- #1
jouvelot
- 53
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Hi everyone,
Weinberg uses spatial translation invariance to derive the momentum operator. But the way he does it puzzles me. Here is an excerpt of the book.
Equation 3.5.1 is the definition of the unitary operator ##U(x)## for translation invariance:
$$U^{-1}(x)XU(x) = X+x,$$ with ##-P/\hbar## as translation generator, while Equation 3.5.6 defines the commutator:
$$[X_i,P_j] = i\hbar\delta_{ij}.$$ What I don't get is how one can "infer" from this equation the momentum operator in Equation 3.5.11, in particular the usual partial derivative.
Note that there is Equation 3.5.8, which states that ##U(x) = exp(-iP.x/\hbar)##, and from which I could fathom Equation 3.5.11, however. Is (3.5.6) just a bogus reference (even though it appears in both the 2013 and 2015 editions of the book)?
Thanks in advance to anyone who can help.
Pierre
Weinberg uses spatial translation invariance to derive the momentum operator. But the way he does it puzzles me. Here is an excerpt of the book.
Equation 3.5.1 is the definition of the unitary operator ##U(x)## for translation invariance:
$$U^{-1}(x)XU(x) = X+x,$$ with ##-P/\hbar## as translation generator, while Equation 3.5.6 defines the commutator:
$$[X_i,P_j] = i\hbar\delta_{ij}.$$ What I don't get is how one can "infer" from this equation the momentum operator in Equation 3.5.11, in particular the usual partial derivative.
Note that there is Equation 3.5.8, which states that ##U(x) = exp(-iP.x/\hbar)##, and from which I could fathom Equation 3.5.11, however. Is (3.5.6) just a bogus reference (even though it appears in both the 2013 and 2015 editions of the book)?
Thanks in advance to anyone who can help.
Pierre