Is $pq+qp$ ever a projection? There is a question over on Mathoverflow to which I added this answer.
Is it an answer? i.e. if $p$ and $q$ are Hilbert space projections, $p=p^2=p^*$ and $q=q^2=q^*$, is $pq+qp$ ever a non-zero Hilbert space projection?
 A: No.
Suppose that $p,q,r:=pq+qp$ are all projections in a vector space,
i.e. $pp=p$, $qq=q$, $(pq+qp)^2 = pq+qp$
(We do not need to require that they are orthogonal or that we are on a hilbert space!).
part 1:
Let $s$, $t$ be arbitrary operators with $sq = tp$.
Then we have
$$
\begin{align}
 sq = tp
 &\implies
 sq = tpp = sqp
 \\ &\implies
 sq = sqq = sqpq
 \\ &\implies
 sq (qp + pq) = sqp + sqpq = 2sq
 \\ &\implies
 2sq = sq (qp+pq) = sq(qp+pq)^2 = 2sq (qp+pq) = 4sq
 \\ &\implies
 sq = 0.
\end{align}
$$
part 2:
By assumption, we have
$(pq+qp)(pq+qp)=pq+qp$,
which is equivalent to
$((pq+qp)p -p)q = (q - (pq+qp)q)p$.
Applying part 1 yields
$((pq+qp)p - p)q=0$.
Then we have
$$
\begin{align}
 pq = (pq+qp) pq = pqpq + qpq
 &\implies
 qpq = pq - pqpq
 \\ &\implies
 qpq = pq - p (pq - pqpq) = pqpq
 \\ &\implies
 pq = (pq+qp) pq = pqpq + qpq = 2qpq
 \\ &\implies
 2qpq = pq = 2q(2qpq ) = 4qpq
 \\ &\implies
 qpq = 0.
\end{align}
$$
By exchanging $p$ and $q$ we also get $pqp =0$.
In summary, we have
$$
pq+qp = (pq + qp)(pq+qp) = pqpq + pqp + qpq + qpqp = 0,
$$
which is what we wanted to show.
(Note: A previous version of this answer was incorrect.
The above uses a large number of elementary
operations and only relies on $rr=r$-type substitution,
which makes me think that there is probably a simpler proof).
