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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 $p_x=-i\hbar\frac{\partial}{\partial x}$?

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  • $\begingroup$ 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. $\endgroup$
    – Anonjohn
    Jan 18, 2021 at 17:14

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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!

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