# Understanding projective measurements as a special case of POVM measurements (“third postulate” in Nielsen and Chuang)

I am working through Nielsen and Chuang's book and am confused about a detail from sections 2.2.3 and 2.2.5.

On page 88 of my copy (section 2.2.5), they write

Projective measurements can be understood as a special case of Postulate 3. Suppose the measurement operators in Postulate 3, in addition to satisfying the completeness relation $$\sum_m M_m^\dagger M_m = I$$ also satisfy the conditions that $$M_m$$ are orthogonal projectors, that is, the $$M_m$$ are Hermitian, and $$M_mM_{m^\prime} = \delta_{m,m^\prime}M_m$$.

It seems to me that they're implying that orthogonal projectors are (1) Hermitian and also (2) satisfy $$M_mM_{m^\prime} = \delta_{m,m^\prime}M_m$$.

My question: My understanding is that a projector is simply an operator which satisfies $$P^2 = P$$, and for projectors to be orthogonal means that the composition of two distinct ones always yields zero, i.e. $$(P_1 \circ P_2)(x) = 0$$ for all $$x$$. But this is all covered by part (2) of the statement alone. So why is (1) necessary?

Edit: Here is a screenshot of their statement of postulate 3  • For completeness, and for those of us who don't have that book, can you also quote Postulate 3? – J. Murray Nov 16 '18 at 18:48
• @J.Murray Done! – Alex Nov 16 '18 at 19:06

You're on the right track. However, if you look closely at what it means for a projector to be orthogonal, then you'll see that an orthogonal projection is necessarily self-adjoint, and vice-versa.

A projector $$P$$ on a vector space $$\mathcal H$$ is an operator which satisfies the following conditions:

1. $$P$$ is idempotent, so $$P\circ P = P$$
2. The entire Hilbert space can be decomposed as a direct sum $$\mathcal H = U\oplus V$$ where $$P(U) = U$$ and $$P(V)=\{0\}$$
3. For all $$u\in U$$, $$P(u) = u$$

For a given $$x\in\mathcal H$$, the decomposition in (2) is given by $$x = u + v$$ where $$u = P(x)$$ and $$v = x-P(x)$$. Note that $$U=Ran(P)$$ and $$V=Ker(P)$$.

When the vector space in question is a Hilbert space, then it's possible to introduce the notion of orthogonality into the picture. We say that a projector $$P$$ is orthogonal if, for any $$u\in Ran(P)$$ and $$v\in Ker(P)$$, we have that $$\langle u,v\rangle = 0$$.

Another way to phrase this is that for any $$x,y\in \mathcal H$$, we have that

$$\langle \underbrace{Px}_{\in U},\underbrace{(I-P)y}_{\in V}\rangle = \langle\underbrace{(I-P)x}_{\in V},\underbrace{Py}_{\in U}\rangle = 0$$

where $$I$$ is the identity operator on $$\mathcal H$$. This immediately implies that

$$\langle Px, y\rangle - \langle Px,Py\rangle = 0$$ and $$\langle x,Py\rangle - \langle Px,Py \rangle= 0$$ which together imply that $$\langle Px, y \rangle = \langle x, Py\rangle$$

We could also have gone the other direction. If we ignore the question of orthogonality and require that $$P$$ is self-adjoint, then

$$\langle Px,(I-P)y\rangle = \langle P\circ P x,(I-P)y \rangle = \langle Px, P\circ(I-P)y\rangle$$ $$= \langle Px, (P-P^2)y\rangle =\langle Px, (P-P)y\rangle = 0$$ The same line of reasoning shows that $$\langle(I-P)x,Py\rangle=0$$, so we see that a projector is orthogonal if and only if it is self-adjoint. Therefore, rather than demanding that our projectors be orthogonal, we can equivalently demand them to be self-adjoint, and the orthogonality follows.

To see where your intuition wasn't quite right, look at what you said.

[...] projectors to be orthogonal means that the composition of two distinct ones always yields zero, i.e. $$(P_1\circ P_2)(x)=0$$ for all $$x$$.

By distinct ones, what I assume you mean is that $$Ran(P_1)\cap Ran(P_2) = \{0\}$$. If $$P_1$$ and $$P_2$$ have that property then certainly $$(P_1 \circ P_2)(x)=0$$, but that's true by definition of what we mean by a projector, and in fact makes no use at all of the notion of orthogonality. It also requires a reference to multiple projectors, which is not good if we're trying to talk about a property of a single projector rather than a family of them.

• I know I'm not supposed to use comments for things like "thanks", but this was a really helpful answer, so I just wanted to say that I appreciate it a lot! – Alex Nov 23 '18 at 2:19

An orthogonal projection is a projection for which the range U and the null space V are orthogonal subspaces.

Thus, orthogonality is a property of a single projection, not of a set of projections, as you state it (some kind of mutual orthogonality) -- so the immediate answer to your question is: "You are using the wrong definition of orthogonal projection".

Immediately afterwards, it is shown that:

A projection is orthogonal if and only if it is self-adjoint.

• I disagree - I think that the text's use of fuzzy language is at fault here. There are two meanings of the qualifier 'orthogonal' in this context, and N&C want both definitions. They just haven't specified it quite clearly enough. (Though the equations do make the requirements completely unambiguous.) – Emilio Pisanty Nov 17 '18 at 8:59
• @EmilioPisanty No. "Orthogonal projector" is a precisely defined mathematical term. You don't get to choose what "orthogonal" might mean here. The mistake might be that it assumes people know that. --- Ironically, "orthogonal projector" is what physicists would usually just call a projector. Non-orthogonal projectors are things like $\sqrt{2}|0\rangle\langle + |$. – Norbert Schuch Nov 17 '18 at 11:39
• Yes, but the qualifier 'orthogonal' as applied to an arbitrary set of (appropriately) bounded operators also refers to the operators being pairwise orthogonal under the Hilbert-Schmidt inner product, $\mathrm{Tr}(A_i^\dagger A_j)=0$. Here N&C want both - they want a set of orthogonal projectors (in the sense you point out) which are also pairwise orthogonal (in the Hilbert-Schmidt sense). I agree that your sense should be the primary reading, but the text is ambiguous enough that OP's confusion is understandable. – Emilio Pisanty Nov 24 '18 at 20:48
• @EmilioPisanty That's the problem with commas, they don't establish a hierarchy. But point is: If you know the term "orthogonal projector" in math, there's only that way to read it. If you don't, you indeed have to guess where the term orthogonal belongs to. (Differently speaking, if you know that "orthogonal projector" is a thing and you would like to express that you are talking about a set of projectors which are orthogonal, you will not phrase it as it is phrased.) – Norbert Schuch Nov 24 '18 at 20:51