The purpose of this post is to explain how the Landau levels of a nonrelativistic charged quantum particle moving in a magnetic field arise. Classically, charged particles spiral along \(\textbf B\)-field lines at the cyclotron frequency \(\omega_B=\frac{qB}{m}\), and indeed this foreshadows the quantum mechanics of Landau levels.
Review of Classical Electromagnetism: Newton, Lagrange, Hamilton
Recall that a charged particle \(q\in\textbf R\) moving in an electromagnetic field \(\textbf E,\textbf B\) classically experiences a Lorentz force \(\textbf F_{\textbf E,\textbf B}\) given in the Newtonian framework by:
\[\textbf F_{\textbf E,\textbf B}=q\textbf E+q\textbf v\times\textbf B\]
By contrast, the Lagrangian \(\mathcal L\) of such a classical charged particle moving through an electromagnetic field \(\textbf E,\textbf B\) cannot ironically be written down in terms of \(\textbf E\) and \(\textbf B\)! Instead, we have to revert to their gauge potentials \(\phi,\textbf A\) defined by \(\textbf E=-\frac{\partial\phi}{\partial\textbf x}-\frac{\partial\textbf A}{\partial t}\) and \(\textbf B=\frac{\partial}{\partial\textbf x}\times\textbf A\). Although \(\phi\) and \(\textbf A\) are typically introduced as mere mathematical “substitutions” that automatically solve the \(2\) source-free Maxwell equations (Gauss’s law for magnetic fields and the Faraday-Lenz law of electromagnetic induction), their role is actually much deeper than this innocuous origin suggests (this feels similar to the story of the canonical partition function \(Z\) in statistical mechanics initially being introduced as a mere normalization factor in the Boltzmann distribution but then turning out to be much more significant). So in terms of the gauge potentials \(\phi,\textbf A\), the Lagrangian \(\mathcal L\) of the charge \(q\) moving through these gauge fields is:
\[\mathcal L=\frac{1}{2}m|\dot{\textbf x}|^2+q\dot{\textbf x}\cdot\textbf A(\textbf x)-q\phi(\textbf x)\]
Note that although \(\mathcal L\) is not gauge-invariant, it merely accumulates a total time derivative \(-q\dot{\Gamma}\) when \(A^{\mu}\mapsto A^{\mu}+\partial^{\mu}\Gamma\) which doesn’t affect the variation of the action functional \(S:=\int\mathcal Ldt\) between fixed configuration space endpoints and so doesn’t affect the equations of motion. One can check that the generalized momentum \(\textbf p\) conjugate to the position \(\textbf x\) is not the usual dynamical momentum \(\textbf p\neq m\dot{\textbf x}\), but rather \(\textbf p=\frac{\partial\mathcal L}{\partial\dot{\textbf x}}=m\dot{\textbf x}+q\textbf A(\textbf x)\) includes an additional magnetic contribution \(q\textbf A(\textbf x)\). It is not gauge-invariant (unlike the dynamical momentum which is trivially gauge-invariant). The Euler-Lagrange equations \(\dot{\textbf p}=\frac{\partial \mathcal L}{\partial\textbf x}\) can be checked to reproduce to the Newtonian Lorentz force above (the one slightly nontrivial vector calculus identity one needs is that \(\dot{\textbf x}\times\left(\frac{\partial}{\partial\textbf x}\times\textbf A\right)=\left(\left(\frac{\partial\textbf A}{\partial\textbf x}\right)^T-\frac{\partial\textbf A}{\partial\textbf x}\right)\dot{\textbf x}\)).
