Quantum Mechanics: Transformation Matrix

[Robben]

Homework Statement

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.

Homework Equations

$|R\rangle = \frac{1}{\sqrt{2}}\left(|x\rangle+i|y\rangle\right)$

$|L\rangle = \frac{1}{\sqrt{2}}\left(|x\rangle-i|y\rangle\right)$

The Attempt at a Solution

Not sure exactly what the question is asking.

Is it asking to use $|x\rangle$ and $|y\rangle$ as a basis to find the transformation matrix that transforms $|x\rangle$ and $|y\rangle$ to $|R\rangle$ and $|L\rangle$?

If that is what it's asking, then would this be correct:

$\mathbb{S}=\left[{\begin{array}{cc} \langle R|x\rangle & \langle L|y\rangle \\ \langle R|x\rangle & \langle L|y\rangle \\\end{array}}\right]?$

 

[TSny]

Hello and welcome to PF!

I think you have the right idea but your matrix for $\mathbb{S}$ is not correct (It has two identical columns).

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 subsrcipt 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 $$\binom{c}{d}_{\!RL}$$ I believe the question is asking you to find a matrix $\mathbb{S}$ that will transform any $\binom{a}{b}_{\!xy}$ into the corresponding $\binom{c}{d}_{\!RL}$.

If, for example, the polarization state is $|\psi \rangle = \frac{1}{\sqrt{3}}|x \rangle + \sqrt{\frac{2}{3}}e^{i\pi/4}|y \rangle$, then if you apply the matrix $\mathbb{S}$ to
\begin{pmatrix}
\frac{1}{\sqrt{3}} \\\sqrt{\frac{2}{3}}e^{i\pi/4}
\end{pmatrix} you will get the corresponding column matrix that expresses the same state in the $|R \rangle$ and $|L \rangle$ basis.

 

[Robben]

Hello and welcome to PF!

Thank you!

I think you have the right idea but your matrix for $\mathbb{S}$ is not correct (It has two identical columns).

I made a mistake. It should be $\mathbb{S}^{\dagger}$ instead of $\mathbb{S}$.

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 subsrcipt 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 $$\binom{c}{d}_{\!RL}$$ I believe the question is asking you to find a matrix $\mathbb{S}$ that will transform any $\binom{a}{b}_{\!xy}$ into the corresponding $\binom{c}{d}_{\!RL}$.

If, for example, the polarization state is $|\psi \rangle = \frac{1}{\sqrt{3}}|x \rangle + \sqrt{\frac{2}{3}}e^{i\pi/4}|y \rangle$, then if you apply the matrix $\mathbb{S}$ to
\begin{pmatrix}
\frac{1}{\sqrt{3}} \\\sqrt{\frac{2}{3}}e^{i\pi/4}
\end{pmatrix} you will get the corresponding column matrix that expresses the same state in the $|R \rangle$ and $|L \rangle$ basis.

Can you elaborate please?

So we need to find a matrix $\mathbb{S}$ such that $\mathbb{S} \binom{a}{b}_{\!xy} = \binom{c}{d}_{\!RL}$ is true?

 

[TSny]

So we need to find a matrix $\mathbb{S}$ such that $\mathbb{S} \binom{a}{b}_{\!xy} = \binom{c}{d}_{\!RL}$ is true?

Yes, I believe that's what you are asked to do. For example, suppose you take a state that is linearly polarized in the x direction: $\binom{1}{0}_{\!xy}$. After multiplying by $\mathbb{S}$, what column vector should you get?