Quantum Mechanics: Linear and Circular polarization states

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Homework Help Overview

The discussion revolves around evaluating matrix elements related to the action of the angular momentum operator \(\mathbb{\hat J}_z\) on linear and circular polarization states in quantum mechanics. The original poster attempts to express the linear polarization states \(|x\rangle\) and \(|y\rangle\) in terms of the circular polarization states \(|R\rangle\) and \(|L\rangle\).

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the need to determine the action of \(\mathbb{\hat J}_z\) on the states \(|R\rangle\) and \(|L\rangle\). There is a focus on substituting the expressions for \(|x\rangle\) and \(|y\rangle\) into the matrix elements, with some questioning the correctness of the original poster's approach.

Discussion Status

The conversation is ongoing, with participants providing guidance on the next steps needed to evaluate the matrix elements. There is recognition that the original poster has made progress in expressing the states but still needs to clarify the action of \(\mathbb{\hat J}_z\) on the circular states.

Contextual Notes

Participants note the importance of coursework in understanding the operator's action, suggesting that specific information from the coursework may be necessary to proceed effectively.

Robben
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Homework Statement



Evaluate the matrix elements ##
{\mathbb S}=\left( \begin{array}{cc} \langle x|\mathbb{\hat J}_z|x\rangle& \langle x|\mathbb{\hat J}_z|y\rangle\\
\langle y|\mathbb{\hat J}_z|x\rangle &\langle y|\mathbb{\hat J}_z|y\rangle\end{array}\right)## by expressing the linear polarization states ##|x\rangle## and ##|y\rangle## in terms of the circular polarization states ##|R\rangle## and ##|L\rangle.##

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



I worked out ##|x\rangle## and ##|y\rangle## and got:

##|x\rangle=\frac{1}{2}\left(|R\rangle+|L\rangle\right)##
##|y\rangle = \frac{-i}{\sqrt{2}}\left(|R\rangle-|L\rangle\right)##.

To get ##\mathbb{S},## do I just work out:

##
{\mathbb S}=\left( \begin{array}{cc}
\langle x \mid \mathbb{I} \ \mathbb{\hat J}_z \mathbb{I} \mid x \rangle
& \langle x \mid \mathbb{I} \ \mathbb{\hat J}_z \mathbb{I} \mid y \rangle \\
\langle y \mid \mathbb{I} \ \mathbb{\hat J}_z \mathbb{I} \mid x \rangle & \langle y \mid \mathbb{I} \ \mathbb{\hat J}_z \mathbb{I} \mid y \rangle \end{array}\right) =
\left( \begin{array}{cc}
\langle x \mid \pm z\rangle \langle \pm z | \ \mathbb{\hat J}_z |\pm z\rangle \langle \pm z \mid x \rangle
& \langle x \mid \pm z\rangle \langle \pm z | \ \mathbb{\hat J}_z |\pm z\rangle \langle \pm z \mid y \rangle \\
\langle y \mid \pm z\rangle \langle \pm z | \ \mathbb{\hat J}_z |\pm z\rangle \langle \pm z \mid x \rangle & \langle y \mid \pm z\rangle \langle \pm z | \ \mathbb{\hat J}_z |\pm z\rangle \langle \pm z \mid y \rangle \end{array}\right) \ ?
##
 
Last edited:
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Hi. You need to find out from your coursework what the action of Jz is on |R> and |L>, that's the whole point of using them instead of |x> and |y>...
 
Goddar said:
Hi. You need to find out from your coursework what the action of Jz is on |R> and |L>, that's the whole point of using them instead of |x> and |y>...

I am not understanding. So what I did was wrong?

The question asks to express ##|x\rangle## and ##|y\rangle## in terms of ##|R\rangle## and ##|L\rangle##.
 
That you did. But then you need to plug these relations in your matrix instead of what you did in the last part, which is what your question was about right?
 
So I did do it correctly and all I have to do is plug in the equations. Thank you!
 
Yes but then you'll still have to evaluate the matrix elements! And for that you'll need to know what <R| Jz|R> gives, for example...
Good luck.
 

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