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 am trying to calculate atom - light (EM field) interaction Hamiltonian, and the results I get seem to me rather unphysical - I get some nonzero matrix elements which should not be there. Please, can you help me by pointing out errors in my calculation / giving advice?

This is how I get matrix elements of atom - light interaction Hamiltonian:

Atom - light interaction is treated semiclassicaly. Light polarization convention is adopted from the book "Optically polarized atoms. Understanding Light-Atom interactions" by Auzinsh, Budker & Rochester. In short, $\pi$ polarization = electric field vector oscilates along z-axis; $\sigma^\pm$ correspond to electric field vector rotating in xy plane while the light propagates in positive direction of z axis: for $\sigma^+$ electric field rotates counter-clockwise as viewed form positive direction of z axis, while for $\sigma^-$ electric field rotates clockwise.

Vectors are described using their components in the spherical basis (see sec. 3.1.2 of the book by Auzinsh et al.). Covariant basis vectors are related to cartesian basis $\hat{x},\hat{y},\hat{z}$ in the following way (Auzinsh et al. eqn. 3.8a-c):

$\hat{\epsilon}_1=-\frac{1}{\sqrt{2}}\left( \hat{x} + i\hat{y}\right)\\ \hat{\epsilon}_0=\hat{z}\\ \hat{\epsilon}_{-1}=\frac{1}{\sqrt{2}}\left( \hat{x} - i\hat{y}\right)$

Components of a vector $\mathbf{v}$ in this basis are contravariant quantities (Auzinsh et al. eqn. 3.16a-c):

$v^1=-\frac{1}{\sqrt{2}}\left( v_x - iv_y\right)\\ v^0=v_z\\ v^{-1}=\frac{1}{\sqrt{2}}\left( v_x + iv_y\right)$

Vector dot product in the spherical basis: $\mathbf{a}\cdot\mathbf{b}\sum_{q=-1}^{1}{a_q b^q}$.

Relation between components of $\mathbf{v}$ and $\mathbf{v}^*$ (conjugate vector) is (Auzinsh et al. eqns D.36, D.41):

$\left(v^*\right)^q=\left(-1\right)^q\left(v^{-q}\right)^*$

For simplicity, model an atom as a two level system, with total electronic angular momenta of both levels $J_g=J_e=\frac{1}{2}$. Therefore each energy level is composed of two degenerate quantum states $m_J=\pm\frac{1}{2}$. In such system $\pi$ (linearly) polarized light induces both $\Delta m_{J} = 0$ transitions. Circular polarizations however should induce just one transition each: $|g,m_J=-\frac{1}{2}\rangle \leftrightarrow |e,m_J=\frac{1}{2}\rangle$ for $\sigma^+$ polarization and $|g,m_J=\frac{1}{2}\rangle \leftrightarrow |e,m_J=-\frac{1}{2}\rangle$ for $\sigma^-$ polarization.

The electric part of the light is (spatial variation neglected):

$\mathbf{E}=\frac{E_0}{2}\left(\mathbf{v}e^{-i\omega t}+\mathbf{v}^*e^{i\omega t}\right)$

Here $E_0$ is amplitude of the field, $\mathbf{v}$ is polarization vector, $\omega$ is angular frequency.

Now, energy of an electric dipole in electric field is:

$\hat{H}=\hat{\mathbf{d}}\cdot\mathbf{E}$

Here $\hat{\mathbf{d}} = -e\hat{\mathbf{r}}$.

$\hat{H}=\sum_{q}\hat{d}_q E^q = \frac{E_0}{2}\sum_{q}\hat{d}_q \left(v^qe^{-i\omega t}+\left(v^*\right)^qe^{i\omega t}\right)$

When calculating matrix elements, values of $\hat{d}_q$ are obtained using Wigner-Eckart theorem (book by Auzinsh et al., eqn 3.121):

$\langle\eta^\prime,m_J^\prime|\hat{d}_q|\eta,m_J\rangle = \left(-1\right)^{J^\prime-m_J^\prime} \langle\eta^\prime\parallel d\parallel\eta\rangle \left( \begin{array}{ccc} J^\prime & 1 & J \\ -m_J^\prime & q & m_J \end{array} \right)$

