Question 6.2: Find the parity-transformed angular momentum operator L'=Π†L...

Since L=r×p , Equation 6.23 [\hat {Q}^′  (\hat{r},\hat{p} )=\hat{\Pi }^{\dagger }\hat{Q} (\hat{r},\hat{p} )\hat{\Pi }=\hat{Q} (-\hat{r},-\hat{p} )] tells us that

\hat{L} ^′  =\hat{\Pi }^{\dagger } \hat{L} \hat{\Pi } =\hat{r} ^′ \times \hat{p} ^′ =(-\hat{r}) \times (-\hat{p})=\hat{r}\times \hat{p}=\hat{L}     (6.24)

We have a special name for vectors like \hat{L} , that are even under parity. We call them pseudovectors, since they don’t change sign under parity the way “true” vectors, such as  \hat{r} or \hat{p} , do. Similarly, scalars that are odd under parity are called pseudoscalars, since they do not behave under parity the way that “true” scalars (such as \hat{r} \cdot \hat{r}   which is even under parity) do. See Problem 6.9. Note: The labels scalar and vector describe how the operators behave under rotations; we will define these terms carefully in the next section. “True” vectors and pseudovectors behave the same way under a rotation—they are both vectors.