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Why is (5.302) an approximation instead of an equality?
Let ##T## be the operator ##\frac{p_x^2}{2m}##.
By the law of indices, we should have ##e^{-\frac{i}{\hbar}(T+V)\Delta t}=e^{-\frac{i}{\hbar}T\Delta t} e^{-\frac{i}{\hbar}V\Delta t}## exactly. Is it because ##T## and ##V## do not commute? So the correct equation should be ##e^{-\frac{i}{\hbar}(T+V)\Delta t}=\frac{1}{2}[e^{-\frac{i}{\hbar}T\Delta t} e^{-\frac{i}{\hbar}V\Delta t}+e^{-\frac{i}{\hbar}V\Delta t} e^{-\frac{i}{\hbar}T\Delta t}]##?
But if this is so, shouldn't (5.302) be correct up to terms of order ##\Delta t## instead of order ##(\Delta t)^2## as claimed by the book?
Let ##T## be the operator ##\frac{p_x^2}{2m}##.
By the law of indices, we should have ##e^{-\frac{i}{\hbar}(T+V)\Delta t}=e^{-\frac{i}{\hbar}T\Delta t} e^{-\frac{i}{\hbar}V\Delta t}## exactly. Is it because ##T## and ##V## do not commute? So the correct equation should be ##e^{-\frac{i}{\hbar}(T+V)\Delta t}=\frac{1}{2}[e^{-\frac{i}{\hbar}T\Delta t} e^{-\frac{i}{\hbar}V\Delta t}+e^{-\frac{i}{\hbar}V\Delta t} e^{-\frac{i}{\hbar}T\Delta t}]##?
But if this is so, shouldn't (5.302) be correct up to terms of order ##\Delta t## instead of order ##(\Delta t)^2## as claimed by the book?
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