$P,Q$ are Hermitian operators with only non-negative eigenvalues, then tr($PQ$) = 0 $\implies$ that $PQ = 0$.

Question

So far I can prove:

suppose there is a basis set $\{|i>\}$ that diagonalize $P$ with eigenvalue $p_i$, then tr($PQ$) = $\sum_i<i|PQ|i>$ = $\sum_ip_i<i|Q|i>$.

Since $P, Q$ are Hermitian with non-negative eigenvalues, both $p_i$ and $<i|Q|i>$ is non-negative...

Now there are at least two possibilities to make tr() = $0$.

i) all $p_i$ = $0$, then of course $PQ = 0$

ii) all $<i|Q|i>$ = $0$, then I can't prove $PQ = 0$...

iii) $p_i$, $<i|Q|i>$ alternates to be $0$, $PQ$ = ???

I think they should be a very easy problem... But I am just stuck somehow..

Answer 1

Here is a fairly direct argument.

You have $$ 0=\operatorname{tr}(PQ)=\operatorname{tr}(P^{1/2}P^{1/2}Q)=\operatorname{tr}(P^{1/2}QP^{1/2}). $$ As $Q\geq0$, you have $P^{1/2}QP^{1/2}=P^{1/2}Q(P^{1/2})^*\geq0$. And the trace is faithful, so $P^{1/2}QP^{1/2}=0$. Then $$ 0=P^{1/2}QP^{1/2}=P^{1/2}Q^{1/2}Q^{1/2}P^{1/2}=(Q^{1/2}P^{1/2})^*Q^{1/2}P^{1/2}=0. $$ So $Q^{1/2}P^{1/2}=0$. Now multiply by $Q^{1/2}$ on the left and by $P^{1/2}$ on the right.