Determine the matrix representation for an operator written as an outer product

Question

Suppose $|v_{i} \rangle$ is an orthonormal basis for an inner product space $V$. What is the matrix representation for the operator $|v_{j}\rangle \langle v_{k}|$, with respect to the $|v_{i}\rangle$ basis?

Any hint with this is appreciated.

Answer 1

We have to take $|v_j\rangle\langle v_k|$, apply it to a basis element, write it again as a combination of the basis, and put the coefficients in columns. That is to say, the matrix $(a_{i\ell})$ we're looking for satisfies $|v_j\rangle\langle v_k| (|v_\ell\rangle) = \sum_i a_{i\ell} |v_i\rangle$. We have $$|v_j\rangle\langle v_k|(|v_\ell\rangle) = \langle v_\ell|v_k\rangle\langle v_k| = \delta_{\ell k}\langle v_k| = \langle v_\ell| = \sum_i\delta_{i\ell}| v_i\rangle,$$so the matrix is the identity. This is why in general we:

• have $(f\otimes v)(w) = f(w)v$;
• identify $V^*\cong V$ using the inner product.
• may see this operator as $|v_j\rangle\otimes \langle v_k|$.

Answer 2

i wasnt able to write a comment under the comment by @Joppy which i believe is right.

the i-th column of the matrix A is the operator applied to $|v_i\rangle$, so it is $|v_j\rangle\langle v_k|v_i\rangle=\langle v_k|v_i\rangle|v_j\rangle=\delta_{ik}|v_j\rangle$

so everywhere else is $0$ except for the k-th column, and that column is equal to $|v_j\rangle$, which has a $1$ in the j-th entry (j-th entry of the column vector, so j-th row) and $0$ everywhere else. so it is $1$ in $(j, k)$ entry and $0$ everywhere else.