# How Translation Operator is defined?

Here, In Shankar's (2nd edition, p-283) QM book, The translation operator is given by

$$T(\epsilon) = I - \frac{i\epsilon}{\hbar}G \tag{11.2.13}$$

Similar In Sakurai (Revised edition 1994 p-45), he wrote about equation as following,

“We now demonstrate that if we take the infinitesimal translation operator to be”

$$\mathscr{T( \vec{dx'})} = 1 - i \vec{K} d \vec{x'} \tag{1.6.20}$$

I do know that this translation operator does satisfy properties such as

1. $$\mathscr{T^\dagger ( \vec{dx'})}\mathscr{T( \vec{dx'})} = 1$$

2. $$\mathscr{T( \vec{dx''})}\mathscr{T( \vec{dx'})} = \mathscr{T( \vec{dx''} +\vec{dx'})}$$

Here my question is that, do we assume that $$T(\epsilon) = I - \frac{i\epsilon}{\hbar}G$$? Or there is some math behind, to take such equation. As In Sakurai he assumed the equation, and then he satisfied the properties required for an operator.

• what are $G$ and $K$? Are both of them the momentum operator? – user2723984 May 25 '20 at 7:06
• @user2723984 Yes, but they are defined in different ways, In Shankar, it's straightforward, G is momentum operator, But in Sakurai, K is defined by $p/\hbar$ , they both are same, just definition is different. – anbhadane May 25 '20 at 7:10

The translation operator in $$L^2(\mathbb{R}^d)$$, i.e. the continuous linear operator such that, for a.e. $$x\in\mathbb{R}^d$$: $$(T(x_0)\psi)(x)=\psi(x-x_0)$$ is given by $$T(x_0)= e^{-i x_0 \hat{p}}$$ where $$\hat{p}$$ is the momentum operator $$\hat{p}=-i\partial_x\;.$$ This is standard to prove: consider the dense subset of rapidly decreasing functions $$\mathscr{S}(\mathbb{R}^d)\subset L^2(\mathbb{R}^d)$$, then for all $$\varphi\in\mathscr{S}$$ by Taylor expansion, $$\varphi(x-a)=\varphi(x)- a\cdot \nabla_x \varphi(x) +a^2 \Delta_x \varphi(x) +\dotsc= (e^{-ia \hat{p}}\varphi)(x)\; ,$$ where the last identification is made using the definitions above and the series expansion of the exponential. Therefore, for rapidly decreasing functions, $$\varphi(x-x_0)=(T(x_0)\varphi)(x)\;.$$ To extend the proof to all square-integrable functions a density argument is used (that may be too advanced to discuss here in details).
If the translation $$x_0=\varepsilon$$ is very small, one could omit the higher order terms in the series expansion of the exponential, at least when acting on the dense subset of rapidly decreasing wavefunctions, thus leading to $$T(\varepsilon)= 1-i\varepsilon \hat{p}+ O(\varepsilon^2)\;.$$