Find the operator \hat{x}^{'} obtained by applying a translation through a distance a to the operator \hat{x} . That is, what is the action of \hat{x}^{'} , as defined by Equation 6.6 [\hat{Q}'=\hat{T}^{\dagger } \hat{Q}\hat{T}], on an arbitrary f(x) ?
Using the definition of \hat{x}^{'} (Equation 6.6) and a test function f(x) we have
\hat{x}^{′} f(x)=\hat{T} ^{\dagger }(a)\hat{x} \hat{T}(a) f(x)
and since \hat{T} ^{\dagger }(a)=\hat{T} (-a) (Equation 6.4 [\hat{T} (a)^{-1}=\hat{T} (-a)=\hat{T} (a)^{\dagger }]),
\hat{x}^{′} f(x)=\hat{T} (-a)\hat{x} \hat{T}(a)f(x)
From Equation 6.1 [\hat{T} (a)\psi (x)=\psi ^{′}(x)=\psi (x-a)]
\hat{x}^{′} f(x)=\hat{T} (-a)x f(x-a)
and from Equation 6.1 again, \hat{T} (-a)\left[xf(x-a)\right] =(x+a)f(x) , so
\hat{x}^{′} f(x)=(x+a) f(x)
Finally we may read off the operator
\hat{x}^{′} =\hat{x} +a (6.7)
As expected, Equation 6.7 corresponds to shifting the origin of our coordinates to the left by a so that positions in these transformed coordinates are greater by a than in the untransformed coordinates.