Finally, the classical Hamiltonian \(H\) is just the Legendre transform of the Lagrangian from the velocity \(\dot{\textbf x}\mapsto\textbf p\) to the generalized momentum:
\[H:=\dot{\textbf x}\cdot\textbf p-\mathcal L=\frac{1}{2}m|\dot{\textbf x}|^2+q\phi(\textbf x)=\frac{1}{2m}|\textbf p-q\textbf A(\textbf x)|^2+q\phi(\textbf x)\]
In particular, notice that when \(H\) is written in terms of \(\dot{\textbf x}\) it just looks like a charge \(q\) moving through an electric field only, reflecting that the magnetic power is always zero \(P_{\textbf B}=0\). However, when written in terms of the generalized momentum \(\textbf p\) (again, I emphasize that this is not the dynamical momentum \(m\dot{\textbf x}\)!), it certainly doesn’t look like that anymore, and in general this prescription \(\textbf p\mapsto\textbf p-q\textbf A(\textbf x)\) in the kinetic energy of the Hamiltonian is called minimal coupling. Similar to the Lagrangian \(\mathcal L\), the Hamiltonian \(H\) changes by a partial time derivative \(q\frac{\partial\Gamma}{\partial t}\) under a gauge transformation (as the velocity \(\dot{\textbf x}=\textbf p-q\textbf A(\textbf x)\) is gauge invariant) which still leaves Hamilton’s equations of motion \(\dot{\textbf x}=(\textbf p-q\textbf A(\textbf x))/m\) and \(\dot{\textbf p}=q(\textbf E+\dot{\textbf x}\times\textbf B)\) gauge invariant (for instance note that \(\frac{\partial\Gamma}{\partial\textbf p}=\textbf 0\)).
Moving Onto Quantum Mechanics
Just as the classical Hamiltonian \(H\) was inevitably written in terms of the gauge potentials \(\phi,\textbf A\), so the corresponding Hamiltonian observable \(H\) in quantum mechanics can also only be written with \(\phi,\textbf A\). For a charge \(q\), it is formally identical to the classical Hamiltonian:
\[H=\frac{|\textbf P-q\textbf A(\textbf X)|^2}{2m}+q\phi(\textbf X)\]
where note that \(|\textbf P-q\textbf A(\textbf X)|^2=|\textbf P|^2+q^2|\textbf A(\textbf X)|^2-q(\textbf P\cdot\textbf A(\textbf X)+\textbf A(\textbf X)\cdot\textbf P)\). We then still have to solve for the dynamics via the usual Schrodinger equation \(i\hbar\dot{|\psi\rangle}=H|\psi\rangle\) using the electromagnetic Hamiltonian \(H\) above.
Suppose we insist that the Schrodinger equation be gauge covariant under the gauge transformations of the potentials \(\phi\mapsto\phi’=\phi+\frac{\partial\Gamma}{\partial t}\) and \(\textbf A\mapsto\textbf A’=\textbf A-\frac{\partial\Gamma}{\partial\textbf x}\). The novelty is that the state \(|\psi\rangle\mapsto |\psi’\rangle=U_{\Gamma}|\psi\rangle\) will have to transform too under the gauge transformations (which we postulate occurs via some linear operator \(U_{\Gamma}\)) in order to have any hope of maintaining gauge covariance. Thus, the Hamiltonian becomes \(H\mapsto H’=H+q\frac{\partial\Gamma}{\partial t}\) and we demand by gauge covariance that \(i\hbar\dot{|\psi’\rangle}=H’|\psi’\rangle\) in addition to \(i\hbar\dot{|\psi\rangle}=H|\psi\rangle\). With these two equations in mind, one can show that \(U_{\Gamma}\) satisfies the differential equation:
\[i\hbar\frac{\partial U_{\Gamma}}{\partial t}=[H,U_{\Gamma}]+q\frac{\partial\Gamma}{\partial t}U_{\Gamma}\]
If we assume for a moment that \([H,U_{\Gamma}]=0\) commute, then the resultant ODE is easy to integrate and gives:
\[U_{\Gamma}(\textbf X,t)=e^{-iq\Gamma(\textbf X,t)/\hbar}\]
which is indeed a local unitary phase as it was suggestively written all along. This suggests a new interpretation of the gauge \(\Gamma\), namely it is a generator of translations in charge space \(q\in\textbf R\), adding more or less charge to the system. The assumption earlier that \([H,U_{\Gamma}]=0\) amounts to conservation of electric charge?
