$\epsilon_{ijk}x_j p_k$ under orthogonal transformation including those involving parity

Question

The easiest way to see that angular momentum $\vec{L}=\vec{r}\times\vec{p}$ is an axial vector is to note that $\vec{r}\to -\vec{r}$ and $\vec{p}\to-\vec{p}$ under parity. But this should mean that every component of $\vec{L}$ must remain unchanged under parity. For example, $L_x=(yp_z-zp_y)$ does not change sign because $y\to -y, p_z\to-p_z, z\to-z$ and $p_y\to -p_y$ under parity.

How can I prove this for a all components in index notation?

I started with the index notation $L_i=\epsilon_{ijk}x_jp_k$. First I want to find out how it transforms under an orthogonal transformation $A$. I know that $x^\prime_i=A_{ij}x_j$ and $p^\prime_i=A_{ij}p_j$ and $\epsilon^\prime_{ijk}=\det(A)\epsilon_{ijk}$ where $\det(A)=-1$ when the orthogonal transformation includes parity. I tricky matter is that $j$ and $k$ are summed over in the expression for $L_i$.

$\circ$ Is it possible to proceed from the expression $\epsilon_{ijk}A_jB_k$ and work out its transformation under orthogonal transformations?

Answer 1

Forgetting that $A$ is orthogonal for the moment, its determinant is defined by$$\det A\cdot\epsilon_{glm}=\epsilon_{hjk}A_{hg}A_{jl}A_{km}.$$Multiplying both sides by $A_{ig}=A_{gi}^T$ for orthogonal $A$,$$\det A\cdot A_{ig}\epsilon_{glm}=\epsilon_{hjk}\delta_{hi}A_{jl}A_{km}=\epsilon_{ijk}A_{jl}A_{km}.$$Thus$$\epsilon_{ijk}(Ax)_j(Ap)_k=\epsilon_{ijk}A_{jl}A_{km}x_lp_m=\det A\cdot A_{ig}\epsilon_{glm}x_lp_m.$$These are the $i$th components of $(Ax)\times(Ap)=\det A\cdot A(x\times p)$. (In fact, for invertible $A$ we get the generalization $(Ax)\times(Ap)=\det A\cdot (A^{-1})^T(x\times p)$, which makes sense if you count powers of $A$.)

Answer 2

I would post my attempt. If anything wrong someone will point it out.

Let us start with the following expression $$\epsilon^\prime_{ijk} B^\prime_l C^\prime_m=A_{ip}A_{jq}A_{kr}A_{ls}A_{mt}B_sC_t\\ (\det{A})\epsilon_{ijk} B^\prime_l C^\prime_m=A_{ip}A_{jq}A_{kr}A_{ls}A_{mt}B_sC_t.$$ Multiplying both sides by $\delta_{jl}\delta_{km}$, $$(\det{A})\epsilon_{ijk} B^\prime_l C^\prime_m\delta_{jl}\delta_{km}=A_{ip}A_{jq}A_{kr}A_{ls}A_{mt}B_sC_t\delta_{jl}\delta_{km}\\=A_{ip}(A^T_{qj}I_{jl}A_{ls})(A^T_{rk}I_{km}A_{mt})\epsilon_{pqr}B_sC_t\\=A_{ip}\delta_{qs}\delta_{rt}\epsilon_{pqr}B_sC_t\\=A_{ip}\epsilon_{pqr}B_qC_r$$ Thus, we reach at the same result in the other answer if both sides are multiplied by $\det A$ and $(\det A)^2=1$ used.

Note But it is not clear why $(\vec{B}\times\vec{C})^\prime_i\neq \epsilon_{ijk}^\prime B^\prime_j C^\prime_k$ but$(\vec{B}\times\vec{C})^\prime_i=\epsilon_{ijk} B^\prime_j C^\prime_k$. In other words, why is $(\vec{B}\times\vec{C})^\prime=(\vec{B}^\prime\times\vec{C}^\prime)$? I leave it for someone else to comment on.