Physics Asked by Butane on December 26, 2020
In the book "Modern Quantum Mechanics" (by J.J. Sakurai and Jim Napolitano) page 44, infinitesimal Translation operator is given:
$mathscr{J}left(d mathbf{x}^{prime}right)left|mathbf{x}^{prime}rightrangle=left|mathbf{x}^{prime}+d mathbf{x}^{prime}rightrangle$ after this, book mentioned properties of the operator as being unitary and adding: $mathscr{J}left(d mathbf{x}^{prime prime}right) mathscr{J}left(d mathbf{x}^{prime}right)=mathscr{J}left(d mathbf{x}^{prime}+d mathbf{x}^{prime prime}right)$ where I have confused is that:
$mathbf{x} mathscr{J}left(d mathbf{x}^{prime}right)left|mathbf{x}^{prime}rightrangle=mathbf{x}left|mathbf{x}^{prime}+d mathbf{x}^{prime}rightrangle=left(mathbf{x}^{prime}+d mathbf{x}^{prime}right)left|mathbf{x}^{prime}+d mathbf{x}^{prime}rightrangle$
The left hand side of the equation is so weird to me because middle one is directly driven from the first equation but to find a meaning at the right hand side of the above equation, I thought it might be true because $d mathbf{x}^{prime}$ is infinitesimally small but where is $mathbf{x}^{prime}$came from because initially it was $mathbf{x}$ at the right hand side of the last equation.
I'm not sure I completely understood your question, but perhaps it will become clear if I explain what the last equation means. The left hand side of the equation means:
By definition, translating the state $|mathbf{x}'rangle$ by $text{d}mathbf{x}'$ gives you the state $|mathbf{x}' + text{d}mathbf{x}'rangle$, and also by definition (since it is now a state of definite position with eigenvalue $mathbf{x}'+ text{d}mathbf{x}'$), when you act on this state with the position operator you get $$mathbf{hat{x}} |mathbf{x}' + text{d}mathbf{x}'rangle = (mathbf{x}' + text{d}mathbf{x}')|mathbf{x}' + text{d}mathbf{x}'rangle.$$
Does this help?
Correct answer by Philip on December 26, 2020
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