is the steady state ones; those when nothing changes or something is only slowly changing. For example, an atom under light forces can cause the atom to move, but at every point in time the atom is nearly in equilibrium with the field.

A second aspect that can be discussed is the spectrum of the scattered light, as we have seen with the Mollow triplet, in which the frequencies and linewidths of the observed spectra are obtained from the optical bloch equations.

The dipole operator

Mechanical forces will come from ${\vec {d}}{\vec {\nabla }}E$, where the dipole operator is

${\vec {d}}={\vec {d}}_{ab}(|b\rangle \langle a|+|a\rangle \langle b|)$

This is related to the density matrix through

${\begin{array}{rcl}\langle d\rangle &=&Tr(\sigma d)={\vec {d}}_{ab}(\sigma _{ab}+\sigma _{ba})\\&=&{\vec {d}}_{ab}(\sigma _{ab}e^{i\omega _{L}t}+\sigma _{ba}e^{-i\omega _{L}t})\\&=&2{\vec {d}}_{ab}(u\cos \omega _{L}t-v\sin \omega _{L}t)\,.\end{array}}$

Note that the $u$ component is in phase with the laser excitation and the $v$ component is in quadrature with the laser. Physically, to recapilate, the atomic dipole moment has two components which oscillate: one in phase, and the other in quadrature with an incident electromagnetic field.

Which part of an oscillator absorbs energy? Is it the component in phase with an excitation or out of phase? The out of phase component is what absorbs energy. Thus, the $v$ component absorbs photons, and the $u$ component, which is in phase, can redistribute photons but does not contribute to absorption. This redistribution can lead to momentum changes, which thus translate to light forces. So both will be important for us to consider.

The optical force has thus been separated to two, distinct categories. One is a dissipative force and the other is a reactive force.

Absorbed energy and power

The energy per unit time absorbed by an atom from the light field is

${\frac {dW}{dt}}=q{\cal {E}}_{0}\cos \omega _{L}t\,{\frac {dr}{dt}}\,,$

representing a dipole $q\,dr$ being driven by an electric field. Recall $u$ and $v$ are off-diagonal elements of the atomic density matrix, known as coherences. Component $u$ is in phase with the driving electric field, and $v$ is in quadrature (90 degrees out of phase) with the driving field. Only the $v$ component results in absorption. Averaging, over a single cycle, we get

$\left\langle {\frac {dW}{dt}}\right\rangle =\hbar \omega _{L}\Omega v$

giving that the number of photons absorbed on average is

$\left\langle {\frac {dN}{dt}}\right\rangle =\Omega v\,.$

Now back to the OBE. From this equation,

${\dot {\sigma }}_{ee}=\Omega v-\Gamma \sigma _{ee}$

we note that (since ${\dot {\sigma }}_{ee}=0$ in steady state)

$\left\langle {\frac {dN}{dt}}\right\rangle =\Gamma \sigma _{ee}\,,$

which shows that the excited state decays with rate $\Gamma$. That is nice to see, since it means the optical Bloch equations are consistent with our expected classical understanding of energy loss and absorption.

Light forces: radiation

An electric field can exert a force on an atom because it has an electric dipole moment. This moment can be in-phase with the field, or out-of-phase with the field. As we know classically from a simple harmonic oscillator, a driving force in phase with the oscillation does not contribute energy to the oscillator, whereas the drive out of phase with the oscillator does transfer energy. We will now identify how this happens with light driving an atom, in terms of the $u$ and $v$ components of the Bloch vector. The force is

${\begin{array}{rcl}{\vec {F}}&=&{\dot {\vec {p}}}\\&=&-{\frac {\partial H}{\partial R}}\\&=&\sum _{j=x,y,z}d_{j}\nabla _{R}\left({E_{ej}({\vec {R}},t)+E_{\perp j}({\vec {R}})}\right)\,,\end{array}}$

where the subscript $e$ on $E$ emphasizes that the electric field is applied externally (the c-number field obtained after cannonical transformation to obtain the semiclassical picture we're using), and the subscript $\perp$ distinguishes the vacuum field (the residual quantum fluctuation fields from the coherent state). For more about this, read Exercise #17 in the back of API -- this shows a canonical transform in which the electric field can be transformed, such that a coherent state field can be represented by a classical c-number field, plus fluctuations from the vacuum.

