From e40e9cbff323eab1706910765fe26aba33d72d05 Mon Sep 17 00:00:00 2001
From: Maciej Topyla <m.m.topyla@student.tudelft.nl>
Date: Sun, 4 Sep 2022 11:19:42 +0000
Subject: [PATCH] Trying out new admonitions
---
src/3_vector_spaces.md | 29 ++++++++++++++---------------
1 file changed, 14 insertions(+), 15 deletions(-)
diff --git a/src/3_vector_spaces.md b/src/3_vector_spaces.md
index ec31a6a..582c9f7 100644
--- a/src/3_vector_spaces.md
+++ b/src/3_vector_spaces.md
@@ -68,33 +68,32 @@ $$\vec{a}_1 = (1, 0, 0) \, ,\qquad \vec{a}_2 = (0, 1, 0)\, ,\qquad \vec{a}_3 = (
We can consider one example in the two-dimensional real vector space $\mathbb{R}$, namely the $(x,y)$ coordinate plane, shown below.
<figure markdown>
- { width="90%" }
+ 
<figcaption></figcaption>
</figure>
-In this figure, you can see how the same vector $\vec{v}$ can be expressed in two different bases. In the first one (left panel), the Cartesian basis, its components are $\vec{v}=(2,2)$. But in the second basis (right panel), the components are different, being instead $\vec{v}=(2.4 ,0.8)$,
-though the magnitude and direction of the vector itself remain unchanged.
+In this figure, you can see how the same vector $\vec{v}$ can be expressed in two different bases. In the first one (left panel), the Cartesian basis is used and its components are $\vec{v}=(2,2)$. But in the second basis (right panel), the components are different, namely $\vec{v}=(2.4 ,0.8)$, while the magnitude and direction of the vector remain unchanged.
-For many problems, both in mathematics and in physics, the appropiate choice of the vector space basis will significantly facilitate
-its solution.
+For many problems, both in mathematics and in physics, the appropiate choice of the vector space basis may significantly simplify the
+solution proces.
## 3.2. Properties of a vector space
-You might be familiar with the concept that one can perform a number of **operations** between vectors. Some important operations that are relevant in this course are are:
+You might be already familiar with the concept of performing a number of various **operations** between vectors, so in this course, let us review some essential operations that are relevant to start working with quantum mechanics:
-- **Addition**: I can add two vectors to produce a third vector, $\vec{a} + \vec{b}= \vec{c}$.
- As with scalar addition, also vectors satisfy the commutative property, $\vec{a} + \vec{b} = \vec{b} + \vec{a}$.
+!!! info "Addition"
+ I can add two vectors to produce a third vector, $$\vec{a} + \vec{b}= \vec{c}$$.
+ As with scalar addition, also vectors satisfy the commutative property, $$\vec{a} + \vec{b} = \vec{b} + \vec{a}$$.
Vector addition can be carried out in terms of their components,
$$ \vec{c} = \vec{a} + \vec{b} = (a_1 + b_1, a_2 + b_2, \ldots, a_n + b_n) = (c_1, c_2, \ldots, c_n) \, .$$
-- **Scalar multiplication**: I can multiply a vector by a scalar number (either real
-or complex) to produce another vector, $\vec{c} = \lambda \vec{a}$.
-Addition and scalar multiplication of vectors are both *associative* and *distributive*, so the following
-relations hold
+!!! info "Scalar multiplication"
+ I can multiply a vector by a scalar number (either real or complex) to produce another vector, $$\vec{c} = \lambda \vec{a}$$.
+ Addition and scalar multiplication of vectors are both *associative* and *distributive*, so the following relations hold
-1. $(\lambda \mu) \vec{a} = \lambda (\mu \vec{a}) = \mu (\lambda \vec{a})$
-2. $\lambda (\vec{a} + \vec{b}) = \lambda \vec{a} + \lambda \vec{b}$
-3. $(\lambda + \mu)\vec{a} = \lambda \vec{a} +\mu \vec{a}$
+ 1. $(\lambda \mu) \vec{a} = \lambda (\mu \vec{a}) = \mu (\lambda \vec{a})$
+ 2. $\lambda (\vec{a} + \vec{b}) = \lambda \vec{a} + \lambda \vec{b}$
+ 3. $(\lambda + \mu)\vec{a} = \lambda \vec{a} +\mu \vec{a}$
- **Vector product**: in addition to multiplying a vector by a scalar, as mentioned above, one can also multiply two vectors among them.
There are two types of vector productions, one where the end result is a scalar (so just a number) and
--
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