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:
- \(\langle v,v\rangle\geq 0\)
- \(\langle v,v\rangle = 0\) iff \(|v\rangle=0\)
- \(\langle x,y\rangle=\langle y,x\rangle\) for all \(x,y\in\mathcal{V}\)
- \(\langle\alpha x+\beta y,z\rangle = \alpha\langle x,z\rangle + \beta\langle y,z\rangle\) 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 \(\sum_i |e_i\rangle\langle e_i|=\mathbb{I}\).