Determine a $2\times 2$ matrix $\mathbb{S}$ that can be used to transform a column vector representing a photon polarization state using the linear polarization vectors $|x\rangle$ and $|y\rangle$ as a basis to one using the circular polarization vectors $|R\rangle$ and $|L\rangle$ as a basis.
Relevant equation:
$|R\rangle = \frac{1}{\sqrt{2}}\left(|x\rangle+i|y\rangle\right)$ and $|L\rangle = \frac{1}{\sqrt{2}}\left(|x\rangle-i|y\rangle\right).$
I am not sure on how to go about doing this problem?
I know that an arbitrary polarization state, $|\psi \rangle = a|x \rangle + b|y \rangle,$ can be written as a column matrix $\binom{a}{b}_{\!xy}$ where the subscript tells us which basis we're using.
The same state of polarization could be expressed in terms of the $|R \rangle$ and $|L \rangle$ basis states: $|\psi \rangle = c|R \rangle + d|L \rangle$. Or, as $\binom{c}{d}_{\!RL},$ so I believe the question is asking to find a matrix $\mathbb{S}$ that will transform any $\binom{a}{b}_{\!xy}$ into the corresponding $\binom{c}{d}_{\!RL}.$
But how do I do that?
Hint: have a look where $|x\rangle$ and $|y\rangle$ are transformed.
One has $$ |x\rangle=\frac{1}{\sqrt{2}}(|R\rangle+|L\rangle)\quad \text{and}\quad |y\rangle=\frac{-i}{\sqrt{2}}(|R\rangle-|L\rangle)$$ which gives $${\mathbb S}=\left( \begin{array}{cc} \frac{1}{\sqrt{2}}&\frac{-i}{\sqrt{2}}\\ \frac{1}{\sqrt{2}} &\frac{i}{\sqrt{2}}\end{array}\right)=\frac{1}{\sqrt{2}}\left( \begin{array}{cc} 1&-i\\ 1 &i\end{array}\right). $$
Explanation
I can explain in the language of linear algebra. One has a two dimensional vector space spanned by two vectors, in your case $|x\rangle$ and $|y\rangle$ (this is a basis of the space). It means that every vector, say $|\psi\rangle$, is of the form $|\psi\rangle =a|x\rangle+b|y\rangle$. Now, let $|R\rangle, |L\rangle$ be another basis of the space and we want to express $|\psi\rangle$ in this basis. Then we look how $|x\rangle$ and $|y\rangle$ are expresed in this new basis. For instance, $|x\rangle=\alpha |R\rangle+\gamma |L\rangle$ and $|y\rangle=\beta |R\rangle+\delta |L\rangle$. Then the corresponding transformation has matrix representation $$ {\mathbb S}=\left(\begin{array}{cc} \alpha & \beta\\ \gamma & \delta \end{array}\right). $$ Note that $|x\rangle=\left(\begin{array}{c} 1\\ 0\end{array}\right)_{xy}$ and $|y\rangle=\left(\begin{array}{c} 0\\ 1\end{array}\right)_{xy}$. Hence $$ {\mathbb S}|x\rangle=\left(\begin{array}{cc} \alpha & \beta\\ \gamma & \delta \end{array}\right)\left(\begin{array}{c} 1\\ 0\end{array}\right)_{xy}=\left(\begin{array}{c} \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}\end{array}\right)_{RL} $$ and $$ {\mathbb S}|y\rangle=\left(\begin{array}{cc} \alpha & \beta\\ \gamma & \delta \end{array}\right)\left(\begin{array}{c} 0\\ 1\end{array}\right)_{xy}=\left(\begin{array}{c} \frac{-i}{\sqrt{2}}\\ \frac{i}{\sqrt{2}}\end{array}\right)_{RL} $$ Because of linearity of matrix multiplication one has for a general $|\psi\rangle=a|x\rangle+b|y\rangle$: $$ {\mathbb S}|\psi\rangle=\left(\begin{array}{cc} \alpha & \beta\\ \gamma & \delta \end{array}\right)\left(\begin{array}{c} a\\ b\end{array}\right)_{xy}= a{\mathbb S}|x\rangle+b{\mathbb S}|y\rangle=\left(\begin{array}{c} \frac{a-bi}{\sqrt{2}}\\ \frac{a+bi}{\sqrt{2}}\end{array}\right)_{RL}.$$