💼 Case: consider a function $f$ to be a vector, then $f(x)$ is a component of the vector and $x$ plays the role of an index.
🗒️ Note: vector spaces of functions are function spaces $f\in \mathcal F$
<aside> <img src="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/3cfbf1b7-25ee-44a3-ab04-828ecbb7a51a/inner_product.png" alt="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/3cfbf1b7-25ee-44a3-ab04-828ecbb7a51a/inner_product.png" width="40px" /> Inner product of function space: $f$ and $g$ are functions defined in the interval $x\in[a, b]$, the inner product is
$$ \lang f|g\rang =\int^b_a\overline{f(x)} g(x)\,\text d \mu(x) $$
where $\text d \mu (x)$ is the measure of a function space.
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💃 Example: in quantum $\text d\mu(x)=\text dx$ and the inner product of two wavefunction $\psi (x), \phi(x)$
$$ \lang \psi|\phi\rang =\int^b_a\overline{\psi(x)} \phi(x)\,\text dx $$
🗒️ Note: $\text d\mu(x)=\text dx$ is most common choice, but ex: Hermitian: $\text d\mu(x)=e^{-x^2}\text d\mu(x)$
Magnitude of a function (Euclidian norm)
$$ ||f||^2=\lang f|f\rang =\int^b_a|f(x)|^2\,\text d \mu (x) \ge 0 $$
🗒️ Notes:
Two functions are orthogonal if $\lang f|g \rang=0$
🗒️ Note: for $\lang f|g\rang ^{+\infin}{-\infin} \ne \pm \infin$ then $\lim{\mu(x)\to\pm\infin}(f,g)=0$
🗒️ Note: for the rest of the notes $\text d\mu(x)=\text dx$
$f\in\mathcal F$ defined in $x\in[a,b]$ can be represented by the basis vectors $\{u_n(x),x\in[a,b]\}^\infin_{n=1}$
$$ \begin{aligned} f(x)&=\sum^\infin_{n=1}f_n u_n(x) \\ \text{where} \quad f_n&=\lang u_n|f\rang =\int^b_a \overline{u_n(x)}f(x)\,\text dx
\end{aligned} $$
🗒️ Note: we can make the vector nature more explicit $|f\rang =\sum^\infin_{n=1}f_n |u_n \rang$
💃 Example: In the case of a wavefunction of a quantum oscillator $u_n =A_nH_n(x)e^{-x^2/2}$, where $H_n(x)$ are Hermite polynomials and $A_n$ is normalisation.
The general wavefunction $\psi=\sum^\infin_{n=1} \alpha_n u_n(x)$ is defined on the basis $u_n(x)$ where $a_n$ is the probability amplitude
💼 Case: consider a set of periodic functions on $[-\pi,\pi]$.
The set of plane waves $u_n(x)=e^{inx}/\sqrt{2\pi}$ can be used as an orthonormal basis:
$$ \lang u_n|u_m\rang = \frac{1}{2\pi} \int^\pi_{-\pi}e^{i(n-m)x}\,\text dx=\left \{ \begin{matrix} 1 & \text{for} & n=m\\ 0 & \text{for} & n\ne m \end{matrix} \right . $$
🗒️ Note: for a set of basis functions to be valid it requires a completeness relation
💃 Example: $f,g\in\mathcal F$.
Basis $\{u_n\}^\infin_{n=1}$ are complete and orthonormal for $\mathcal F$
$$ f(x)=\sum^\infin_{n=1}f_nu_n(x) \qquad g(x)=\sum^\infin_{n=1}g_nu_n(x) $$
with $f_n=\lang u_n |f\rang$ and $g_n=\lang u_n |g_n\rang$
Taking the inner product
$$ \lang f|g\rang =\sum^\infin_{n=1}\overline{f_n}g_n=\sum^\infin_{n=1}\lang f|u_n\rang \lang u_n |g\rang = \lang f|\left ( \sum ^\infin_{n=1}|u_n\rang \lang u_n| \right )|g\rang $$
which implies
$$ \sum ^\infin_{n=1}|u_n\rang \lang u_n|=\hat {\bold 1} $$
🗒️ Note: this is equivalent to
$$ \sum^\infin_{n=1}u_n(x) \overline{u_n(y)} =\delta(x-y) $$
where $\delta(x)$ is Dirac-$\delta$ ( discrete case: $(\hat {\bold 1}){ij}=\delta{ij}$ where $\delta_{ij}$ is Kronecker-$\delta$)
