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Let us assume you have a 3D bulk periodic crystal which has inversion symmetry e.g. $r\rightarrow -r$. Assume we are considering spinful operators with $S=1/2$.

Now let us imagine cutting a surface of this, lets say the surface $(x, y, 0)$. This surface should inherit the inversion symmetry of the bulk, e.g. it should be invariant under $(x,y,0)\rightarrow (-x,-y,0)$.

Now, according to the Wikipedia page for the parity operator (https://en.wikipedia.org/wiki/Parity_(physics)),

"In a two-dimensional plane, a simultaneous flip of all coordinates in sign is not a parity transformation; it is the same as a 180°-rotation. "

I can believe that this is not parity as e.g. this question points out that 2D parity would imply that spin is odd under parity. However, I cannot quite justify that this should be a $180°$ rotation.

The surface symmetry $(x,y,0)\rightarrow (-x,-y,0)$ looks a lot like a $180°$-rotation, except that a $180°$-rotation would also act on spins as $e^{i\pi S}$, e.g. it would give a factor of $+i$ for spin up and a factor $-i$ for spin down.

However, I would expect this factor to be absent on the surface, as there is no such factor in the bulk symmetry which the surface symmetry is inherited from (e.g. the bulk parity does not act on spin, so the surface symmetry should equally not act on spin).

So what is this effective symmetry? Is it a spinless $180°$-rotation in an otherwise spinful system? And if so, does anyone have an intuitive physical picture why this rotation does not give the usual factors for the spins?

Or should this symmetry actually act on spin?

Any help is greatly appreciated!

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The Wikipedia article is actually a bit misleading because it implicitly refers to a spinless system. In case where spin is involved you are right. It is a bit confusing to make this sort of dimensional reasoning with spin because it intrinsically lives in 3D space.

To make thing more formal recall that from group structure : $$i = r_z(180) \times \sigma_z $$

On 2D spatial coordinates, the action of the reflection with respect to the $z$ axis is identity ($\sigma_z \sim e$), so for the operators we have $$R_{z, space} (180) = \Pi_{space}$$ whereas the action on the spin degree freedom of the inversion operator is identity ($i \sim e$) because spin is an angular momentum. So $$R_{z, spin} (180) = \sigma^{-1}_{z,spin}$$

We conclude that for the spatial degrees of freedom in 2D, rotation by 180 is indeed equal to reflection. The same does not hold for spin.

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  • $\begingroup$ Thank you for the reply! I am somewhat confused by your last equation. Do you mean this as a counter example which does not hold in this situation? So in this concrete situation, there is no action on spin but only a flip of sign of the coordinates, like I suspected? $\endgroup$
    – G.Lang
    May 28, 2020 at 19:24
  • $\begingroup$ Yes exactly I mean it as a counter example! Rotation by 180 is not equivalent to inversion, as demonstrated by the last equation, but this is only the case when spins are considered. $\endgroup$
    – user140255
    May 29, 2020 at 17:19
  • $\begingroup$ Thank you so much for your answer and the clarification! $\endgroup$
    – G.Lang
    May 30, 2020 at 16:20

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