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Physics
Quantum Physics
Ballentine on construction of the (Galilean) symmetry generators
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[QUOTE="EE18, post: 6862805, member: 701792"] Per your advice (from private messages) to me, I've begun a reread of Ballentine and am back here. I'm at the point where I need to use the Jacobi identity but I end up obtaining ##e^{i\textbf{v}\cdot \textbf{G}} V_\alpha e^{-i\textbf{v}\cdot \textbf{G}} = V_\alpha + [\textbf{v}\cdot \textbf{G},V_\alpha] = V_\alpha + [v_j G_j,i[H,Q_\alpha]] \implies v_\alpha I = iv_j([G_j,Q_\alpha],H] - \delta_{\alpha,j}I)## but it's not clear how to go further from here. It seems to me like to obtain (3.50) Ballentine uses perhaps a completely different argument that has nothing to do with (3.49). Perhaps Ballentine is saying that the action at ##t=0## on the position basis is of ##e^{i\textbf{v}\cdot \textbf{G}}|\textbf{x}\rangle = \textbf{x}|\textbf{x}\rangle## so that ##Q = Q'## under this transformation and thus we get that the two commute, i.e. 3.50? Edit: I think I am correct and that there is no connection between 3.49 and 3.50 since 3.49 is used later to prove 3.58. [/QUOTE]
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Ballentine on construction of the (Galilean) symmetry generators
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