Classical Harmonic Oscillator in a Magnetic Field Consider a 3D  charged Harmonic Oscillator which is placed in a homogenous magnetic field along the z direction. The equations of motion in plane are then given as:
$$
m\ddot{x}=kx+a\dot{y},$$
$$
m\ddot{y}=ky-a\dot{x},
$$
where $a=qB_z$.
Because I really knew no other way forward, I assumed:
$$
y=A\exp(i\omega t)+B\exp(-i \omega t)
$$
$$
x=C\exp(i\omega t)+D\exp(-i\omega t).
$$
My first question is this, is this Ansatz even valid or is there a better one?
I comforted myself, that both dimensions should have the same $\omega$ because mass, charge and magnetic field are the same for both dimensions, is this really valid though?
After some arithmetic I reached:
$$
\omega_{1,2,3,4}=\pm\sqrt{\frac{a^2}{2m^2} \pm \sqrt{\frac{a^2}{4m^2}-\frac
{k^2}{m^2}}}
$$
First of all I am uncertain if all four solutions are physical further more, if $\frac{a^2}{4m^2}<\frac{k^2}{m^2} $ discriminant has an imaginary part, hence the motion is damped.
But I recall, that the Lorenz-Force does no work, can it still damp? And if $\frac{a^2}{4m^2}<\frac{k^2}{m^2} $ there should be no damping, is this physical? Or did I make a mistake?
 A: To decouple your equations, you can perform the change of variable
$$z=x+iy$$
so that
$$m\ddot z=kz-ia\dot z$$
Solutions are exponential, $z=e^{rt}$ with $mr^2=k-iar$. Finally, $x$ and $y$ are given by the real and imaginary parts of $z(t)$.
A: This is a system of linear differential equations with constant coefficients, which can be solved by many approaches. Perhaps, it is worth pointing out that, since the order of the system is $4$, one may potentially have four linearly independent solutions (not two as the OP suggests.) However, the ansatz $x,y\propto e^{i\omega t}$ is a possible method of solution.
Apart from this small clarifications, this is a homework problem. I strongly suggest that you do some reading (beyond wikipedia) on linear ODE with constant coefficients and systems of such equations - it is part of the obligatory background for a physicist.
A: The Ansatz is valid. The real requirement is that both $x$ and $y$ have to be real. Clearly you cannot achieve that at all times if the frequencies are different. Second, the reality condition also imposes relations between A and B (and C and D).
The fact that you got four solutions is of course a consequence that you assumed you have four independent variables, so by itself it does not necessarily mean the solution is wrong (the outer $\pm$ is more about phase than actual frequency). However what does trouble, is that your solution does not reproduce the trivial answer when you put $B=0$. You definitely made a mistake somewhere. Also the dimensionality is inconsistent: you essentially add $a^2/m^2$ with $a/m$ inside the larger square root. This also indicates there is a mistake somewhere.
In general when you come up with an expression it is always good to perform these simple sanity checks:

*

*check limiting cases where you already know what the answer should be

*check if dimensions between different terms match

