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Prove that the angular momentum operator is hermitian

  1. Feb 28, 2013 #1
    Greetings,

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

    [itex]\vec{L}=\vec{r}\times\vec{p}[/itex]

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

    [itex]\vec{L''}=\vec{p''}\times\vec{r''}[/itex]

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

    [itex]\vec{L''}=\vec{p}\times\vec{r}[/itex]

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

    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
     
  2. jcsd
  3. Feb 28, 2013 #2

    Bill_K

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    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.
     
  4. Feb 28, 2013 #3
    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.
     
  5. Feb 28, 2013 #4

    stevendaryl

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