We now make some approximations. First, the vacuum field is even at ${\vec {R}}$ (the position of the atom), and thus does not contribute any force. Thus,

${\begin{array}{rcl}{\vec {F}}&\approx &\sum _{j=x,y,z}d_{j}\nabla _{R}\left({E_{ej}({\vec {R}},t)}\right)\,.\end{array}}$

Next, we assume the atom is sufficiently localized (in position space; this assumption is not valid when considering cooling to the single photon recoil limit, because at that point the cooled atoms become delocalized to more than an optical wavelength) such that we may replace the operator ${\vec {R}}$ with ${\langle }{\vec {R}}\rangle ={\vec {r}}_{G}$, a c-number. That is, we evaluate the electric field at the center of the atomic wavefunction. This gives us

${\begin{array}{rcl}{\vec {F}}&\approx &\sum _{j=x,y,z}d_{j}\nabla _{R}E_{ej}({\vec {r}}_{G},t)\,.\end{array}}$

This is a simple expression for light forces which we shall now expand on, by inserting steady state solutions for the optical Bloch equation. There are two different timescales involved here. The external motion of an atom occurs on a timescale of $\hbar /E_{recoil}$, whereas the internal states evolve with time $\Gamma ^{-1}$. These two are very well separated for atomic sodium, and most alkali atoms. But for certain atoms, like metastable helium, they are not, so one must be careful (such systems have really not been studied much in the community yet).

We will be considering forces when velocity $v\approx 0$ and $r=0$. We assume that the electric field is given by a traveling or standing wave,

${\vec {E}}_{e}({\vec {r}},t)={\hat {e}}{\mathcal {E}}_{0}({\vec {r}})\cos \left[{\omega _{L}t+\phi ({\vec {r}})}\right]\,,$

where ${\hat {e}}$ is the polarization. Assume $\phi (0)=0$. The gradient of this is

$\nabla E_{ej}=e_{j}\left[{\cos \omega _{l}t\,\nabla {\mathcal {E}}_{0}-\sin \omega _{L}t{\mathcal {E}}_{0}\nabla \phi }\right]\,.$

This is one part of the expression. The other comes from the atomic dipole moment, given by the optical Bloch equations. We have

$\langle d_{j}\rangle =2d_{ab}\left[{u_{st}\cos \omega _{L}t-v_{st}\sin \omega _{L}t}\right]\,.$

The subscript "$st$" emphasizes these are steady state values. Integrating over one cycle of the radiation field (such that cross terms with $\cos \times \sin$ go away) gives us

${\begin{array}{rcl}{\vec {F}}={\hat {e}}\cdot {\vec {d}}_{ab}\left[{u_{st}\nabla {\mathcal {E}}_{0}+v_{st}{\mathcal {E}}_{0}\nabla \phi }\right].\,.\end{array}}$

The two terms of this can be identified as the reactive force, and the dissipative force. Specifically, the reactive term $\sim u_{st}\nabla {\mathcal {E}}_{0}$, and the dissipative term $\sim v_{st}{\mathcal {E}}_{0}\nabla \phi$. We may usefully re-express this in terms of the Rabi frequency

$\Omega =-{\vec {d}}_{ab}\cdot {\hat {e}}{\mathcal {E}}_{0}/\hbar \,,$

in terms of which the reactive force can be expressed using

${\vec {\alpha }}={\frac {\nabla \Omega }{\Omega }}\,,$

giving

${\vec {F}}_{react}=-\hbar \Omega u_{st}{\vec {\alpha }}\,.$

Similarly, the dissipative force can be expressed using

${\vec {\beta }}=\nabla \phi ,,$

giving

${\vec {F}}_{diss}=-\hbar \Omega v_{st}{\vec {\beta }}\,.$

It is useful to consider situations where we have just the dissipative, or just the reactive forces.