Here $\langle\eta^\prime\parallel d\parallel\eta\rangle$ is reduced dipole matrix element, and $\left( \begin{array}{ccc} J^\prime & 1 & J \\ -m_J^\prime & q & m_J \end{array} \right)$ is Wigner 3-j symbol. Therefore, matrix elements of the Hamiltonian are: $\langle\eta^\prime,m_J^\prime|\hat{H}|\eta,m_J\rangle = \frac{E_0}{2} \langle\eta^\prime\parallel d\parallel\eta\rangle \left(-1\right)^{J^\prime-m_J^\prime} \sum_{q} \left( \begin{array}{ccc} J^\prime & 1 & J \\ -m_J^\prime & q & m_J \end{array} \right) \left(v^qe^{-i\omega t}+\left(v^*\right)^qe^{i\omega t}\right)$

Now consider atom interacting with $\sigma^+$ polarized light. In the spherical basis components of polarization vector are: $v^1=1,v^0=0,v^{-1}=0$, and components of its conjugate vector are: $\left(v^*\right)^1=0,\left(v^*\right)^0=0,\left(v^*\right)^{-1}=-1$. Matrix elements of the interaction Hamiltonian are:

$\langle\eta^\prime,m_J^\prime|\hat{H}_{\sigma^+}|\eta,m_J\rangle = \frac{E_0}{2} \langle\eta^\prime\parallel d\parallel\eta\rangle \left(-1\right)^{J^\prime-m_J^\prime} \left[ \left( \begin{array}{ccc} J^\prime & 1 & J \\ -m_J^\prime & 1 & m_J \end{array} \right) e^{-i\omega t} - \left( \begin{array}{ccc} J^\prime & 1 & J \\ -m_J^\prime & -1 & m_J \end{array} \right) e^{i\omega t}\right]$

For the transition $|g,m_J=-\frac{1}{2}\rangle \leftrightarrow |e,m_J=\frac{1}{2}\rangle$ everything seems to be in order. Nonzero values of matrix element $\langle g,m_J=-\frac{1}{2}|\hat{H}_{\sigma^+}|e,m_J=\frac{1}{2}\rangle$ and its conjugate transpose $\langle e,m_J=\frac{1}{2}|\hat{H}_{\sigma^+}|g,m_J=-\frac{1}{2}\rangle$ are provided by terms containing $e^{i \omega t}$ and $e^{-i \omega t}$ accordingly.

However, nonzero matrix elements corresponding to transition $|g,m_J=\frac{1}{2}\rangle \leftrightarrow |e,m_J=-\frac{1}{2}\rangle$ appear in the Hamiltonian too!

$\langle g,\frac{1}{2}|\hat{H}_{\sigma^+}|e,-\frac{1}{2}\rangle = \frac{E_0}{2} \langle\eta^\prime\parallel d\parallel\eta\rangle \left(-1\right)^{\frac{1}{2}-\frac{1}{2}} \left[ \left( \begin{array}{ccc} \frac{1}{2} & 1 & \frac{1}{2} \\ -\frac{1}{2} & 1 & -\frac{1}{2} \end{array} \right) e^{-i\omega t} -\left( \begin{array}{ccc} \frac{1}{2} & 1 & \frac{1}{2} \\ -\frac{1}{2} & -1 & -\frac{1}{2} \end{array} \right) e^{i\omega t}\right]=\\ =-\frac{E_0}{2} \langle\eta^\prime\parallel d\parallel\eta\rangle \left[ -\frac{1}{\sqrt{3}} e^{-i\omega t} -0 e^{i\omega t}\right] \neq 0$

which should not be the case! Currently the only "solution" I can think of is to treat the light field quantum mechanically - in that case photon creation/annihilation operators will take care of these matrix elements. However I would like to treat my problem semiclassically.

