# Nielsen and Chuang give 4 nomenclatures / definitions for projective measurements - are they equal?

What this question is about (the 4 nomenclatures in short):

Nielsen, M.A. and Chuang, I.L. state in their famous book (see http://dx.doi.org/10.1017/CBO9780511976667, p. 87 ff.) that there are 4 alternative definitions (or variants, nomenclatures) of projective measurements.

1)) Usually they are defined using an observable (a hermition operator) on the state space with a spectral decomposition $$M= \sum_m mP_m,$$ where $$m$$ are the eigenvalues and $$P_m$$ the projectors onto the eigenspace of $$M$$ with eigenvalues $$m$$. This can be found on page 87 and 88.

2)) Now, the authors also mention that measurements alternatively (with an alternative nomenclature) are defined by giving a basis $$|m\rangle$$ - which is nothing else but giving $$P_m$$. This can be found on page 88-89.

3)) The third variant apparently is to give a complete set of orthogonal projectors $$P_m$$ that satisfy the completeness relation $$\sum_m P_m=I$$ and $$P_mP_{m'}=\delta_{m,m'}P_m.\tag{1}$$ This can be found on page 88-89.

4)) Moving on they also say that with $$\vec{v}\in\mathbb{R}^3$$ and $$\vec{v}\cdot\vec{\sigma}=v_1\sigma_X+v_2\sigma_Y+v_3\sigma_Z$$ one can define an observable (with eigenvalues $$\pm 1$$), that corresponds to a measurements "along the $$\vec{v}$$ axis" (which is a historic artifact regarding spins). This can be found on page 90.

The question itself:

How do these nomenclatures / definitions connect? Are they the equal? What is your take on the following thoughts of mine?

My thoughts:

Thought 1: Firstly, what's strange about 1)) in comparison to 2)), 3)) and 4)) is that suddenly the eigenvalues don't seem to be important. This is the first difference I notice, although there seems to be a more relevant one...

Thought 2: Namely, that in 2)) one has more demands on the $$P_m$$'s (as stated in 2)) above) - they need to be a) a complete set b) of orthonormal projectors and c) obey (1)- It seems odd that 1)) doesn't demand the completeness relation to be true. Truly, on page 88 Nielsen and Chuang do say that 1)) is a special version of a more general postulate, that indeed demands completeness relation to be true. Why don't they then include this in the definition on page 87? Are all these demands taken care of by saying that in 1)) $$M$$ is hermitian?

Thought 3: Lastly, I see a difference between 3)), 4)) and 1)), 2)). 3)) and 4)) seem to be using a basis of the state space. To see that 4)) uses a basis one just needs to use the usual bloch-sphere representation to understand, that $$\vec{v}$$ gives an axis in the blochsphere, which can be associated with 2 antiparallel arrows - 2 basis-vectors of the state space. In contrast, 1)) and 2)) don't explicitly say that there is a basis involved. It might be included inmplicitly from what already was topic of thought 3: Maybe the 4 demands on $$P_m$$ (mentioned in Thought 2) already determine, that $$P_m$$ looks like $$P_m=|\psi_m\rangle\langle\psi_m|$$ for a basis $$\{|\psi_m\rangle\}$$ of the state space.

Thought 4: In my opinion 2)) and 1)) cannot be equal. If you look at 2 qubits and just measure the first one - using the observable $$\sigma_z\otimes I$$, you surely won't be using a basis of the state space.

These are notsomuch definitions of a projective measurement but ingredients that you need to do calculations with them. A projective measurement is described by a list of eigenvalues and projectors that correspond to those eigenvalues. Given that, and given a state $$|\psi \rangle$$, you can calculate probabilities corresponding to particular eigenvalues.

(1) You can specify a set of projectors and eigenvalues by just giving a hermitian operator. The projectors and eigenvalues specified are the ones you get by diagonalizing that operator.

(2) gives a measurement basis $$|m \rangle$$. Remembering that when there is only one linearly independent eigenvector per eigenvalue, we can just write the projector as $$P=|m \rangle\langle m|$$, we have thereby specified the eigenvalues $$m$$ and the projectors $$|m \rangle \langle m |$$. Usually, in this notation, $$m$$ is the eigenvalue itself. But if it is an index $$m=1,2,3, ...$$ then it is also an index for the eigenvalues, which you could then write $$\lambda_1, \lambda_2, ...$$.

