Is there a solution using $\Psi(x,t)$ such that
$$\frac{\partial^2\Psi}{\partial x^2} + \frac{x^2}{1+t^2}\Psi = \frac{\partial \Psi}{\partial t}$$
?
This is a follow-up question to this question which solves $\frac{\partial^2\Psi}{\partial x^2} + \frac{1}{1+t^2}\Psi = \frac{\partial \Psi}{\partial t}$ using separation of variables method $\Psi = X(x)T(t)$.
Unfortunately, the separation of variables method does not seem to work for solving this modified form of $\frac{x^2}{1+t^2}\Psi$?
Suppose that $\Psi$ is of the form:
$$\Psi = a(t)\Psi_1 + b(t)\Psi_2$$
where $\Psi_1, \Psi_2$ have the property that
$$\boxed{ \frac{d^2\Psi_1}{dx^2} = E_1\Psi_1\\ \frac{d^2\Psi_2}{dx^2} = E_2\Psi_2}$$
$$\boxed{ \Psi_1 = \Psi_3e^{-iE_1t}\\ \Psi_2 = \Psi_4e^{-iE_2t}}$$
$$\boxed{\int_{-\infty}^\infty \Psi_3 \Psi_4 dx = 0\\\int_{-\infty}^\infty \Psi_3 \Psi_3 dx = 1\\\int_{-\infty}^\infty \Psi_4 \Psi_4 dx = 1}$$
Then
$$\frac{d^2\Psi}{dx^2} + \frac{x^2}{1+t^2}\Psi = i\frac{d\Psi}{dt}$$
$$a\frac{d^2\Psi_1}{dx^2} + b\frac{d^2\Psi_2}{dx^2}+ a\frac{x^2}{1+t^2}\Psi_1 + b\frac{x^2}{1+t^2}\Psi_2 = i\Psi_3\frac{d}{dt}(a(t)e^{-iE_1t}) + i\Psi_4\frac{d}{dt}(b(t)e^{-iE_2t})$$
$$\color{red}{aE_1\Psi_1} + \color{red}{b E_2 \Psi_2} + a\frac{x^2}{1+t^2}\Psi_1 + b\frac{x^2}{1+t^2}\Psi_2 = i\Psi_1(\frac{da}{dt} - \color{red}{iE_1a}) + i\Psi_2(\frac{db}{dt} - \color{red}{iE_2a})$$
$$a\frac{x^2}{1+t^2}\Psi_1 + b\frac{x^2}{1+t^2}\Psi_2 = i\Psi_1\frac{da}{dt} + i\Psi_2\frac{db}{dt}$$
Multiplying all sides by $\Psi_3$ and integrating over $x$ to infinity:
$$ e^{-iE_1t}\int \Psi_3 a\frac{x^2}{1+t^2}\Psi_3 + e^{-iE_2t} \int\Psi_3 b\frac{x^2}{1+t^2}\Psi_4 = i \frac{da}{dt} e^{-iE_1t} \color{orange}{(\int \Psi_3\Psi_3)} + i\frac{db}{dt}e^{-iE_2t} \color{orange}{(\int \Psi_3\Psi_4)}$$
Hence
$$ e^{-iE_1t}\int \Psi_3 a\frac{x^2}{1+t^2}\Psi_3 + e^{-iE_2t} \int\Psi_3 b\frac{x^2}{1+t^2}\Psi_4 = i \frac{da}{dt} e^{-iE_1t} $$
so
$$ \frac{a(t)}{1+t^2}\int x^2 \Psi_3 \Psi_3 + e^{-i(E_2-E_1)t}\frac{b(t)}{1+t^2} \int x^2 \Psi_3 \Psi_4 = i \frac{da}{dt} $$
Similarly,
$$ e^{i(E_2-E_1)t}\frac{a(t)}{1+t^2} \int x^2 \Psi_4 \Psi_3 + \frac{b(t)}{1+t^2} \int x^2\Psi_4 \Psi_4 = i \frac{db}{dt} $$
Assuming further that
$$ \boxed{ \int_{-\infty}^\infty x^2 \Psi_3 \Psi_3 dx = 0 \\\int_{-\infty}^\infty x^2 \Psi_4 \Psi_4 dx = 0\\\int_{-\infty}^\infty x^2 \Psi_3 \Psi_4 dx = A\\\int_{-\infty}^\infty x^2 \Psi_4 \Psi_3 dx = B}$$
$$\boxed{a_{\{t=0\}} = 1\\b_{\{t=0\}} = 0}$$
we obtain
$$ \frac{da}{dt} = -iA b(t)e^{-i(E_2-E_1)t}\left(\frac{1}{1+t^2}\right) $$
$$ \frac{db}{dt} = -iB a(t)e^{i(E_2-E_1)t}\left(\frac{1}{1+t^2}\right) $$
This coupled first order differential equations behave in quite stable oscillations over time, eg. $|b(t)|^2$:
Replacing the one-off disturbance to periodic $V(t)=\cos(kt)$ (analytically derived here), $|b(t)|^2$:
Reversing initial condition such that $a_{\{t=0\}}=0, b_{\{t=0\}} = 1, |b(t)|^2$:
Therefore given a time varying contraint, the probability of a transition from $\Psi_1$ to $\Psi_2$ necessarily oscillates higher and lower over time (even for $t<0$) called Rabi cycle. Measurement at specified times create true random binary number of specified probability.
