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.

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.
• Hmmm. So, based on your comment, it seems to me then that my calculation in the way that I carried out is correct, due to the initial state of the particle being $\left|S_z^+\right>$ and the magnetic field being aligned with it. Classically, however, would not the direction of the moment change direction? The potential energy of the particle classically is: $U=-\vec{\mu}\cdot\mathbf{B}$, and to minimize this one would have $\mu$ align with $\mathbf{B}$. Commented Nov 1, 2018 at 6:03
• @T.Zaborniak The classical torque is what I wrote in my answer, and if you take the dot product of $\vec{N}=d\vec{S}/dt$ with the direction $\hat{z}$ of $\vec{B}$, you can see that $S_{z}$ is independent of time.
• I still don't get your classical explanation exactly. I'm thinking that in writing $S_z$ you mean $L_z$, where $L_z$ is the classical angular momentum in the $z$-direction, and by $\vec{S}$ you mean the classical angular momentum. That being the case, isn't it necessary that $L_z$ change when $t$ goes from $t<0$ to $t>0$ to have $|d\vec{L}/dt|$ remain a constant? The magnitude of $\vec{L}$ obviously would have to stay the same, but because it is $L_z\hat{\mathbf{z}}$, its direction would need to change... Commented Nov 1, 2018 at 6:31