1
$\begingroup$

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?

$\endgroup$

1 Answer 1

0
$\begingroup$

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 $\hat L^2$.)

$\endgroup$
7
  • $\begingroup$ 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
    Jan 2 at 2:12
  • $\begingroup$ @jnez71 Other questions here and here might help regarding how to measure a given observable. $\endgroup$
    – nanoman
    Jan 3 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
    Jan 3 at 4:15
  • $\begingroup$ @jnez71 Unfortunately no up-votes have arrived -- would you consider accepting? $\endgroup$
    – nanoman
    Mar 14 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
    Mar 14 at 17:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.