# What motivates the trial solution of $\left[-\frac{\hbar^2\nabla^2}{2m}+\frac{e^2y^2B^2}{2m}-\frac{i\hbar eyB}{m}\right]\psi(x,y,z)=E\psi(x,y,z).$?

The time-independent Schrodinger equation for the problem of charged particles in an uniform magnetic field $${\vec B}=B{\hat k}$$, in the Coulomb gauge $${\vec A}=(-yB,0,0)$$, reduces to the following differential equation $$\left[-\frac{\hbar^2\nabla^2}{2m}+\frac{e^2y^2B^2}{2m}-\frac{i\hbar eyB}{m}\frac{\partial}{\partial x}\right]\psi(x,y,z)=E\psi(x,y,z).$$ It can be solved by assuming a trial solution of the form $$\psi(x,y,z)=f(y)\exp [i(k_xx + k_zz)].$$

What motivates this trial solution? I can guess the part $$\exp(ik_zz)$$. What motivates the plane wave part $$\exp(ik_xx)$$ when the Hamiltonian couples $$y$$ with $$p_x=-i\hbar\frac{\partial}{\partial x}$$?

• Notice that this can't be a physical, since the we could flip the roles of $x$ and $y$ by switching gauges. The reason is purely mathematical: every differential equation of the form $(\frac{\partial^2}{\partial x^2}+\frac{\partial}{\partial x}+c)f(x)=0$ always has $e^{ikx}$ as a solution as you are probably aware from the damped oscillator problems in mechanics. – Anonjohn Jan 18 at 17:14

The (not entirely) physical reason to try that ansatz is the fact that the Hamiltonian commutes with both $$p_x$$ and $$p_z$$, and they obviously commute between themselves: $$\begin{cases} [p_x, H]=0\\ [p_z, H]=0\\ [p_x, p_z]=0 \end{cases}$$ hence $$\{p_x,p_z,H\}$$ is a set of commuting observables. For what concerns $$\hat{x}$$ and $$\hat{z}$$ directions, we then expect the eigenfunctions of $$H$$ to be plain waves. You can easily see that $$[p_y,H]\neq 0$$, as $$H$$ contains explicitly $$y$$, so you have to include an unknown function $$f(y)$$ in the trial solution.
Anyway, with a different choice of Gauge, for instance $$\vec{A}=(0,Bx,0)$$, you can switch the roles of $$x,y,z$$. Note that this affects only the wave function, and not the energy levels of the system, as one expects from the fact that the choice of $$\vec{A}$$ is arbitrary. I hope this can be of some help!