Dual Space
Definition
The Dual Space for a vector space \(V\) is the set of linear functions \(L: V\to \mathbb{F}\).
Theorem
In finite dimensions, vector spaces of equal dimensions are isomorphic. (I.e. the vector space and its dual space are isomorphic)
Theorem
In infinite dimensions, this is not quite true. In infinite dimensions, we can distinguish linear maps by whether they are bounded or unbounded. So we have two types of dual spaces:
- \(\mathcal{V}^*\) is the set of all bounded linear maps on \(\mathcal{V}\). (Topological Dual)
- \(\mathcal{V}^\times\) is the set of all linear maps on \(\mathcal{V}\). (Algebraic Dual) (Dirac Delta Function, position ket, momentum ket, etc. are in this set)
Thus, in infinite dimensions a vector space and its topological dual space is isomorphic, but a vector space \(\subset\) algebraic dual space.
Definition of Dual Basis from Vector Basis
Let \(\{|i\rangle\}\) be a basis for \(\mathcal{V}\), then define the dual basis \(\{\langle i|\}\) for \(\mathcal{V}^*\) by \(\langle i|j \rangle = \delta_j^i\) (extended by linearity).
Proof. Homework #1 (Prove Span and Linearity of Dual Basis).
Using the Dual Basis
Given a dual basis, \(\langle i|j\rangle = \delta^i_j\),
- \(|a\rangle = \sum_k a^k |k\rangle\). Let \(\langle i|\in\mathcal{V}^*\) be a basis vector. Then \(\langle i|a\rangle = \langle i|a^k|k\rangle = a^i\). (N.B. Einstein Summation Notation, note that \(i,a\) are fixed thus \(k\) is a dummy index)
- \(A|j\rangle = A^i_j|i\rangle\). Thus, \(A^i_j=\langle i|A|j\rangle\).
- \(|a\rangle = \sum_k a^k |k\rangle\) thus \(\langle a| = \sum_k \langle k|a_k = \langle k|a_k\)
- Since we can write \(\langle f|f_i\langle i|\), \(f_i\equiv \langle f|i\rangle\), \(\langle f|a\rangle = f_i\rangle i|a\rangle = f_ia^i=\langle f|a\rangle\).
Comparing the Algebraic and Topological Duals
\(\mathcal{V}^* \subset \mathcal{V}^X\)