Consider a particle in a potential $V(x)$ represented by a definite energy wave function, sollution of the corresponding time-independent Shrodinger equation given by $$\Psi_n(x,t)=\psi_n(x)e^{-i\omega_nt}$$where $E_n=\hbar \omega_n$
Using the energy operator $$\hat{E}_n=i\hbar \frac{\partial}{\partial t}$$

Question

Consider a particle in a potential $V(x)$ represented by a definite energy wave function, sollution of the corresponding time-independent Shrodinger equation given by $$\Psi_n(x,t)=\psi_n(x)e^{-i\omega_nt}$$where $E_n=\hbar \omega_n$ 
Using the energy operator $$\hat{E}_n=i\hbar \frac{\partial}{\partial t}$$

richyw · Answer

a) obtain the expectation value of E, E;

richyw · Answer

This is an assignment question so please don't just give me the answer. But some help would be very much appreciated!

anonymous · Answer

hate quantum . . .

richyw · Answer

I am just getting started on it (obviously!)

fwizbang · Answer

Is the wavefunction in an eigenstate of energy?

richyw · Answer

to be honest. I don't even know what that means. I found out the answer once I knew that$$\int^\infty_{-\infty}\Psi^*(x)\Psi(x)=1$$

fwizbang · Answer

Certainly one way to get the answer is by direct calculation. the expectation value of any operator is

$$< O(t) > =\int\limits_{-\infty}^{\infty}\psi^*(x,t)O \psi(x,t) dx$$

where O is the operator. 

Another method(in this case at least) is to notice that the wvfct is an energy eiegenstate, which means that
$$\hat E \Psi = \hbar \omega_n  \Psi$$

which means that for this wvfct, the only possible result of a measurement of E
would be  $$ \hbar\omega_n $$.