(3) All projectors will automatically follow these relations, and they can also be used to define what a projector is. Usually one also includes the fact that they must be hermitian $$P_m^\dagger = P_m$$. Similar to (2), if $$m$$ is just an index here, and not the value of the eigenvalue itself, then it is also an index for the eigenvalues $$\lambda_m$$. Even if the eigenvalues are not specified, they are part of the measurement scheme in Quantum Mechanics.

(4) This is just a concrete example of (1) with $$M=\vec{v} \cdot \vec{\sigma}$$.

One does not have more demands on the $$P_m$$ in (2), as you can prove that any set of projectors coming from (1) diagonalization of an operator (whether hermitian or not) or (2) an orthonormal basis $$|m \rangle$$ will always be complete and orthonormal.
Thought 3: Lastly, I see a difference between 3)), 4)) and 1)), 2)). 3)) and 4)) seem to be using a basis of the state space. To see that 4)) uses a basis one just needs to use the usual bloch-sphere representation to understand, that v⃗ gives an axis in the blochsphere, which can be associated with 2 antiparallel arrows - 2 basis-vectors of the state space. In contrast, 1)) and 2)) don't explicitly say that there is a basis involved. It might be included inmplicitly from what already was topic of thought 3: Maybe the 4 demands on Pm (mentioned in Thought 2) already determine, that Pm looks like Pm=|ψm⟩⟨ψm| for a basis ${|\psi_m\rangle} of the state space. While (1) and (2) don't specify a basis, this is not essential. As long as we have defined our projectors - whether it is through a basis or some diagonalizable operator - we have the ingredients to calculate probabilities corresponding to measurement outcomes. The axis $$\vec{v}$$ on the Bloch sphere is not relevant to this, it is only the fact that you have a hermitian operator in (4). As for your guess that we might always have $$P_m = |m \rangle \langle m|$$ for some basis $$|m \rangle$$, this is not true in general, because sometimes there is more than one eigenvector corresponding to a particular eigenvalue. In that case, you just add the projectors of equal eigenvalues together: $$P_0 = |m=0 \rangle \langle m=0 |$$ $$...$$ $$P_5 = |m=5,\text{eigenvector 1} \rangle \langle m=5,\text{eigenvector 1}| + |m=5,\text{eigenvector 2} \rangle \langle m=5,\text{eigenvector 2}|$$ So for that reason, you do need to know which eigenvalues each basis vector corresponds to, in order to construct the projectors when given a measurement basis. Or at least you need to know that all eigenvalues are not "degenerate" (multiple $$E\vec{V}$$s). Thought 4: In my opinion 2)) and 1)) cannot be equal. If you look at 2 qubits and just measure the first one - using the observable σz⊗I, you surely won't be using a basis of the state space. Not sure what you mean here. If you are looking at a 2-qubit system, you should also give a basis corresponding to two particles per eigenvector. This might mean that $$m$$ stands for two eigenvalues, one corresponding to each particle. All in all I recommend you review your fundamentals. Take a look at the postulates in chapter 4 of Shankar's Principles of Quantum Mechanics. Here at the beginning he lists the basic assumptions of the theory. That will be your foundation for all things quantum mechanical. • Please verify for me if the following two statements are correct: 2) is equal to 1) only if m is an index. If not, one "forgets" about eigenvalues that are degenerate. 4) is not equal (but "less") than 1) since it doesn't cover every possible$M$. Aug 18, 2022 at 14:58 • Comment #1 reply: Actually (2) is equal to (1) only if m is the eigenvalue itself. Because then you know which eigenvectors correspond to degenerate eigenvalues. If you just list the eigenvectors you wouldn't know. (4) is not a definition of measurement at all, it is just an example of (1) for a particular example of$M$, so yes it is less than 1. Aug 18, 2022 at 15:11 • Comment #2 reply: The reason$M$is taken as hermitian is that for every measurement, there is an operator corresponding to it, which is always hermitian. For example the operators for position, momentum, spin in z direction, all have corresponding operators. So in the context of a real-life measurement, though you could find the probabilities if you are given the projectors/eigenvalues, you have this hermitian operator at your disposal which gives them anyway. Aug 18, 2022 at 15:15 • Reply to "Comment 1 thread": Yes, giving a basis with$m$being the eigenvalue as the label allows for$|m\rangle$to have degenerate eigenvalues, you just have to add another