Physics Asked on May 1, 2021
Often times one simply postulates that the momentum operator is the generator of translations or follows the reasoning here, for example. I am working through Sakurai and Napolitano’s Modern Quantum Mechanics 3rd Ed, and below is the part I want clarified (copied from 2nd Ed.) When he says, "We are led to speculate," I do not see why. The two expressions
$$ vec xcdotvec P,~~quadtext{and}qquad 1-ihat{vec K}cdot dvec x ~~,$$
seem quite different to me. I can look at these two expressions and, without the suggestion in the book, I am not drawn to speculate in this manner. What connection suggests this speculation?
I suspect what is confusing you is the apparent typo in (1.6.29) where the p is supposed to be P to yield the first of the equations (1.6.28): it is a type 2 generating function, after all, F(x,P), with ${mathbf X}=partial F/partial {mathbf P}$, in our case $= {mathbf x} +d {mathbf x} $. So, as he reminds you, F is a machine consisting of the identity plus a piece that ads a small increment $d {mathbf x} $ to ${mathbf x}$ and nobody else, as ${mathbf P}=partial F/partial {mathbf x}={mathbf p}$, so, basically, a gradient w.r.t. x.
This is meant to evoke the standard linear QM shift operator of Lagrange, a mere rewriting of Taylor's series around the constant dx, $$ e^{id{mathbf x}cdot {mathbf K}}f({mathbf x})= f({mathbf x}+d{mathbf x}) approx ( I +id{mathbf x}cdot {mathbf K} )f({mathbf x}) = f({mathbf x})+ d{mathbf x}cdot partial f/partial {mathbf x}, $$ the increment term also involving a gradient.
The evocation and the name are not that necessary, really; if you were a Martian, the commutation relations of K with x dictating representation thereof with a gradient $-inabla$ would tell you everything you need. But the two S's remind you that you should expect to observe this shadow-dance between QM and classical mechanics, and K should strongly evoke the classical phase-space variable p, so it's good mental hygiene to call it momentum.
The minus sign you have is because S defines his translation on kets, which transform with a minus sign w.r.t. bras, and functions are basically bras, $f(x) =langle x | frangle$. I could have done the whole thing with the minus sign and the kets, without reminding you that you are really discussing a bland Lagrange shift, in pompous language, after all.
Correct answer by Cosmas Zachos on May 1, 2021
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