# Thread: Find matrix elements and expansion for operator in position representation

1. ## Find matrix elements and expansion for operator in position representation

In 1-D let TL be an operator defined on the position eigenstates |x> such that TL|x>=|x+L>.

Find the matrix elements TL(x,x')=<x'|TL|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. TL shifts the state of the system along the positive x axis. Is TL diagonal in the position representation?

I have TL(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

2. ## Re: Find matrix elements and expansion for operator in position representation

Originally Posted by oldguynewstudent
In 1-D let TL be an operator defined on the position eigenstates |x> such that TL|x>=|x+L>.

Find the matrix elements TL(x,x')=<x'|TL|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. TL shifts the state of the system along the positive x axis. Is TL diagonal in the position representation?

I have TL(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
Just noticed a mistake in what I wrote earlier the kets in the first term should have a prime in them

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

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