Eg.
Suppose for the harmonic oscillator
$$\Psi_0 = \left(\frac{1}{\pi}\right)^{\frac{1}{4}}e^{-\frac{x^2}{2}}e^{iE_0t}$$ $$\Psi_2 = \left(\frac{1}{\pi}\right)^{\frac{1}{4}}\left(\frac{1}{2}\right)^{\frac{1}{2}}(2x^2-1)e^{-\frac{x^2}{2}}e^{iE_2t}$$ $$H = \frac{x^2}{1+t^2}$$ $$E_0 = \frac{1}{2}\hbar \omega$$ $$E_2 = \frac{5}{2}\hbar \omega$$
Then
$$\int_{x=-\infty}^\infty \int_{t=-\infty}^\infty \bar \Psi_0 H \Psi_2 dtdx$$
$$=\left(\frac{1}{\sqrt{2\pi}}\right)\int_{t=-\infty}^\infty \frac{e^{i(-E_0+E_2)t}}{1+t^2} \left(\int_{x=-\infty}^\infty (2x^4 e^{-x^2} - x^2e^{-x^2})dx \right)dt$$
$$=\left(\frac{1}{\sqrt{2\pi}}\right)\int_{t=-\infty}^\infty \frac{e^{i(-E_0+E_2) t}}{1+t^2} \left[\frac{3}{2}\sqrt\pi - \frac{1}{2}\sqrt\pi\right] dt$$
$$=\left(\frac{1}{\sqrt2}\right)\int_{t=-\infty}^\infty \frac{e^{i2\hbar \omega t}}{1+t^2}dt$$
$$=\left(\frac{1}{\sqrt2}\right)\pi e^{-2\hbar \omega}$$
since
$$\boxed {\int_{-\infty}^\infty x^2e^{-x^2} = \frac{1}{2}\sqrt \pi}$$
$$\boxed {\int_{-\infty}^\infty x^4e^{-x^2} = \frac{3}{4}\sqrt \pi}$$
$$\boxed {\int_{-\infty}^\infty \frac{e^{ikt}}{1+t^2} = \pi e^{-|k|}}$$
by Gaussian integration here, here, and by contour integration here.
Extending to the 1+1D Dirac constraint equation:
$$ \boxed{i\begin{bmatrix} 1&0&0&0 \\ 0&1&0&0 \\ 0&0&-1&0 \\ 0&0&0&-1\end{bmatrix} \frac{d\Psi}{dt} + i\begin{bmatrix} 0&0&0&1 \\ 0&0&1&0 \\ 0&-1&0&0 \\ -1&0&0&0\end{bmatrix} \frac{d\Psi}{dx} \color{red}{-A_0\begin{bmatrix} 1&0&0&0 \\ 0&1&0&0 \\ 0&0&-1&0 \\ 0&0&0&-1\end{bmatrix}\Psi} = m\Psi}$$
Suppose that $\Psi$ is of the form:
$$\Psi = a(t)\Psi_1 + b(t)\Psi_2$$
where $\Psi_1, \Psi_2$ have the property that
$$\boxed{ i\gamma_x\frac{d\Psi_1}{dx} = m_1\Psi_1 - i\gamma_t\frac{d\Psi_1}{dt}\\ i\gamma_x\frac{d\Psi_2}{dx} = m_2\Psi_2 - i\gamma_t\frac{d\Psi_2}{dt}}$$
$$\boxed{ \Psi_1 = \Psi_3e^{-iE_1t}\\ \Psi_2 = \Psi_4e^{-iE_2t}}$$
$$\boxed{\int_{-\infty}^\infty \bar \Psi_3 \gamma_t \Psi_4 dx = 0\\\int_{-\infty}^\infty \bar \Psi_4 \gamma_t \Psi_3 dx = 0\\\int_{-\infty}^\infty \bar \Psi_3 \gamma_t \Psi_3 dx = 1\\\int_{-\infty}^\infty \bar \Psi_4 \gamma_t \Psi_4 dx = 1}$$
$$\boxed{ A_0 = A(t)A_x }$$
Then
$$i\gamma_t \frac{d\Psi}{dt} + i\gamma_x \frac{d\Psi}{dx} \color{red}{- A_0\gamma_t\Psi} = m\Psi$$
$$i\gamma_t \frac{d}{dt}(a(t)\Psi_1 + b(t)\Psi_2) + i\gamma_x \frac{d}{dx}(a(t)\Psi_1 + b(t)\Psi_2) - A_0\gamma_t(a(t)\Psi_1 - b(t)\Psi_2)= m_1a(t)\Psi_1 + m_2b(t)\Psi_2$$
