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Commutation of operator

  1. Oct 2, 2007 #1
    Let the translation operator be:

    [tex]F (\textbf {l} ) = exp \left( \frac{-i \textbf{p} \cdot \textbf{l}}{\hbar} \right) [/tex]

    where p is the momentum operator and l is some finite spatial displacement

    I need to find [tex] [x_i , F (\textbf {l} )] [/tex]

    let me start with a fundamental commutation relation:

    [tex] [x_i , G ( \textbf{p} ) ]=i \hbar \frac{ \partial G}{\partial p_i} [/tex]

    also let me define:

    [tex]\textbf{p} = (p_i,p_j,p_k) [/tex]
    [tex]\textbf{l} = (\Delta x, \Delta y, \Delta z)[/tex]

    we expand F in a taylor series:

    [tex]F (\textbf {l} ) = \sum_{n=0}^{\infty} \frac{ \left( \frac{ -i \textbf{p } \cdot \textbf{ l}}{\hbar} \right) ^n}{n!} [/tex]

    [tex]= \sum_{n=0}^{\infty} \frac{ \left( \frac{-i p_i \Delta x}{\hbar} \right) ^n}{n!} \frac{ \left( \frac{-i p_j \Delta y}{\hbar} \right) ^n}{n!} \frac{ \left(\frac{ -i p_k \Delta z}{\hbar} \right) ^n}{n!}[/tex]

    [tex]= \sum_{n=0}^{\infty} \frac{ \left(\frac{ -i \Delta x}{\hbar} \right) ^n}{n!} \frac{ \left(\frac{ -i \Delta y}{\hbar} \right) ^n}{n!} \frac{ \left(\frac{ -i \Delta z}{\hbar} \right) ^n}{n!} p_i^n p_j^n p_k^n[/tex]

    we now apply:

    [tex] [x_i , G ( \textbf{p} ) ]=i \hbar \frac{ \partial G}{\partial p_i} [/tex]

    which leaves us with:

    [tex]= \sum_{n=0}^{\infty} \frac{ \left(\frac{ -i \Delta x}{\hbar} \right) ^n}{n!} \frac{\left(\frac{ -i \Delta y}{\hbar} \right) ^n}{n!} \frac{ \left(\frac{ -i \Delta z}{\hbar} \right) ^n}{n!} n i \hbar p_i^{n-1} p_j^n p_k^n[/tex]

    [tex][x_i , G ( \textbf{p} ) ]=\frac{n i \hbar}{p_i} F (\textbf {l} ) [/tex]

    wondering if this looks okay.
     
    Last edited: Oct 2, 2007
  2. jcsd
  3. Oct 3, 2007 #2

    dextercioby

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    It's not okay. Compute this commutator

    [tex] \left[x_{1},e^{\frac{p_{1}l_{1}+p_{2}l_{2}+p_{3}l_{3}}{i\hbar}} \right] [/tex]
     
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