Orthonormal Bases

An orthonormal basis is a basis made of unit vectors that are mutually orthogonal. It is the coordinate system in which inner products, norms, projections, probabilities, and matrix elements take their cleanest finite-dimensional form.

For infinite-dimensional Hilbert spaces, where expansions converge as limits and completeness must be stated carefully, see Completeness and Orthonormal Bases.

Helpful background. bases and coordinates for expanding a vector in a chosen basis; inner products for norms, conjugation, and orthogonality.

Orthogonal unit vectors

Let $\mathcal H$ be a finite-dimensional complex inner-product space. A basis

$$\mathcal B = \{\lvert e_1\rangle,\ldots,\lvert e_n\rangle\}$$

is orthonormal if

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

This equation combines two statements:

$$\lVert e_i\rVert=1$$

for every $i$, and

$$\langle e_i\vert e_j\rangle=0 \qquad \text{when } i\ne j.$$

The word “orthogonal” means mutually perpendicular with respect to the inner product. The word “normal” means unit norm. “Orthonormal” means both.

Expansion Coefficients

If $\mathcal B$ is orthonormal, every vector $\lvert\psi\rangle\in\mathcal H$ has a unique expansion

$$\lvert\psi\rangle = \sum_i c_i\lvert e_i\rangle.$$

Taking the inner product with $\lvert e_j\rangle$ gives

$$\langle e_j\vert\psi\rangle = \sum_i c_i\langle e_j\vert e_i\rangle = \sum_i c_i\delta_{ji} = c_j.$$

Thus the coordinates in an orthonormal basis are

$$c_i = \langle e_i\vert\psi\rangle.$$

This formula is one of the main reasons orthonormal bases are so useful: no inverse Gram matrix is needed.

Completeness Relation

The identity operator can be written

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

This is the finite-dimensional completeness relation. Applying it to a vector gives

$$I\lvert\psi\rangle = \sum_i \lvert e_i\rangle \langle e_i\vert\psi\rangle = \lvert\psi\rangle.$$

Each term $\lvert e_i\rangle\langle e_i\rvert$ is the rank-one projector onto the direction $\lvert e_i\rangle$. The sum of all these mutually orthogonal projectors is the identity.

Norms and Inner Products in Components

If

$$\lvert\psi\rangle = \sum_i c_i\lvert e_i\rangle, \qquad \lvert\phi\rangle = \sum_i d_i\lvert e_i\rangle,$$

then

$$\langle\phi\vert\psi\rangle = \sum_i d_i^*c_i.$$

The norm becomes

$$\lVert\psi\rVert^2 = \sum_i \lvert c_i\rvert^2.$$

For a normalized quantum state,

$$\sum_i \lvert c_i\rvert^2=1.$$

This simple squared-modulus formula assumes an orthonormal basis. In a nonorthonormal basis, the Gram matrix appears.

Projection Onto a Subspace

If $\{\lvert u_1\rangle,\ldots,\lvert u_k\rangle\}$ is an orthonormal basis for a subspace $U\subset\mathcal H$, the orthogonal projector onto $U$ is

$$P_U = \sum_{a=1}^k \lvert u_a\rangle\langle u_a\rvert.$$

For any vector $\lvert\psi\rangle$, the projected vector is

$$P_U\lvert\psi\rangle = \sum_{a=1}^k \lvert u_a\rangle \langle u_a\vert\psi\rangle.$$

The squared norm $\lVert P_U\psi\rVert^2$ is the total weight of $\psi$ in the subspace $U$. This is the linear algebra behind degenerate projective measurements.

Constructing an orthonormal basis

The Gram–Schmidt process turns independent vectors into orthonormal ones by subtracting the components already represented. For two independent vectors $v_1,v_2$, set

$$e_1=\frac{v_1}{\lVert v_1\rVert},\qquad w_2=v_2-e_1\langle e_1|v_2\rangle,\qquad e_2=\frac{w_2}{\lVert w_2\rVert}.$$

The subtraction makes $w_2$ orthogonal to $e_1$; independence ensures $w_2\ne0$. For $v_1=(1,0)^T$ and $v_2=(1,1)^T$, it gives $e_1=(1,0)^T$ and $e_2=(0,1)^T$. Repeat the same subtraction against every previously constructed $e_j$ for longer lists. In floating-point calculations, nearly dependent inputs need more stable orthogonalization; see Conditioning and Stability.