$$i\gamma_t \left(\Psi_3\frac{d}{dt}(a(t)e^{-iE_1t}) + \Psi_4\frac{d}{dt}(b(t)e^{-iE_2t}) \right) + i\gamma_x \left(a(t)\frac{d}{dx}(\Psi_1) + b(t)\frac{d}{dx}(\Psi_2)\right) - A_0a(t)\gamma_t\Psi_1 - A_0b(t)\gamma_t\Psi_2 = m_1a(t)\Psi_1 + m_2b(t)\Psi_2$$ and so $$i\gamma_t \left(\frac{da}{dt}\Psi_1 - iE_1a(t)\Psi_1 + \frac{db}{dt}\Psi_2 -iE_2b\Psi_2 \right) + i\gamma_x \left( a(t)m_1\Psi_1 - ia(t)\gamma_t\frac{d\Psi_1}{dt} + b(t)m_2\Psi_2 - ib(t)\gamma_t\frac{d\Psi_2}{dt}\right) - A_0a(t)\gamma_t\Psi_1 - A_0b(t)\gamma_t\Psi_2 = m_1a(t)\Psi_1+ m_2b(t)\Psi_2$$
Multiplying by $\bar \Psi_3$ and integrating $\int_{-\infty}^\infty dx$:
$$ i\frac{da}{dt}e^{-iE_1t}\color{orange}{\int\bar \Psi_3\gamma_t \Psi_3} + E_1a(t)e^{-iE_1t}\color{orange}{\int\bar \Psi_3\gamma_t \Psi_3} + i\frac{db}{dt}e^{-iE_2t}\color{orange}{\int\bar \Psi_3\gamma_t\Psi_4} +E_2b(t)e^{-iE_2t}\color{orange}{\int\bar \Psi_3\gamma_t \Psi_4} + im_1a(t)\int \bar \Psi_3\gamma_x\Psi_1 - iE_1a(t)\int \bar \Psi_3\gamma_x\gamma_t\Psi_1 + im_2b(t)\int \bar \Psi_3\gamma_x\Psi_2 - iE_2b(t)\int \bar \Psi_3\gamma_x\gamma_t\Psi_2 - A(t)a(t) e^{-iE_1t}\int A_x\bar \Psi_3 \gamma_t\Psi_3 - A(t)b(t)e^{-iE_2t}\int A_x\bar \Psi_3 \gamma_t\Psi_4 = m_1a(t) \int \bar \Psi_3\Psi_1 + m_2b(t)\int \bar \Psi_3\Psi_2$$ and so $$ i\frac{da}{dt} + E_1a(t) + im_1a(t)\int \bar \Psi_3\gamma_x\Psi_3 - iE_1a(t)\int \bar \Psi_3\gamma_x\gamma_t\Psi_3 + im_2b(t)e^{-i(E_2-E_1)t}\int \bar \Psi_3\gamma_x\Psi_4 - iE_2b(t)e^{-i(E_2-E_1)t}\int \bar \Psi_3\gamma_x\gamma_t\Psi_4 - A(t)a(t) \int A_x\bar \Psi_3 \gamma_t\Psi_3 - A(t)b(t)e^{-i(E_2-E_1)t}\int A_x\bar \Psi_3 \gamma_t\Psi_4 = m_1a(t) \int \bar \Psi_3\Psi_3+ m_2b(t)e^{-i(E_2-E_1)t}\int \bar \Psi_3\Psi_4$$
Therefore,
$$\boxed{ \frac{da}{dt} = a(t)\left( E_1\color{blue}{\int \bar \Psi_3\gamma_x\gamma_t\Psi_3} - m_1\color{blue}{\int \bar \Psi_3\gamma_x\Psi_3} +iE_1- im_1 \color{blue}{\int \bar \Psi_3\Psi_3} \color{red}{- i A(t)} \color{blue}{\int A_x\bar \Psi_3 \gamma_t\Psi_3} \right) + b(t)e^{-i(E_2-E_1)t}\left( E_2\color{blue}{\int \bar \Psi_3\gamma_x\gamma_t\Psi_4} - m_2\color{blue}{\int \bar \Psi_3\gamma_x\Psi_4} -i m_2\color{blue}{\int \bar \Psi_3\Psi_4} \color{red}{- iA(t)}\color{blue}{\int A_x\bar \Psi_3 \gamma_t\Psi_4} \right) }$$
or
$$ \boxed {\frac{da}{dt} = k_1(t)a(t) + k_2(t) e^{-i(E_2-E_1)t}b(t)}$$
and similarly,
