Suppose our plan is to measure experimentally the position $(x,y)$ in the plane and the momentum $(p_{x}, p_{y})$ of a quantum particle. Assuming the canonical commutation relation between $x$ and $p_{x}$, we will bypass so to speak the law by performing the following sequence of measurements: $$ x \rightarrow p_{y}\rightarrow p_{x} \rightarrow y$$ Since, for each successive measurement the commutators vanish i.e $[x,p_y]=[p_y,p_x]=[p_x, y]=0$, we can make the measurement of the above mentioned quantities without affecting the system. Is this a viable way to measure the position and momentum simultaneously?


Even if measurement of $x$ puts your system in a state with definite $x$, as soon as you measure $p_x$ this will no longer be the case. The same is true about $p_y$ and $y$. So no, you still cannot use this to produce a state with definite $x$ and $p_x$.

Of course there exist states with definite $(x,p_y)$, definite $(p_y,p_x)$, or definite $(p_x,y)$, but that doesn't help us here. There do not exist any states in QM where both $\Delta x$ and $\Delta p$ are both $0$, as this would violate the HUP. There is no way around this.

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  • $\begingroup$ but the operators are commuting so the operators in each step are compatible. $\endgroup$ – user11937 Jun 29 at 20:33
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    $\begingroup$ @user11937 That doesn't matter. The fact that $[x,p_x]\neq0$ means that there exist no states with definite $x$ and $p_x$, no matter how you do the measurements. $\endgroup$ – BioPhysicist Jun 29 at 20:41

I think you might misunderstand the difference between the commutation (or not) of observables and measurement of those observables.

If two observables do not commute, it means that there is no state that is an eigenstate of both observables. For example, there is no state vector $|x,p_x\rangle$ such that

$$X|x,p_x\rangle = x|x,p_x\rangle$$


$$P_x|x,p_x\rangle = p_x|x,p_x\rangle$$

Thus, the problem isn't a measurement problem, the problem is that such a state (with definite $x$ position and definite $x$ momentum) does not exist.

Now, a postulate of QM is that, just after a measurement of an observable, the state of the particle (system) is an eigenstate of that observable.

In your example, you begin with a state vector $|x,p_y\rangle$ that satisfies

$$X|x,p_y\rangle = x|x,p_y\rangle$$


$$P_y|x,p_y\rangle = p_y|x,p_y\rangle$$

But the state represented by this state vector is not an eigenstate of $P_x$ thus, upon measurement of $P_x$, the state vector is no longer $|x,p_y\rangle$.

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  • $\begingroup$ this is what I don' t understand: If I measure $x$ then obviously $p_{x}$ is affected. But $[p_{y},x]=0$ so we can subsequently measure $p_{y}$ after $x$ because they are in the same eigenstate. But because $[p_{y}, p_{x}]=0$ and thus it follows that $p_{y}$ has to also be in the same eigenstate as $p_{x}$ because they commute. It follows now, that we can both measure $p_{y}$ with absolute precision because it is in the same eigenstate as $x$ and we cannot measure it because it is in the same eigenstates as $p_{x}$. $\endgroup$ – user11937 Jun 29 at 22:04
  • $\begingroup$ I could make my above comment into a question if you don't mind $\endgroup$ – user11937 Jun 29 at 22:12
  • $\begingroup$ @user11937, sorry, but I don't follow your reasoning at all. Maybe it will be clear to me what you're trying to say later. $\endgroup$ – Alfred Centauri Jun 29 at 22:32
  • $\begingroup$ I made a question out of my comment physics.stackexchange.com/questions/562731/… $\endgroup$ – user11937 Jun 29 at 22:49

