# Commutation of operator

1. Oct 2, 2007

### indigojoker

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 $$[x_i , F (\textbf {l} )]$$

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

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.

Last edited: Oct 2, 2007
2. Oct 3, 2007

### dextercioby

It's not okay. Compute this commutator

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