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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.

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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.

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  • $\begingroup$ 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 $\mathbf{B}$. $\endgroup$ Commented Nov 1, 2018 at 6:03
  • $\begingroup$ @T.Zaborniak Yes, it's the fact that the field and the initial spin direction are aligned that leads to the spin not changing. (There is something a odd about the expression for the time-dependent wave function in the question though; the absolute value of the upper component of the spinor should be 1 for all time.) $\endgroup$ Commented Nov 1, 2018 at 6:08
  • $\begingroup$ @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. $\endgroup$ Commented Nov 1, 2018 at 6:12
  • $\begingroup$ 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... $\endgroup$ Commented Nov 1, 2018 at 6:31

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