Commutator of position and momentum

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The commutator of position and momentum, specifically ##[p_x, r]## where ##r=(x,y,z)##, can be expanded by calculating the individual commutators of ##p_x## with each position operator: ##[p_x, x]##, ##[p_x, y]##, and ##[p_x, z]##. This results in the expression ##[\hat{p}_x, \mathbf{\hat{r}}] = ([\hat{p}_x, \hat{x}], [\hat{p}_x, \hat{y}], [\hat{p}_x, \hat{z}])##. The vector operator ##\mathbf{\hat{r}}## comprises the three position operators, and their relationships follow the same principles as vector components. Thus, the expansion of the commutator reflects the inherent structure of these operators. Understanding this relationship is crucial for further analysis in quantum mechanics.
Kara386
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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?
 
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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?

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}])##
 
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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}])##
Ah, thank you. :)
 
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