under parity transformation in 1D and 3D, we know that a parity transformation takes $\vec{r}\mapsto-\vec{r}$ and $\vec{p}\mapsto-\vec{p}$.
For the 3D case, the above means that the orbital angular momentum $\vec{L}=\vec{r}\times\vec{p}$ is invariant under a parity transformation. Since spin and orbital angular momentum satisfy the same commutation relations, also spin does not change under a parity transformation in 1D and 3D.

On the other hand, in the 2D case, we know that only one component of $\vec{r}$ and $\vec{p}$ flips its sign. So, in this case, $L_z=x\ p_y-y\ p_x$ changes sign under a parity transformation. But, now there are no other components of the angular momentum, so it makes no sense to talk about the commutation relations that the orbital angular momentum operators must satisfy, so I cannot use the same argument as above (for the commutation relations) to conclude that the spin must also change sign.

So, what does happen with the way that spin transforms in 2D under a parity transformation?


If we use the definition of angular momentum as generator of rotations, regardless of the dimensions of the space, a vector operator $\textbf{V}$ under rotations must satisfy
\begin{equation} U^{\dagger}(R)\textbf{V}_a U(R)=R_{ab}\textbf{V}_b \end{equation} Where ${R}_{ab}$ is a D dimensional rotation matrix and $U(R)$ is its representation in the Hilbert space where the operator $\textbf{V}$ lives. Let's consider an infinitesimal rotation, $R_{ab}=\delta_{ab}+\epsilon K_{ab}$ where $\epsilon$ is a small parameter and $K^T=-K$. The unitary operator $U(R)$, for such an infinitesimal operator must take the form \begin{equation} U(R)= \textbf{1}+\frac{i\epsilon}{2}K_{ab} \textbf{J}_{ab} + O(\epsilon^2) \end{equation} Observe that since U must be unitary, all the compotents of $\textbf{J}_{ab}$ must be hermitian and since $K$ is antisymmetric, $\textbf{J}_{ab}$ can be taken antisymmetric too. Putting this form of $U(R)$ into the first equation, we find the commutation relations: \begin{equation} i\left[\textbf{V}_c, \textbf{J}_{ab}\right]=\delta_{ac}\textbf{V}_b - \delta_{bc}\textbf{V}_a \end{equation} and this formula does not depend on the number of components of $\textbf{V}$. Computing \begin{equation} U^\dagger(R)\textbf{J}_{ab}U(R) \end{equation} you can show that $\textbf{J}$ behaves as a tensor under rotations.

In D-dimension, the angular momentum (i.e. the generator of rotations) is therefore a D dimensional rank 2 antisymmetric tensor.

In 2 dimensions such an object has just 1 independent component as you noticed and, as is well know, in 3 dimensions $\textbf{J}$ has three independent components and can be reduced to an axial vector.

Returning to our question, in the specific case of D=2, the commutation relations written above are \begin{equation} i\left[\textbf{X},\textbf{J}_{12}\right]=\textbf{Y} \end{equation} \begin{equation} i\left[\textbf{Y},\textbf{J}_{12}\right]=\textbf{-X} \end{equation} Using this commutation relations you can show that $\textbf{J}$ changes sign under a parity transformation.

Let's flip the X axis leaving Y unaltered: \begin{equation} \frac{1}{i}\textbf{Y} =\frac{1}{i}\pi^\dagger \textbf{Y} \pi = \pi^\dagger \left[\textbf{X},\textbf{J}_{12}\right]\pi=\pi^\dagger \textbf{X}\pi \pi^\dagger \textbf{J}_{12}\pi - \pi^\dagger \textbf{J}_{12}\pi \pi^\dagger \textbf{X}\pi = - \left[\textbf{X}, \pi^\dagger \textbf{J}_{12} \pi\right] \end{equation} Which implies \begin{equation} \pi^\dagger \textbf{J}_{12} \pi = - \textbf{J}_{12} \end{equation} The same can be shown for $\textbf{J}_{21}$. Since $\textbf{J}$ changes sign under a parity transformation and, as you noticed, $\textbf{L}$ changes sign too, $\textbf{S}=\textbf{J} - \textbf{L}$ is odd under parity.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.