Jones-Vector Representation

The Jones vector approach to describing optics makes life easier by completely describing the phase and amplitude of the polarization ellipse in a single column vector. The most general vector is given as:

\begin{equation} J = \begin{pmatrix} A_xe^{i\delta_x} \\ A_ye^{i\delta_y} \end{pmatrix} \end{equation}

which we can recognize comes directly from our complex envelope representation before! Obviously, since life takes place in real space, we must take the real part of the Jones vector in order to describe the actual propagation i.e. $Re[J_xe^{i\omega{t}}] = Re[A_xe^{i\omega{t} + \delta_x}]$.

We can thus consider the case when $A_x = A_y = 1$ and that the phase difference between the two waves is $\delta = \delta_y – \delta_x = m\pi$. As we saw before, this phase difference corresponds to linearly polarized light! We can arbitrarily set $\delta_x = 0$ and $\delta_y = \pi$ such that we have the Jones vector:

\begin{equation} J = \begin{pmatrix} cos(0) + isin(0) \\ cos(\pi) + isin(\pi) \end{pmatrix} = \begin{pmatrix} 1 \\ -1 \end{pmatrix} \end{equation}

Now let’s consider the case when $A_x = A_y = 1$ but the phase difference is now $\delta = m\pi/2$. This then corresponds to circularly polarized light, but let’s see what the Jones vector yields:

\begin{equation} J = \begin{pmatrix} cos(0) + isin(0) \\ cos(\pi/2) + isin(\pi/2) \end{pmatrix} = \begin{pmatrix} 1 \\ \frac{1}{\sqrt{2}}(1 + i) \end{pmatrix} \end{equation}

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