Landau levels: Hamiltonian with ladder operators

In summary, the conversation discusses the application of intuition from the simple harmonic oscillator to the Landau level problem. The Hamiltonian for the Landau level problem is compared to the Hamiltonian of a harmonic oscillator, and the relevant part for diagonalization is identified as the non-commutativity of p_x and x. The conversation also notes that the frequency in the Landau level problem is related to the classical motion of a particle in a magnetic field.
  • #1
Garlic
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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
 
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  • #2
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 [itex][a,a^{\dagger}] = 1[/itex], 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 [itex]p_y[/itex] and [itex]p_z[/itex] 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 [itex]p_x[/itex] and [itex]x[/itex] 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 [itex]x[/itex] coordinate but with a shifted equilibrium position, and a frequency [itex]\omega = eB/m[/itex]. (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?
 
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1. What are Landau levels?

Landau levels are a phenomenon that occurs when a charged particle, such as an electron, is placed in a magnetic field. The energy levels of the particle become quantized, forming discrete energy states known as Landau levels.

2. What is the Hamiltonian in relation to Landau levels?

The Hamiltonian is a mathematical operator that represents the total energy of a system, including its potential and kinetic energy. In the context of Landau levels, the Hamiltonian is used to describe the energy levels of a charged particle in a magnetic field.

3. What are ladder operators?

Ladder operators are mathematical tools used to describe the energy states of a system, such as Landau levels. They are used to raise or lower the energy of a particle by a certain amount, and can also be used to calculate the transition probabilities between different energy levels.

4. How are Landau levels experimentally observed?

Landau levels can be observed through a variety of experimental techniques, such as angle-resolved photoemission spectroscopy (ARPES) and cyclotron resonance. These techniques involve measuring the energy and momentum of particles in a magnetic field to identify the discrete energy levels.

5. What is the significance of Landau levels in physics?

Landau levels have significant implications in many areas of physics, including condensed matter physics and quantum mechanics. They provide a fundamental understanding of how particles behave in a magnetic field and have practical applications in fields such as electronics and materials science.

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