# Finding the energy levels of an electron in a plane perpendicular to a uniform magnetic field

Suppose we have an electron, mass $m$, charge $-e$, moving in a plane perpendicular to a uniform magnetic field $\vec{B}=(0,0,B)$. Let $\vec{x}=(x_1,x_2,0)$ be its position and $P_i,X_i$ be the position and momentum operators.

The electron has Hamiltonian

$H=\frac{1}{2m}((P_1-\frac{1}{2}eBX_2)^2+(P_2+\frac{1}{2}eBX_1)^2)$

How can I show that this is analogous to the one dimensional harmonic oscillator and then use this fact to describe its energy levels?

I have attempted to expand out the Hamiltonian and found:

$(\frac{P^2_1}{2m}+ \frac{1}{2} m (\frac{eB}{2m}))^2X^2_1+(\frac{P^2_2}{2m}+\frac{1}{2}m(\frac{eB}{2m})^2)X^2_2+\frac{eB}{2m}(X_1P_2-P_1X_2)$

This looks very similar to the 2D harmonic oscillator, if anyone can help/point out where I am wrong I'd much appreciate it!

• Looks like you have an error in the second formula there. I think there should only be one close parenthesis just before $X_1^2$. And perhaps you don't want the open paren in from of $\frac{P_2^2}{2m}$ either. – dmckee Nov 20 '12 at 14:47

## 1 Answer

Starting from the canonical momenta:

$\Pi_1= P_1 -\frac{1}{2}eBX_2$ and $\Pi_2= P_2 +\frac{1}{2}eBX_1$,

We get

$[\Pi_1, \Pi_2] = ieB\hbar$

Thus the operators:

$a = \frac{1}{\sqrt{2eB\hbar}}(\Pi_1+i\Pi_2)$ and

$a^{\dagger} = \frac{1}{\sqrt{2eB\hbar}}(\Pi_1-i\Pi_2)$

satisfy the canonical commutation relation $[a, a^{\dagger}] = 1$

Making the substitution in the Hamiltonian we obtain:

$H =\hbar \omega (a a^{\dagger}+\frac{1}{2})$, with $\omega=\frac{eB}{m}$.

The difference from the Harmonic oscillator lies in the fact that now the energy levels are infinitely degenerate for example, the ground states equation:

$a\Psi(X_1,X_2) = 0$

has an infinite number of solutions and any function of the form:

$\Psi = f(X_1+iX_2) \mathrm{exp}(-eB(X_1^2+X_2^2)/4\hbar)$

is a ground state. These are the lowest Landau levels.

• I'm sorry, this is very good, but would you mind explaining in a little more detail how this gives us the energy levels? – Freeman Nov 20 '12 at 16:40
• What is the signifiance of the canonical commutation relation? – Freeman Nov 20 '12 at 16:47
• The Hamiltonian expressed in terms of the creation and annihilation operators has exactly the form of the harmonic oscillator Hamiltonian. Thus the energy levels are exactly equal to those of the harmonic oscillator, however with infinite degeneracy per level (The harmonic oscillator energy levels are nondegenerate). The advantage of using this method is that it allows an algebraic solution of the energy levels (i.e. without solving differential equations), please see the quantum harmonic oscillator Wikipedia page: en.wikipedia.org/wiki/Quantum_harmonic_oscillator – David Bar Moshe Nov 20 '12 at 16:58
• These creation annihilation operators, are they $a,a^{\dagger}$, I don't see how $H =\hbar \omega (a a^{\dagger}+\frac{1}{2})$? don't we want to find it in terms of $X$ and $P$ to compare to the harmonic oscillator? Sorry for being slow.. it's been a very long day! – Freeman Nov 20 '12 at 17:04
• If you substitute the expressions of $a$ and $a^{\dagger}$ given in the answer in the Hamiltonian $H=\hbar\omega(a a^{\dagger}+\frac{1}{2})$ you get the Hamiltonian you started with expressed in terms of $X$ and $P$. – David Bar Moshe Nov 20 '12 at 17:16