# What is the distribution for a function of different quantum observables?

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?

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 $$\hat L^2$$.)

• 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? Jan 2 at 2:12
• @jnez71 Other questions here and here might help regarding how to measure a given observable. Jan 3 at 2:13
• 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! Jan 3 at 4:15
• @jnez71 Unfortunately no up-votes have arrived -- would you consider accepting? Mar 14 at 9:06
• 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. Mar 14 at 17:59