$$\boxed{ \frac{db}{dt} = a(t)e^{i(E_2-E_1)t}\left(E_1\color{blue}{\int \bar \Psi_4\gamma_x\gamma_t\Psi_3} - m_1\color{blue}{\int \bar \Psi_4\gamma_x\Psi_3} - im_1 \color{blue}{\int \bar \Psi_4\Psi_3} \color{red}{- i A(t)} \color{blue}{\int A_x\bar \Psi_4 \gamma_t\Psi_3} \right) + b(t) \left( E_2\color{blue}{\int \bar \Psi_4\gamma_x\gamma_t\Psi_4} - m_2\color{blue}{\int \bar \Psi_4\gamma_x\Psi_4} +iE_2 - i m_2\color{blue}{\int \bar \Psi_4\Psi_4} \color{red}{- iA(t)}\color{blue}{\int A_x\bar \Psi_4 \gamma_t\Psi_4} \right) }$$
or
$$ \boxed {\frac{db}{dt} = k_3(t)e^{i(E_2-E_1)t}a(t) + k_4(t) b(t)}$$
Assuming that most of the integrals evaluate to $0$, and $A(t)A(x) = \left(\frac{1}{1+t^2}\right)(x^2)$, the equations may reduce to
$$ \frac{da}{dt} = -iC e^{-i(E_2-E_1)t}b(t)\left( \frac{1}{1+t^2}\right)$$
$$ \frac{db}{dt} = -iD e^{i(E_2-E_1)t}a(t)\left( \frac{1}{1+t^2}\right)$$
which gives an oscillating $a(t), b(t)$ solution as in the Schrodinger constraint equation.
Extending to the 3 dimensional spherical coordinates:
$$ \boxed{i\gamma_t \frac{d\Psi}{dt} + i\gamma_r \frac{d\Psi}{dr} + i\gamma_\theta \frac{d\Psi}{d\theta} + i\gamma_\phi \frac{d\Psi}{d\phi} \color{red}{-A_0\gamma_t\Psi} = m\Psi}$$
Suppose that $\Psi$ is of the form:
$$\Psi = a(t)\Psi_1 + b(t)\Psi_2$$
where $\Psi_1, \Psi_2$ have the property that
$$\boxed{ \Psi_1 = \Psi_3e^{-iE_1t}\\ \Psi_2 = \Psi_4e^{-iE_2t}}$$
$$\boxed{\int_{r=0}^\infty \int_{\theta=0}^{\pi} \int_{\phi=0}^{2\pi} \bar \Psi_3 \gamma_t \Psi_4 dx = 0\\\int_{r=0}^\infty \int_{\theta=0}^{\pi} \int_{\phi=0}^{2\pi} \bar \Psi_4 \gamma_t \Psi_3 dx = 0\\\int_{r=0}^\infty \int_{\theta=0}^{\pi} \int_{\phi=0}^{2\pi} \bar \Psi_3 \gamma_t \Psi_3 dx = 1\\\int_{r=0}^\infty \int_{\theta=0}^{\pi} \int_{\phi=0}^{2\pi} \bar \Psi_4 \gamma_t \Psi_4 dx = 1}$$
$$\boxed{ A_0 = A(t)A_r }$$
Then
$$i\gamma_t \frac{d\Psi}{dt} + i\gamma_r \frac{d\Psi}{dr} + i\gamma_\theta \frac{d\Psi}{d\theta} + i\gamma_\phi \frac{d\Psi}{d\phi} \color{red}{- A_0\gamma_t\Psi} = m\Psi$$
$$i\gamma_t \frac{d}{dt}(a(t)\Psi_1 + b(t)\Psi_2) + i\gamma_r \frac{d}{dr}(a(t)\Psi_1 + b(t)\Psi_2) + i\gamma_\theta \frac{d}{d\theta}(a(t)\Psi_1 + b(t)\Psi_2) + i\gamma_\phi \frac{d}{d\phi}(a(t)\Psi_1 + b(t)\Psi_2) \color{red}{- A_0\gamma_t\Psi} = (m_1a(t)\Psi_1 + m_2b(t)\Psi_2)$$
$$i\gamma_t \left(\Psi_3\frac{d}{dt}(a(t)e^{-iE_1t}) + \Psi_4\frac{d}{dt}(b(t)e^{-iE_2t}) \right) + i\gamma_r \left(a(t)\frac{d}{dr}(\Psi_1) + b(t)\frac{d}{dr}(\Psi_2)\right) + i\gamma_\theta \left(a(t)\frac{d}{d\theta}(\Psi_1) + b(t)\frac{d}{d\theta}(\Psi_2)\right)+ i\gamma_\phi \left(a(t)\frac{d}{d\phi}(\Psi_1) + b(t)\frac{d}{d\phi}(\Psi_2)\right) - A_0a(t)\gamma_t\Psi_1 - A_0b(t)\gamma_t\Psi_2 = m_1a(t)\Psi_1 + m_2b(t)\Psi_2$$
