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?