The commutator of position and momentum, specifically ##[\hat{p}_x, \mathbf{\hat{r}}]##, can be expanded into its components as ##([\hat{p}_x, \hat{x}], [\hat{p}_x, \hat{y}], [\hat{p}_x, \hat{z}])##. This indicates that the position operator vector ##\mathbf{\hat{r}}## consists of the individual position operators ##\hat{x}##, ##\hat{y}##, and ##\hat{z}##, which are treated separately when calculating the commutators with the momentum operator ##\hat{p}_x##. The discussion clarifies that the expansion of the commutator follows the vector operator representation, confirming the relationship between the components of the vector and their respective operators.
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Kara386 said:How would ##[p_x, r]## be expanded? Where ##r=(x,y,z)##, the position operators. Do you do the commutators of ##p_x## with ##x, y,z## individually? So ##[p_x,x]+[p_x,y+p_x,z]## for example?
Ah, thank you. :)PeroK said:More generally, a vector operator such as ##\mathbf{\hat{r}}## represents three operators ##(\hat{x}, \hat{y}, \hat{z})##, related in the same way as the components of a vector.
In this case, essentially by definition:
##[\hat{p}_x, \mathbf{\hat{r}}] = ([\hat{p}_x, \hat{x}], [\hat{p}_x, \hat{y}], [\hat{p}_x, \hat{z}])##