$$i\gamma_t \left(\frac{da}{dt}\Psi_1 - iE_1a(t)\Psi_1 + \frac{db}{dt}\Psi_2 -iE_2b\Psi_2 \right)+ ia(t) \gamma_r \frac{d\Psi_1}{dr} + ib(t) \gamma_r \frac{d\Psi_2}{dr} + a(t)i\gamma_\theta \frac{d\Psi_1}{d\theta} + b(t)i\gamma_\theta \frac{d\Psi_2}{d\theta}+ a(t)i\gamma_\phi \frac{d\Psi_1}{d\phi} + b(t)i\gamma_\phi \frac{d\Psi_2}{d\phi} - A_0a(t)\gamma_t\Psi_1 - A_0b(t)\gamma_t\Psi_2 = m_1a(t)\Psi_1 + m_2b(t)\Psi_2$$
Multiplying by $\bar \Psi_3$ and integrating $\int_{r=0}^\infty \int_{\theta=0}^{\pi} \int_{\phi=0}^{2\pi} dr d\theta d\phi$:
$$ i\frac{da}{dt}\color{orange}{\int \int \int \bar \Psi_3 \gamma_t \Psi_1} + E_1a(t)\color{orange}{\int \int \int \bar \Psi_3 \gamma_t \Psi_1} + i\frac{db}{dt} \color{orange}{\int \int \int \bar \Psi_3 \gamma_t\Psi_2} +E_2b(t)\color{orange}{\int \int \int \bar \Psi_3 \gamma_t \Psi_2} + ia(t) \int \int \int \bar \Psi_3\gamma_r \frac{d\Psi_1}{dr} + ib(t) \int \int \int \bar \Psi_3\gamma_r\frac{d\Psi_2}{dr} + a(t)i\int \int \int \bar \Psi_3\gamma_\theta \frac{d\Psi_1}{d\theta} + b(t)i\int \int \int \bar \Psi_3\gamma_\theta \frac{d\Psi_2}{d\theta}+ a(t)i\int \int \int \bar \Psi_3\gamma_\phi \frac{d\Psi_1}{d\phi} + b(t)i\int \int \int \bar \Psi_3\gamma_\phi \frac{d\Psi_2}{d\phi} - A(t)a(t)\int \int \int A_r \bar \Psi_3\gamma_t\Psi_1 - A(t)b(t)\int \int \int A_r \bar \Psi_3\gamma_t\Psi_2 = m_1a(t)\int \int \int \bar \Psi_3\Psi_1 + m_2b(t)\int \int \int \bar \Psi_3\Psi_2$$
and so
$$ i\frac{da}{dt} + E_1a(t) + ia(t) \int \int \int \bar \Psi_3\gamma_r \frac{d\Psi_3}{dr} + ib(t) e^{-i(E_2-E_1)t} \int \int \int \bar \Psi_3\gamma_r\frac{d\Psi_4}{dr} + a(t)i\int \int \int \bar \Psi_3\gamma_\theta \frac{d\Psi_3}{d\theta} + ib(t)e^{-i(E_2-E_1)t}\int \int \int \bar \Psi_3\gamma_\theta \frac{d\Psi_4}{d\theta}+ a(t)i\int \int \int \bar \Psi_3\gamma_\phi \frac{d\Psi_3}{d\phi} + ib(t)e^{-i(E_2-E_1)t}\int \int \int \bar \Psi_3\gamma_\phi \frac{d\Psi_4}{d\phi} - a(t)A(t)\int \int \int A_r \bar \Psi_3\gamma_t\Psi_1 - b(t)A(t)\int \int \int A_r \bar \Psi_3\gamma_t\Psi_2 = m_1a(t)\int \int \int \bar \Psi_3\Psi_3 + m_2b(t)e^{-i(E_2-E_1)t}\int \int \int \bar \Psi_3\Psi_4$$
Therefore,
