Inner Product in Hilbert Space

Gino Prasad

05/08/2023

https://www.youtube.com/watch?v=3N2vN76E-QA&ab_channel=QuantumSense

Inner product definition

A hilbert space H is defined such that every convergent limit is also in H.

Let $u, v$ be two complex numbers

By properties of the hilbert space, $u \cdot v \in H$

Let $u = a + bi$

Let $v = c + di$

If b, d equal 0, then we will use the euclidean distance.

We want to preserve euclidean distance for real numbers while still being able to provide a valid distance function for complex numbers

Ideally we would want the below equations to be true

$(u, v) = |u| |v| cos(\theta)$

$|u| = |i u| = \sqrt{(u, u)}$

$(u, av + w) = a(u, v) + (u, w)$

if $(u, u) = 1$

$(iu, iu) = 1$

$i(iu, u) = 1$

Therefore$ (iu, u) = -(u, iu)$

Using $(u, v) = (v, u)^*$ fixes this

$i^* = -i$

$i(iu, u) = i (u, iu)^* = -i^2 (u, u)^* = 1$

This condition also makes it so that $(u, u)$ is a real number since $(u, u) = (u, u)^*$

Given orthonormal basis $E$:

Let $u$ be in the span of the basis

$u = \sum_i c_i E_i$

$(E_j, u) = (E_j, \sum_i c_i E_i)$

Since inner product is linear in second term

$(E_j, u) = \sum_i (E_j, c_i E_i)$

$(E_j, u) = \sum_i c_i (E_j, E_i)$

Using Kronecker delta,

$\delta_{ij} = 1$ when $i = j$; 0 otherwise

$(E_j, u) = \sum_i c_i \delta_{ij}$

$(E_j, u) = c_j \delta_{jj} = c_j$

Inner Product of Two Vectors in Orthonormal Basis

$(u, v) = (\sum_i c_i E_i, \sum_j d_j E_j)$

$= \sum_j d_j (\sum_i c_i E_i, E_j)$

$= \sum_j d_j (E_j, \sum_i c_i E_i)^*$

$= \sum_j \sum_i d_j c_i^* (E_j, E_i)^*$

$= \sum_j \sum_i d_j c_i^* (E_i, E_j)$

Using kronecker delta

$= \sum_j \sum_i d_j c_i^* \delta_{ij}$

$= \sum_i c_i^* d_i$

if c_i are real

$= \sum_i c_i d_i$