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diff --git a/source/know/concept/landau-quantization/index.md b/source/know/concept/landau-quantization/index.md index 82ea86e..592d266 100644 --- a/source/know/concept/landau-quantization/index.md +++ b/source/know/concept/landau-quantization/index.md @@ -8,23 +8,23 @@ categories: layout: "concept" --- -When a particle with charge $q$ is moving in a homogeneous +When a particle with charge $$q$$ is moving in a homogeneous [magnetic field](/know/concept/magnetic-field/), quantum mechanics decrees that its allowed energies split into degenerate discrete **Landau levels**, a phenomenon known as **Landau quantization**. -Starting from the Hamiltonian $\hat{H}$ for a particle with mass $m$ -in a vector potential $\vec{A}(\hat{Q})$: +Starting from the Hamiltonian $$\hat{H}$$ for a particle with mass $$m$$ +in a vector potential $$\vec{A}(\hat{Q})$$: $$\begin{aligned} \hat{H} &= \frac{1}{2 m} \big( \hat{p} - q \vec{A} \big)^2 \end{aligned}$$ -We choose $\vec{A} = (- \hat{y} B, 0, 0)$, -yielding a magnetic field $\vec{B} = \nabla \times \vec{A}$ -pointing in the $z$-direction with strength $B$. +We choose $$\vec{A} = (- \hat{y} B, 0, 0)$$, +yielding a magnetic field $$\vec{B} = \nabla \times \vec{A}$$ +pointing in the $$z$$-direction with strength $$B$$. The Hamiltonian becomes: $$\begin{aligned} @@ -32,18 +32,18 @@ $$\begin{aligned} &= \frac{\big( \hat{p}_x - q B \hat{y} \big)^2}{2 m} + \frac{\hat{p}_y^2}{2 m} + \frac{\hat{p}_z^2}{2 m} \end{aligned}$$ -The only position operator occurring in $\hat{H}$ is $\hat{y}$, -so $[\hat{H}, \hat{p}_x] = [\hat{H}, \hat{p}_z] = 0$. -Because $\hat{p}_z$ appears in an unmodified kinetic energy term, -and the corresponding $\hat{z}$ does not occur at all, -the particle has completely free motion in the $z$-direction. -Likewise, because $\hat{x}$ does not occur in $\hat{H}$, -we can replace $\hat{p}_x$ by its eigenvalue $\hbar k_x$, -although the motion is not free, due to $q B \hat{y}$. +The only position operator occurring in $$\hat{H}$$ is $$\hat{y}$$, +so $$[\hat{H}, \hat{p}_x] = [\hat{H}, \hat{p}_z] = 0$$. +Because $$\hat{p}_z$$ appears in an unmodified kinetic energy term, +and the corresponding $$\hat{z}$$ does not occur at all, +the particle has completely free motion in the $$z$$-direction. +Likewise, because $$\hat{x}$$ does not occur in $$\hat{H}$$, +we can replace $$\hat{p}_x$$ by its eigenvalue $$\hbar k_x$$, +although the motion is not free, due to $$q B \hat{y}$$. -Based on the absence of $\hat{x}$ and $\hat{z}$, -we make the following ansatz for the wavefunction $\Psi$: -a plane wave in the $x$ and $z$ directions, multiplied by an unknown $\phi(y)$: +Based on the absence of $$\hat{x}$$ and $$\hat{z}$$, +we make the following ansatz for the wavefunction $$\Psi$$: +a plane wave in the $$x$$ and $$z$$ directions, multiplied by an unknown $$\phi(y)$$: $$\begin{aligned} \Psi(x, y, z) @@ -51,15 +51,15 @@ $$\begin{aligned} \end{aligned}$$ Inserting this into the time-independent Schrödinger equation gives, -after dividing out the plane wave exponential $\exp(i k_x x + i k_z z)$: +after dividing out the plane wave exponential $$\exp(i k_x x + i k_z z)$$: $$\begin{aligned} E \phi &= \frac{1}{2 m} \Big( (\hbar k_x - q B y)^2 + \hat{p}_y^2 + \hbar^2 k_z^2 \Big) \phi \end{aligned}$$ -By defining the cyclotron frequency $\omega_c \equiv q B / m$ and rearranging, -we can turn this into a 1D quantum harmonic oscillator in $y$, +By defining the cyclotron frequency $$\omega_c \equiv q B / m$$ and rearranging, +we can turn this into a 1D quantum harmonic oscillator in $$y$$, with a couple of extra terms: $$\begin{aligned} @@ -67,8 +67,8 @@ $$\begin{aligned} &= \bigg( \frac{1}{2} m \omega_c^2 \Big( y - \frac{\hbar k_x}{m \omega_c} \Big)^2 + \frac{\hat{p}_y^2}{2 m} \bigg) \phi \end{aligned}$$ -The potential minimum is shifted by $y_0 = \hbar k_x / (m \omega_c)$, -and a plane wave in $z$ contributes to the energy $E$. +The potential minimum is shifted by $$y_0 = \hbar k_x / (m \omega_c)$$, +and a plane wave in $$z$$ contributes to the energy $$E$$. In any case, the energy levels of this type of system are well-known: $$\begin{aligned} @@ -77,29 +77,29 @@ $$\begin{aligned} } \end{aligned}$$ -And $\Psi_n$ is then as follows, -where $\phi$ is the known quantum harmonic oscillator solution: +And $$\Psi_n$$ is then as follows, +where $$\phi$$ is the known quantum harmonic oscillator solution: $$\begin{aligned} \Psi_n(x, y, z) = \phi_n(y - y_0) \exp(i k_x x + i k_z z) \end{aligned}$$ -Note that this wave function contains $k_x$ (also inside $y_0$), -but $k_x$ is absent from the energy $E_n$. +Note that this wave function contains $$k_x$$ (also inside $$y_0$$), +but $$k_x$$ is absent from the energy $$E_n$$. This implies degeneracy: -assuming periodic boundary conditions $\Psi(x\!+\!L_x) = \Psi(x)$, -then $k_x$ can take values of the form $2 \pi n / L_x$, for $n \in \mathbb{Z}$. +assuming periodic boundary conditions $$\Psi(x\!+\!L_x) = \Psi(x)$$, +then $$k_x$$ can take values of the form $$2 \pi n / L_x$$, for $$n \in \mathbb{Z}$$. -However, $k_x$ also occurs in the definition of $y_0$, so the degeneracy -is finite, since $y_0$ must still lie inside the system, -or, more formally, $y_0 \in [0, L_y]$: +However, $$k_x$$ also occurs in the definition of $$y_0$$, so the degeneracy +is finite, since $$y_0$$ must still lie inside the system, +or, more formally, $$y_0 \in [0, L_y]$$: $$\begin{aligned} 0 \le y_0 = \frac{\hbar k_x}{m \omega_c} = \frac{\hbar 2 \pi n}{q B L_x} \le L_y \end{aligned}$$ -Isolating this for $n$, we find the following upper bound of the degeneracy: +Isolating this for $$n$$, we find the following upper bound of the degeneracy: $$\begin{aligned} \boxed{ @@ -108,12 +108,12 @@ $$\begin{aligned} } \end{aligned}$$ -Where $A \equiv L_x L_y$ is the area of the confinement in the $(x,y)$-plane. -Evidently, the degeneracy of each level increases with larger $B$, -but since $\omega_c = q B / m$, the energy gap between each level increases too. +Where $$A \equiv L_x L_y$$ is the area of the confinement in the $$(x,y)$$-plane. +Evidently, the degeneracy of each level increases with larger $$B$$, +but since $$\omega_c = q B / m$$, the energy gap between each level increases too. In other words: the [density of states](/know/concept/density-of-states/) is a constant with respect to the energy, -but the states get distributed across the $E_n$ differently depending on $B$. +but the states get distributed across the $$E_n$$ differently depending on $$B$$. |