# Expanding a translation operator

1. Nov 17, 2012

### Woodles

I'm trying to understand the construction of the T(ε) operator and why it is equal to I-iεG/hbar.

The textbook I'm using (Shankar) talks defines the translation operator with the phase factor:

$T(ε)\left|x\right\rangle=e^{i \epsilon g(x)/\hbar}\left|x+\epsilon\right\rangle$

and translationational invariance

$\langleψ| H|ψ\rangle=\langle ψ_\epsilon| H|ψ_\epsilon\rangle$

The book then says

"To derive the conservation law that goes with the above equation, we must first construct the operator T(e) explicitly. Since ε=0 correspons to no translation, we may expand T(ε) to order (ε) as
$I-\frac{i ε}{\hbar} G$

Why is this so? How can you find an equation for only T without it acting on anything?

2. Nov 17, 2012

### The_Duck

It's less an equation for T than a definition of G. There's some operator T(a), which depends on a, such that multiplying any state by T(a) translates that state by a distance a. Shankar assumes that we can expand the function T(a) as a power series in a. The zeroth order term must be the identity operator, because T(0) is the identity operator. The first order term Shankar calls (-i/hbar * G), which is just a definition of the operator G

Last edited: Nov 17, 2012
3. Nov 18, 2012

### dextercioby

T(ε) is an uniparametric group of unitary operators on a Hilbert space. Its generator G is a self-adjoint operator acting on the same space (dense everywhere subset of it). Mathematically, this is covered through Stone's theorem and its reverse.

The + or - sign when linearizing the exponential is a convention for the so-called active vs passive view of symmetries. He chose - which IIRC stands for the passive view of looking at space translations.