Bases and Coordinates

A basis is a linearly independent spanning set, and coordinates are the scalars used to write a vector in that basis. The vector is abstract; its coordinate column is a basis-dependent representation.

This distinction is one of the most important pieces of mathematical hygiene in quantum mechanics. A state vector, an operator, or a wavefunction may look different in different representations even when the underlying object has not changed.

Bases

Let $V$ be a vector space over a field $\mathbb F$, such as $\mathbb R$ or $\mathbb C$. A finite list

$$\mathcal B=(e_1,\ldots,e_n)$$

is a basis of $V$ if:

• it spans $V$, meaning every $v\in V$ can be written as a linear combination of the $e_i$;
• it is linearly independent, meaning no basis vector is redundant.

Together these conditions imply that every vector has a unique expansion

$$v = \sum_{i=1}^n c^i e_i.$$

The numbers $c^i$ are the coordinates of $v$ in the basis $\mathcal B$.

The index placement convention used here is summarized in Index Notation and Summation Conventions.

Coordinates Are Not the Vector

Once a basis is chosen, the coordinate column of $v$ is written

$$[v]_{\mathcal B} = \begin{pmatrix} c^1\\ \vdots\\ c^n \end{pmatrix}.$$

The map

$$C_{\mathcal B}:V\to\mathbb F^n, \qquad C_{\mathcal B}(v)=[v]_{\mathcal B},$$

is a linear isomorphism. It depends on the basis. Changing the basis changes $C_{\mathcal B}$ and therefore changes the coordinate column.

The abstract vector $v$ does not change when its coordinates are rewritten.

Standard Coordinates

For $V=\mathbb C^n$, the standard basis is

$$e_1= \begin{pmatrix} 1\\ 0\\ \vdots\\ 0 \end{pmatrix}, \quad e_2= \begin{pmatrix} 0\\ 1\\ \vdots\\ 0 \end{pmatrix}, \quad \ldots, \quad e_n= \begin{pmatrix} 0\\ 0\\ \vdots\\ 1 \end{pmatrix}.$$

With this basis, a vector and its coordinate column look identical. That visual convenience is also the source of a common mistake: in another basis, the same vector can have a different column.

Orthonormal Bases

If $V$ has an inner product, a basis $\mathcal B=(e_1,\ldots,e_n)$ is orthonormal when

$$\langle e_i\vert e_j\rangle = \delta_{ij}.$$

In an orthonormal basis, coordinates are obtained by inner products. The detailed finite-dimensional treatment is Orthonormal Bases.

$$c^i = \langle e_i\vert v\rangle.$$

Thus

$$v = \sum_i e_i\,\langle e_i\vert v\rangle.$$

In Dirac notation this becomes the familiar resolution of the identity,

$$I = \sum_i \lvert e_i\rangle\langle e_i\rvert,$$

but the coordinate idea is already present in ordinary linear algebra.

Change of Coordinates

Let $\mathcal B=(e_1,\ldots,e_n)$ and $\mathcal F=(f_1,\ldots,f_n)$ be two bases of the same vector space. The same vector can be written as

$$v = \sum_i c^i e_i = \sum_a d^a f_a.$$

The columns $[v]_{\mathcal B}$ and $[v]_{\mathcal F}$ are different coordinate descriptions of the same vector. A change-of-coordinates matrix translates between them.

For orthonormal bases in a complex inner-product space, the overlap matrix

$$S_{ai} = \langle f_a\vert e_i\rangle$$

gives

$$d^a = \sum_i S_{ai}c^i.$$

The detailed coordinate transformation law is treated in Change of Basis. The physics-oriented passive convention is treated in Change of Basis. The invariant idea is simpler: coordinates change, the vector does not.

Worked Example

In $\mathbb R^2$, take

$$e_1= \begin{pmatrix} 1\\ 1 \end{pmatrix}, \qquad e_2= \begin{pmatrix} 1\\ -1 \end{pmatrix}.$$

These vectors form a basis. Write

$$v= \begin{pmatrix} 3\\ 1 \end{pmatrix} = a e_1+b e_2.$$

Then

$$a+b=3, \qquad a-b=1.$$

Solving gives $a=2$ and $b=1$, so

$$[v]_{\mathcal B} = \begin{pmatrix} 2\\ 1 \end{pmatrix}.$$

In the standard basis, the coordinate column is $(3,1)^T$. In the basis $\mathcal B=(e_1,e_2)$, the coordinate column is $(2,1)^T$. Both columns describe the same vector.

Why Quantum Mechanics Needs This

Quantum mechanics constantly changes representations. A spin state can be written in the $z$-basis or the $x$-basis. A wavefunction can be written in position or momentum representation. An operator can be written as a matrix in an energy basis or a differential operator in position representation.

The basis is part of the representation. Probabilities and expectation values are invariant only when states, operators, and measurement projectors are translated consistently.

Common Mistakes

• Treating a coordinate column as the vector itself.
• Calling a vector a superposition without naming the basis.
• Forgetting that a coordinate map depends on an ordered basis.
• Changing state coordinates without changing operator matrices consistently.
• Assuming formulas from the standard basis remain visually unchanged in every basis.

Cross-Links

• Vector Spaces and Dual Spaces
• Linear Maps
• Index Notation and Summation Conventions
• Orthonormal Bases
• Change of Basis
• Matrices as Linear Maps
• Inner Products
• Finite-Dimensional Hilbert Spaces
• Bases and Representations
• Change of Basis
• Operator Representations

References

• S. Axler, Linear Algebra Done Right, 3rd ed., Springer, 2015.
• G. Strang, Linear Algebra and Its Applications, 4th ed., Brooks/Cole, 2006.
• P. R. Halmos, Finite-Dimensional Vector Spaces, 2nd ed., Springer, 1974.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.

Exercises

1. In $\mathbb R^2$, let $e_1=(1,1)^T$ and $e_2=(1,-1)^T$. Find the coordinates of $(5,1)^T$ in the basis $(e_1,e_2)$.

Solution

Write

$$\begin{pmatrix} 5\\ 1 \end{pmatrix} = a \begin{pmatrix} 1\\ 1 \end{pmatrix} + b \begin{pmatrix} 1\\ -1 \end{pmatrix}.$$

Then $a+b=5$ and $a-b=1$, so $a=3$ and $b=2$. The coordinate column is

$$\begin{pmatrix} 3\\ 2 \end{pmatrix}.$$

2. Why is an ordered basis needed for coordinates?

Solution

The coordinate column records coefficients in a specific order. If the basis order is swapped, the same vector has the same coefficients attached to the same basis vectors, but they appear in different positions in the column.

3. In an orthonormal basis $\{e_i\}$, what is the coordinate $c^i$ of $v$ along $e_i$?

Solution

The coordinate is

$$c^i=\langle e_i\vert v\rangle.$$

This formula uses orthonormality.