# Inner Product Space

## Definition

An inner product on a vector space $$\mathcal{V}$$ is an operation, $$\langle\cdot,\cdot\rangle:\mathcal{V}\times\mathcal{V}\to\mathbb{R}$$, on $$\mathcal{V}$$ that assigns, to each pair of vectors $$v,w$$ a real number $$\langle v,w\rangle$$ satsifying the follwoing conditions:

1. $$\langle v,v\rangle\geq 0$$
2. $$\langle v,v\rangle = 0$$ iff $$|v\rangle=0$$
3. $$\langle x,y\rangle=\langle y,x\rangle$$ for all $$x,y\in\mathcal{V}$$
4. $$\langle\alpha x+\beta y,z\rangle = \alpha\langle x,z\rangle + \beta\langle y,z\rangle$$ for all $$x,y,z\in\mathcal{V}$$ and all scalars $$\alpha, \beta\in\mathbb{F}$$.

An inner product space is defined as a vector space with an inner product: ( $$\mathcal{V}, (\cdot,\cdot)$$ ).

### Definition from class

Inner Product often it is useful to introduce a rule that maps pairs of vectors to numbers ( $$\mathbb{C}$$ or $$\mathbb{R}$$ ) with the properties:

• $$(|a\rangle,|b\rangle) = (|b\rangle,|a\rangle)^*$$
• The inner product is linear in the second argument, sesquilinear in the first argument
• The only way to get zero with an inner product of a vector with itself is only if the vector is zero

## Defintion - Orthogonality

$$|a\rangle$$ and $$|b\rangle$$ are orthogonal iff $$(|a\rangle,|b\rangle) = 0$$.

### Definition - Orthonormality

A basis is orthonormal if $$(|i\rangle,|j\rangle)=\delta^i_j$$.

## Definition - 2-Norm

The inner product defines a norm on $$\mathcal{V}$$.

• The 2-norm is given as $$\sqrt{(|a\rangle,|a\rangle)}=\sqrt{\langle a|a\rangle} = |||a\rangle||$$

## Properties of IPS

For an orthonormal basis, $$(|a\rangle,|b\rangle) = (\sum_{i}a^i|i\rangle,\sum_jb^j|j\rangle) = \sum_{ij}a_ib^j(|i\rangle,|j\rangle) = \sum_{ij}a_ib^j\delta^i_j=\sum_i a_ib^i=\sum_i a^{i*}b^i$$.

### Cauchy-Schwarz Inequality

$$|(|v\rangle,|w\rangle)\leq |||v\rangle||\cdot|||w\rangle||$$

### Using Dual Basis

In an ON basis, $$(|j\rangle,|a\rangle)=\langle j|a\rangle= \langle j|\sum_ia^i|i\rangle = a^j$$ gets the components of $$|a\rangle$$.

Similarly for $$A^i_j=\langle i|A|j\rangle = (|i\rangle,A|j\rangle)$$

### Relating the Dual Basis with the Inner Product

Suppose we have an inner product space $$\mathcal{V}$$. For any $$|f\rangle\in\mathcal{V}$$ we can always define the Functional $$\langle f|$$ by $$\langle f|v\rangle (|f\rangle,|v\rangle), \forall|v\rangle\in\mathcal{V}$$.

Thus, the existence of an inner product defines a natural map from the topological dual space and the vector space.

## Classes of Operators

### Aside

For any unitary operator $$U$$ [ $$U^\dagger = U^{-1}$$ ], there exists a Hermitian operator, $$G$$ [ $$G^\dagger = G$$ ], for which $$U=\exp(iG)$$. $$G$$ is called a generator of the Unitary symmetry.

Proof. Let $$G$$ be Hermitian. Then $$\exp(iG)^\dagger = \exp(-iG)$$. So, $$\exp(-iG)\exp(iG)=\exp(iG)\exp(-iG)=\exp(0)=\mathbb{I}$$. Hence $$\exp(iG)$$ is Unitary.

## Orthogonality of Operators

$$Q^TQ = \mathbb{I}$$.

## Separable

### Definition

An inner product space is separable if there exists a sequence of vectors $$\{|v_i\rangle \}$$ such that no (non-trivial) $$|v\rangle\in V$$ is orthogonal to all the $$|v_i\rangle$$. Such a set is called complete’’. An ON basis is a complete set of ON vectors iff $$|v\rangle=\sum_i |e_i\rangle\langle e_i|v\rangle$$ and $$\sum_i |e_i\rangle\langle e_i|=\mathbb{I}$$.

Created: 2024-05-30 Thu 21:16

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