Exercise 2.10

Suppose \(V\) is an inner product space with orthonormal basis \(|v_i\rangle\), and \(A = |v_j\rangle\langle v_k|\) is an operator in \(V\). From eq. (2.25) in the book, we could conclude that the outer product representation for a linear operator \(A:V\to V\) is given by

(1)\[ A = \sum_{m,n}\langle v_n |A|v_m\rangle |v_n\rangle\langle v_m| \]

where \(|v_m\rangle\) is the \(m-\)th basis for vector space \(V\) and \(\langle v_n|A|v_m\rangle\) is the matrix element \(A_{nm}\) of \(A\). Since \(A = |v_j\rangle\langle v_k|\), from eq. (1) we should have

(2)\[\begin{split} \langle v_j |A|v_k\rangle = A_{nm}=\begin{cases} 1 & n=j, m=k \\ 0 & {\rm otherwise} \end{cases} \end{split}\]

Thus, the matrix representation for operator \(|v_j\rangle\langle v_k|\) is a matrix with only the element in the \(j-\)th row and \(k-\)th column as \(1\), and all other elements are \(0\).