Hermitian conjugation identity

1. Jul 12, 2017

Dyatlov

1. The problem statement, all variables and given/known data
$(\hat A \times \hat B)^*=-\hat B^* \times \hat A^*$
Note that $*$ signifies the dagger symbol.

2. Relevant equations
$(\hat A \times \hat B)=-(\hat B \times \hat A)+ \epsilon_{ijk} [a_j,b_k]$

3. The attempt at a solution
Using as example $R$ and $P$ operators:
$(\hat R \times \hat P)^*_i=-(\hat P \times \hat R)^*_i+ \epsilon_{ijk} [Y,P_z]$
$(\hat R \times \hat P)^*_i=-(\hat P \times \hat R)^*_i$
$(\hat R \times \hat P)^*=-\hat P^* \times \hat R^*$

2. Jul 12, 2017

strangerep

Start instead from a component-wise definition:
$$(A \times B)_i ~=~ \epsilon_{ijk} A_j B_k$$ (where the usual summation convention applies to repeated indices).

3. Jul 13, 2017

Dyatlov

Since $\epsilon_{ijk}$ is antisymmetric then we have
$\epsilon_{ijk}A_jB_k=A_jB_k-A_kB_j$
$A_jB_k-A_kB_j=-(B_jA_k-B_kA_j)$
$(A \times B)_i=-(B \times A)_i$
Since A and B are Hermitian the same equlity holds for their self-adjoint counterparts.

Last edited: Jul 13, 2017
4. Jul 13, 2017

strangerep

That equation does not make sense. On the LHS, $i$ is a free index, but $j,k$ are dummy summation indices. However, on your RHS both $j$ and $k$ are free indices, and there's no $i$ at all. Both left and right hand sides of such an equation must have exactly the same free indices.

The LHS uses a version of the summation convention. It is short for $$\sum_{j,k} \epsilon_{ijk}A_jB_k$$
This is wrong if $B_j$ and $A_k$ don't commute (which is presumably the case here, since the problem statement didn't specify commutativity). So you can't blithely interchange $A$ and $B$ like that.

Your original problem statement doesn't say that $A,B$ are Hermitian.

But that doesn't solve the problem as stated.

Start with this: $$\left( \sum_{j,k} \epsilon_{ijk}A_jB_k \right)^\dagger ~=~ \dots\,? \dots$$ Hint: for arbitrary operators $X,Y$, what is $(XY)^\dagger$ ?

5. Jul 14, 2017

Dyatlov

Thanks for the replies.
The title mentions that I am solving the identity for Hermitian operators.
I know that $\epsilon_{ijk}A_jB_k$ is a sum over j and k, with $j,k=1,2,3$.
$(\epsilon_{ijk}A_jB_k)^\dagger=(\epsilon_{ijk}B^\dagger_kA^\dagger_j)=-(B^\dagger \times A^\dagger)_i$
Therefore:
$(A \times B)^\dagger_i=-(B^\dagger \times A^\dagger)_i$

6. Jul 14, 2017

strangerep

No it doesn't -- it mentions "Hermitian conjugation", which is an operation which can be performed on any operator.

Anyway, I take it you're now happy with the solution.

7. Jul 15, 2017