In this case, we can consider the state $|\psi\rangle_{xy}$ in the $xy$ plane to be a tensor product of two one-dimensional states, $|\psi\rangle_x$ and $|\psi\rangle_y$. These states are independent: measuring one will not affect the other. Let's conduct your measurements in sequence. First, we have the state $$ |\psi\rangle_x\otimes |\psi\rangle_y $$ We then conduct a $y$ position measurement, causing the $y$ ket to collapse to a position eigenket $y = q$. The state is now $$ |\psi\rangle_x\otimes |y = q\rangle_y $$ Next, we measure the $x$ momentum, getting $p_x = m$, and putting our state in the form $$ |p_x = m\rangle_x\otimes |y = q\rangle_y $$ Next, we measure the $y$ momentum. It's true that this measurement commutes with our $p_x$ measurement: our ket $|\psi\rangle_{xy}$ is really the tensor product $|\psi\rangle_x\otimes |\psi\rangle_y$, and measuring the momentum of $|\psi\rangle_x$ should have no effect on the momentum of $|\psi\rangle_y$. But that doesn't mean that the measurement doesn't affect the state $|\psi\rangle_y$. The measurement has the same effect that it would on any state: collapsing it to an eigenstate. After the measurement, our state becomes $$ |p_x = m\rangle_x\otimes |p_y = n\rangle_y $$ for some measured $y$ momentum $n$. The $y$ position is lost, and we now have only the momentum.

The reason for this is that independent kets don't care about what the other ket is doing. Sure, from your perspective, you conducted the measurements $x, p_y, p_x, y$, each of which individually commute. But from the perspective of the $y$ ket, you only conducted two measurements: $y$ and $p_y$, which clearly don't commute. So simultaneous eigenstates are impossible.

Answering your question from another angle: while your suggested measurements commute adjacently $$ xp_yp_xy = p_yxp_xy = xp_xp_yy = xp_yyp_x $$ they do not commute in general $$ xp_yp_xy \ne yp_yp_xx $$ indicating that a simultaneous eigenstate of the four operators is impossible.

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[...] by performing the following sequence of measurements: $x \rightarrow p_y \rightarrow p_x \rightarrow y $ [...]

Corresponding to these four stages of the sequence of measurements we can consider four generally distinct states (sets of observational data), $\psi_1, \psi_2, \psi_3$ and $\psi_4$, such that for any arbitrary state $\phi$ holds

$$ \langle \phi | \hat x \, \psi_1 \rangle \, \langle \psi_1 | \psi_1 \rangle = \langle \phi | \psi_1 \rangle \, \langle \psi_1 | \hat x \, \psi_1 \rangle, \text{ and abbreviating } x := \frac{\langle \psi_1 | \, \hat x \, \psi_1 \rangle}{\langle \psi_1 | \psi_1 \rangle}, $$

$$ \langle \phi | \hat p_y \, \psi_2 \rangle \, \langle \psi_2 | \psi_2 \rangle = \langle \phi | \psi_2 \rangle \, \langle \psi_2 | \hat p_y \, \psi_2 \rangle, \text{ and abbreviating } p_y := \frac{\langle \psi_2 | \, \hat p_y \, \psi_2 \rangle}{\langle \psi_2 | \psi_2 \rangle}, $$

$$ \langle \phi | \hat p_x \, \psi_3 \rangle \, \langle \psi_3 | \psi_3 \rangle = \langle \phi | \psi_3 \rangle \, \langle \psi_3 | \hat p_x \, \psi_3 \rangle, \text{ and abbreviating } p_x := \frac{\langle \psi_3 | \, \hat p_x \, \psi_3 \rangle}{\langle \psi_3 | \psi_3 \rangle}, $$

$$ \langle \phi | \hat y \, \psi_4 \rangle \, \langle \psi_4 | \psi_4 \rangle = \langle \phi | \psi_4 \rangle \, \langle \psi_4 | \hat y \, \psi_4 \rangle, \text{ and abbreviating } y := \frac{\langle \psi_4 | \, \hat y \, \psi_4 \rangle}{\langle \psi_4 | \psi_4 \rangle}. $$

[...] the commutators vanish i.e $[x, p_y] = [p_y, p_x] = [p_x, y] = 0$ [...]

