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

Question

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.

Answer 1

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 $M=\vec{v} \cdot \vec{\sigma}$.

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Addressing your thoughts in particular:

Thought 1 I have addressed in the answer text itself. Feel free to ask in the comments if you have a question.

Thought 2: Namely, that in 2)) one has more demands on the Pm'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?

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.

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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.

Answer 2

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.