3  Inner Product Spaces

3.1 Notations for Inner Products

Inner products are variously denoted

\[\langle\cdot|\cdot\rangle;\quad\langle\cdot,\cdot\rangle;\quad\left(\cdot,\cdot\right);\quad\left(\cdot|\cdot\right).\]

I will principally use the first of these in order to connect to Dirac notation familiar from quantum mechanics. I will sometimes use the second.

3.2 Definition of an Inner Product Space

An Inner Product Space \(\mathcal{I}\) is a vector space \(\mathcal{V}\) over the field \(\mathcal{F}=\mathbb{R}\text{ or }\mathbb{C}\) together with a binary function

\[\left\langle \cdot|\cdot\right\rangle :V\times V\rightarrow F\,\,\text{(the \emph{inner product})}\] obeying the following axioms.

\(\forall|a{\rangle},|b{\rangle},|c{\rangle}{\in}V;\,\,\alpha,\beta\in F\):

  • [I1]: \(\langle a|b\rangle=\left(\langle b|a\rangle\right)^{*}\) (conjugate symmetry)

  • [I2]: \(\langle c|\left(\alpha|a\rangle+\beta|b\rangle\right)=\alpha\langle c|a\rangle+\beta\langle c|b\rangle\) (linearity in the second argument)

  • [I3]: \(\langle a|a\rangle\ge0\), & \(\left(\langle a|a\rangle=0\right)\iff\left(|a\rangle=|0\rangle\right)\) (positive definiteness).

Equivalently, an Inner product space is an ordered pair \(\mathcal{I}=\left(\mathcal{V},\left\langle \cdot|\cdot\right\rangle \right)\) obeying these axioms.

NB in the mathematics literature, linearity in the first argument is often used instead. The results are equivalent, but a consistent choice must be made.

3.3 Examples of Inner Product Spaces

  • \(\mathbb{C}\), with \(\langle x|y\rangle=x^{*}y\).

  • complex functions \(f\left(x\right)\), with \(\langle f|g\rangle=\int f^{*}\left(x\right)g\left(x\right)\text{d}x\).

  • \(\mathbb{C}^{N}\) with the Euclidean inner product:

\[|x\rangle=\left(\begin{array}{c} x_{1}\\ x_{2}\\ \vdots\\ x_{N} \end{array}\right),\quad\langle x|y\rangle=\sum^{N}_{i=1}x^{*}_{i}y_{i}.\]

3.4 Properties of Inner Product Spaces

3.4.1 Lemma [\(l\)I1]: \(\langle a|0\rangle=\langle0|a\rangle=0\)

Proof:

\[\text{\textbf{[V4]}}:\quad\exists|0\rangle\in V||a\rangle+|0\rangle=|a\rangle\forall|a\rangle\in V.\]

Choose \(|a\rangle=|0\rangle\). Therefore

\[|0\rangle=|0\rangle+|0\rangle.\]

Create inner products with \(\langle b|\):

\[\langle b|0\rangle=\langle b|\left(|0\rangle+|0\rangle\right)\]

Now use [I2]:

\[\langle b|0\rangle=\langle b|0\rangle+\langle b|0\rangle\]

and subtract \(\langle b|0\rangle\) from both sides:

\[0=\langle b|0\rangle.\]

For the other case, use [I1]:

\[0=\left(\langle0|b\rangle\right)^{*}\]

and take the complex conjugate of both sides:

\[0=\langle0|b\rangle.\]

QED.

3.4.2 Lemma [\(l\)I2]: \(\langle x|x\rangle\in\mathbb{R}\).

Proof: from [I1],

\[\langle x|y\rangle=\left(\langle y|x\rangle\right)^{*}\]

and therefore

\[\langle x|x\rangle=\left(\langle x|x\rangle\right)^{*}\]

QED.