$$\boxed{\frac{da}{dt} = -im_1a(t)\int \int \int \bar \Psi_3\Psi_3 -i m_2b(t)e^{-i(E_2-E_1)t}\int \int \int \bar \Psi_3\Psi_4 +i E_1a(t) - a(t) \int \int \int \bar \Psi_3\gamma_r \frac{d\Psi_3}{dr} - b(t) e^{-i(E_2-E_1)t} \int \int \int \bar \Psi_3\gamma_r\frac{d\Psi_4}{dr} - a(t)\int \int \int \bar \Psi_3\gamma_\theta \frac{d\Psi_3}{d\theta} - b(t)e^{-i(E_2-E_1)t}\int \int \int \bar \Psi_3\gamma_\theta \frac{d\Psi_4}{d\theta} - a(t)\int \int \int \bar \Psi_3\gamma_\phi \frac{d\Psi_3}{d\phi} - b(t)e^{-i(E_2-E_1)t}\int \int \int \bar \Psi_3\gamma_\phi \frac{d\Psi_4}{d\phi} -i a(t)A(t)\int \int \int A_r \bar \Psi_3\gamma_t\Psi_3 -i b(t)A(t)e^{-i(E_2-E_1)t}\int \int \int A_r \bar \Psi_3\gamma_t\Psi_4 }$$
Similarly,
$$\boxed{\frac{db}{dt} = -im_1a(t)e^{i(E_2-E_1)t}\int \int \int \bar \Psi_4\Psi_3 -i m_2b(t)\int \int \int \bar \Psi_4\Psi_4 + i E_2b(t) - a(t) e^{i(E_2-E_1)t}\int \int \int \bar \Psi_4\gamma_r \frac{d\Psi_3}{dr} - b(t) \int \int \int \bar \Psi_4\gamma_r\frac{d\Psi_4}{dr} - a(t)e^{i(E_2-E_1)t}\int \int \int \bar \Psi_4\gamma_\theta \frac{d\Psi_3}{d\theta} - b(t)\int \int \int \bar \Psi_4\gamma_\theta \frac{d\Psi_4}{d\theta} - a(t)e^{i(E_2-E_1)t}\int \int \int \bar \Psi_4\gamma_\phi \frac{d\Psi_3}{d\phi} - b(t)\int \int \int \bar \Psi_4\gamma_\phi \frac{d\Psi_4}{d\phi} -i a(t)A(t)e^{i(E_2-E_1)t}\int \int \int A_r \bar \Psi_4\gamma_t\Psi_3 -i b(t)A(t)\int \int \int A_r \bar \Psi_4\gamma_t\Psi_4 }$$
Eg.
Using approximate hydrogen ground state solutions
$$ \Psi_1 = \begin{bmatrix} e^{-r} e^{-iE_1t}\\ 0 \\ i \cos\theta e^{-r}e^{-iE_1t}\\ i \sin\theta e^{i\phi}e^{-r}e^{-iE_1t} \end{bmatrix}, \Psi_2 = \begin{bmatrix} 0 \\ e^{-r}e^{-iE_2t} \\ i \sin\theta e^{i\phi}e^{-r} e^{-iE_2t} \\ - i \cos\theta e^{-r} e^{-iE_2t}\end{bmatrix}$$
and
$$ \gamma_t = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1\end{bmatrix} $$ $$ \gamma_r = \begin{bmatrix} 0 & 0 & \cos \theta & \sin \theta e^{-i\phi} \\ 0 & 0 & sin \theta e^{i\phi} & -\cos \theta \\ -\cos \theta & -\sin \theta e^{-i\phi} & 0 & 0 \\ -\sin \theta e^{i\phi} & \cos \theta & 0 & 0\end{bmatrix} $$ $$ \gamma_\theta = \frac{1}{r}\begin{bmatrix} 0 & 0 & -\sin \theta & \cos \theta e^{-i\phi} \\ 0 & 0 & \cos\theta e^{i\phi} & \sin\theta \\ \sin\theta & -\cos\theta e^{-i\phi} & 0 & 0 \\ -\cos\theta e^{i\phi} & -\sin\theta & 0 & 0\end{bmatrix} $$ $$ \gamma_\phi = \frac{1}{r\sin\theta}\begin{bmatrix} 0 & 0 & 0 & -ie^{-i\phi} \\ 0 & 0 & ie^{i\phi} & 0 \\ 0 & ie^{-i\phi} & 0 & 0 \\ -ie^{i\phi} & 0 & 0 & 0\end{bmatrix}$$
we obtain (triple integral $r$ from $0$ to $30$, $A_r = \frac{1}{r}$, unverified):
Therefore,
$$\boxed{ 25\frac{da}{dt} = -9im_1a(t) + 25iE_1a(t) - 120ia(t) + 3a(t) - 95ia(t)A(t) \\ 25\frac{db}{dt} = -9im_2b(t) + 25iE_2b(t) - 42ib(t) - 95ib(t)A(t)}$$
which are uncoupled and thus non-oscillating.
