The smallest wave packet in the lowest Landau level

Introduction

Exercise 12.5 from Modern Condensed Matter Physics (Girvin and Yang, 2019) asks to construct a Gaussian wave packet in the lowest Landau level in the Landau gauge, such that it is localized as closely as possible around some point $\mbf R\ceq\p{R_x,R_y}$.

Actually, we can prove that the smallest wave packet is a Gaussian wave packet. Here is the derivation.

The problem

First, for readers who are not familiar with the Landau levels, here is a brief introduction. For an electron confined in the $xy$ plane under a magnetic field $\mbf B=B\bhat z$, its Hamiltonian is $$H=\fr1{2m_e}\p{p_x^2+\p{p_y-\fr{eB}cx}^2}$$ under the Landau gauge $\mbf A=Bx\bhat y$. Its eigenstates in the position representation are $$\fc{\psi_{nk}}{x,y}=\e^{\i ky}\fc{H_n}{\fr xl-kl} \e^{-\p{x-kl^2}^2/2l^2}$$ labeled by $n\in\bN$ and $k\in\bR$, where $H_n$ is the Hermite polynomial of degree $n$ and $l\ceq\sqrt{\hbar c/eB}$. States with the same $n$ are degenerate in energy ($E_n=\p{n+1/2}\hbar eB/m_ec$) and make up the $n$th Landau level. The Landau level with $n=0$ is called the lowest Landau level.

The problem, now, is this optimization problem: $$\begin{align*} \min_{a_k}\quad&\mel{\Psi}{x^2+y^2}{\Psi}\\ \st\quad&\braket{\Psi}{\Psi}=1,\\ &\mel{\Psi}{x}{\Psi}=R_x,\\ &\mel{\Psi}{y}{\Psi}=R_y \end{align*}$$ (optimizing $\a{x^2+y^2}$ is equivalent to optimizing $\sgm_x^2+\sgm_y^2$ because $\a x$ and $\a y$ are both fixed), where $\ket\Psi$ is defined as the state whose position representation is $$\fc\Psi{x,y}=\int\d k\,a_k\e^{\i ky}\e^{-\p{x-kl^2}^2/2l^2}.$$

The solution

Consider the moment-generating function $$\begin{align*} \fc M{u,v}&\ceq\mel{\Psi}{\e^{ux+vy}}{\Psi}\\ &=\iint\d x\d y\,\e^{ux+vy} \int\d k\,a_k^*\e^{-\i ky}\e^{-\fr1{2l^2}\p{x-kl^2}^2} \int\d k'\,a_{k'}\e^{\i k'y}\e^{-\fr1{2l^2}\p{x-k'l^2}^2}\\ &=\iint\d k\d k'\,a_k^*a_{k'}\int\d x\,\e^{ ux-\fr1{2l^2}\p{x-kl^2}^2-\fr1{2l^2}\p{x-k'l^2}^2 }\underbrace{\int\d y\,\fc\exp{vy+\i\p{k'-k}y}}_{2\pi\fc\dlt{k'-k-\i v}}\\ &=2\pi\int\d k\,a_k^*a_{k+\i v}\underbrace{\int\d x\,\fc\exp{ ux-\fr1{2l^2}\p{x-kl^2}^2-\fr1{2l^2}\p{x-\p{k+\i v}l^2}^2 }}_{l\sqrt\pi\fc\exp{\fr14l^2\p{4ku+u^2+2\i uv+v^2}}}\\ &=2\pi^{3/2}l\fc\exp{\fr14l^2\p{u^2+2\i uv+v^2}} \int\d k\,a_k^*a_{k+\i v}\e^{kl^2u}\\ &=2\pi^{3/2}l\int\d k\,a_k^*\left( a_k+kl^2a_ku+\i a_k'v+\fr14l^2\p{1+2k^2l^2}a_ku^2 \right.\\&\qquad\qquad\qquad\qquad\left. {}+\fr14\p{l^2a_k-2a_k''}v^2 +\fr\i2l^2\p{a_k+2ka_k'}uv+\cdots \right), \end{align*}$$ where $a_k'\ceq\d a_k/\d k$ and $a_k''\ceq\d^2a_k/\d k^2$. On the other hand, we have $$\fc M{u,v}=\mel{\Psi}{1+ux+uy+\fr12u^2x^2+\fr12v^2y^2+uvxy+\cdots}{\Psi}.$$ Compare the expansion coefficients, and we have $$\begin{align*} \braket{\Psi}{\Psi}&=2\pi^{3/2}l\int\d k\,a_k^*a_k,\\ \mel{\Psi}{x}{\Psi}&=2\pi^{3/2}l^3\int\d k\,a_k^*ka_k,\\ \mel{\Psi}{y}{\Psi}&=2\i\pi^{3/2}l\int\d k\,a_k^*a_k',\\ \mel{\Psi}{x^2}{\Psi}&=\fr12\pi^{3/2}l^3\int\d k\,a_k^*\p{1+2k^2l^2}a_k,\\ \mel{\Psi}{y^2}{\Psi}&=\fr12\pi^{3/2}l\int\d k\,a_k^*\p{l^2-2a_k''}a_k. \end{align*}$$

