Landau levels: Hamiltonian with ladder operators

[Garlic]

Homework Statement

Landau levels in quantum mechanics

In short:
take the hamiltonian $$ H=\vec{Π}^{\,2} $$ with $$ \vec{Π}= \vec{p} + e \vec{A(x)} $$ and $$ A(x)=xB \hat{e}_y $$
and bring it in the form
$H = \frac{ Π^2_3}{2m} + \hbar \omega_c ( a^{ \dagger } a + \frac{1}{2}$

I have brought the Hamiltionian in this form:
$H= \frac{1}{2m} (Π^2_3 + Π_{+} Π_{-} +e \hbar B)$

There is a point where I don't understand the given solutions.
They just say the ladder operators are defined as this:
$a^{ \dagger }/a:= \frac{1}{ \sqrt{ 2 e \hbar B }} (Π_1 \pm iΠ_2)$

I really don't understand how does this definition just fall off the sky. I've looked at many lesson notes from different universities, but they just defined it like this.
But I don't know how to derive this from the regular ladder operator formulas.

PS: there was a note from MIT,saying that the ladder operators of the total angular momentum are analogues for the $$ a/a^{\dagger} $$ ladder operators. Maybe the solution involves just replacing an operator with another?

Relevant Equations

$$ a= \frac{1}{ \sqrt{2} } (\frac{x}{x_0} +i \frac{ x_0 p }{ \hbar } $$

Dear PF,
I hope I've formulated my question understandable enough.
Thank you for your time,
Garli

 

[king vitamin]

The trick is to use your intuition from the usual harmonic oscillator. I assume you have probably seen the simple harmonic oscillator in your course if you are already doing Landau levels; to review, your Hamiltonian is
$$
H = \frac{p^2}{2m} + \frac{1}{2} m \omega^2 x^2.
$$
The solution comes from defining
$$
x = \sqrt{\frac{\hbar}{2 m \omega}} \left( a + a^{\dagger} \right), \qquad p = i \sqrt{\frac{\hbar m \omega}{2}} \left( a - a^{\dagger} \right),
$$
after which the Hamiltonian becomes
$$
H = \hbar \omega \left( a^{\dagger} a + \frac{1}{2} \right).
$$
Then using the commutation relation [a,a^{\dagger}] = 1, you can find the spectrum of this Hamiltonian.

How does this relate to the Landau level problem? Well, let's write out the full Hamiltonian:
$$
H = \frac{1}{2m} p_x^2 + \frac{1}{2m} \left( p_y + e x B \right)^2 + \frac{1}{2m} p_z^2.
$$
The key point to notice is that p_y and p_z commute with each other and everything else in this Hamiltonian, so that the relevant part needed to diagonalize it is entirely in the fact that p_x and x do not commute. In fact, rewriting it as
$$
H = \frac{1}{2m} p_x^2 + \frac{1}{2} \left( \frac{e B}{m} \right) \left( x + \frac{p_y}{eB} \right)^2 + \frac{1}{2m} p_z^2,
$$
it looks just like a harmonic oscillator in the x coordinate but with a shifted equilibrium position, and a frequency \omega = eB/m. (Is this frequency a familiar quantity, say from the motion of a classical particle in a magnetic field?)

Can you see how to apply the harmonic oscillator solution to this Hamiltonian now?