Vector Spaces, Bases, and Linear Combinations

A vector space over a field $\mathbb{F}$ is a set $V$ equipped with addition and scalar multiplication satisfying eight axioms: associativity and commutativity of addition, existence of a zero vector, additive inverses, and distributive and compatibility laws for scalars. For quantum mechanics the field is almost always $\mathbb{C}$.

Examples include $\mathbb{C}^n$, the space of n-component column vectors; the space of complex-valued functions on a set; and the space of polynomials with complex coefficients. Each example has its own notion of vector, but they all obey the same abstract rules.

A subspace $W$ of $V$ is a subset that is itself a vector space under the same operations. The set of solutions to a homogeneous linear equation and the span of any collection of vectors are subspaces. Subspaces are the building blocks for decomposing a vector space into simpler pieces.

The abstract viewpoint frees us from coordinates. A quantum state $|\psi\rangle$ exists as an abstract vector in a Hilbert space; choosing a basis merely gives a concrete list of numbers. This separation between object and representation is one of the great advantages of linear algebra.

A finite set of vectors $\{\mathbf{v}_1, \dots, \mathbf{v}_k\}$ is linearly independent if the only scalars $c_i$ satisfying $c_1\mathbf{v}_1 + \cdots + c_k\mathbf{v}_k = \mathbf{0}$ are $c_1 = \cdots = c_k = 0$. In other words, no vector in the set can be written as a combination of the others.

The span of a set $S$ is the collection of all finite linear combinations of vectors from $S$. It is the smallest subspace containing $S$. If $\operatorname{span}(S) = V$, we say $S$ is a spanning set for $V$. A basis is both independent and spanning.

Linear independence guarantees that the representation of a vector in terms of a basis is unique. Spanning guarantees that every vector has such a representation. Together these two properties make bases powerful tools for computation and for proofs.

Geometrically, two nonzero vectors are independent unless they lie on the same line through the origin. In $\mathbb{C}^3$, three vectors are independent if they do not all lie in the same plane. This intuition carries over to higher dimensions even though visualization fails.

A basis of a vector space $V$ is a linearly independent spanning set. The dimension of $V$ is the number of vectors in any basis; this number is well defined because all bases of a finite-dimensional space have the same cardinality. We write $\dim(V) = n$.

The standard basis of $\mathbb{C}^n$ consists of the vectors $\mathbf{e}_j$ with a 1 in the j-th position and 0 elsewhere. Every vector $\mathbf{v} = (v_1, \dots, v_n)^T$ can be written uniquely as $\sum_j v_j \mathbf{e}_j$. The coordinates $v_j$ are the components of $\mathbf{v}$ in this basis.

An n-dimensional complex vector space is isomorphic to $\mathbb{C}^n$, but the isomorphism depends on the chosen basis. This means that abstract results can be proved once in $\mathbb{C}^n$ and then transferred back to any n-dimensional space, provided the basis is specified.

In quantum mechanics, the dimension of the state space is the number of linearly independent outcomes of a measurement. A qubit lives in a two-dimensional Hilbert space, so every pure state can be described by two complex amplitudes relative to a chosen computational basis.

Once a basis $\{\mathbf{e}_j\}$ is fixed, every vector $\mathbf{v}$ corresponds to a column vector of its coordinates. We write this column as $[\mathbf{v}]_\mathcal{B} = (v_1, \dots, v_n)^T$. Addition and scalar multiplication are performed componentwise.

The same abstract vector has different column representations in different bases. This is not a contradiction: the column vector is a coordinate artifact, while the abstract vector is basis independent. It is essential to keep track of which basis is being used.

Linear maps also become concrete as matrices once a basis is chosen. If $T: V \to W$ and bases for $V$ and $W$ are fixed, the j-th column of the matrix $[T]$ contains the coordinates of $T(\mathbf{e}_j)$ in the basis of $W$. Matrix multiplication then implements composition of linear maps.

In Dirac notation, a ket $|\psi\rangle$ is an abstract vector. Choosing a basis $\{|j\rangle\}$ gives a column vector of amplitudes $\psi_j = \langle j|\psi\rangle$. The abstract equation $|\psi\rangle = \sum_j \psi_j |j\rangle$ becomes the familiar column-vector equation $\psi = (\psi_1, \dots, \psi_n)^T$.

Let $\mathcal{B} = \{\mathbf{e}_j\}$ and $\mathcal{B}^\prime = \{\mathbf{e}_j^\prime\}$ be two bases of the same vector space $V$. The change-of-basis matrix $P$ has columns $[\mathbf{e}_j^\prime]_\mathcal{B}$, i.e. the coordinates of the new basis vectors expressed in the old basis. Then $[\mathbf{v}]_\mathcal{B} = P [\mathbf{v}]_\mathcal{B}^\prime$.

Because $P$ is invertible, the coordinate vectors in the two bases are related by $[\mathbf{v}]_\mathcal{B}^\prime = P^{-1} [\mathbf{v}]_\mathcal{B}$. A linear operator $A$ is represented by $[A]_\mathcal{B}$ in the old basis and by $[A]_\mathcal{B}^\prime = P^{-1} [A]_\mathcal{B} P$ in the new basis.

The matrices $[A]_\mathcal{B}$ and $[A]_\mathcal{B}^\prime$ are called similar. Similarity preserves determinant, trace, eigenvalues, and rank, because these quantities are properties of the abstract operator rather than of any particular coordinate system.

In quantum mechanics, changing basis corresponds to changing which observable is diagonal. The operator itself does not change; only the numbers used to describe it change. A judicious choice of basis can turn an operator into a diagonal matrix, which simplifies spectral analysis enormously.

Consider the vector space $\mathbb{C}^2$ with standard basis $\mathcal{B} = \{|0\rangle, |1\rangle\}$ and a rotated basis $\mathcal{B}^\prime = \{|+\rangle, |-\rangle\}$, where $|\pm\rangle = (|0\rangle \pm |1\rangle)/\sqrt{2}$. Express the state $|\psi\rangle = \alpha |0\rangle + \beta |1\rangle$ in the primed basis.

Step 1 -- Invert the basis relations. We have $|0\rangle = (|+\rangle + |-\rangle)/\sqrt{2}$ and $|1\rangle = (|+\rangle - |-\rangle)/\sqrt{2}$. These equations follow by adding and subtracting the definitions of $|\pm\rangle$.

Step 2 -- Substitute. Then $|\psi\rangle = \alpha (|+\rangle + |-\rangle)/\sqrt{2} + \beta (|+\rangle - |-\rangle)/\sqrt{2} = ((\alpha+\beta)/\sqrt{2}) |+\rangle + ((\alpha-\beta)/\sqrt{2}) |-\rangle$.

Step 3 -- Identify new coordinates. Thus $\psi_+^\prime = (\alpha+\beta)/\sqrt{2}$ and $\psi_-^\prime = (\alpha-\beta)/\sqrt{2}$. For $\alpha=1$, $\beta=0$ the state is $|0\rangle$, which has equal amplitudes $\frac{1}{\sqrt{2}}$ in both $|+\rangle$ and $|-\rangle$.