indigojoker
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Let the translation operator be:
F (\textbf {l} ) = exp \left( \frac{-i \textbf{p} \cdot \textbf{l}}{\hbar} \right)
where p is the momentum operator and l is some finite spatial displacement
I need to find [x_i , F (\textbf {l} )]
let me start with a fundamental commutation relation:
[x_i , G ( \textbf{p} ) ]=i \hbar \frac{ \partial G}{\partial p_i}
also let me define:
\textbf{p} = (p_i,p_j,p_k)
\textbf{l} = (\Delta x, \Delta y, \Delta z)
we expand F in a taylor series:
F (\textbf {l} ) = \sum_{n=0}^{\infty} \frac{ \left( \frac{ -i \textbf{p } \cdot \textbf{ l}}{\hbar} \right) ^n}{n!}
= \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!}
= \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
we now apply:
[x_i , G ( \textbf{p} ) ]=i \hbar \frac{ \partial G}{\partial p_i}
which leaves us with:
= \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
[x_i , G ( \textbf{p} ) ]=\frac{n i \hbar}{p_i} F (\textbf {l} )
wondering if this looks okay.
F (\textbf {l} ) = exp \left( \frac{-i \textbf{p} \cdot \textbf{l}}{\hbar} \right)
where p is the momentum operator and l is some finite spatial displacement
I need to find [x_i , F (\textbf {l} )]
let me start with a fundamental commutation relation:
[x_i , G ( \textbf{p} ) ]=i \hbar \frac{ \partial G}{\partial p_i}
also let me define:
\textbf{p} = (p_i,p_j,p_k)
\textbf{l} = (\Delta x, \Delta y, \Delta z)
we expand F in a taylor series:
F (\textbf {l} ) = \sum_{n=0}^{\infty} \frac{ \left( \frac{ -i \textbf{p } \cdot \textbf{ l}}{\hbar} \right) ^n}{n!}
= \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!}
= \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
we now apply:
[x_i , G ( \textbf{p} ) ]=i \hbar \frac{ \partial G}{\partial p_i}
which leaves us with:
= \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
[x_i , G ( \textbf{p} ) ]=\frac{n i \hbar}{p_i} F (\textbf {l} )
wondering if this looks okay.
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