The Landau Problem

The problem of electrons confined to a plane pierced by a perpendicular magnetic field is sometimes referred to as the Landau problem. Understanding this quantum mechanical problem is a prerequisite for understanding the physics of the integer and fractional quantum Hall effects.

Consider the Hamiltonian \begin{equation} \label{eq:landau-problem-hamiltonian} H = \frac{1}{2m} \left( \mathbf{p} - e \mathbf{A} \right)^ {2} \ , \end{equation} where $ \mathbf{p} $ and $ m $ are the momentum and (effective) mass of the electrons, and $ \mathbf{A} $ is the gauge potential corresponding to a magnetic field in the symmetric gauge: \begin{equation} \mathbf{A} = \frac{B}{2} \left( - x _ {2} , x _ {1} , 0 \right) \ , \end{equation} so that $ \mathbf{B} = \nabla \times \mathbf{A} = B \,\widehat{\mathbf{z} } $. Notice in particular that since $ A _ {z} = 0 $, the dynamics along the $ z $-axis can be separated out and is free. We will pay no further attention to this fact, and use Latin indices $ i, j, \cdots \in \left\lbrace 1,2 \right\rbrace $.

It is useful to define the operators \begin{equation} \Pi _ {i} = p _ {i} - e A _ {i} \ , \end{equation} whose commutations relations are \begin{equation} \left[ \Pi _ {i} , \Pi _ {j} \right] = i \epsilon _ {ij} e B \ , \end{equation} where $ \epsilon _ {ij} $ is the totally antisymmetric two-dimensional Levi-Civita symbol. (It is useful, incidentally, to write $ A _ {i} = -\frac{B}{2} \epsilon _ {ij} x _ {j} $ sometimes.) Our conventions completely specified by $ \epsilon _ {12} = +1 $.

Let us also remark on the dimensions here: a simple exercise in dimensional analysis will tell us that, in natural units (where $ \hbar =1 $), the quantity $ eB $ has dimensions of inverse length squared. We use the symbol $ \ell _ {\text{B}} ^ {2} = \left( eB \right)^ {-1} $, which will be a characteristic length scale for our problem.

Consider now the following operators constructed out of the $ \Pi _ {i} $: \begin{equation} a = \frac{1}{\sqrt{2 e B} } \left( \Pi _ {1} + i \Pi _ {2} \right) \quad \text{and} \quad a ^ {\dagger } = \frac{1}{\sqrt{2eB} } \left( \Pi _ {1} - i \Pi _ {2} \right) \ , \end{equation} which satisfy the familar commutation relations \begin{equation} \left[ a, a ^ {\dagger } \right] = 1 \ , \end{equation} and so the Hamiltonian in \eqref{eq:landau-problem-hamiltonian} is equivalently expressed as \begin{equation} H = \frac{eB}{m} \left( a ^ {\dagger } a + \frac{1}{2} \right) \ . \end{equation}

Similar to our earlier discussions of the harmonic oscillator, the ground state $ \left\vert 0 \right\rangle $ of this problem can be defined by the equation $ a \left\vert 0 \right\rangle = 0 $, and excited states are constructed by successively acting on the ground state with $ a ^ {\dagger } $: \begin{equation} \left\vert n \right\rangle = \frac{1}{\sqrt{n!} } \left( a ^ {\dagger } \right)^ {n} \left\vert 0 \right\rangle \ , \end{equation} whose energy is given by \begin{equation} E _ {n} = \frac{eB}{m} \left( n + \frac{1}{2} \right) \quad \text{for} \quad n \in \mathbb{N} \ . \end{equation} These are called Landau levels, and the state $ n = 0 $ is called the lowest Landau level, sometimes abbreviated LLL.

We now introduce the operators \begin{equation} M _ {i} = p _ {i} + e A _ {i} \ ,
\end{equation} which looks a lot like the $ \Pi _ {i} $ operator, but notice the crucial sign difference! This might seem like an odd object to consider until one computes \begin{equation} \left[ \Pi _ {i} , M _ {j} \right] = 0 \ , \end{equation} due to the antisymmetry of the Levi-Civita symbol. From this, it follows that \begin{equation} \label{eq:M-commutes-H} \left[ M _ {i} , H \right] = 0 \ , \end{equation} and therefore that eigenstates of the $ M _ {i} $ operators are eigenstates of the Hamiltonian, and further, that acting on the states $ \left\vert n \right\rangle $ that we constructed earlier with the $ M _ {i} $ lead to degenerate states.

