Spin-1/2 in a magnetic field

[Orbor]

Problem
A spin-1/2 is placed in a magnetic field with both x and z-components so that its
Hamiltonian is $H=-b_x \sigma^x-b_z\sigma^z$, where $\sigma^x$ and $\sigma^z$ are the Pauli matrices. The real constants $b_x$ and $b_z$ have units of energy, and account for both the magnetic field components and coupling constants between the spin and the magnetic field.

Consider that the spin-component along the z-axis of the spin-1/2 is known to be $+\hbar /2$ at $t = 0$.
What is the probability that the spin component along the z-axis at time $t ≥ 0$ will be measured to be $-\hbar /2$?

Solution
Time-evolution of the initial state yields $\vert \psi(t)\rangle=e^{-i \hat H t/ \hbar}\vert \uparrow_z \rangle$, hence the probability of measuring $-\hbar /2$ is $\vert \langle \downarrow_z \vert \psi(t) \rangle \vert ^2=0$.

Is this correct or am I missing something important here?

 

[blue_leaf77]

Solution
Time-evolution of the initial state yields $\vert \psi(t)\rangle=e^{-i \hat H t/ \hbar}\vert \uparrow_z \rangle$, hence the probability of measuring $-\hbar /2$ is $\vert \langle \downarrow_z \vert \psi(t) \rangle \vert ^2=0$.

Is this correct or am I missing something important here?

$\vert \uparrow_z \rangle$ and $\vert \downarrow_z \rangle$ are not the eigenstates of the problem's Hamiltonian. You may probably want to know how $\vert \uparrow_z \rangle$ and $\vert \downarrow_z \rangle$ expand into the basis made of the eigenstates of the Hamiltonian.