The relevant commutators are of course not between the result values (of the individual stages of measurements), but between the measurement operators (which are applied, stage by stage); i.e.

$$ [\hat x, \hat p_y] = [\hat p_y, \hat p_x] = [\hat p_x, \hat y] = 0,$$


$$ [\hat x, \hat p_x] = [\hat y, \hat p_y] = \mathbf i \, \hbar \neq 0.$$

Since, for each successive measurement the commutators vanish [...]

The vanishing commutators have to do with compatibility of measurement operators, and thus with the possibility of interchanging the order of measurement stages. For instance, from $ [\hat x, \hat p_y] = 0 $ we have for state $\psi_2$ and any arbitrary state $\phi$:

$$ \langle \phi | \, (\hat p_y \, \hat x) \, \psi_2 \rangle = \langle \phi | \, (\hat x \, \hat p_y) \, \psi_2 \rangle = p_y \, \langle \phi | \, \hat x \, \psi_2 \rangle.$$

This, by itself, does neither guarantee, nor does it rule out, that

$$\langle \phi | \, \hat x \, \psi_2 \rangle \, \langle \psi_2 | \psi_2 \rangle = \langle \phi | \psi_2 \rangle \, \langle \psi_2 | \, \hat x \, \psi_2 \rangle \tag{*}.$$

Accordingly, we can distinguish two cases:
Either condition $(\ast)$ fails and there is no specific $x$-value attributable to state $\psi_2$ at all -- then the supposed plan apparently fails already at measurement stage 2.

Or condition $(\ast)$ holds, and we continue to consider measurement stage 3; i.e. state $\psi_3$ with $\langle \phi | \hat p_x \, \psi_3 \rangle \, \langle \psi_3 | \psi_3 \rangle = \langle \phi | \psi_3 \rangle \, \langle \psi_3 | \hat p_x \, \psi_3 \rangle$ from above.
But the application of operator $\hat x$ to such a state $\psi_3$ is undefined (or, at least, doesn't result in any arbitrary state $\phi$ at all)! For, suppose to the contrary that we could specify $\phi \equiv \hat x \, \psi_3$. This yields (due to the measurement operators $\hat x$ and $\hat p_x$ being self-adjoint):

$$\langle \psi_3 | (\hat x \, \hat p_x) \, \psi_3 \rangle \, \langle \psi_3 | \psi_3 \rangle = \langle \hat x \, \psi_3 | \hat p_x \, \psi_3 \rangle \, \langle \psi_3 | \psi_3 \rangle = \langle \hat x \, \psi_3 | \psi_3 \rangle \, \langle \psi_3 | \hat p_x \, \psi_3 \rangle = \langle \psi_3 | \hat x \, \psi_3 \rangle \, \langle \hat p_x \, \psi_3 | \psi_3 \rangle = \langle \psi_3 | (\hat p_x \, \hat x) \, \psi_3 \rangle \, \langle \psi_3 | \psi_3 \rangle.$$

But on the other hand with the above commutator holds:

$$\langle \psi_3 | (\hat x \, \hat p_x) \, \psi_3 \rangle \, \langle \psi_3 | \psi_3 \rangle = \langle \psi_3 | (\hat p_x \, \hat x) \, \psi_3 \rangle \, \langle \psi_3 | \psi_3 \rangle + \mathbf i \, \hbar \, (\langle \psi_3 | \psi_3 \rangle)^2.$$

Consequently: $ \mathbf i \, \hbar \, (\langle \psi_3 | \psi_3 \rangle)^2 = 0$, i.e. contrary to stage 3 of the measurement sequence having been carried out and corresponding relevant observational data $\psi_3$ having been obtained at all.
Thus, the supposed plan fails in any case:
If a sequence of measurements has been carried out involving application of conjugate measurement operators then the values thus obtained are not together characterizing the same state.

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