# If two commuting observables do not share eigenspaces, can't one measurement affect the other due to an extra wave function collapse?

When we learn quantum mechanics, we are told that the only way to extract information from a system is to conduct measurements. We are told that if two observables commute then performing one measurement does not affect the outcome of the other.

This is how we are told measurements work: if $a$ is an eigenvalue of $\hat A$, then if we perform the measurement corresponding to $\hat A$, the probability of measuring $a$ is $\langle \psi| \hat P_a | \psi \rangle$ where $\hat P_a$ is the projection operator onto the $a$-eigenspace. $| \psi \rangle$ then collapses into the state $\frac{\hat P_a | \psi \rangle}{\sqrt{\langle \psi | \hat P_a |\psi \rangle}}$.

What if, however, we have two operators that commute but do not share eigenspaces? Take, for example, the two operators

$\hat A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{pmatrix}, \hat B = \begin{pmatrix} 3 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{pmatrix}.$

Certainly, $[\hat A, \hat B] = 0$. However, these operators do not share eigenspaces for any eigenvalues. Let's say we conduct the measurement $\hat B$ right before we conduct the measurement $\hat A$. When we measure $\hat B$, we will measure an eigenvalue $b$ with probability $\langle \psi| \hat P_b | \psi \rangle$ and the state will collapse into $\frac{\hat P_b | \psi \rangle}{\sqrt{\langle \psi | \hat P_b |\psi \rangle}}$. The probability to measure $a$ after this is just $\frac{\langle \psi | \hat P_a \hat P_b | \psi \rangle }{\langle \psi | \hat P_b | \psi \rangle }$. Afterwards, the state will collapse into $\frac{ \hat P_a \hat P_b | \psi \rangle}{\sqrt{\langle \psi | \hat P_a \hat P_b |\psi \rangle}}$.

Now, the probability of measuring some eigenvalue $a$ does not change depending on whether or not we measured $\hat B$ beforehand. This can be easily checked, as the probability of measuring an eigenvalue $a$ is just, summing over all possible $b$ measurements,

$\sum_b \langle \psi | \hat P_b| \psi\rangle \frac{\langle \psi | \hat P_a \hat P_b | \psi \rangle }{\langle \psi | \hat P_b | \psi \rangle } = \sum_b \langle \psi | \hat P_a \hat P_b | \psi \rangle = \langle \psi | \hat P_a | \psi \rangle$

which is the same as before. Indeed, measuring $\hat B$ before $\hat A$ does not affect the probability of measuring any particular $a$, which is what we were told in class.

HOWEVER, the collapsed state after the $\hat A$ measurement is proportional to $\hat P_a \hat P_b |\psi \rangle$, NOT proportional to $\hat P_a | \psi \rangle$. If $\hat A$ and $\hat B$ share no eigenspaces, then this will always be a different state then if we had conducted no $\hat B$ measurement at all. In other words, the fact that we conducted the $\hat B$ measurement lives on in the fact that our state has collapsed further than it would have otherwise. How do we know that we won't be able to detect this? For example, maybe we could wait some time, let the system evolve in some way, and then pick up some interference effect. How does this extra collapse of the wave function affect causality in QFT with regards to commuting space-like separated local operators? And why is this never talked about in introductory QM books?

Of course they share an eigenspace. Indeed the vectors $$\vert 1\rangle=\left(\begin{array}{c} 1 \\ 0 \\0 \end{array}\right) \qquad \vert 2\rangle=\left(\begin{array}{c} 0 \\ 1 \\0 \end{array}\right) \qquad \vert 3\rangle=\left(\begin{array}{c} 0\\ 0 \\1 \end{array}\right)$$ are each common eigenvectors of both operators. The fact they do not have the same eigenvalues does not affect the fact they are common eigenstates.