Parity Operator in 2D: Understanding Transformation & Spin

[Malamala]

Hello! What is the 2D (acting in spin space) representation of the parity operator. In principle we can make it a diagonal matrix with the right transformation and given that $P^2=1$ the matrix would be diag(1,1) or diag(1,-1). However spin shouldn't change under parity and using that it seems like the diag(1,1) is the right representation. But that doesn't make much sense to me, as in this case it looks like all objects in 2D have the same parity. Can someone help me with this? Thank you!

 

[jambaugh]

Here is my understanding of the mathematical aspects of it. I don't know what conventions are in common use.

Let's see... rotations in 2-dim is represented by the group SO(2) the corresponding spin group is U(1). Its vector representation is 1-(complex)-dimensional, with generator $\boldsymbol{i}$. Now, parity is not a part of the spin group, it is not a "special" orthogonal transformation but rather general orthogonal group transformations, #O(2) it's "spinor" correspondent is the pin group, the group generated by the vector and bivector elements of the respective Clifford algebra. This is (in one formulation) the Quaternions. Let $\boldsymbol{j}, \boldsymbol{k}$ express, respectively, inversion in the x and y directions of the plane and then $\boldsymbol{i} = \boldsymbol{jk}$ generates rotation in the x-y plane. (Note I'm mixing around the associations so the bivector element is $\boldsymbol{i}$ the generator of the U(1) spin group. This will also match up with the alternative formulation below.)

Now we get to business. Parity transforms LH coordinates to RH coordinates (or when actively applied changes the LH vs RH orientation of an object.) In our odd 3 dimensional world, this can be achieved by inverting all three (and odd number of) spatial directions. Thus parity is usually expressed by inversion in the x, y, and z directions simultaneously. It is more ambiguous in 2 dimensions. We can effect parity change either by inverting the x-direction or the y-direction. Since the dimension is even inverting both effects a 180-degree rotation with no reversal of orientation.

However in either case (or some rotated version thereof) the adjoint action on the rotation generator is to change its sign: $\boldsymbol{jij^{-1} = -\boldsymbol{i}$ and similar. So when we revert back to the spin group, parity will manifest as complex conjugation in the 1-dim complex representation of U(1).

There is an alternative formulation of the Pin group analogous to the Marajona spinor representation of 3+1 relativistic spin. We can allow that the generator of inversions square to +1 hence:
$\gamma_x = \sigma_3, \gamma_y = \sigma_1, \text{ and } \boldsymbol{i} = \gamma_x\gamma_y = i\sigma_2$. Thence:
\boldsymbol{i} = \left( \begin{array}{cc} 0 & 1\\ -1 & 0\end{array}\right)
(if I got the signs correct.)

Now it looks like just another version of the SO(2) rotation matrix but it is not. A planar rotation of a vector through angle $\theta$ will correspond to a half-angle rotation $e^{i\theta/2}$.

Now we will again have two parity inverting operations. One effected by $\gamma_x = \sigma_3$ which in this real 2-dim spinor representation will invert the 2nd component e.g. corresponding to complex conjugation and the other effected by $\gamma_y = \sigma_1$ will swap the two components and will thus correspond to complex conjugation followed by multiplication by $\boldsymbol{i}$.

You can equate these with two with the operators P and -P and which you choose to associate with which would be a matter of convention.