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

  1. #1
    Senior Member oldguynewstudent's Avatar
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    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” <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')= \int dx^{'} \int dx |x+L> x \delta (x^{'} - x) <x| = \int dx |x+L> x <x|

    <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
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  2. #2
    Senior Member oldguynewstudent's Avatar
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    Re: Find matrix elements and expansion for operator in position representation

    Quote Originally Posted by oldguynewstudent View Post
    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” <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')= \int dx^{'} \int dx |x+L> x \delta (x^{'} - x) <x| = \int dx |x+L> x <x|

    <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')= \int dx^{'} \int dx |x^{'}+L> x \delta (x^{'} - x) <x| = \int dx |x+L> x <x|

    <x|T_L|\psi >=\int d\psi^{'} \int dx |\psi^{'}+L> x \delta (\psi^{'}+L - x) <x|= \psi (x-L)
    Last edited by oldguynewstudent; Mar 27th 2015 at 10:18 AM. Reason: mistake
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