Suppose we have a quantum mechanical particle prepared in a pure state $\psi$, and an apparatus that can measure the orbital angular momentum of the particle along a specified orthogonal axis ($x$, $y$, or $z$). First we measure its angular momentum along $x$, then along $y$, and then along $z$.

We repeat this experimental procedure for a huge number $n$ of identically prepared particles and record the results. So for each particle $k \in \{1,2,\ldots,n\}$ we have three real values we'll call $L_{x_k}$, $L_{y_k}$, and $L_{z_k}$. Finally, we make a histogram / empirical probability distribution of the squared-magnitude $L_k^2 := L_{x_k}^2 + L_{y_k}^2 + L_{z_k}^2\ \ \forall k$.

My questions are:

  1. Will the histogram of the squared-magnitude depend on the order that the angular momentum components were measured in? E.g. if we had instead first measured along $y$, then $x$, then $z$.
  2. Will the histogram of the squared-magnitude match the Born probability distribution of $\psi$ expressed on the eigenbasis of the operator $\hat{L^2}$? Or does the latter only correspond to an apparatus that can measure the squared-magnitude "directly"? (As opposed to via a deterministic function of the individual axis measurements).
  3. In general, if one has an observable $\hat{Q}$ that can be expressed as a function of other "constituent" observables $\hat{Q} = f(\hat{T}_1, \hat{T}_2, \ldots)$, does the Born distribution of $\psi$ expressed on the eigenbasis of $\hat{Q}$ match the probabilistic pushforward through $f$ of the Born distributions for the individual experiments $\{\hat{T}_1, \hat{T}_2, \ldots\}$?

I would be surprised if the answer to #3 is "yes" (with a corresponding "no" to #1 and "yes" to #2) because it allows handling quantum mechanical predictions with classical probability theory in a way that I never see done: monte carlo the Born distributions of the individual observables $\{\hat{T}_1, \hat{T}_2, \ldots\}$ and pass the results through $f$ deterministically to compute statistics for $\hat{Q}$ (as opposed to ever finding the eigenbasis of $\hat{Q}$). But I don't really know quantum mechanics "in practice" so perhaps this is done all the time?


1 Answer 1


tl;dr 1. Yes 2. No 3. No

During each measurement, the quantum state changes (collapses) to the observed eigenstate -- or in the case of a degenerate eigenvalue, projects onto the observed eigenspace. So your sequential measurements are operating on different states.

In the case of observables $\hat T_i$ that mutually commute, there is a common eigenbasis for all the observables, including $\hat Q$, so measuring the $\hat T_i$ sequentially just performs the collapse in stages (projections), leading to the same distribution for $Q$. It doesn't generally lead to the same state as measuring $\hat Q$ directly, though, because the latter would preserve any superposition within an eigenspace of $\hat Q$ that is collapsed by measuring the $\hat T_i$.

But the observables $\hat L_x,\hat L_y,\hat L_z$ do not commute. Measurements of each operate in different eigenbases. This is what makes quantum mechanics behave differently from classical probability. For example, if the particle starts in a state with $L^2 = 2$, then each measurement of $\hat L_x,\hat L_y,\hat L_z$ can give values $\{-1,0,1\}$. Thus, your "measurement" $L^2_k$ can be as low as $0$ or as high as $3$, whereas the actual $L^2$ is always $2$. (Indeed, the state remains in this eigenspace of $\hat L^2$ during the measurements because $\hat L_x,\hat L_y,\hat L_z$ each individually commute with $\hat L^2$.)

  • $\begingroup$ This makes sense to me, thanks. Quantum experiment design must be very tricky; it seems like you have to "be careful what you look at." A machine that spits out $L^2$ by computing $L_x^2 + L_y^2 + L_z^2$ from three axial measurements (in some order) is fundamentally different than a machine that measures $L^2$ without ever observing the individual components. Interesting. I feel like it'd be easy to not realize that a certain observation was indirectly made by my apparatus, like the information is there but unused as an intermediate signal of some sort. Do you agree? $\endgroup$
    – jnez71
    Commented Jan 2, 2022 at 2:12
  • $\begingroup$ @jnez71 Other questions here and here might help regarding how to measure a given observable. $\endgroup$
    – nanoman
    Commented Jan 3, 2022 at 2:13
  • $\begingroup$ Hm, the answers there weren't super useful in my opinion. Regardless, this answer seems good. Hoping to see a few up-votes before accepting though in case we're both missing something; gotta mitigate my confirmation bias. Thanks! $\endgroup$
    – jnez71
    Commented Jan 3, 2022 at 4:15
  • $\begingroup$ @jnez71 Unfortunately no up-votes have arrived -- would you consider accepting? $\endgroup$
    – nanoman
    Commented Mar 14, 2022 at 9:06
  • $\begingroup$ Not too keen on it; perhaps a bounty is in order.. One question about your answer's last paragraph: if $L^2$ commutes with e.g. $L_x$ (so they share an eigenbasis) and the particle is in an eigenstate of $L^2$ (with e.g. eigenvalue $2$) then isn't it also in an eigenstate of $L_x$ since they share an eigenbasis? I.e. the particle being in an eigenstate of $L^2$, owing to the commutation with each of $L_x$, $L_y$, and $L_z$ individually, seems to imply that the particle is also in an eigenstate of each of $L_x$, $L_y$, and $L_z$, yielding no variance in any of their measurements. $\endgroup$
    – jnez71
    Commented Mar 14, 2022 at 17:59

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