Exercise 2.31

Suppose \(V\) is a vector space with orthonormal basis \(\{|i\rangle_{V}\}\) and \(A:V\to V\) is a \(m\times m\) positive operator. Suppose \(W\) is a vector space with orthonormal basis \(\{|j\rangle_{W}\}\) and \(B:W\to W\) is a \(n\times n\) positive operator. Then the tensor product of \(A\) and \(B\) is an operator \(A\otimes B : V\otimes W\to V\otimes W\), and the basis in vector space \(V\otimes W\) is \(\{|i\rangle_{V}\otimes|j\rangle_{W}\}\). To prove operator \(A\otimes B\) is positive, we need to prove for arbitrary vector \(|\psi\rangle \in V\otimes W\), we have

(1)\[ \langle \psi|A\otimes B |\psi\rangle \geq 0, \text{where }|\psi\rangle = \sum_{i,j} c_{ij} |i\rangle_{V}\otimes|j\rangle_{W} \]

We could compute \(\langle \psi|A\otimes B|\psi\rangle \) directly as

(2)\[\begin{split} \begin{align} \langle \psi|A\otimes B|\psi\rangle &= \left(\sum_{i',j'} c^{*}_{i'j'} \langle i'|_{V}\otimes\langle j'|_{W}\right)A\otimes B \left(\sum_{i,j} c_{ij} |i\rangle_{V}\otimes|j\rangle_{W}\right) \\ &= \sum_{i,i',j,j'} c^{*}_{i'j'}c_{ij} \langle i'|A|i\rangle \langle j'|B|j\rangle \\ &= \sum_{i,j} |c_{ij}|^2 \langle i|A|i\rangle \langle j|B|j\rangle \\ \end{align} \end{split}\]

Note that \(A\) and \(B\) are positive operator, then for basis vector \(|i\rangle_{V}\) and \(|j\rangle_W\), we should always have \(\langle i|A|i\rangle\geq 0\) and \(\langle j|B|j\rangle\geq 0\). Also, since for any coefficient \(c_{ij}\), we have \(|c_{ij}|^2 \geq0\), for eq. (2) we should have

(3)\[ \langle \psi|A\otimes B|\psi\rangle = \sum_{i,j} |c_{ij}|^2 \langle i|A|i\rangle \langle j|B|j\rangle \geq 0 \]

Then we could conclude that, \(\langle \psi|A\otimes B|\psi\rangle \geq 0 \) for arbitrary state \(|\psi\rangle \in V\otimes W\), and the tensor product of two positive operators is positive.