Radiation-pressure force

Consider an atom in a traveling wave,

${\vec {E}}_{e}={\hat {e}}{\mathcal {E}}_{0}\cos(\omega _{L}t-{\vec {k}}_{L}\cdot {\vec {r}})\,.$

For this, ${\vec {\alpha }}=0$, whereas ${\vec {\beta }}=-{\vec {k}}_{L}$. This gives

${\begin{array}{rcl}{\vec {F}}_{diss}=\Omega v_{st}\hbar {\vec {k}}_{L}\,.\end{array}}$

Now recalling that $v=r_{y}/2$ we can use our steady state solution from Solutions of the optical Bloch equations to find

${\vec {F}}_{diss}=\hbar {\vec {k}}_{L}{\frac {\Gamma }{2}}{\frac {\Omega ^{2}/2}{(\omega _{L}-\omega _{0})^{2}+\Omega ^{2}/2+\Gamma ^{2}/4}}\,.$

This is an expression we saw right at the start of the class, argued intuitively, and now derived rigorously. The maximum value is

${\vec {F}}_{diss,max}=\hbar {\vec {k}}_{L}{\frac {\Gamma }{2}}\,.$

It can also be written in the more intuitive form reminiscent of what we saw for the absorbed power subsection above

${\begin{array}{rcl}{\vec {F}}_{diss}&=&\Gamma \sigma _{ee}^{st}\hbar {\vec {k}}_{L}\\&=&\left\langle {\frac {dN}{dt}}\right\rangle \hbar {\vec {k}}_{L}\,.\end{array}}$

This clearly shows that the dissipative force is simply the rate of photon scattering events times the momentum per photon.

Moving atoms & friction force

Laser cooling requires us to include moving atoms in the picture. The simplest case is as follows. Let us go into a moving frame,

$\omega _{L}\rightarrow \omega _{L}-{\vec {k}}_{L}\cdot {\vec {v}}_{0}\,,$

now including the Doppler shift. Let $\theta$ be the angle between ${\vec {k}}_{L}$ and ${\vec {v}}_{0}$. Substituting into the above expression, and taylor expanding around $v_{0}=0$, we get

$F(v_{0})=F(v_{0}=0)+\alpha _{1}v_{0}\cdots \,,$

where the coefficient $\alpha _{1}$ is

$\alpha _{1}=\hbar k_{L}^{2}\cos(\theta )~{\frac {2s}{(1+s)^{2}}}~{\frac {2\delta }{\Gamma }}{\frac {1}{1+{\frac {4\delta ^{2}}{\Gamma ^{2}}}}}\,.$

We'll use this result when looking at optical molasses in the low-intensity limit, in which case we can make the approximation $s\ll 1$ to give

$\alpha _{1}\approx \hbar k_{L}^{2}\cos(\theta )~2s~{\frac {2\delta }{\Gamma }}{\frac {1}{1+{\frac {4\delta ^{2}}{\Gamma ^{2}}}}}\,.$

Note that here $\alpha _{1}$ is given for the case of one traveling wave, not two counterpropagating beams.

Reaction forces

The much more interesting case is atoms in a standing wave. Earlier, we argued that this case could be viewed as the sum of two traveling wave interactions; we then took the force as the sum of the two dissipative forces. Now, we shall take a different viewpoint. Let the standing wave be

${\vec {E}}_{e}(r,t)={\hat {e}}_{x}{\mathcal {E}}_{0}\cos k_{L}z\,\cos \omega _{L}t\,.$

The dissipative force for this standing wave is actually zero, because there is no phase, and $\beta$ is the derivative of the phase. There is actually only a reaction force from a standing wave. What is happening physically is there is interference happening, with processes such as an atom absorbing a photon from one beam, and emitting into the other beam. Let us see what the resulting force is. The reactive force comes from $u_{st}$. We get

${\vec {F}}_{react}={\frac {-\hbar (\omega _{L}-\omega _{0})}{4}}{\frac {\nabla (\Omega ^{2})}{(\omega _{L}-\omega _{0})^{2}+\Gamma ^{2}/4+\Omega ^{2}/2}}\,.$

What is the connection between the two traveling, counter-propagating fields $E_{1}$ and $E_{2}$, and the standing waves? Consider this phasor diagram:

File:Light forces from steady-state solutions-lf-phasor

The reactive force comes from $u$, the component in phase with the electric field. From the classical harmonic oscillator, we know that if the motion is between $0$ and $180$ degrees we absorb energy from the field, and if it between $180$ and $360$ degrees, we deliver energy to the field. The above phasor diagram shows that $u$ is $45$ degrees from $E_{1}$, but $-45$ degrees from $E_{2}$. This indicates that the atom is absorbing energy from one field, and delivering it to the other beam, and these physics are at the heart of stimulated forces. Let us take this situation to an extreme, to better understand whether the scenario is best described as being two traveling waves, or one standing wave. Arrange the standing wave such that atoms are moving with velocity $v_{0}$:

