Inner Products and Norms
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You should be comfortable with vector spaces, linear maps, and the concept of a basis from the previous lecture. Familiarity with complex numbers and their conjugates is essential for the complex case.
Why this matters
Inner products provide the geometric structure that allows us to measure length, angle, and orthogonality in abstract vector spaces.
In physics, the ability to define a "dot product" is not merely a computational convenience; it is the foundation of probability amplitudes in quantum mechanics and energy conservation in classical systems. Without an inner product, a vector space is just a collection of elements with addition and scaling rules. With an inner product, it becomes a geometric space where we can ask meaningful questions about distance, projection, and orthogonality. This lecture establishes the axioms that define this structure and introduces the norm, which quantifies the "size" of a vector.
The Geometric Intuition
Imagine you are working in a space where vectors are not arrows in 3D space, but perhaps wavefunctions or abstract states. To make sense of these objects, we need a way to measure how "close" two states are or how "large" a state is. The inner product generalizes the familiar dot product from high school physics. In \(\mathbb{R}^n\), the dot product \(\mathbf{u} \cdot \mathbf{v}\) gives a scalar that depends on the magnitudes of the vectors and the angle between them. In an abstract vector space \(V\), we define a function \(\langle u, v \rangle\) that behaves similarly. It must be linear in one argument and conjugate-linear in the other (for complex spaces), and it must be positive-definite, meaning \(\langle v, v \rangle\) is always a non-negative real number, zero only if \(v\) is the zero vector. This positive-definiteness is crucial because it allows us to define a length, or norm, via \(\|v\| = \sqrt{\langle v, v \rangle}\). This turns the algebraic structure into a geometric one, enabling concepts like orthogonality (\(\langle u, v \rangle = 0\)) and projection.
Axioms of the Inner Product
Let \(V\) be a vector space over the field \(\mathbb{F}\) (where \(\mathbb{F}\) is either \(\mathbb{R}\) or \(\mathbb{C}\)). An inner product is a map \(\langle \cdot, \cdot \rangle : V \times V \to \mathbb{F}\) satisfying the following axioms for all \(u, v, w \in V\) and \(\alpha \in \mathbb{F}\):
- Conjugate Symmetry: \(\langle u, v \rangle = \overline{\langle v, u \rangle}\). In the real case, this reduces to symmetry: \(\langle u, v \rangle = \langle v, u \rangle\).
- Linearity in the First Argument: \(\langle \alpha u + v, w \rangle = \alpha \langle u, w \rangle + \langle v, w \rangle\). Note that in the complex case, the second argument is conjugate-linear: \(\langle u, \alpha v \rangle = \overline{\alpha} \langle u, v \rangle\).
- Positive Definiteness: \(\langle v, v \rangle \ge 0\), and \(\langle v, v \rangle = 0\) if and only if \(v = 0\).
These axioms ensure that the inner product behaves consistently with our geometric intuition. For example, in \(\mathbb{C}^n\), the standard inner product is \(\langle u, v \rangle = \sum_{i=1}^n u_i \overline{v_i}\). The conjugation on the second argument is vital to ensure that \(\langle v, v \rangle = \sum |v_i|^2\) is real and non-negative.
The Induced Norm
Given an inner product, we define the norm (or length) of a vector \(v\) as:
This definition satisfies the three axioms of a norm: positivity, homogeneity, and the triangle inequality. Positivity and homogeneity follow directly from the inner product axioms. The triangle inequality, \(\|u + v\| \le \|u\| + \|v\|\), is a non-trivial result that relies on the Cauchy–Schwarz inequality. This inequality is the cornerstone of inner product spaces, linking the algebraic structure of the inner product to the geometric property of distance.
The norm allows us to define the distance between two vectors as \(d(u, v) = \|u - v\|\). This metric structure is essential for defining convergence, continuity, and completeness, which are prerequisites for Hilbert spaces in quantum mechanics.
Orthogonality and Orthonormal Bases
Two vectors \(u\) and \(v\) are said to be orthogonal if \(\langle u, v \rangle = 0\). A set of vectors \(\{e_1, e_2, \dots, e_n\}\) is orthonormal if \(\langle e_i, e_j \rangle = \delta_{ij}\), where \(\delta_{ij}\) is the Kronecker delta. Orthonormal bases are particularly useful because they simplify the representation of vectors. If \(\{e_i\}\) is an orthonormal basis, then any vector \(v\) can be written as \(v = \sum_{i=1}^n \langle v, e_i \rangle e_i\) (with the convention above, linear in the first argument; physicists write the same coefficient as \(\langle e_i | v \rangle\)). The coefficients \(\langle v, e_i \rangle\) are the projections of \(v\) onto the basis vectors.
