
#1
Jan2512, 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 = hbar 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) )? 



#2
Jan2512, 05:08 PM

P: 2,059

Functions of operators are defined from the power series for those functions. So, because HE>=EE>, and H^2E>=H(EE>)=E(HE>)=E^2E>, etc., as long as these power series converge, then you will have f(H)E>=f(E)E>.




#3
Jan2612, 03:02 AM

P: 31

Thank you matterwave for the clear explanation




#4
Jan2612, 04:09 AM

P: 640

Functions with operator valued arguments acting on eigenstates 



#5
Jan2812, 09:00 AM

P: 31

Yes, makes sense, thanks for pointing it out torquil.



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