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I would really appreciate your help with this one!

I've been looking at a simple problem, linearly polarized excitation scheme in a two-level atom, [itex]J_g=J_e=1[/itex]. At first orientation of coordinate system is such that light is polarized along [itex]y[/itex] direction and propogates in [itex]z[/itex] direction. Therefore in Zeeman basis it is viewed as superposition of both left and right polarized light at equal amounts. Using electric dipole approximation and RWA, and assuming reduced dipole matrix element and amplitude of electric field are equal to unity I get the total Hamiltonian in form

[tex] H_a=\left(

\begin{array}{cccccc}

\Delta \hbar & 0 & 0 & 0 & -\frac{1}{4 \sqrt{3}} & 0 \\

0 & \Delta \hbar & 0 & \frac{1}{4 \sqrt{3}} & 0 & -\frac{1}{4 \sqrt{3}} \\

0 & 0 & \Delta \hbar & 0 & \frac{1}{4 \sqrt{3}} & 0 \\

0 & \frac{1}{4 \sqrt{3}} & 0 & 0 & 0 & 0 \\

-\frac{1}{4 \sqrt{3}} & 0 & \frac{1}{4 \sqrt{3}} & 0 & 0 & 0 \\

0 & -\frac{1}{4 \sqrt{3}} & 0 & 0 & 0 & 0

\end{array}

\right)[/tex]

Here [itex]\Delta[/itex] represents detuning from transition resonance. Note that I have arranged magnetic sublevels in increasing order of [itex]m_J[/itex] - matrix element in the upper left corner corresponds to [itex]\langle J_g,m_J=-1| \hat{H}_b | J_g,m_J=-1\rangle[/itex]

An alternate approach suggests that I choose such coordinate system, in which the light beam propagates in [itex]x^\prime y^\prime[/itex] plane and is polarized along [itex]z^\prime[/itex] axis. In this case I would get the following Hamiltonian matrix:

[tex]H_b=\left(

\begin{array}{cccccc}

\Delta \hbar & 0 & 0 & -\frac{1}{2 \sqrt{6}} & 0 & 0 \\

0 & \Delta \hbar & 0 & 0 & 0 & 0 \\

0 & 0 & \Delta \hbar & 0 & 0 & \frac{1}{2 \sqrt{6}} \\

-\frac{1}{2 \sqrt{6}} & 0 & 0 & 0 & 0 & 0 \\

0 & 0 & 0 & 0 & 0 & 0 \\

0 & 0 & \frac{1}{2 \sqrt{6}} & 0 & 0 & 0

\end{array}

\right)[/tex]

Change of coordinate system [itex]xyz\rightarrow x^\prime y^\prime z^\prime[/itex] should be equal to coordinate rotation at Euler angles [itex]\alpha=0, \beta=\pi/2,\gamma=0[/itex]. Therefore a proper Wigner D matrix [itex]W_D[/itex] would allow me to perform the following operation:

[tex]W_D^{-1}H_a W_D = H_b[/tex]

Here Wigner D matrix for two level scheme can be written as

[tex]\left(

\begin{array}{cc}

W_g & 0 \\

0 & W_e

\end{array}

\right)[/tex]

With

[tex]W_g=W_e=\left(

\begin{array}{ccc}

\frac{1}{2} & -\frac{1}{\sqrt{2}} & \frac{1}{2} \\

\frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2}} \\

\frac{1}{2} & \frac{1}{\sqrt{2}} & \frac{1}{2}

\end{array}

\right)[/tex]

The problem is that I get

[tex]W_D^{-1}H_a W_D = H_a[/tex]

and not [itex] H_b [/itex]! Even more, for Wigner D matrix with [itex]\alpha=0, \beta=-\pi/2,\gamma=0[/itex] I get almost correct result, which would be

[tex]W_D^{\prime -1}H_b W_D^{\prime} = H_a[/tex]

but with incorrect signs for off-diagonal elements.

Can anyone suggest where I made a mistake?

Thanks!

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# Wigner D matrix

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