Can the spin of a spin-$1/2$ particle flip in a time-varying magnetic field? Given that the state of the particle at time $t=-\infty$ is $\left|S_z^+\right>$, a magnetic field of the form $\mathbf{B}=B \tanh(t/\tau) \hat{\mathbf{z}}$, the Hamiltonian is $H=-\vec{\mu}\cdot\mathbf{B}$, where $\vec{\mu}=-\gamma \mathbf{S}$, how do you find the probability that the particle is in the state $\left|S_z^-\right>$ at time $t$?
I've approached the problem so far by setting up the Schödinger equation using the above definition of the Hamiltonian to find, by representing the state of the particle $\left|\Psi,t\right>$ as the column vector:
$$
\left|\Psi,t\right>=\begin{pmatrix}\psi_+\\\psi_-\\\end{pmatrix},
$$
and solving:
$$
i\hbar\frac{\partial}{\partial t}\left|\Psi,t\right>=\hat{H}\left|\Psi,t\right>.
$$
In doing so, with $H$ being, more precisely:
$$
\hat{H}=\gamma B\frac
{\hbar}{2}\tanh{\Big(\frac{t}{\tau}}\Big)\begin{pmatrix}1&0\\0 &-1\\\end{pmatrix},$$
I've found the general solution to be, taking into account the initial condition:
$$
\left|\Psi,t\right>=\begin{pmatrix}C_1 e^{\frac{-i\gamma B\tau}{2}}\cosh{\Big(\frac{t}{\tau}\Big)}\\0\\\end{pmatrix}.
$$
The thing is that I don't know if I've answered my own question correctly, as intuitively it seems to me that upon the flipping of the direction of the magnetic field at $t=0$, the state could or would change to $\left|S_z^-\right>$, meaning that the probability of finding the particle at time $t$ in state $\left|S_z^+\right>$ would not always be one.
Further, by operating the Hamiltonian on the original state, one finds that original state multiplied by a constant (if the Hamiltonian is evaluated at $t=-\infty$), meaning (I think) that it exists in a stationary state. 
I would appreciate either affirmation of my result, or a pointing out that it is incorrect, and suggestions as to how I should re-tackle the problem if indeed my current solution is incorrect.
 A: You can certainly flip a spin using a magnetic field.  It does not even have to be time dependent; a time-independent field will work, provided it is pointing in the right direction.  But the direction is key; you cannot change the $z$ projection of the spin with a magnetic field that points entirely in the $z$-direction.
A single spin-$\frac{1}{2}$ particle behaves essentially like a classical magnetic moment.  If it is exposed to a magnetic field in the $z$-direction, the spin will precess around the $z$-axis.  (Classically, this is a consequence of the torque being $\vec{N}=\vec{\mu}\times\vec{B}$.)  If the spin does not point precisely along the $\pm z$-axis, the $x$- and $y$-components of $\vec{S}$ will change; however, the $z$-component will remain constant. This is what you have found with your calculation.  If the magnetic field pointed had a component in the $xy$-plane, your initial $\vec{S}$ would precess around a different direction, and the $z$-projections of the spin would change in the way you were hoping to see.
