I Which operator for reflection in quantum mechanics?

  • Thread starter Amentia
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Hello,

I know we have the parity operator for inversion in quantum mechanics and for rotations we have the exponentials of the angular momentum/spin operators. But what if I want to write the operator that represent a reflection for example just switching y to -y, the matrix in real space being:

$$\begin{pmatrix}
1 & 0 & 0\\
0 & -1 & 0\\
0 & 0 & 1
\end{pmatrix}=
\begin{pmatrix}
-1 & 0 & 0\\
0 & -1 & 0\\
0 & 0 & -1
\end{pmatrix}
\begin{pmatrix}
-1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & -1
\end{pmatrix}
$$ ?

Is it possible to write it as a composition of rotations and/or parity operators? Or is there already an operator for this kind of transformation like partial parity... ? I could not find it on the internet or book chapters.

Thank you for any hints about that.
 
In case it is unclear, I will take an example that is of interest to me. I wanted to ask if I can write an operator for reflection along y as:
$$\Pi e^{-\frac{i}{\hbar}\hat{L}_{y}\pi}\otimes e^{-\frac{i}{\hbar}\hat{S}_{y}\pi} = \Pi e^{-\frac{i}{\hbar}\hat{J}_{y}\pi}$$
when I want to perform a reflecion on a spin 3/2 particle and $$\Pi$$ is the parity operator while L acts on angular momentum for l=1 and S on spin 1/2 while J is the operator when I add them.

And I know that we have:
$$\Pi |l,m\rangle = (-1)^{l}|l,m\rangle$$

but I do not know if this operator acts in a simple way on spin 1/2 states. I believe also it should commute wih the rotation operators. Sorry that was actually two questions into one but if my assumptions that I can use the parity operator to represent a reflection is wrong, this second question was not really necessary...
 

A. Neumaier

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One can write any reflection ##R## as a product of the rotation ##Q=RR_0^{-1}## and another given reflection ##R_0##, since the rotation group ##SO(3)## has index 2 in the full orthogonal group ##O(3)## generated by rotations and reflections.
 
Thank you for your answer but I am not sure to understand. You are saying I could write my reflection as $$R=QR_{0},$$ but it implies writing another reflection while my question is about how to write such a reflection, either directly or as a composition of other operators which are not reflections. Or did you mean something else?
 

A. Neumaier

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Thank you for your answer but I am not sure to understand. You are saying I could write my reflection as $$R=QR_{0},$$ but it implies writing another reflection while my question is about how to write such a reflection, either directly or as a composition of other operators which are not reflections. Or did you mean something else?
You can take the parity as the reflection ##R_0##.
 
Ok, so assuming the parity does not act on spin 1/2, I assume my first equation is correct? I could say, calling ##R_{y}## my reflection operator:

$$R_{y}|\frac{3}{2},\frac{3}{2}\rangle = e^{-\frac{i}{\hbar}\hat{J}_{y}\pi}\Pi|\frac{3}{2},\frac{3}{2}\rangle \\

R_{y}|\frac{3}{2},\frac{3}{2}\rangle = e^{-\frac{i}{\hbar}\hat{L}_{y}\pi}\Pi\otimes e^{-\frac{i}{\hbar}\hat{S}_{y}\pi} |1,1\rangle\otimes|\frac{1}{2},\frac{1}{2}\rangle \\
R_{y}|\frac{3}{2},\frac{3}{2}\rangle = e^{-\frac{i}{\hbar}\hat{L}_{y}\pi}\Pi|1,1\rangle\otimes e^{-\frac{i}{\hbar}\hat{S}_{y}\pi}|\frac{1}{2},\frac{1}{2}\rangle \\
R_{y}|\frac{3}{2},\frac{3}{2}\rangle = -e^{-\frac{i}{\hbar}\hat{L}_{y}\pi}|1,1\rangle\otimes e^{-\frac{i}{\hbar}\hat{S}_{y}\pi}|\frac{1}{2},\frac{1}{2}\rangle
$$

And then just proceed by developping the rotation operator with the Pauli matrices and matrices for angular momentum equal to 1?
 

A. Neumaier

Science Advisor
Insights Author
6,052
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I didn't have time to look closely at your example. I only answered to your post #1. My statement holds for any dimension, though, also for tensor products.
 
All right, so it should work for me, thank you again!
 

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