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**oldguynewstudent** In 1-D let T_{L} be an operator defined on the position eigenstates |x> such that T_{L}|x>=|x+L>.

Find the matrix elements T_{L}(x,x^{'})=<x^{'}|T_{L}|x> and construct an explicit expansion for this operator in the position representation. Show that “in the position representation” $\displaystyle <x|T_L|\psi >= \psi (x-L)$, i.e. T_{L }shifts the state of the system along the positive x axis. Is T_{L }diagonal in the position representation?

I have T_{L}(x,x^{'})=$\displaystyle \int dx^{'} \int dx |x+L> x \delta (x^{'} - x) <x| = \int dx |x+L> x <x|$

$\displaystyle <x|T_L|\psi >=\int d\psi \int dx |\psi+L> x \delta (\psi+L - x) <x|= \psi (x-L)$

Here is where I really get confused. Should the delta be reversed? Could someone please help me understand this better?

Thanks