There is another way to see why the above is the correct gauge transformation of states, even if its derivation was a little handwavy. Specifically, if one defines the four-covector operator \(\mathcal D:=\frac{\partial}{\partial X}+\frac{iq}{\hbar}A\) with corresponding covariant components \(\mathcal D_{\mu}=\partial_{\mu}+\frac{iqA_{\mu}}{\hbar}\) for spacetime index \(\mu=0,1,2,3\), then the Schrodinger equation looks like that for a free particle (descending from kets to wavefunctions \(|\psi\rangle\mapsto\psi(\textbf x,t)\)):
\[i\hbar c\mathcal D_0\psi=-\frac{\hbar^2}{2m}\mathcal D^2\psi\]
where \(\mathcal D^2:=\mathcal D_1^2+\mathcal D_2^2+\mathcal D_3^2\). Now then, check (using the product rule and \([A_{\mu},U_{\Gamma}]=0\)) that each \(\mathcal D_{\mu}\psi\mapsto U_{\Gamma}\mathcal D_{\mu}\psi\) under the gauge transformation \(\Gamma\) so the Schrodinger equation is thus gauge invariant since \(D_{\mu}U_{\Gamma}=U_{\Gamma}D_{\mu}\) as operators (though the Schrodinger equation is not Lorentz covariant! That’s a topic for another time).
Landau Levels
Assume \(\textbf E=\textbf 0\) so that there is only a uniform magnetic field \(\textbf B=B\hat{\textbf k}\) along the \(z\)-axis with \(B>0\). Of course, there are many possible magnetic vector potentials \(\textbf A\) one could use here related by gauge transformations. One option is the Landau gauge \(\textbf A_{\text{Landau}}:=Bx\hat{\textbf j}\) (think fluid mechanically to see why \(B\hat{\textbf k}=\frac{\partial}{\partial\textbf x}\times\textbf A_{\text{Landau}}\)). Then the Hamiltonian becomes:
\[H=\frac{P_1^2+(P_2-qBX)^2+P_3^2}{2m}\]
In particular, because the Hamiltonian \([H,P_2]=[H,P_3]=0\) is translationally invariant in the \(y\) and \(z\)-directions, postulate the Cartesian separation-of-variables ansatz \(|\psi\rangle=|\tilde\psi\rangle\otimes|k_2\rangle\otimes|k_3\rangle\). Then the eigenequation \(H|\psi\rangle=E|\psi\rangle\) for the Hamiltonian \(H\) effectively reduces to (not being too careful about the domains of various operators):
\[\left(\frac{P_1^2}{2m}+\frac{q^2B^2}{2m}\left(X-\frac{\hbar k_2}{qB}\right)^2\right)|\tilde{\psi}\rangle=\left(E-\frac{\hbar^2k_3^2}{2m}\right)|\tilde{\psi}\rangle\]
Thus, in the Landau gauge \(\textbf A_{\text{Landau}}\), a quantum-mechanical charged particle moving through a uniform magnetic field along the \(z\)-axis looks like a one-dimensional quantum harmonic oscillator along the \(x\)-axis oscillating at the classical cyclotron frequency \(\omega_B=qB/m\) with center displaced away from the origin at \((k_2\ell_B^2,0,0)\) with \(\ell_B:=\sqrt{\hbar/qB}\) the characteristic length scale of any quantum harmonic oscillator (if \(q<0\) then just take \(|q|\)), in this context called the magnetic length (e.g. for a proton or electron in \(B=1\text{ T}\), it is \(\ell_B=25\text{ nm}\)). The corresponding ladder of quantum harmonic oscillator energies are the Landau energy levels of this charge, quantized in the usual way:
\[E_{n,k_3}=(n+1/2)\hbar\omega_B+\frac{\hbar^2k_3^2}{2m}\]
with corresponding position space wavefunctions \(\psi_{n,k_2,k_3}(x,y,z)\propto e^{ik_2y+ik_3z}H_n\left(\frac{x-k_2\ell_B^2}{\ell_B}\right)e^{-(x-k_2\ell_B^2)^2/2\ell_B^2}\) (although, unlike the Landau levels, these wavefunctions are not gauge invariant so don’t pay too much attention to them). As this formula makes explicit, unlike in a typical one-dimensional quantum harmonic oscillator where each energy level is non-degenerate, here each Landau level exhibits a mind-boggling amount of degeneracy since the energies \(E_{n,k_3}\) doesn’t depend on the particle’s momentum \(k_2\) in the \(y\)-direction despite the fact that wavefunctions \(\psi_{n,k_2,k_3}\) evidently do. We can understand this graphically by plotting the energy levels \(E_{n,k_3=0}\) as a function of the degenerate momentum \(k_2\):
