Vector spaces - Physics 162

Definition¶

Quantum mechanics involves complex vector spaces, but in time we will also want to consider real vector spaces, and they are also vector spaces you may be more familiar with, so we will start by defining them together. Consider a field $\mathbb{F}$ which for us will always be $\mathbb{R}$ or $\mathbb{C}$ . The basic idea of a field (defined in any decent abstract algebra textbook, or Wikipedia is an honorable source) is that the operations of addition, multiplication, subtraction, and division have the you are used to for real numbers. (Other important examples of fields include the rational numbers and integers modulo $p$ .) A vector space over $\mathbb{F}$ , called “real” and “complex” vector spaces for $F = \mathbb{R},\mathbb{C}$ respectively, is a set $V$ of elements $\ket{v}$ with the following properties:

Vector addition. For all $\ket{v},\ket{w}\in V$ there is a notion of addition $\ket{v} + \ket{w} \in V$ wuth the following properties:

Addition is commutative: $\ket{v} + \ket{w} = \ket{w} + \ket{v}$ .
Addition is associative. If there is a third vector $\ket{y} \in V$ ,

(\ket{v} + \ket{w}) + \ket{y} = \ket{v} + (\ket{w} + \ket{y})

(1)

a zero vector $\ket{0}\in V$ exists such that $\ket{v} + \ket{0} = \ket{v}$ .

Scalar multiplication. For all $a \in \mathbb{F}$ , $\ket{v} \in V$ , there is a notion of scalar multiplication such that $a\ket{v} \in V$ with the following properties:

Multiplication is associative: For all $b \in \mathbb{F}$ , $a(b\ket{v}) = (ab)\ket{v}$ , where $ab$ is the standard mjultiplication of numbers in $\mathbb{F}$ .
$1\ket{v} = \ket{v}$ .
$0 \ket{v} = \ket{0}$ .

Distributive properties

For all $a\in \mathbb{F}$ , $\ket{v},\ket{w} \in V$ ,

a\left(\ket{v} + \ket{w}\right) = a \ket{v} + b \ket{w}

(2)

for all $a,b \in \mathbb{F}$ , $\ket{v} \in V$ ,

(a + b)\ket{v} = a\ket{v} + b \ket{v}

(3)

From these rules we can also deduce the existence of an additive inverse: for every $\ket{v} \in V$ , there exists a vector $\ket{-v} \in V$ such that $\ket{v} + \ket{-v} = \ket{0}$ . This can be seen by construction: set $\ket{-v} = (-1) \ket{v}$ . Then

\ket{v} + \ket{-v} = \ket{v} + (-1) \ket{v} = (1 + (-1) \ket{v} = 0 \ket{v} = \ket{0}

(4)

Note that I have not yet introduced any notion of the length of a vector, of whether two vectors are orthogonal, and so on. As we will see, these require some additional structure.

Examples¶

There are a number of more and less familiar examples.

$\mathbb{C}^n$ , the space of $n$ -component column vectors

\ket{v} = \begin{pmatrix} c_1 \\ c_2 \\ \vdots \\ c_n \end{pmatrix}

(5)

with $c_k \in \mathbb{C}$ . We define vector addition and scalar multiplication in the usual way:

\begin{pmatrix} c_1 \\ c_2 \\ \vdots \\ c_n \end{pmatrix} + \begin{pmatrix} d_1 \\ d_2 \\ \vdots \\ d_n \end{pmatrix} = \begin{pmatrix} c_1+ d_1 \\ c_2+d_2 \\ \vdots \\ c_n+d_n \end{pmatrix}

(6)

a \begin{pmatrix} c_1 \\ c_2 \\ \vdots \\ c_n \end{pmatrix} = \begin{pmatrix} a c_1 \\ a c_2 \\ \vdots \\ a c_n \end{pmatrix}

(7)

for any $a \in \mathbb{C}$ .

Note we can do ths same with $c_k, d_k \in \mathbb{R}$ : then we have a real vector space. Here the zero vector is defined by $c_k = 0$ .

The space of $n\times n$ complex-valued matrices $M_n(\mathbb{C})$ . Addition and scalar multiplication are just matrix addition and scalar miltiplication (for $M \in M_n$ , $aM$ is elementwise multiplication by $a$ .)
Degree- $n$ polynomials over $\mathbb{C}$ :

\ket{a_0,\ldots a_n} = a_0 + a_1 x + a_2 x^2 + \ldots a_n x^n

(8)

with addition and scalar multiplication working in the standard way. Note that this is clearly equivalent to $\mathbb{C}^n$ . Note also that there is no reason for $n$ to be finite — we could work with the space of all polynomials.

