Probability of measurement of a time evolved two-particle spinor in QM Let's say we have a ket describing the spin of two particles. Let's say also they are in a specific state $\left|\downarrow\downarrow\right\rangle=\left|\downarrow\right\rangle_{(1)}  \otimes\left|\downarrow\right\rangle_{(2)}$, and let this be a state where they both have spin down, so produce $-\hbar/2$ when measuring on them with $S_i^z$. And let's say when using the Hamiltonian on this state it produces an eigenvalue $\lambda$ and that the hamiltonian is time-independent. Then I can write
$$\left|\chi(t)\right\rangle=e^{-i(\lambda/\hbar)t}\left|\downarrow\downarrow\right\rangle. $$
How do I now find the probability of measuring for example $+\hbar/2$ with $S_2^z$ on this time-evolved state?. All attempts I just get zero.
 A: Lets take some general eigenstate of some time independent Hamiltonian $|E\rangle$, with eigenvalue $E$, and some observable $A$. Now after some time $t$ $|E\rangle$ will evolve to a state $e^{-i H t/\hbar}|E\rangle = e^{-iEt/\hbar}|E\rangle$ so the expectation of $A$ on the time evolved $|E\rangle$ will be
\begin{align}
\langle E |e^{i H t/\hbar} A e^{-i H t/\hbar}|E\rangle &= \langle E |e^{i E t/\hbar} A e^{-i E t/\hbar}|E\rangle\\
&=\langle E | A |E\rangle\;.
\end{align}
So for any eigenstate of the Hamiltonian the expectation of any observable will not change. For this reason eigenstates of the Hamiltonian are called stationary states; their time evolution is trivial and does not change any observable. In your case the initial chance of observing the system in the $+\hbar/2$ state was zeros and as it is a stationary state that does not change.
It is important to note, however, that the time evolution of eigenstates of the Hamiltonian does have important effects. In particular it effects how eigenstates with different eigenvalues interfere with each other. Indeed as eigenstates of the Hamiltonian form a complete basis the time evolution of any state can be described in terms of the time evolution of eigenstates of the Hamiltonian.
