In physics vectors are often represented using ket-vectors so a vector represented like

$$ \vec v=v_x\hat e_x+v_y\hat e_y+v_z\hat e_z $$

in ket notation is given by

$$ \ket v=v_x\ket{e_x}+v_y\ket{e_y}+v_z\ket{e_z} $$

Keep in mind that vector spaces are extremely general. Any object that obeys the following rules can be treated as a vector space and all the results associated with vector spaces will follow as well.

<aside> 💡 Vector Space: a space, $V$, formed by a set of objects, vectors $\ket v$, obeying

• Closed Under Addition: $\ket v,\ket u\in V\to\ket v+\ket u\in V$

Vector addition is defined and stays in $V$

• Closed Under Scalar Multiplication: given a scalar $c\in \Complex$ or someother field and $\ket v\in V$ then $c\ket V\in V$

Scalar multiplication is defined and stays in $V$

• Zero Vector: There exists a zero vector, $\ket 0\in V$, such that $\ket v+\ket 0=\ket v$ and $0\ket v=\ket0$ for any $\ket v\in V$
• For any $\ket v\in V$ there exists a vector $\ket u\in V$ such that $\ket v+\ket u=\ket 0\to\ket v=-\ket u$

Vector addition and scalar multiplication must follow these properties

1. Communative: $\ket v+\ket u=\ket u+\ket v$
2. Associative: $(\ket v+\ket u)+\ket w=\ket v+(\ket u+\ket w)$
3. Identity: $1\cdot\ket v=\ket v$
4. Associative: $a(b\ket v)=(ab)\ket v$
5. Distributive: $(a+b)\ket v=a\ket v+b\ket v$
6. Distributive: $a(\ket v+\ket u)=a\ket v+a\ket u$

These last two rules mean that scalar multiplication has a bilinear form in that it is linear in each argument.

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Currently our vector space is only equipped with vector addition and scalar multiplication. We don’t have a way to multiply two vectors. There are several ways to define products. An algebra is a vector space over a field that has a product that takes two vectors and outputs another. A common operator that satisfies the properties required is the cross product in $\R^2$ and $\R^3$ and the commutator with matrices. However there’s another special product that returns a scalar called the inner product. Vector spaces with an inner product are called inner-product spaces

Inner Product Spaces

<aside> 💡 Inner Product: Can be defined for real and complex fields Takes two input vectors and gives a scalar.

Consider $u,v\in V$, the inner product $(v,u)=c$ where $c$ is a scalar

The inner product is not always communative:

$$ (v,u)=\overline{(u,v)}\\ (v,u)=c~~~(u,v)=c^*

$$

The inner product is a sesquilinear form

1. Linear in second input: $(v,a_1u_1+a_2u_2)=a_1(v,u_1)+a_2(v,u_2)$

2. Antilinear in first input: $(a_1v_1+a_2v_2,u)=a_1^(v_1,u)+a_2^(v_2,u)$

   This is due to the first property and the linearity in the second input

   $$ (a_1v_1+a_2v_2,u)=\overline{(u,a_1v_1+a_2v_2)}=\overline{a_1(u,v_1)}+ \overline{a_2(u,v_2)}\\ =a_1^(v_1,u)+a_2^(v_2,u) $$

The inner product $(v,v)$ gives a real “magnitude” and is positive definite

$$ (v,v)>0~~~(v,v)=c=c^*\to\text{real} $$

$(v,v)=0$ only if $v=0$

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• Parrallelogram Identity: All inner product norms induced by an inner product satisfy this identity

  $$ ||x+y||^2+||x-y||^2=2||x||^2+2||y||^2 $$

Dual Spaces

Using an inner product we can define a “dual” vector to every vector in a vector space $V$.

<aside> 💡 Dual Space: For any vector $v\in V$, the function ${\tilde v(\cdot)}=(v,\cdot)$ takes a vector and returns a scalar. So if we input the vector $u\in V$ we get $\tilde v(u)=(v,u)=\text{scalar}$.

This function forms a new vector space $\tilde V$

• $\tilde v+\tilde u=(v,\cdot)+(u,\cdot)=(v+u,\cdot)=\tilde{(v+u)}\in\tilde V$where we used the antilinearity property
• $c\tilde v=c(v,\cdot)=(c^*v,\cdot)\in \tilde V$
• $\tilde 0=(0,\cdot)$
• $\tilde v+\tilde u=(v,\cdot)+(u,\cdot)=(0,\cdot)\to(v+u,\cdot)=(0,\cdot)\to u=-v$ Since this property holds in $V$ it also holds in $\tilde V$

Every vector in $V$ has a corresponding dual vector in $\tilde V$

We can think of the inner product as taking a vector from $V$ with a dual vector from $\tilde V$ and giving a scalar: $(u,v)=\tilde u(v)$

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In Dirac notation ket vectors $\ket v$ are vectors from our vector space $V$ while bra vectors $\bra v$ are dual vectors from our dual vector space $\tilde V$

$$ u\to\ket u\\ \tilde v=(v,\cdot)\to\bra v $$

A bra (dual vector) acts on a ket (vector) and gives a scalar (inner product)

Basis Vectors

<aside> 💡 A set of vectors $\{\ket {e_1},\ket{e_2},\dots,\ket{e_N}\}$ form a basis of our vector space $V$ if

1. They are linearly independent

   This means no one vector can be written in terms of the other. This means the only solution to the equation:

   $$ \sum_{n=0}^N c_n\ket {e_n}=0 $$

   Is the trivial solution $c_1=c_2=\dots=c_N=0$

2. All vectors can be written as a superposition of them, i.e. they form a complete set

Any vector $\ket v\in V$ can be written as

$$ \ket v = \sum_{n=0}^Nv_n\ket{e_n} $$

where $v_n$ are scalar components for each basis vector and N is the dimension of the space. The dimension is defined as the number of linearly independent vectors required to span the vector space

If there is an inner product, then every vector has a dual. The dual of the basis vectors form a basis in the dual space

$$ \ket {e_n}\to\bra{e_n}\\ \ket v=\sum v_n\ket{e_n}\\ \bra v=\sum \bra{v_ne_n}=\sum v_n^*\bra{e_n} $$

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• Othornormal Bases: A basis is considered orthornormal is the inner product of basis vectors forms a kronecker-delta function meaning all basis vectors are orthogonal and normalized

  $$ \braket{e_n|e_{n'}}=\delta_{nn'} $$