Complex functions on an interval: let $x \in [0,1]$ . The set of all functions $\psi(x)$ forms a vector space under the standard addition and scalar multiplication of functions if we choose the right boundary conditions. These boundary conditions yield vector spaces:

Dirichlet $\psi(0) = \psi(1) = 0$ .
Neumann $\psi'(0) = \psi'(1) = 0$
Periodic $\psi(0) = \psi(1)$ (so $\psi$ is a function on a circle).

However, the boundary condition $\psi(0) = a$ , $\psi(1) = b$ for nonzero $a,b \in \mathbb{C}$ is not a vector space under standard addition of functions: the sum of two such functions does not satisfy the required boundary conditions and so is not in $V$ .

Complex square-integrable functions on $\mathbb{R}$ : that is, functions $\psi(x)$ for $x\in \mathbb{R}$ such that

\int_{-\infty}^{\infty} dx |\psi(x)|^2 < \infty

(9)

Subspaces¶

A set $M \subset V$ is a vector subspace if it is a vector space under the same laws for addition and scalar multiplication. A standard example is any plane through the origin, such as $V = \mathbb{C}^3$ ,

M = \left\{ \begin{pmatrix} c_1 \\ c_2 \\ 0 \end{pmatrix} \forall c_i \in \mathbb{C} \right\}

(10)

Similarly, any complex line through the origin, defined as the set of vectors

a \begin{pmatrix} c_1 \\ c_2 \\ \vdots \\ c_n \end{pmatrix}

(11)

for fixed $c_k\in \mathbb{C}^n$ and all $a \in \mathbb{C}^n$ .

A counterexample is any complex line that does not run through the origin, defined as the set of all vectors of the form

a \begin{pmatrix} c_1 \\ c_2 \\ \vdots \\ c_n \end{pmatrix} + \begin{pmatrix} d_1 \\ d_2 \\ \vdots \\ d_n \end{pmatrix}

(12)

with $c_k,d_k$ fixed and the same for all vectors in this space, and $a$ any complex number. It is clear that the sum of two vectors is not in this set, if there is at least one $d_k \neq 0$ .

Linear independence¶

Definition. A set of vectors $\ket{v_1},\ldots,\ket{v_m} \in V$ is linearly independent if

\sum_{i = 1}^m a_k \ket{v_k} = 0 \Leftrightarrow a_k = 0\ \forall\ k = 1,\ldots,n

(13)

Let us give some examples.

$\mathbb{C}^3$ . These vectors are linearly independent:

\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\ ; \ \ \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}\ ; \ \ \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}

(14)

Similarly these are linearly independent:

\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}\ ; \ \ \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}\ ; \ \ \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}

(15)

However, these three are not:

\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}\ ; \ \ \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}\ ; \ \begin{pmatrix} 1/2 \\ 1/2 \\ 1 \end{pmatrix}

(16)

because

\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} + \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} - 2 \begin{pmatrix} 1/2 \\ 1/2 \\ 1 \end{pmatrix} = 0

(17)

In the space of functions on the interval $[0,1]$ satisfying periodic boundary conditions, the vectors

\ket{n} = \sin\left(\frac{n\pi x}{L}\right)

(18)

are linearly independent. Similarly, for $n$ th order polynomials, the monomials $\ket{k} = x^k$ are a linearly independent set of $n$ th order polynomials.

Dimension of a vector space¶

Definition: the dimension of a vector space $V$ is the maximum number of linearly independent vectors in $V$ . Any such maximal collection is called a basis.

Theorem: Given a basis $\ket{k}$ , $k = 1,\ldots,n$ , then for any vector $\ket{v}$ there is a unique set of complex numbers $a_{k = 1,\ldots,n}$ such that

\sum_{k = 1}^n a_k \ket{k}

(19)

Proof. Assume the contrary, that

\ket{v} = \sum_{k = 1}^n a_k \ket{k} = \sum_{k = 1}^n b_k \ket{k}

(20)

for $a_k,b_k \in \mathbb{C}$ different numbers. The point is that if this is true,

\ket{0} = \ket{v} - \ket{v} = \sum_{k = 1}^n a_k \ket{k} - \sum_{k = 1}^n b_k \ket{k} = \sum_{k = 1}^n (a_k + b_k)\ket{k}

(21)

but this cannot be zero of $a_k,b_k$ differ in any way, so we have a contradiction to our supposition. Thus $a_k = b_k$ .