File:Light forces from steady-state solutions-lf-standingwave

Is the radiation force experienced by this atom a reactive, or dissipative force? Above, we showed rigorously that $\beta =0$. However, intuitively, the above scenario disagrees with that picture, because an atom at $v_{0}$ arguably absorbes only from one beam, with which it is in resonance, and not the other. Let us return to the phasor diagram of the field:

File:Light forces from steady-state solutions-lf-phasor2

On this diagram, we now also include the $v$ component, which rapidly averages away to zero in the mathematical expressions above. The sum of these two, $u+v\sim \langle d\rangle _{total}$, points along one field, say $E_{2}$. This total moment only takes energy out of $E_{1}$, agreeing with the intuitive picture above. If things are done carefully, the two pictures thus agree. In the standing wave picture, $u$ takes a photon from $E_{1}$, and puts it into $E_{2}$. $v$ is responsible for spontaneous scattering, which scatters photons from both fields $E_{1}$ and $E_{2}$, so the net is reduction of two photons from $E_{1}$, and none from $E_{2}$. In the traveling wave picture, $E_{2}$, say, is completely out of resonance, whereas $E_{1}$ is responsible for all the dissipative force. Two photons are thus taken out of $E_{1}$, agreeing with the standing wave picture. Thus, the interpretation of these forces on atoms in a standing wave is ambiguous -- it can either be understood as being a spontaneous force, or not, depending on picture, and preference.

Reactive force magnitudes

Let us go back to discussing the physics of the reactive force expression

${\vec {F}}_{react}={\frac {-\hbar (\omega _{L}-\omega _{0})}{4}}{\frac {\nabla (\Omega ^{2})}{(\omega _{L}-\omega _{0})^{2}+\Gamma ^{2}/4+\Omega ^{2}/2}}\,.$

We can write this as a conservative force, such that ${\vec {F}}=-\nabla U$, where

$U={\frac {\hbar (\omega _{L}-\omega _{0})}{2}}\ln \left[{1+{\frac {\Omega ^{2}/2}{(\omega _{L}-\omega )^{2}+\Gamma ^{2}/4}}}\right]\,.$

This is a very useful observation for laser cooling. The absolute magnitiude of this force is useful to know. Given $\Omega ^{2}$, the magnitude of the force $|F_{react}|$ is maximum when $|\omega _{L}-\omega |\sim |\Omega |$. The maximum, for this optimal detuning, is

${\rm {max}}|F_{react}|\sim \hbar {\frac {\nabla (\Omega )^{2}}{\Omega }}\sim \hbar \nabla \Omega \sim \hbar k_{L}\Omega \,.$

This makes intuitive sense; the strongest force you can get from an optical field is the photon momentum times the Rabi frequency. In each cycle, the atom absorbs a photon from one beam, and emits it into the other beam, giving it $2\hbar k_{L}$ per cycle.

Reactive forces at finite velocity

If the atom is traveling with velocity $v_{0}$ through a standing wave, what is $\alpha$ for $F_{react}=-\alpha v_{0}$? To derive the answer, we cannot take $u$ to be the steady state value $u_{st}$, because the atom is moving rapidly across changes in the electric field. The atom can only respond on a timescale $\Gamma ^{-1}$, so it has some memory, resulting in $u\sim u_{st}({\vec {r}}-{\vec {v}}_{0}\Delta t)$. Thus allows one to obtain the friction coefficient $\alpha$ for an atom in a standing wave. In the weak intensity limit, $\alpha =2\alpha _{TW}$, where $\alpha _{TW}$ is the traveling wave result. However, in the strong intensity limit, things change considerably. The upshot is that blue detuned light may be needed to cool atoms, in a standing wave. See the experiment by A. Aspect, PRL 1986.

References

• see API pp. 370 - 378
• Further reading on friction force in a standing wave (not covered in class)
  • C. Cohen-Tannoudji, Les Houches 1990, pp. 34-35
  • J.P. Gordon and A. Ashkin, PRA 21, 1606 (1980)