Edit: Today I had a chance to ask this question to a guest professor from St. Petersburg State University. He convinced me that there is no error in my calculation. The matrix elements I was worried about are actually the counter-rotating (negative frequency) terms. Application of the rotating wave approximation allows one to get rid of them. See the answer by emarti for more detailed explanation.

share|improve this question
    
You have an error somewhere, almost certainly in the calculation of the last two 3j symbols. In your last equation the initial matrix element is time-dependent but always real but, while $\langle\eta'||d||\eta\rangle$ may be complex, the combination $\langle\eta'||d||\eta\rangle e^{-i\omega t}$ will be complex and nonzero most of the time. –  Emilio Pisanty Jan 9 '13 at 18:16
1  
Thank you for your input, Emilio! I am quite convinced that I am calculating 3j symbols correctly. Values of reduced matrix elements $\langle \eta^\prime\parallel d\parallel\eta\rangle$ should be real. Initial matrix elements would be real only if the corresponding electric dipole operator matrix elements were real, which I believe is not the case. The Wigner-Eckart theorem tells us that for $\Delta m=\pm 1$ transition the only nonzero component of $\hat{\mathbf{d}}$ is for either $q=+1$ or $-1$. And this corresponds to complex $x,y$ components in the Cartesian basis. –  Arturs C. Jan 10 '13 at 21:35
add comment

1 Answer

up vote 2 down vote accepted

I think you're very close to the right understanding with the following statement:

Currently the only "solution" I can think of is to treat the light field quantum mechanically - in that case photon creation/annihilation operators will take care of these matrix elements. However I would like to treat my problem semiclassically.

You have not included the atom's resonant frequencies $\omega_n$. However, you correctly noticed that, quantum mechanically, the $e^{\pm i \omega t}$ terms correspond to annihilation and creation operators of photons, and so only one of the will contribute to the transition. The way to do this semiclasically is the rotating wave approximation. I won't go through it (there are plenty of excellent texts*), but the steps will look as follows.

Your time-dependent Hamiltonian is something like $$H = - \vec d \cdot \vec E \cos \omega t$$

You can define a rotating frame by $\psi_R = R(t) \psi$, where $R(t)$ is a time-dependent unitary matrix that gets rid of the time-dependence of your Hamiltonian. For a two-level system, $R = exp(i \sigma_z \omega t/2)$. \begin{eqnarray*} i \hbar \dot \psi & = & H \psi \\ i \hbar R \dot \psi & = & R H R^\dagger R \psi \\ i \hbar \dot \psi_R & = & (R H R^\dagger + i \hbar \dot R R^\dagger) \psi_R \\ i \hbar \dot \psi_R & = & H_R \psi_R \qquad H_R = R H R^\dagger + i \hbar \dot R R^\dagger \end{eqnarray*}

This transformation will turn your Hamiltonian into an (almost) time-independent Hamiltonian of $\psi_R$. The 'almost' is that you get terms like $e^{i (\omega - \omega_0)}$, which are low frequency and you keep, and terms like $e^{-i (\omega + \omega_0)}$, which are high frequency and you neglect. The non-zero terms you are worrying about are called the counter-rotating terms. There are situations where they do matter, for instance when the detuning is very large (for our optical trap, the counter-rotating terms contribute 30% to the final calculation).

By the way, that's a very good book you're working from. The problem you're working on is very important, and one that many of us atomic physicists have had to tackle for our research.

[*]: Atomic Physics by Chris Foot and Optical Resonances and Two-Level Atoms by L. Allen and J. H. Eberly are the two next to me. Cohen-Tannoudji must have a section on it.

share|improve this answer
    
Thank you for the nice explanation! The rotating wave approximation (when applicable) indeed helps to get rid of the counter-rotating terms. Having looked into books that deal with linearly polarized almost-resonant light, I didn't expect to find counter-rotating terms as separate matrix elements, and this eventually led me to the faulty assumption that these terms should not appear in the Hamiltonian at all. –  Arturs C. Jan 11 '13 at 21:26
    
The rotating wave approximation can be confusing. I originally learned it in the context of a spin-1/2 particle in a magnetic field, where the coordinates of the spin and real space match. It's a little more confusing in general problems, where the real space and spin space coordinate systems are nontrivially related. –  emarti Jan 12 '13 at 5:40
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.