From 0a2b52f618c8a76a2f2a0e29223b5289c0225634 Mon Sep 17 00:00:00 2001
From: Jacky Zhao <j.zhao2k19@gmail.com>
Date: Sun, 24 Mar 2024 22:50:38 +0000
Subject: [PATCH] simpler katex fix
---
docs/features/Latex.md | 11 +++++++++++
1 files changed, 11 insertions(+), 0 deletions(-)
diff --git a/docs/features/Latex.md b/docs/features/Latex.md
index b2bdb2d..fdc9d27 100644
--- a/docs/features/Latex.md
+++ b/docs/features/Latex.md
@@ -39,6 +39,17 @@
\end{bmatrix}
$$
+$$
+\begin{array}{rll}
+E \psi &= H\psi & \text{Expanding the Hamiltonian Operator} \\
+&= -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} \psi + \frac{1}{2}m\omega x^2 \psi & \text{Using the ansatz $\psi(x) = e^{-kx^2}f(x)$, hoping to cancel the $x^2$ term} \\
+&= -\frac{\hbar^2}{2m} [4k^2x^2f(x)+2(-2kx)f'(x) + f''(x)]e^{-kx^2} + \frac{1}{2}m\omega x^2 f(x)e^{-kx^2} &\text{Removing the $e^{-kx^2}$ term from both sides} \\
+& \Downarrow \\
+Ef(x) &= -\frac{\hbar^2}{2m} [4k^2x^2f(x)-4kxf'(x) + f''(x)] + \frac{1}{2}m\omega x^2 f(x) & \text{Choosing $k=\frac{im}{2}\sqrt{\frac{\omega}{\hbar}}$ to cancel the $x^2$ term, via $-\frac{\hbar^2}{2m}4k^2=\frac{1}{2}m \omega$} \\
+&= -\frac{\hbar^2}{2m} [-4kxf'(x) + f''(x)] \\
+\end{array}
+$$
+
> [!warn]
> Due to limitations in the [underlying parsing library](https://github.com/remarkjs/remark-math), block math in Quartz requires the `$$` delimiters to be on newlines like above.
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