# Show that a set of projectors summing to the identity implies mutually orthogonal projectors

The general setting is the study of positive operator measures in quantum mechanics, instead of the projector operator measures. Going from the PVM to the POVM is just saying that our bunch of operators are not pairwise orthogonal. SO the question is:

In a matrix algebra or the bounded operators on a Hilbert space, how to prove that if a bunch of projectors sums to the identity [or their sum is less than or equal to the identity], they must be mutually orthogonal ?

Can we prove this when we only know that

• a projector is an operator $$p$$ such that $$p = p^* = p^2$$
• a projector is always positive, i.e.,

$$\forall\,\,{ \lvert \psi \rangle} \qquad 0 \leq \langle\psi \lvert P \lvert \psi \rangle.$$

I cannot find a proof that if $$p, q$$ are (positive) projectors then $$pq$$ is again positive.

We know that in general, if two operators commute and are positive, then their product is again positive. But how the constraint by the identity forces them to be orthogonal in our setting ?

My book on quantum mechanics says, $$\sum_j p_j \leq \text{Id} \iff \forall{ i \neq j} \quad p_i \leq \text{Id} - p_j$$ without proof unfortunately.

• – glS Nov 30 '19 at 19:45

Let us assume that there are two projections (the general case works similarly). So assume we have $p_1, p_2$ with $p_1+p_2\leq id$. Let $H_1$ be the space on which $p_1$ projection and $H_1^\perp$ be the orthogonal complement, similarly for $H_2, H_2^\perp$. We want to show that $H_2\subset H_1^\perp$. Let us assume not. So choose $v\in H_2\setminus H_1^\perp$ with $\|v\|_2=1$ where $\|\cdot\|_2$ denotes the norm in the Hilbert space. Now we can create an orthonormal basis for the hilbert space call it $\{v_1, v_2, ...\}$, where $v=v_1$. Then $p_2(v)=v$ and $p_1(v)$=$a_1v_1+a_2v_2+\cdots$. Without loss of generality we can assume that $a>0$, if not we can multiply $v$ by a scalar with absolute value one so that this is true. Now since the $v_i$ are orthogonal we get

$\|p_1(v)+p_2(v)\|^2_2=\|(1+a)v_1\|_2^2+\sum_{i=2}^\infty\|a_2v_2\|_2^2>1$. This contradicts that $p_1+p_2\leq id$.

One can prove the claim without using that the $$p$$ are operators, but just that they are projections in a C$$^*$$-algebra. Also, no equality is needed, only that they are below the identity.

Suppose $$\sum_{k=1}^Np_k\leq Id.$$ Fix indices $$i$$ and $$j$$ with $$i\ne j$$. We have $$\sum_{k\ne i}p_k\leq Id-p_i.$$Then $$0\leq p_ip_jp_i\leq p_i\left(\sum_{k\ne i} p_k\right)p_i\leq p_i(Id-p_i)p_i=0.$$ So $$p_ip_jp_i=0$$. But then $$0=p_ip_jp_i=p_ip_j^2p_i=(p_jp_i)^*p_jp_i,$$ and then $$p_jp_i=0$$.

• is there a typo somewhere? I don't understand how you go from the second to the third equation – glS Aug 27 '19 at 17:26
• No typo. If $a,b$ are selfadjoint with $a\leq b$, then $c^*ac\leq c^*bc$. – Martin Argerami Aug 27 '19 at 17:27
• if you refer to the third equation, then I understand the first inequality because you are using $p_j\le\sum_{k\neq i}p_k$. But then using your second equation I would get $p_i(\sum_{k\neq i}p_k)p_i\le p_i(\mathrm{Id}-p_j)p_i$, while you have an equality instead of $\le$, and $p_i$ instead of $p_j$ – glS Aug 27 '19 at 17:31
• When I wrote "no typo", I somehow thought I would have to eat my words. Yes, there was a typo in the second equation. – Martin Argerami Aug 27 '19 at 17:51
• right, that makes more sense. Still, there should be a $\le$ between third and fourth term in the third equation, right (even though I know in this case it doesn't matter, as the whole thing turns out to be zero). Anyway, can't you prove that step more easily by just writing $0\le p_i p_j p_i\le p_i(\mathrm{Id}-p_i)p_i=0$, using the inequality given in the OP, $p_j\le\mathrm{Id}-p_i$? – glS Aug 27 '19 at 17:58