Extending to magnetic vector potential:
$$ L_{QED} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} + i\bar \Psi \gamma_t \frac{d\Psi}{dt} + i\bar \Psi \gamma_r\frac{d\Psi}{dr} + i \bar \Psi \gamma_\theta \frac{d\Psi}{d\theta} + i\bar \Psi \gamma_\phi \frac{d\Psi}{d\phi} - m\bar \Psi \Psi - q\bar\Psi \gamma_t A_t \Psi -q\bar\Psi \gamma_r A_r \Psi - q\bar\Psi \gamma_\theta A_\theta \Psi - q\bar\Psi \gamma_\phi A_\phi \Psi$$
Taking Euler-Lagrange procedure with respect to $\bar \Psi$:
$$ \boxed{i \gamma_t \frac{d\Psi}{dt} + i \gamma_r\frac{d\Psi}{dr} + i \gamma_\theta \frac{d\Psi}{d\theta} + i \gamma_\phi \frac{d\Psi}{d\phi} - q \gamma_t A_t \Psi - q \gamma_r A_r \Psi - q \gamma_\theta A_\theta \Psi - q \gamma_\phi A_\phi \Psi = m \Psi}$$
Suppose $\Psi$ changes into
$$ \Psi \rightarrow e^{iqa(t)}e^{iqb(r)}e^{iqc(\theta)}e^{iqd(\phi)}\Psi$$
and $A_t, A_r, A_\theta, A_\phi$ changes into
$$ A_t \rightarrow A_t - \frac{da(t)}{dt}$$ $$ A_r \rightarrow A_r - \frac{db(r)}{dr}$$ $$ A_\theta \rightarrow A_\theta - \frac{dc(\theta)}{dr}$$ $$ A_\phi \rightarrow A_\phi - \frac{dd(\phi)}{dr}$$
Then
$$i \gamma_t \frac{d}{dt}\left(e^{iqa(t)}e^{iqb(r)}e^{iqc(\theta)}e^{iqd(\phi)}\Psi\right) + i \gamma_r\frac{d}{dr}\left(e^{iqa(t)}e^{iqb(r)}e^{iqc(\theta)}e^{iqd(\phi)}\Psi\right) + i \gamma_\theta \frac{d}{d\theta}\left(e^{iqa(t)}e^{iqb(r)}e^{iqc(\theta)}e^{iqd(\phi)}\Psi\right) + i \gamma_\phi \frac{d}{d\phi}\left(e^{iqa(t)}e^{iqb(r)}e^{iqc(\theta)}e^{iqd(\phi)}\Psi\right) - q \gamma_t (A_t- \frac{da(t)}{dt})\left(e^{iqa(t)}e^{iqb(r)}e^{iqc(\theta)}e^{iqd(\phi)}\Psi\right) - q \gamma_r (A_r- \frac{db(r)}{dr}) \left(e^{iqa(t)}e^{iqb(r)}e^{iqc(\theta)}e^{iqd(\phi)}\Psi\right) - q \gamma_\theta (A_\theta- \frac{dc(\theta)}{d\theta}) \left(e^{iqa(t)}e^{iqb(r)}e^{iqc(\theta)}e^{iqd(\phi)}\Psi\right) - q \gamma_\phi (A_\phi- \frac{dd(\phi)}{d\phi}) \left(e^{iqa(t)}e^{iqb(r)}e^{iqc(\theta)}e^{iqd(\phi)}\Psi\right) = m e^{iqa(t)}e^{iqb(r)}e^{iqc(\theta)}e^{iqd(\phi)}\Psi $$
and