Define $\fc\vphi k\ceq a_k\sqrt{2\pi^{3/2}l}$. Define fictitious position and momentum operators acting on $\vphi$ as $$\Xi\vphi:k\mapsto k\fc\vphi k,\quad \Pi\vphi:k\mapsto-\i\fc{\vphi'}k.$$ Using the constraints of the original optimization problem and abusing the bra–ket notation on $\vphi$, we have $$\braket{\vphi}{\vphi}=1,\quad\mel\vphi\Xi\vphi=\fr{R_x}{l^2},\quad \mel\vphi\Pi\vphi=-R_y.$$ The objective function then becomes $$\mel{\Psi}{x^2+y^2}{\Psi}=\fr12l^2+\mel{\vphi}{\mcal H}{\vphi},$$ where $\mcal H\ceq \Pi^2/2+l^4\Xi^2/2$ is a fictitious Hamiltonian, which is the Hamiltonian of a harmonic oscillator with mass $1$ and angular frequency $\omg\ceq l^2$.

The optimization problem can now be re-stated in terms of $\ket\vphi$ as $$\begin{align*} \min_{\ket\vphi}\quad&\mel\vphi{\mcal H}{\vphi}\\ \st\quad&\braket\vphi\vphi=1,\quad\mel\vphi\Xi\vphi=R_x/\omg,\quad\mel\vphi\Pi\vphi=-R_y. \end{align*}$$ Physically, this means that we want to find the state of a harmonic oscillator with the given expectation values of position and momentum and the lowest energy. To find it, we can use Hisenberg’s uncertainty principle: $$\begin{align*} \a{\mcal H}&=\fr12\a{\Pi^2}+\fr12\omg^2\a{\Xi^2}\\ &=\fr12\p{\a\Pi^2+\sgm_\Pi^2}+\fr12\omg^2\p{\a{\Xi^2}+\sgm_\Xi^2}\\ &=\fr12\sgm_\Pi^2+\fr12\omg^2\sgm_\Xi^2+\fr12R_y^2+\fr12 R_x^2\\ &\ge\omg\sgm_\Pi\sgm_\Xi+\fr12R^2 \ge\fr12\omg+\fr12R^2. \end{align*}$$ The equality in the first “$\ge$” is achieved when $\sgm_\Pi=\omg\sgm_\Xi$, and that in the second “$\ge$” is achieved when the uncertainty principle is saturated. As we know from quantum mechanics, the coherent state of a harmonic oscillator satisfies both conditions. The wavefunction of this state is $$\fc\vphi k=\p{\fr\omg\pi}^{1/4} \fc\exp{-\fr12\omg\p{k-\fr{R_x}{\omg}}^2-\i R_yk}.$$ Express the final result in terms of $a_k$: $$a_k=\fr1{\sqrt2\pi}\e^{-\i kR_y}\e^{-\fr1{2l^2}\p{R_x-kl^2}^2}.$$ We may work out the integral to get the wave function of the wave packet: $$\fc{\Psi}{x,y}=\fr1{\sqrt{2\pi}l}\fc\exp{-\fr1{4l^2}\p{ \p{x-R_x}^2+\p{y-R_y}^2-2\i\p{x+R_x}\p{y-R_y} }}.$$ This is a Gaussian wave packet centered at $\mbf R$ with covariance matrix $\opc{Diag}{l^2,l^2}$.

Further problems

The optimal wave packet is indeed Gaussian. This makes me curious about whether this is a coincidence or not.

Another thing worth noting is that this result is actually the Dirac delta wave function peaking at $\mbf R$ projected into the lowest Landau level. This was actually my first idea to solve the problem. I was like: well, isn’t the Dirac delta the smallest possible wave packet by all means? If the basis is complete, I can surely combine them into a Dirac delta, and it would be very easy to work out $a_k$ in this case. Then, I was like: nah, merely a single Landau level is not complete, so I cannot do that anyway. I then did not even bother to proceed with this approach and went on to trying other methods. It turns out that this approach is actually correct—at least it gives the same result as the correct approach.