To actually compute the degeneracy \(\text{dim}\ker(H-E_{n,k_3=0})\) of the \(n\)-th Landau level, it is necessary to discretize \(k_2\) (otherwise all Landau levels would just be infinitely degenerate). To do this, we imagine the particle is part of a physical sample of dimensions \(L_1,L_2\) along the \(x,y\)-axes (assume henceforth \(k_3=0\) so the particle is confined to the \(xy\)-plane). Looking again at the form of the wavefunctions \(\psi_{n,k_2}\), we see that this quantizes the degenerate momentum \(k_2\in\frac{2\pi}{L_2}\textbf Z\). And because the equilibrium position of the one-dimensional quantum harmonic oscillator in the \(x\)-direction lies at \(x=k_2\ell_B^2\) but \(x\in[0,L_1]\), this clearly implies \(k_2\in[0,L_1/\ell_B^2]\). So the degeneracy of each Landau level is:
\[\text{dim}\ker(H-E_{n,k_3=0})=\frac{L_1/\ell_B^2}{2\pi/L_2}=\frac{L_1L_2}{2\pi\ell_B^2}=\frac{\Phi_{\textbf B}}{\Phi_{\textbf B,0}}\]
where the magnetic flux through the square box is \(\Phi_{\textbf B}:=BL_1L_2\) and the quantum of magnetic flux is \(\Phi_{\textbf B,0}:=h/q\); for a proton/electron it is \(\Phi_{\textbf B,0}=\). The fact that the quantum of magnetic flux \(\Phi_{\textbf B,0}\) is so small explains the mind-boggling degeneracy of Landau levels (and underlies such exotic phenomena as the integer/fractional quantum Hall effect).
Symmetric Gauge Instead of Landau Gauge
Earlier we worked with the Landau gauge \(\textbf A_{\text{Landau}}=Bx\hat{\textbf j}\), however there is something a bit unsatisfactory about it, namely that it doesn’t respect the \(SO(2)\) rotational isotropy of the magnetic field \(\textbf B=B\hat{\textbf k}\) in the \(xy\)-plane (nor does it respect translational symmetry in the \(x\)-direction). As a sort of tradeoff, one can instead fix the (rotationally) symmetric gauge \(\textbf A_{\text{symmetric}}:=-\frac{1}{2}\textbf x\times\textbf B=\frac{B}{2}(-y\hat{\textbf i}+x\hat{\textbf j})\) which one should check does give rise to \(\partial_{\textbf x}\times\textbf A_{\text{symmetric}}=B\hat{\textbf k}\). This now respects the \(SO(2)\) rotational isotropy (again, think fluid mechanically) but fails to be translationally symmetric in both the \(x\) and \(y\) directions. Nevertheless, the \(SO(2)\) rotational isotropy now guarantees that \([H,L_3]=0\) angular momentum is conserved, and therefore a well-defined quantum number (in addition to the \(z\)-axis momentum \(k_3\) as before). More precisely, to actually prove that \([H,L_3]=0\) one can check that in the symmetric gauge the Hamiltonian \(H\) can be written:
\[H=\frac{(P_1+qBX_2/2)^2+(P_2-qBX_1/2)^2+P_3^2}{2m}=\frac{|\textbf P|^2}{2m}+\frac{1}{8}m\omega_B^2(X_1^2+X_2^2)-\frac{\omega_B}{2}L_3\]
from which \([H,L_3]=0\) follows from the commutation relations \(\textbf L\times\textbf X=i\hbar\textbf X\) and \(\textbf L\times\textbf P=i\hbar\textbf P\). Now we want to solve \(H|\psi\rangle=E|\psi\rangle\) with \(|\psi\rangle=|\tilde{\psi}\rangle\otimes|m_{\ell}\rangle\otimes|k_3\rangle\). This leads to:
\[H|\psi\rangle=E|\psi\rangle\]
(some Wirtinger derivative calculation here I’m not bothered to fill in at the moment for a way to solve the Schrodinger equation).