Expressing a state in Schrödinger picture and time evolution operator

Question

I have two question, regarding the derivation of two expression that I will post below:

If we have $\{ |\phi_{n,\tau}\rangle\}$ the set of eigenvectors of the Hamiltonian of a system. This set, is at the same time a basis.

In all the below expressions, $\tau$ represent additional degeneracies that might exist.

For a state, in an initial time $t_0$ we have:

$$|\Psi,t_0\rangle=\sum_{n,\tau} c_{n,\tau}|\phi_{n,\tau}\rangle$$

For a state, in an arbitrary time $t$:

$$|\Psi,t\rangle=\sum_{n,\tau} e^{-i\frac{E_n(t-t_0)}{\hbar}}|\phi_{n,\tau}\rangle \langle \phi_{n,\tau}|\Psi,t_0\rangle$$

Can the above expression be equalized to:

$$|\Psi,t\rangle= e^{-i\frac{H(t-t_0)}{\hbar}}|\Psi,t_0\rangle$$

If, yes, how? For example if we make the assumption that the initial state is an eigenstate of the Hamiltonian i.e:

$$|\Psi,t_0\rangle=\sum_{\tau} c_{n,\tau}|\phi_{n,\tau}\rangle$$

After some time $t$ passes, the state is:

$$|\Psi,t\rangle=\sum_{\tau} e^{-i\frac{E_n(t-t_0)}{\hbar}}|\phi_{n,\tau}\rangle \langle \phi_{n,\tau}|\Psi,t_0\rangle$$

$$|\Psi,t\rangle=e^{-i\frac{E_n(t-t_0)}{\hbar}} \sum_{\tau} |\phi_{n,\tau}\rangle \langle \phi_{n,\tau}|\Psi,t_0\rangle$$

$$|\Psi,t\rangle=e^{-i\frac{E_n(t-t_0)}{\hbar}}|\Psi,t_0\rangle$$

Here I can understand how we operate. Because the exponential is not dependent from $n$ we can move it out of the summation. But above, how is that possible?

In the last example, we are dealing with stationary states, and $U(t,t_0)=e^{-i\frac{E_n(t-t_0)}{\hbar}}$ is the time evolution operator. My question is:

Only stationary states can be expressed as the product between the time evolution operator and the initial "form" of the stationary state, or the state can be an arbitrary one, like in the first case. Understandably if, the answer to my first question, is yes, then obviously every state, whether it's a stationary state or not, can be expressed as we are expressing a stationary state, meaning: $$|\Psi,t\rangle=U(t,t_0)|\Psi,t_0\rangle$$

Additionally, for when the Hamiltonian is time independent, then are the two follow up expressions the same:

$$U(t,t_0)=e^{-i\frac{E_n(t-t_0)}{\hbar}}$$ and $$U(t,t_0)=e^{-i\frac{H(t-t_0)}{\hbar}}.$$

Answer 1

You should reason in the opposite direction to what you wrote about. If you know that $$H|\phi_{m}\rangle=E_m|\phi_m\rangle,$$ then you know that $$e^{-i H t/\hbar}|\phi_m\rangle=e^{-i E_m t/\hbar}|\phi_m\rangle.$$

Using these two equations you can show that if

$$|\Psi(0)\rangle=\sum_m c_m|\phi_n\rangle,$$ it follows that $$|\Psi(t)\rangle=e^{-iHt/\hbar}|\Psi(0)\rangle=\sum_m c_me^{-iHt/\hbar}|\phi_m\rangle=\sum_m c_me^{-iE_mt/\hbar}|\phi_m\rangle,$$

where $m=\{n,\tau\}$ , $c_m=\langle \phi_m|\Psi(0)\rangle$ and I have chosen $t_0=0$ but you can easily add it back.

Note that $e^{-iHt/\hbar}\neq e^{-i E_m t/\hbar}$, the latter cannot get out of the sum.