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The operator for linear momentum $\mathbf{p}$ and the operator for orbital angular momentum $\mathbf{L}$ ($\mathbf{L} = \mathbf{r} \times \mathbf{p}$) are Hermitian. Is the cross product between $\mathbf{p}$ and $\mathbf{L}$ also Hermitian? If I go component wise and use the commutator bracket relation $[p_i,L_j]=i\hbar \epsilon _{ijm}p_m$, then I get-

$(\mathbf{p} \times \mathbf{L})_i = p_yL_z - p_zL_y$

If I take the hermitian conjugate, I get -

$(\mathbf{p} \times \mathbf{L})^\dagger_i = L_zp_y - L_yp_z$

Now, $[L_i, p_j]$ doesn't commute if $i\neq j$. So, we get that $(\mathbf{p} \times \mathbf{L})_i \neq (\mathbf{p} \times \mathbf{L})^\dagger_i$ in this case. Thus, it's not Hermitian.

But on using the vector triple product relation, I see that it's Hermitian indeed. Can someone explain to me where am I going wrong?

Also, can we state this in general for any such cross product between two Hermitian operators?

The operator for linear momentum $\mathbf{p}$ and the operator for orbital angular momentum $\mathbf{L}$ ($\mathbf{L} = \mathbf{r} \times \mathbf{p}$) are Hermitian. Is the cross product between $\mathbf{p}$ and $\mathbf{L}$ also Hermitian? If I go component wise and use the commutator bracket relation $[p_i,L_j]=i\hbar \epsilon _{ijm}p_m$, then I get $$(\mathbf{p} \times \mathbf{L})_x = p_yL_z - p_zL_y.$$

If I take the hermitian conjugate, I get
$$(\mathbf{p} \times \mathbf{L})^\dagger_x = L_zp_y - L_yp_z.$$

Now, $[L_i, p_j]$ doesn't commute if $i\neq j$. So, we get that $(\mathbf{p} \times \mathbf{L})_i \neq (\mathbf{p} \times \mathbf{L})^\dagger_i$ in this case. Thus, it's not Hermitian.

But on using the vector triple product relation, I see that it's Hermitian indeed. Can someone explain to me where am I going wrong?

Also, can we state this in general for any such cross product between two Hermitian operators?