Let’s try and understand this a little better. A straightforward computation reveals that the $ M _ {i} $ operators satisfy the following commutation relation \begin{equation} \left[ M _ {i} , M _ {j} \right] = -i \epsilon _ {ij} e B \ , \end{equation} so it is clear that the $ M _ {i} $ cannot be simultaneously diagonalised. (Note also, in passing, that the $ M _ {i} $ are momentum-like variables.) Analogous to what we did with the $ \Pi _ {i} $ operators, we can construct the operators \begin{equation} b = \frac{1}{\sqrt{2eB} } \left( M _ {1} - i M _ {2} \right) \quad \text{and} \quad b ^ {\dagger } = \frac{1}{\sqrt{2eB} } \left( M _ {1} + i M _ {2} \right) \ , \end{equation} such that \begin{equation} \left[ b, b ^ {\dagger } \right] = 1 \ . \end{equation} Further, it follows from \eqref{eq:M-commutes-H} that \begin{equation} \left[ b, H \right] = 0 \ . \end{equation} It follows from this that the ground state of the system in question is not specified solely by $ a \left\vert 0 \right\rangle = 0 $, but also by $ b \left\vert 0 \right\rangle = 0 $. We may then construct excited states of the form \begin{equation} \left\vert n,k \right\rangle = \frac{1}{\sqrt{n! k!} } \left( a ^ {\dagger } \right)^ {n} \left( b ^ {\dagger } \right)^ {k} \left\vert 0 \right\rangle \ , \end{equation} and for all $ k \in \mathbb{N} $, the eigenstates of the Hamiltonian in \eqref{eq:landau-problem-hamiltonian} are \begin{equation} H \left\vert n,k \right\rangle = \frac{eB}{m} \left( n+\frac{1}{2} \right)\left\vert n,k \right\rangle \ . \end{equation} Each eigenstate has an infinite degeneracy! In particular, when projecting onto the lowest Landau level (with $ n=0 $), the Hamiltonian is simply \begin{equation} H \big\vert _ {\text{LLL}} = \frac{eB}{2m} \mathbf{1} \ , \end{equation} i.e. proportional to the identity operator.

It is natural to ask, at this stage, what this infinite degeneracy corresponds to. It is helpful to recall that a charged particle in a plane pierced by a magnetic field executes cyclotron motion. This does not, however, specify the center about which the cyclotron motion takes place — that is determined by a constant of integration when solving the classical equations of motion. The arbitrariness of the center about which cyclotron motion is the executed is reflected in the symmetry generated by the $ M _ {i} $ operators. (The fact that the $ M _ {i} $ have dimensions of momentum is a minor itch we will momentarily scratch.)

Indeed, the momentum $ p _ {i} $ of the quantum mechanical particle, written in terms of the operators $ \Pi _ {i} $ and $ M _ {i} $ is simply \begin{equation} p _ {i} = \frac{1}{2} \left( \Pi _ {i} + M _ {i} \right) \ , \end{equation} which may be interpreted as a a decomposition of the momentum into a part governing cyclotron motion $ (\Pi _ {i} ) $ and a part governing the center of motion $ (M _ {i} ) $. Analogously, the position $ x _ {i} $ can also be broken up into parts associated to cyclotron motion and the center of motion. Defining \begin{equation} R _ {i} = - \frac{1}{eB} \epsilon _ {ij} M _ {j} \ , \end{equation} we see that the operator $ R _ {i} $, which controls the center of motion and has dimensions of position, satisfies the commutation relation \begin{equation} \label{eq:R-commutation-relations} \left[ R _ {i} , R _ {j} \right] = - i \ell _ {\text{B}} ^ {2} \epsilon _ {ij} \ . \end{equation}

We have already seen that the degeneracy of each Landau level is infinite, but we can do a little better than that in characterising the degeneracy. From the commutation relations in \eqref{eq:R-commutation-relations}, we see that each center of cyclotron motion occupies an area $ \ell _ {\text{B}} ^ {2} $. (In the Landau problem, the quantity $ \ell _ {\text{B}} ^ {2} $ acts like Planck’s constant, except in real space.) Following the usual heuristic that the number of states is the phase space area in units of Planck’s constant, we find that for the Landau problem the number of degenerate states on a plane of area $ A $ is given by \begin{equation} \label{eq:Landau-degeneracy} N = \frac{A}{2 \pi \ell _ {\text{B}} ^ {2} } \ . \end{equation} This can be rephrased in two equivalent ways. The first is that the degeneracy per unit area is constant, a proportional to $ eB/2 \pi $. The second rewrites \eqref{eq:Landau-degeneracy} in terms of the flux of the magnetic field through the surface $ \Phi = B A $ as \begin{equation} N \times \frac{2 \pi }{e} = \Phi \ , \end{equation} and this suggests that the total flux $ \Phi $ is quantised in integer multiples of the quantum of magnetic flux: \begin{equation} \Phi _ {0} = \frac{2 \pi }{e} \ . \end{equation}