Cross Products and Axial Vectors

Note that the cross product of two vectors behaves like a vector in many ways. Under a parity transformation in which the direction of all three coordinate axes are inverted, a vector will change sign, but the cross product of two vectors will not change sign. It is therefore actually something different from a vector. We call it an axial vector. It turns out this this type of cross product of vectors can only be treated as a vector in three dimensions. In reality it is an antisymmetric tensor.

\bgroup\color{black}$\displaystyle (\vec{x}\times\vec{p})_{ij}=x_i p_j-x_j p_i =...
...pmatrix}0 & L_z & -L_y \\ -L_z & 0 & L_x \\ L_y & -L_x & 0\end{pmatrix} $\egroup

Since there are only three independent numbers in this tensor, it can be cast as a vector.

Lets use the angular momentum as an example. We know that angular momentum is normally defined as \bgroup\color{black}$ \vec{L}=\vec{r}\times\vec{p}$\egroup. Both \bgroup\color{black}$ \vec{r}$\egroup and \bgroup\color{black}$ \vec{p}$\egroup are normal vectors and change sign in the coordinate system undergoes a parity inversion. \bgroup\color{black}$ \vec{L}$\egroup then obviously does not change sign and is an Axial vector. We can write the angular momentum as the axial vector

\bgroup\color{black}$\displaystyle L_k=r_ip_j\epsilon_{ijk} $\egroup

or as the antisymmetric tensor.

\bgroup\color{black}$\displaystyle L_{ij}=r_ip_j-r_jp_i $\egroup

These two are easily related.

\bgroup\color{black}$\displaystyle L_k={1\over 2}L_{ij}\epsilon_{ijk} $\egroup

and

\bgroup\color{black}$\displaystyle L_{ij}=L_k\epsilon_{ijk} $\egroup

We can check this by putting the last two equations together.

\bgroup\color{black}$\displaystyle L_{ij}=L_k\epsilon_{ijk}=({1\over 2}L_{mn}\ep...
...i}\delta_{nj}-\delta_{mj}\delta_{ni})
={1\over 2}(L_{ij}-L_{ji})=L_{ij} $\egroup

Jim Branson 2012-10-21