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Commit d3827df7 authored by Maciej Topyla's avatar Maciej Topyla
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latex fix

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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:
!!! 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, $$\vec{a} + \vec{b} = \vec{b} + \vec{a}$$.
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,
$$ \vec{c} = \vec{a} + \vec{b} = (a_1 + b_1, a_2 + b_2, \ldots, a_n + b_n) = (c_1, c_2, \ldots, c_n) \, .$$
$$ \vec{c} = \vec{a} + \vec{b} = (a_1 + b_1, a_2 + b_2, \ldots, a_n + b_n) = (c_1, c_2, \ldots, c_n).$$
!!! info "Scalar multiplication"
I can multiply a vector by a scalar number (either real or complex) to produce another vector, $$\vec{c} = \lambda \vec{a}$$.
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. $(\lambda + \mu)\vec{a} = \lambda \vec{a} +\mu \vec{a}$
$$1. \vspace{5pt} (\lambda \mu) \vec{a} = \lambda (\mu \vec{a}) = \mu (\lambda \vec{a})$$
$$2. \vspace{5pt} \lambda (\vec{a} + \vec{b}) = \lambda \vec{a} + \lambda \vec{b}$$
$$3. \vspace{5pt} (\lambda + \mu)\vec{a} = \lambda \vec{a} +\mu \vec{a}$$
!!! info "Vector product"
In addition to multiplying a vector by a scalar, as mentioned above, one can also multiply two vectors among them.
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