I Exact dynamics of spin in varying magnetic field

[KDPhysics]

TL;DR

Can one use the Schrodinger picture propagator for a sudden, constant perturbation?

Consider an uncharged particle with spin one-half moving with speed $v$ in a region with magnetic field $\textbf{B}=B\textbf{e}_z$. In a certain length $L$ of the particle's path, there is an additional, weak magnetic field $\textbf{B}_\perp=B_\perp \textbf{e}_x$. Assuming the electron has magnetic moment $\mu$ then
$$
H(t) = H_0 + V(t)
$$
where $H_0=-\mu B \sigma_z$ and
$$
V(t)=\begin{cases}
-\mu B_\perp \sigma_x, \ \text{ for } 0<t<l/v\\
0, \ \text { otherwise}
\end{cases}
$$
Assuming the particle starts out in the $|+\rangle$ state (spin-up along the $z$-axis) then I found using perturbation theory that the probability that the spin flips to $|-\rangle$ after time $t>L/v$ is
$$
P(t>L/v) = \bigg[\frac{B_\perp}{B}\sin\bigg(\frac{\mu B L}{\hbar v}\bigg)\bigg]^2
$$
I am wondering how I could derive the result without assuming that $B_\perp\ll B$?

My first instinct was to use the propagator to evolve the state from $t=0$ to $t=L/v$:
\begin{align}
&e^{-iH(t)t/\hbar} = e^{i\mu\textbf{B}\cdot \boldsymbol{\sigma}t/\hbar} = \cos \bigg(\frac{\mu B't}{\hbar}\bigg)\mathbb{1}+i\sin \bigg(\frac{\mu B't}{\hbar}\bigg)\frac{B\sigma_z+B_\perp \sigma_z}{B'}\\
\implies & |+(t)\rangle = \cos \bigg(\frac{\mu B't}{\hbar}\bigg)|+\rangle+i\sin \bigg(\frac{\mu B't}{\hbar}\bigg)\frac{B|+\rangle+B_\perp |-\rangle}{B'}\\
\implies & |\langle-|+(t)\rangle|^2 = \bigg[\frac{B_\perp}{B'}\sin \bigg(\frac{\mu B't}{\hbar}\bigg)\bigg]^2
\end{align}
where $B'=\sqrt{B^2+B_\perp^2}$. Taking the $B_\perp/B<<1$ limit (perturbative limit) then I recover
$$
P(t>L/v) = \bigg[\frac{B_\perp}{B}\sin\bigg(\frac{\mu B L}{\hbar v}\bigg)\bigg]^2
$$
as desired.

However, I'm not entirely sure if my approach of using the Schrödinger picture propagator $U(t)=e^{-iHt/\hbar}$ is correct. Indeed since $H(t<0)$ does not commute with $H(t>0)$, there is no guarantee that the $|+\rangle$ state at time $t=0^-$ will not immediately jump and transition to some other state at $t=0^+$, implying that assuming the state will be $|+\rangle$ at $t=0^+$ could be wrong. Why did my argument still yield the correct result?

 

[KDPhysics]

It seems like I overlooked the simple fact that the state cannot change during a sudden, $\textit{finite}$ perturbation, so I was right in assuming that the spin would be $|+\rangle$ at $t=0^+$.

To understand why the system's state must be continuous over the sudden perturbation in the Hamiltonian, we can write
\begin{cases}
i\hbar \frac{d}{dt}|\psi(t)\rangle = H_0 |\psi(t)\rangle, \ &t<0\\
i\hbar \frac{d}{dt}|\psi(t)\rangle = (H_0+V) |\psi(t)\rangle, \ &t>0
\end{cases}
Integrating from $0^-=-\tau$ to $0^+=\tau$ where $\tau \rightarrow 0$ then
$$
i\hbar(|\psi(\tau)\rangle-|\psi(-\tau)\rangle) = \int_{-\tau}^0 H_0 |\psi(t)\rangle \ dt + \int_0^\tau (H_0+V)|\psi(t)\rangle \ dt
$$
Since $|\psi(t)\rangle$ has to be continuous over $(-\tau,0)$ and $(0,\tau)$, we can apply the Mean Value Theorem, which states that there must be some $\tau_-\in(-\tau,0)$ and some $\tau_+\in(0,\tau)$ such that
$$
\int_{-\tau}^0 H_0 |\psi(t)\rangle \ dt + \int_0^\tau (H_0+V)|\psi(t)\rangle \ dt = \tau (H_0|\psi(\tau_-)\rangle + (H_0+V)|\psi(\tau_+)\rangle)
$$
Taking the limit as $\tau \rightarrow 0$ we find that
$$
\lim_{\tau \rightarrow 0} (|\psi(\tau)\rangle-|\psi(-\tau)\rangle) = \frac{i}{\hbar}\lim_{\tau \rightarrow 0}\tau[H_0|\psi(\tau_-)\rangle + (H_0+V)|\psi(\tau_+)\rangle] = 0
$$
implying that $|\psi(0^+)\rangle = |\psi(0^-)\rangle$ (assuming $V_0$ is finite).