Evolution of quantum state that cannot exist with accompanying Hamiltonian

Question

So I am studying a certain Hamiltonian that has projection operators in its definition. To keep it simple, suppose our Hilbert space is a one particle system that can be spin up/spin down (excited, non excited state), with Hamiltonian H = ZP , with P projector on non excited state and Z the pauli spin matrix, which just measures the spin in the z-direction.

Now we know that the excited state is not allowed because $H |\uparrow > = 0$ because of the projection operator in the Hamiltonian. But if we do prepare our system in that state and look at the time evolution for a small time interval $\Delta t$:

$ \exp(-iH(\Delta t) | \uparrow > = \textbf{1} |\uparrow> - i\Delta t \cdot H |\uparrow> = |\uparrow>$

and so time evolution just leaves the excited state in the excited state, but I would suspect it should "annihilate" it? Why isn't this so?

Answer 1

If $H= \sigma_z P$, where $$ P= \left[\matrix{0&0\cr 0&1}\right] $$ is the projector on spin down, then $$ H= \left[\matrix{1&0\cr 0 &-1}\right]\left[\matrix{0&0\cr 0&1}\right]= \left[\matrix{0&0\cr 0 &-1}\right]. $$ so the two eigenstates are $(1,0)^T$ with energy zero and $(0,1)^T$ with energy $-1$ and
$$ \exp\{-iHt/\hbar\}= \left[\matrix{0&0\cr 0 &e^{it/\hbar}}\right]. $$

Under time evolution $(1,0)^T\to (1,0)^T$ and $(0,1)^T\to e^{it/\hbar} (0,1)^T$. What is confusing here?

By the way, (Grammar Police warning!) why start your question with the word "So"? Why not simply say "I am studying..." Starting with "So" seems so unnecessary...