Prove that the angular momentum operator is hermitian

[Septim]

Greetings,

My task is to prove that the angular momentum operator is hermitian. I started out as follows:

\vec{L}=\vec{r}\times\vec{p}

Where the above quantities are vector operators. Taking the hermitian conjugate yields

\vec{L''}=\vec{p''}\times\vec{r''}

Here I have used double quotes to represent that the hermitian conjugate of the corresponding quantity.

\vec{L''}=\vec{p}\times\vec{r}

Here the fact that the momentum and position are hermitian operators were used. However
\vec{L''}=\vec{p}\times\vec{r}=-\vec{r}\times\vec{p}=-\vec{L}[

There has to be a flaw somewhere but I was not able to catch it, though I was able to prove that the angular momentum operator is hermitian when inspected component by component. I am yet to understand the error in the above derivation. Any help is appreciated.

Thanks in advance

 

[Bill_K]

L′′ = p′′ × r′′

The Hermitian conjugate reverses the operator order, but you've also reversed the order of the arguments to the cross product, which should bring in a minus sign.

 

[Septim]

Thanks for the reply. So as far as I am concerned you tell me that reversing the arguments of the cross product is superfluous. But how can I denote that I have reversed the orders of the operators without doing that? My point is to demonstrate that I have reversed the order of operators in vector notation.

 

[stevendaryl]

Thanks for the reply. So as far as I am concerned you tell me that reversing the arguments of the cross product is superfluous. But how can I denote that I have reversed the orders of the operators without doing that? My point is to demonstrate that I have reversed the order of operators in vector notation.

In general, if \vec{A} and \vec{B} are operators, then (\vec{A} \times \vec{B})^\dagger = - \vec{B}^\dagger \times \vec{A}^\dagger