Help Solve a Puzzling Two-Level Atom Problem

[Arturs C.]

Hi, everyone!

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, J_g=J_e=1. At first orientation of coordinate system is such that light is polarized along y direction and propogates in z 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
H_a=\left(<br /> \begin{array}{cccccc}<br /> \Delta \hbar &amp; 0 &amp; 0 &amp; 0 &amp; -\frac{1}{4 \sqrt{3}} &amp; 0 \\<br /> 0 &amp; \Delta \hbar &amp; 0 &amp; \frac{1}{4 \sqrt{3}} &amp; 0 &amp; -\frac{1}{4 \sqrt{3}} \\<br /> 0 &amp; 0 &amp; \Delta \hbar &amp; 0 &amp; \frac{1}{4 \sqrt{3}} &amp; 0 \\<br /> 0 &amp; \frac{1}{4 \sqrt{3}} &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> -\frac{1}{4 \sqrt{3}} &amp; 0 &amp; \frac{1}{4 \sqrt{3}} &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; -\frac{1}{4 \sqrt{3}} &amp; 0 &amp; 0 &amp; 0 &amp; 0<br /> \end{array}<br /> \right)
Here \Delta represents detuning from transition resonance. Note that I have arranged magnetic sublevels in increasing order of m_J - matrix element in the upper left corner corresponds to \langle J_g,m_J=-1| \hat{H}_b | J_g,m_J=-1\rangle

An alternate approach suggests that I choose such coordinate system, in which the light beam propagates in x^\prime y^\prime plane and is polarized along z^\prime axis. In this case I would get the following Hamiltonian matrix:
H_b=\left(<br /> \begin{array}{cccccc}<br /> \Delta \hbar &amp; 0 &amp; 0 &amp; -\frac{1}{2 \sqrt{6}} &amp; 0 &amp; 0 \\<br /> 0 &amp; \Delta \hbar &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; \Delta \hbar &amp; 0 &amp; 0 &amp; \frac{1}{2 \sqrt{6}} \\<br /> -\frac{1}{2 \sqrt{6}} &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; \frac{1}{2 \sqrt{6}} &amp; 0 &amp; 0 &amp; 0<br /> \end{array}<br /> \right)
Change of coordinate system xyz\rightarrow x^\prime y^\prime z^\prime should be equal to coordinate rotation at Euler angles \alpha=0, \beta=\pi/2,\gamma=0. Therefore a proper Wigner D matrix W_D would allow me to perform the following operation:
W_D^{-1}H_a W_D = H_b
Here Wigner D matrix for two level scheme can be written as
\left(<br /> \begin{array}{cc}<br /> W_g &amp; 0 \\<br /> 0 &amp; W_e<br /> \end{array}<br /> \right)
With
W_g=W_e=\left(<br /> \begin{array}{ccc}<br /> \frac{1}{2} &amp; -\frac{1}{\sqrt{2}} &amp; \frac{1}{2} \\<br /> \frac{1}{\sqrt{2}} &amp; 0 &amp; -\frac{1}{\sqrt{2}} \\<br /> \frac{1}{2} &amp; \frac{1}{\sqrt{2}} &amp; \frac{1}{2}<br /> \end{array}<br /> \right)
The problem is that I get
W_D^{-1}H_a W_D = H_a
and not H_b! Even more, for Wigner D matrix with \alpha=0, \beta=-\pi/2,\gamma=0 I get almost correct result, which would be
W_D^{\prime -1}H_b W_D^{\prime} = H_a
but with incorrect signs for off-diagonal elements.

Can anyone suggest where I made a mistake?
Thanks!

 

[DrDu]

Shouldn't the upper right block of H_a be hermitian so as to be diagonalizable by W_e?

 

[Arturs C.]

DrDu, thank you for the advice!
Indeed changing signs in the upper right (and corresponding lower left) block of H_a to make it hermitian does solve the problem with diagonalization. Therefore I double checked my formula for finding the interaction matrix elements and I found no error in it!
It turned out the definition of Euler angles I was using to calculate Wigner D matrix was the wrong one :) To rotate my coordinate system properly I should perform rotations in the following order:
1) \alpha around z axis;
2) \beta around y axis;
3) \gamma around z axis;
And in my case \alpha=\pi/2,\ \beta=\pi/2,\ \gamma=0 which gives the following Wigner D matrix subatrices:
W_g=W_e=\left(<br /> \begin{array}{ccc}<br /> -\frac{i}{2} &amp; \frac{i}{\sqrt{2}} &amp; -\frac{i}{2} \\<br /> \frac{1}{\sqrt{2}} &amp; 0 &amp; -\frac{1}{\sqrt{2}} \\<br /> \frac{i}{2} &amp; \frac{i}{\sqrt{2}} &amp; \frac{i}{2}<br /> \end{array}<br /> \right)
When performing transformation of the H_a, it gives:
W_D^{-1}H_a W_D = \left( <br /> \begin{array}{cccccc} <br /> \Delta \hbar &amp; 0 &amp; 0 &amp; -\frac{i}{2 \sqrt{6}} &amp; 0 &amp; 0 \\ <br /> 0 &amp; \Delta \hbar &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\ <br /> 0 &amp; 0 &amp; \Delta \hbar &amp; 0 &amp; 0 &amp; \frac{i}{2 \sqrt{6}} \\ <br /> -\frac{i}{2 \sqrt{6}} &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\ <br /> 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\ <br /> 0 &amp; 0 &amp; \frac{i}{2 \sqrt{6}} &amp; 0 &amp; 0 &amp; 0 <br /> \end{array} <br /> \right)
I was a bit confused at first about why the imaginary unit i shows up here. As most of you may know multiplication with the imaginary unit introduces phase shift by \frac{\pi}{2}. Indeed, if you calculate time dependence of electric fields determined by polarization vectors \varepsilon_{xy} = \left( \frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}}\right) and \varepsilon_{z} = \left( 0,1,0\right) (components in Zeeman basis not Cartesian!), it is easy to conclude that the Euler rotations performed on \varepsilon_{xy} do result in i\,\varepsilon_{z} (which is \varepsilon_{z} shifted in phase by \frac{\pi}{2}). After introducing phase shift in calculations of Hamiltonian H_b I got the very same result as using W_D^{-1}H_a W_D.