Change Between Orthonormal Bases

Two orthonormal bases of the same finite-dimensional Hilbert space are related by a unitary matrix. If

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

is the overlap matrix between old basis $\{\lvert e_i\rangle\}$ and new basis $\{\lvert f_a\rangle\}$, then

$$\sum_i S_{ai}S_{bi}^* = \delta_{ab}.$$

Thus

$$SS^\dagger=I.$$

The mathematical transformation law is developed in Change of Basis. The physics-facing change-of-basis workflow is developed in Change of Basis, while Unitary Operators explains norm and inner-product preservation.

Physical Meaning of Basis Choice

An orthonormal basis can represent a measurement context. If a state is expanded as

$$\lvert\psi\rangle = \sum_i c_i\lvert e_i\rangle,$$

and the measurement projects onto the basis vectors $\lvert e_i\rangle$, then

$$p(i)=\lvert c_i\rvert^2.$$

But “being a superposition” is basis-relative. A vector can be a single basis vector in one orthonormal basis and a superposition in another. The physical content lies in the state, the measurement, and their inner products, not in a component list alone.

For the Core Formalism interpretation, see Bases and Representations and Probability in Different Bases.

Worked Example

In $\mathbb C^2$, define

$$\lvert +\rangle = \frac{1}{\sqrt2} \begin{pmatrix} 1\\ 1 \end{pmatrix}, \qquad \lvert -\rangle = \frac{1}{\sqrt2} \begin{pmatrix} 1\\ -1 \end{pmatrix}.$$

They are normalized:

$$\langle+\vert+\rangle = 1, \qquad \langle-\vert-\rangle = 1,$$

and orthogonal:

$$\langle+\vert-\rangle = \frac12(1-1) = 0.$$

Thus $\{\lvert+\rangle,\lvert-\rangle\}$ is an orthonormal basis.

For

$$\lvert\psi\rangle = \begin{pmatrix} 1\\ 0 \end{pmatrix},$$

the expansion coefficients in this basis are

$$c_+ = \langle+\vert\psi\rangle = \frac{1}{\sqrt2}, \qquad c_- = \langle-\vert\psi\rangle = \frac{1}{\sqrt2}.$$

Therefore

$$\lvert\psi\rangle = \frac{1}{\sqrt2}\lvert+\rangle + \frac{1}{\sqrt2}\lvert-\rangle.$$

Common Mistakes

• Using $c_i=\langle e_i\vert\psi\rangle$ in a basis that is not orthonormal.
• Forgetting to check unit norms as well as orthogonality.
• Treating an orthonormal basis label as a physical outcome before the observable or measurement is specified.
• Confusing a passive basis change with a physical transformation of the state.
• Assuming that a continuous generalized basis works exactly like a finite orthonormal basis.
• Calling a state a superposition without saying relative to which basis.

References

• S. Axler, Linear Algebra Done Right, 3rd ed., Springer, 2015.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.
• P. R. Halmos, Finite-Dimensional Vector Spaces, 2nd ed., Springer, 1974.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• G. Strang, Linear Algebra and Its Applications, 4th ed., Brooks/Cole, 2006.

Exercises

1. Show that the vectors $(1,i)^T/\sqrt2$ and $(1,-i)^T/\sqrt2$ form an orthonormal basis of $\mathbb C^2$.

Solution

Each vector has squared norm

$$\frac{\lvert1\rvert^2+\lvert i\rvert^2}{2} = 1.$$

Their inner product is

$$\frac12 \begin{pmatrix} 1 & -i \end{pmatrix} \begin{pmatrix} 1\\ -i \end{pmatrix} = \frac12(1-1) = 0.$$

Two orthonormal vectors in $\mathbb C^2$ form an orthonormal basis.

2. Let $\{\lvert e_1\rangle,\lvert e_2\rangle,\lvert e_3\rangle\}$ be orthonormal and let $\lvert\psi\rangle=2\lvert e_1\rangle-i\lvert e_3\rangle$. Find $\lVert\psi\rVert^2$.

Solution

In an orthonormal basis, square the moduli of the coefficients:

$$\lVert\psi\rVert^2 = \lvert2\rvert^2+\lvert-i\rvert^2 = 5.$$

3. If $P=\lvert e_1\rangle\langle e_1\rvert+\lvert e_2\rangle\langle e_2\rvert$ for two orthonormal vectors, what subspace does $P$ project onto?

Solution

It projects onto the span of $\lvert e_1\rangle$ and $\lvert e_2\rangle$. Components orthogonal to both basis vectors are removed.