Functions with operator valued arguments acting on eigenstates


by qtm912
Tags: acting, arguments, eigenstates, functions, operator, valued
qtm912
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#1
Jan25-12, 05:02 PM
P: 31
This question concerns the outcome when operator valued functions act on an energy eigenstate. Given an eigenstate at t =0, say |Ej > , I have seen or inferred in some of the literature that the following applies :

exp(-iHt/h) |Ej > = exp(- iEj t/h) |Ej >

Where h = h-bar
Ej is energy eigenstate j
H is the Hamiltonian

I am unclear in particular why we can say that if we apply a function with an operator as the argument to this energy eigenstate, it will return the function with the eigenvalue as the argument times the energy eigenstate. (assuming I have inferred what is in the notes correctly)

Would this then be a general result for all functions or do only certain functions satisfy this relationship. (So for example would it be true for sin(H) instead of exp(-iHt/h) )?
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Matterwave
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#2
Jan25-12, 05:08 PM
P: 2,059
Functions of operators are defined from the power series for those functions. So, because H|E>=E|E>, and H^2|E>=H(E|E>)=E(H|E>)=E^2|E>, etc., as long as these power series converge, then you will have f(H)|E>=f(E)|E>.
qtm912
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#3
Jan26-12, 03:02 AM
P: 31
Thank you matterwave for the clear explanation

torquil
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#4
Jan26-12, 04:09 AM
P: 640

Functions with operator valued arguments acting on eigenstates


Quote Quote by Matterwave View Post
Functions of operators are defined from the power series for those functions. So, because H|E>=E|E>, and H^2|E>=H(E|E>)=E(H|E>)=E^2|E>, etc., as long as these power series converge, then you will have f(H)|E>=f(E)|E>.
Thus, if the operator has eigenvalues that exceed the convergence radius of the expansion, problems will arise.
qtm912
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#5
Jan28-12, 09:00 AM
P: 31
Yes, makes sense, thanks for pointing it out torquil.


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