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?

  • 5
    $\begingroup$ I suspect that Sakurai will answer at least the second half of this question if you keep reading $\endgroup$ Commented Aug 21, 2019 at 13:51
  • $\begingroup$ Does the answer do it? $\endgroup$ Commented Oct 22, 2019 at 15:56

1 Answer 1


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...)

Since translations in different directions commute, you may multiplex this to arbitrarily dimensioned space, as you do, a red herring here.


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