In quantum mechanics, the basis vectors often represent eigenstates of an observable, and the coefficients are probability amplitudes. The norm squared of the coefficient, \(|\langle e_i, v \rangle|^2\), gives the probability of measuring the corresponding eigenvalue. This direct link between inner products and physical probabilities underscores the importance of this mathematical structure.
The derivation
Cauchy–Schwarz and the Triangle Inequality
In any real or complex inner-product space \((V,\langle\cdot,\cdot\rangle)\) with induced norm \(\|x\|=\sqrt{\langle x,x\rangle}\), every pair of vectors \(u,v\in V\) satisfies the Cauchy–Schwarz inequality \(\left|\langle u,v\rangle\right|\le \|u\|\,\|v\|\), with equality if and only if \(u\) and \(v\) are linearly dependent. From this the norm obeys the triangle inequality \(\|u+v\|\le\|u\|+\|v\|\), so \(\|\cdot\|\) is a genuine norm.
A common source of confusion is the placement of the complex conjugate. In physics, the convention is often that the inner product is linear in the second argument and conjugate-linear in the first, whereas in mathematics, it is typically linear in the first and conjugate-linear in the second. Always check the convention used in your text. Mixing conventions can lead to sign errors and incorrect adjoints. For instance, if \(\langle u, v \rangle\) is linear in \(v\), then \(\langle u, \alpha v \rangle = \alpha \langle u, v \rangle\), but if it is linear in \(u\), then \(\langle \alpha u, v \rangle = \alpha \langle u, v \rangle\). Consistency is key. C
Worked example
Problem. Consider the vector space \(\mathbb{C}^2\) with the standard inner product \(\langle u, v \rangle = u_1 \overline{v_1} + u_2 \overline{v_2}\). Let \(u = (1, i)\) and \(v = (1, -i)\). Calculate \(\langle u, v \rangle\), \(\|u\|\), and \(\|v\|\). Verify the Cauchy–Schwarz inequality.
- Calculate the inner product: \[ \langle u, v \rangle = 1 \cdot \overline{1} + i \cdot \overline{-i} = 1 \cdot 1 + i \cdot i = 1 - 1 = 0 \] Thus, \(u\) and \(v\) are orthogonal.
- Calculate the norm of \(u\): \[ \|u\|^2 = \langle u, u \rangle = 1 \cdot \overline{1} + i \cdot \overline{i} = 1 \cdot 1 + i \cdot (-i) = 1 + 1 = 2 \] \[ \|u\| = \sqrt{2} \]
- Calculate the norm of \(v\): \[ \|v\|^2 = \langle v, v \rangle = 1 \cdot \overline{1} + (-i) \cdot \overline{-i} = 1 \cdot 1 + (-i) \cdot i = 1 + 1 = 2 \] \[ \|v\| = \sqrt{2} \]
- Verify Cauchy–Schwarz: \[ |\langle u, v \rangle| = |0| = 0 \] \[ \|u\| \|v\| = \sqrt{2} \cdot \sqrt{2} = 2 \] Since \(0 \le 2\), the inequality holds. Equality does not hold because \(u\) and \(v\) are not linearly dependent (they are orthogonal and non-zero).
Units check. The inner product is dimensionless in this abstract context, but if \(u\) and \(v\) represented physical quantities with units, the inner product would carry the product of their units. The norm would carry the units of the vector components. Here, we work in dimensionless abstract space, so the numerical values are consistent.
Three quick tests
- Is the function \(\langle u, v \rangle = u_1 v_1 + u_2 v_2\) a valid inner product on \(\mathbb{C}^2\)?
Answer
No. It fails conjugate symmetry. For \(u=(1,i)\), \(v=(1,1)\), \(\langle u, v \rangle = 1+i\), but \(\overline{\langle v, u \rangle} = \overline{1+i} = 1-i\). They are not equal. - If \(\langle u, v \rangle = 0\), must \(u\) or \(v\) be the zero vector?
Answer
No. They are orthogonal. For example, in \(\mathbb{R}^2\), \((1,0)\) and \((0,1)\) are orthogonal but non-zero. - Why is positive definiteness required for an inner product?
Answer
To ensure that \(\langle v, v \rangle\) is a non-negative real number, allowing the definition of a real-valued norm \(\|v\|\). Without it, the square root might not be real, and the geometric interpretation of length would fail.
Problem set for PU-104 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.