index saying this is eigenvector with eigenvalue$m=1$number 1, or number 2, as I did in my answer (usually you'd find a more compact notation). Even more common than this is to index it by finding a different hermitian operator, which is simultaneously diagonalizable with eigenvalues$m2$, and for which all vectors have a unique combination of$m_1, m_2$. Then you label the eigenvectors as$|m_1, m_2 \rangle$. Aug 18, 2022 at 16:16 • Reply to "Comment #2 thread": If you define a measurement with basis vectors and corresponding eigenvalues, you can construct a unique hermitian operator with this. But, depending on what you're doing you might never construct it because you don't need to, in order to predict probabilities. Aug 18, 2022 at 16:19 Maybe I can give a more concise answer. Your main confusion seems to be whether the spectral representation of an observable in 1) gives a complete, orthogonal set of projectors. Indeed, it does. After all, the entire Hilbert space is supposed to factorize into a direct sum of the eigenspaces of any Hermitian operator. This is exactly the statement that $$\sum_m P_m = 1$$. Furthermore, the eigenspaces of a Hermitian operator are supposed to be orthogonal: the overlap of two eigenvectors of $$M$$ with different eigenvalues always vanishes. This is equivalent to the statement $$P_m P_{m'} = \delta_{m m'} P_{m'}$$. As for your particular confusions: 1. Indeed, the eigenvalues don't actually matter so much for defining the measurement. What is more important is defining the set of orthogonal projectors $$P_m$$. The eigenvalue $$m$$ simply labels which of the outcomes occur. If you are thinking in terms of measuring an observable $$M = \sum_m m P_m$$, then measuring an observable $$M' = \sum_m f(m) P_m$$ for an invertible function $$f(m)$$ gives an identical physical description of the measurement process, and provides you with all the same information. So the precise values of $$m$$ don't matter, so long as they uniquely label the different eigenspaces. 2. Let's consider your particular example $$M = \sigma^z \otimes I$$. I claim that the measurement of this observable is equivalent to the approach in definition 2). Indeed, your spectral decomposition is $$M = +1 \left( \frac{1 + \sigma^z \otimes I}{2} \right) + (-1) \left( \frac{1 - \sigma^z \otimes I}{2} \right) = +1 P_0 + (-1) P_1$$ The two projectors $$P_0$$ and $$P_1$$ respectively project the first qubit onto the states $$\lvert 0 \rangle$$ and $$\lvert 1 \rangle$$, and they clearly satisfy $$P_0 P_1 = 0$$ and $$P_0 + P_1 = 1$$. Note, however, that the projectors are not rank-one! each of them projects onto a two-dimensional space, since the state of the second qubit is unaffected by the measurement. I suspect this is the source of your confusion: you're thinking of the $$P_m$$'s as if they are equal to $$\lvert m \rangle \langle m \rvert$$, when in reality they may be a sum over several such rank-one projectors. • Regarding your comments on my confusion in thought 1: Could it be also considered this way: "Indeed, the eigenvalues don't actually matter so much for defining the measurement. What is more important is defining ..." (and now my adaption) "a basis and a measurement outcome / a eigenvalue to each basis vector as a label for which outcome occurs (which also includes that eigenvalues may be degenerate and the same eigenvalue can be awarded to several basis vectors) Aug 18, 2022 at 15:07 • You should keep in mind that when you obtain a measurement outcome corresponding to a degenerate eigenvalue, you do not project into a definite state in that eigenspace. So as soon as you start talking about particular basis vectors, you're going down the wrong path. You explicitly do NOT want to choose a full basis, only a set of projectors. – Zack Aug 18, 2022 at 15:37 • I understand. But the measurement could still be characterized by a basis, right? One just needs to keep in mind that when having a degenerate eigenvalue with 2 eigenvectors for example, that if the result of the measurement is that eigenvalue, the state after the measurement not necessarily is either of the two corresponding eigenvectors. Aug 18, 2022 at 17:36 • You could specify a basis if you so chose, but you are then overspecifying the measurement. The choice of basis within each eigenspace is immaterial. This may not seem like a big deal if you're imagining measuring projectors with only rank 1 or 2, but if you plan to measure a single qubit out of many, specifying an entire basis of$2^N\$ states instead of your two projectors becomes entirely the wrong way to think about the measurement.