# Ignoring 2nd order terms in QM definitions of translation and momentum operators

In Sakurai's 'Modern Quantum Mechanics', he defines the infinitesimal translation operator as:

$$\mathcal{J}(d\mathbf{x}')=1-i\mathbf{K}\cdot d\mathbf{x}'$$,

and then he goes on to prove this satisfies some properties the translation operator should have, e.g.:

$$\mathcal{J}(d\mathbf{x}')\mathcal{J}(d\mathbf{x}'')=(1-i\mathbf{K}\cdot d\mathbf{x}')(1-i\mathbf{K}\cdot d\mathbf{x}'')\simeq 1-i\mathbf{K}\cdot(d\mathbf{x}'+d\mathbf{x}'')$$,

ignoring $$(d\mathbf{x}')^2$$ terms. What is the significance of this approximation? (Is it an approximation to ignore squared infinitesimals?) Isn't there another operator that satisfies these properties exactly?

I found somewhere else they defined the infinitesimal translation operator as

$$\mathcal{J}(\delta \mathbf{x})= 1-i\hbar\mathbf{p}\cdot\delta\mathbf{x}+...$$

I'm not sure what is after the ellipsis but I'm guessing $$\mathcal{J}(\delta \mathbf{x})= \sum_n(-i\hbar\mathbf{p}\cdot\delta\mathbf{x})^n$$? Would this satisfy the properties, and also for example the fact that $$p_x=-i\hbar\frac{\partial}{\partial x}$$ in position space?

• I suspect that Sakurai will answer at least the second half of this question if you keep reading – By Symmetry Aug 21 at 13:51
• Does the answer do it? – Cosmas Zachos Oct 22 at 15:56

This is physics speak for the elementary calculus of Lagrange's shift operator effecting Lie-group motion in one dimension, $${\cal J}(a) = e^{a\partial_x}.$$ you may readily verify that $${\cal J}(a) ~ f(x)= f(x+a) = \sum_{n=0}^\infty \frac{a^n}{n!} f^{(n)}(x) = e^{a\partial_x} f(x) ~,$$ the Taylor expansion around x.
It is evident the group is Abelian, and successive translations involve addition of shifts. The coefficient of a "small" Lie group motion motion around the identity (a=0) is the Lie algebra element $$-i{\mathbf K}$$; this is the significance of chucking away higher powers of a.
To apply this to QM, exponentiate your momentum ("Lie's 3rd theorem": generating the group), $${\cal J}(\delta x ) = e^{\delta x ~\partial_x} = e^{\delta x ~ i\hat p/\hbar},$$ so you know exactly what the missing terms are: they have the factorials of the order of the term in their denominator. (You have also upended the prefactors of $$\hat p$$ in your mis-scaled expression...)