# Functions with operator valued arguments acting on eigenstates

1. Jan 25, 2012

### qtm912

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) )?

2. Jan 25, 2012

### Matterwave

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>.

3. Jan 26, 2012

### qtm912

Thank you matterwave for the clear explanation

4. Jan 26, 2012

### torquil

Thus, if the operator has eigenvalues that exceed the convergence radius of the expansion, problems will arise.

5. Jan 28, 2012

### qtm912

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