$$i e^{iqb(r)}e^{iqc(\theta)}e^{iqd(\phi)}\gamma_t \left(e^{iqa(t)}\frac{d\Psi}{dt} + iq\frac{da}{dt}e^{iqa(t)}\Psi \right) + i e^{iqa(t)}e^{iqc(\theta)}e^{iqd(\phi)}\gamma_r\left(e^{iqb(r)}\frac{d\Psi}{dr} + iq\frac{db}{dr}e^{iqb(r)}\Psi \right) + i e^{iqa(t)}e^{iqb(r)}e^{iqd(\phi)} \gamma_\theta \left(e^{iqc(\theta)}\frac{d\Psi}{d\theta} + iq\frac{dc}{d\theta}e^{iqc(\theta)}\Psi \right) + i e^{iqa(t)}e^{iqb(r)}e^{iqc(\theta)}\gamma_\phi \left(e^{-iqd(\phi)}\frac{d\Psi}{d\phi} + iq\frac{dd}{d\phi}e^{iqd(\phi)}\Psi \right) - \left(e^{iqa(t)}e^{iqb(r)}e^{iqc(\theta)}e^{iqd(\phi)}\right)q \gamma_t (A_t - \frac{da}{dt})\Psi - \left(e^{iqa(t)}e^{iqb(r)}e^{iqc(\theta)}e^{iqd(\phi)}\right)q \gamma_r (A_r - \frac{db}{dr}) \Psi - \left(e^{iqa(t)}e^{iqb(r)}e^{iqc(\theta)}e^{iqd(\phi)}\right)q \gamma_\theta (A_\theta - \frac{dc}{d\theta})\Psi - \left(e^{iqa(t)}e^{iqb(r)}e^{iqc(\theta)}e^{iqd(\phi)}\right)q \gamma_\phi (A_\phi - \frac{dd}{d\theta}) \Psi = m e^{iqa(t)}e^{iqb(r)}e^{iqc(\theta)}e^{iqd(\phi)}\Psi$$ and $$=e^\kappa i\gamma_t \frac{d\Psi}{dt} \color{orange}{- e^\kappa\gamma_tq\frac{da}{dt}\Psi} + e^\kappa i\gamma_r \frac{d\Psi}{dr} \color{orange}{- e^\kappa\gamma_rq\frac{db}{dr}\Psi} + e^\kappa i\gamma_\theta \frac{d\Psi}{d\theta} \color{orange}{- e^\kappa\gamma_\theta q\frac{dc}{d\theta}\Psi} + e^\kappa i\gamma_\phi \frac{d\Psi}{d\phi} \color{orange}{- e^\kappa\gamma_\phi q\frac{dd}{d\phi}\Psi} - e^\kappa q \gamma_t A_t\Psi \color{orange}{+ e^\kappa q \gamma_t \frac{da}{dt}\Psi} - e^\kappa q \gamma_r A_r\Psi \color{orange}{+ e^\kappa q \gamma_r \frac{db}{dr}\Psi} - e^\kappa q \gamma_\theta A_\theta\Psi \color{orange}{+ e^\kappa q \gamma_\theta \frac{dc}{d\theta}\Psi} - e^\kappa q \gamma_\phi A_\phi\Psi \color{orange}{+ e^\kappa q \gamma_\phi \frac{dd}{d\phi}\Psi}= m e^\kappa \Psi$$ so $$e^\kappa \left( i\gamma_t \frac{d\Psi}{dt} + i\gamma_r \frac{d\Psi}{dr} + i\gamma_\theta \frac{d\Psi}{d\theta} + i\gamma_\phi \frac{d\Psi}{d\phi} - q \gamma_t A_t\Psi - q \gamma_r A_r\Psi - q \gamma_\theta A_\theta\Psi - q \gamma_\phi A_\phi\Psi\right)= e^\kappa (m\Psi)$$
$$\boxed{i\gamma_t \frac{d\Psi}{dt} + i\gamma_r \frac{d\Psi}{dr} + i\gamma_\theta \frac{d\Psi}{d\theta} + i\gamma_\phi \frac{d\Psi}{d\phi} - q \gamma_t A_t\Psi - q \gamma_r A_r\Psi - q \gamma_\theta A_\theta\Psi - q \gamma_\phi A_\phi\Psi= m \Psi}$$
The cancellation of $e^\kappa$ leaves the above equation unchanged when $\Psi$ changes to $e^\kappa \Psi$.