How Does a Spin-1/2 Particle Behave in a Magnetic Field with X and Z Components?

[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.