Measurement interpretation - do individual operators get applied?

Question

The description of measurement operators in Nielsen and Chuang is as follows:

Quantum measurements are described by a collection $\{M_m\}$ of measurement operators. These are operators acting on the state space of the system being measured. The index $m$ refers to the measurement outcomes that may occur in the experiment. If the state of the quantum system is $|\psi \rangle$ immediately before the measurement then the probability that result $m$ occurs is $$p(m) = \langle \psi | M_m^{\dagger} M_m |\psi \rangle,$$ and the state of the system after the measurement is $$\frac{M_m | \psi \rangle }{ \sqrt{\langle \psi | M_m^{\dagger} M_m |\psi \rangle} }.$$

In order to make a measurement, I'm confused as to whether an individual operator $M_m$ is applied, or if the whole set $\{M_m\}$ is somehow applied.

This confusion stems from: Suppose we have $|\psi \rangle = \frac{1}{\sqrt{2}}|0 \rangle + \frac{1}{\sqrt{2}}|1 \rangle$ and measurement operators $\{M_0 = |0 \rangle \langle 0 |,M_1 = |1 \rangle \langle 1 |\}$. If I can choose to apply $M_0$, then I get outcome 0 with probability $p(0) = 1/2$ and the state after measurement is $\frac{M_0 |\psi\rangle}{1/\sqrt{2}} = |0 \rangle.$ So I don't know how to reconcile the fact that there is only a 1/2 chance of getting outcome 0, yet the system is in a pure state $|0\rangle$ afterwards. Does it somehow make sense that $M_0$ gives me outcome 1 yet leaves the system in state $|0\rangle$, or do I need to somehow imagine that the entire set $\{M_0, M_1\}$ gets applied? Or is there some other interpretation?

It's also possible that I am confusing the measurement outcomes $0,1$ with the basis states $|0 \rangle, |1 \rangle$?

Answer 1

Here is how I think about measurements. The state $|\psi\rangle$ is in a superposition of being $|0\rangle$ or $|1\rangle$ simultaneously. If the set of measurement operators $\{M_m\}$ consists of $|0\rangle \langle 0|, |1\rangle \langle 1|$, then you are asking the following question:

Is the system in the $|0\rangle$ state or in the $|1\rangle$ state?

Of course, what you would observe, let it be $|0\rangle$ or $|1\rangle$ is probabilistic. This measurement destroys the superposition and each subsequent measurement by this set of operators would give you a deterministic answer.

Now, to answer your question, it seems you are indeed confusing the outcomes and measurement basis. They are the same things. Getting an 'outcome 0' would mean you have observed $M_0$ in the measurement apparatus. Observing $M_0$ immediately implies that the system is now in $|0\rangle$. The other outcome is similar. It also does not make sense that you would get outcome '1' if you observe $M_0$.

Answer 2

A measurement operation includes all of the operators in the set $\{M_m\}$; we do not have control over which operator gets applied. This makes sense if we consider a number of relevant examples.

1. Suppose we have a Stern-Gerlach device that takes our qubit and sends it on one path (call it path $0$) if it has spin up and on another path (call it path $1$) if it has spin down. Our measurement process in the spin up / spin down basis is equivalent to asking "where is the qubit?" If we measure the qubit to be on path $0$, meaning we obtain result $0$, the measurement update rule tells us that the qubit must be spin up. If we obtain result $1$, the rule tells us that the qubit must be spin down. Here, $0$ and $1$, as well as "spin up" and "spin down," are merely labels; what's important is that we associate the correct operator $M_m$ with the correct path. Who chooses which operator to apply? Quantum mechanics does! The overall state of the qubit dictates the probability with which we will measure each outcome.

2. Suppose we have some harmonic oscillator mode, so that we can express our state in, eg, the photon-number basis as $$|\psi\rangle=\sum_{n=0}^\infty c_n |n\rangle.$$ We know from Born's rule that the probability of measuring the state to have $n$ photons is $|c_n|^2$. How can we describe this measurement? We ask: how many photons does the state have? Our measurement operators are the photon-number projectors $M_m=|m\rangle\langle m|$. If our measurement result is some number of photons $m$, then operator $M_m$ gets applied to the state, which makes the state evolve as $|\psi\rangle\to |m\rangle$. Everything is again consistent: there are lots of possible measurement outcomes, each occurs with some probability dictated by the state, and the measurement update rule tells us what happens to the state given that we find a particular measurement outcome. We cannot simply choose to use only a single operator $M_m$, because that would imply our measurement succeeded in producing state $|m\rangle$ with probability unity, which goes against Born's rule. To choose only a single operator we would need to use some postselection such that we ignore all of the cases in which we do not get our desired measurement result.

3. Maybe the easiest question to understand is "where is the particle?" If we have a state with wavefunction given by $\psi(x)$, we know that there's some probability $|\psi(x)|^2 dx$ of finding the particle to be within the range $dx$. We can write a set of measurement operators such as $M_m=|x\rangle\langle x| dx_m$, where $dx_m$ is some range of positions labeled by $m$: if we measure result $m$, the update rule tells us the particle must be localized within $dx_m$. We cannot simply force the particle to be in one desired location, so the measurement has to take into account all possible location outcomes. This is why the measurement operators have to be normalized; in this case, that would be because $\int |x\rangle\langle x|dx=\mathbb{I}$. The position example is probably the easiest to understand conceptually but has a few mathematical pitfalls with delta functions and position eigenstates, so to be mathematically rigorous it is easier need to consider ranges of positions as we did here.