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I propose the following thought experiment:

Suppose we have a beam of identically prepared electrons that is splits into two. The first goes through detector A that detects the $x+y$ where $x$ is the coordinate along x direction and $y$ is the coordinate along the $y$ direction. Then, we measure the difference of the momenta of the electrons in the $x$ and $y$ directions i.e. $p_{x}-p_{y}$. Then, according to the postulates of quantum mechanics, we can measure the both quantities to arbitrary precision since $$ [x+y, p_{x}-p_{y}]=[x,p_{x}]-[y,p_{y}]=0$$ The second beam of electrons is subjected to a similar measurement by a detector B but this time we measure $x-y$ and then measure the sum of momenta i.e. $p_{x}+p_{y}$. Then, again we can measure $x-y$ and $p_{x}+p_{y}$ to arbitrary precision because $$ [x-y, p_{x}+p_{y}]=[x,p_{x}]-[y,p_{y}]=0$$ Then, adding the results of the measurements we have $(x+y)+(x-y)=2x$ and then $(p_{x}-p_{y})+(p_{x}-p_{y})=2p_{x}$. Both of which, $x$ and $p_{x}$ can be measured to arbitrary precision thus violating the uncertainty principle. If on the other hand, we carry out this experiment and find that we are not able to measure the above quantities to arbitrary precision then it follows that the postulates of quantum mechanics do not correctly predict the outcome of the experiment in the sense that the commutator vanishes but we can't measure the quantities to arbitrary precision.

Does this mean that the postulates of quantum mechanics are inconsistent? (I certainly don't hope so!)

Qmechanic
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user11937
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  • Related – Tobias Fünke Oct 16 '21 at 11:01
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    A beam of identically prepared electrons that splits in two seems like a contradiction to me, perhaps a misunderstanding of what "identically prepared" means – garyp Oct 16 '21 at 13:32
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    Note if you perform a $\pi/4$ rotation $(x,y)\mapsto (u,v)$, $x+y=u\sqrt{2}$ and $p_x-p_y=p_v\sqrt{2}$, so you may as well replace those expressions by $[x,p_y]=0$ and $[y,p_x]=0$. The extra algebra serves only to obfuscate your question. – J. Murray Oct 16 '21 at 19:11
  • @J.Murray I don’t see how you rotate a scalar quantity. The quantity is the magnitude in the x direction plus the magnitude in the y direction which happens to be scalar quantity. – user11937 Oct 16 '21 at 23:29
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    @user11937 I’m saying if you rotate the $(x,y)$ axes by $\pi/4$ to get new axes $(u,v)$, then the quantity $x+y$ simply becomes $u\sqrt 2$ while $x-y$ becomes $v\sqrt 2$, and similarly for the momenta. Adding $x+y$ doesn’t give you some new thing - it’s just position measured along a different axis. – J. Murray Oct 16 '21 at 23:35
  • @J.Murray look if I have detector that measures $x+y=1+\sigma$ where $\sigma$ is the uncertainty in the measurement. Then $y= 1-x+\sigma$ now you tell me how your coordinate transformation reduces the straight line with intercept at 1 and slope equal to -1 to your new (u,v) axis. – user11937 Oct 16 '21 at 23:51
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    @user11937 I’m afraid I don’t really understand what you’re asking. All I’m saying is that measuring $x+y$ and $p_x-p_y$ is exactly the same (besides the factor of $\sqrt 2$) as measuring $x$ and $p_y$ with a detector which is oriented at 45 degrees with respect to the original one, so the addition and subtraction of various quantities does not have any influence on the heart of your question. – J. Murray Oct 17 '21 at 01:28
  • @J.Murray if you are going to do a coordinate transformation you should also do a proper coordinate transformation on the p_u and p_v operators. If you do coordinate transformation for p_u using the chain rule you get p_u = p_x + p_y and for p_v = p_x - p_y. And if you were careful in reading the question, you would notice that after measuring the u coordinate we measure p_x - p_y which happens to be p_v. So I don’t see what the problem is. – user11937 Oct 17 '21 at 03:54
  • @user11937 Again, there’s no problem per se - but you may as well just say you’re measuring $x$ and $p_y$ on one beam and $y$ and $p_x$ on the other. You therefore claim that you’re simultaneously measuring $x$ and $p_x$ - which makes very little sense to me because you’re measuring different beams. In any case, using the unrotated coordinates does nothing to elucidate the problem, and serves only as an algebraic smokescreen which obscures the real question, whether you realize it or not. – J. Murray Oct 17 '21 at 04:00
  • @J.Murray I am not understanding what the purpose of coordinate transformation is in answering this problem. Are you saying that the measurements of an experiment are independent of reference frame? And if the result of an experiment holds in one frame it automatically holds in another? – user11937 Oct 17 '21 at 04:34
  • @user11937 I’m saying that your setup could be rephrased as “I split the beam into two and measure $(x,p_y)$ on one beam and $(y,p_x)$ on the other, which seems to give a simultaneous measurement of $x$ and $p_x$.” The addition and subtraction serves no purpose other than to make the question somewhat more opaque; that being said, there are several good answers already. – J. Murray Oct 17 '21 at 04:40
  • @J.Murray again I don't see what coordinate transformation seeks to accomplish. Its like saying we don't need the complicated formulas for the measurement of spin because if we measure the spin in its rotating reference frame it will always be 0. – user11937 Oct 17 '21 at 22:06
  • @user11937 Obviously my point is not getting across. In your response to Mark’s answer, you argue that your proposal is fundamentally different from the one he suggests. My point is that they are exactly the same, and the fact that you think otherwise suggests that you’re being confused by your choice of coordinates. Regardless, it’s clear that I’m not being effective so I’ll stop. – J. Murray Oct 17 '21 at 22:56
  • @J.Murray All you are saying is that the commutator of u and p_v is 0 in the new coordinates. Then you say that I should have stated that I am measuring u and p_v in the new coordinates instead as if that changes anything about my question. Are you implying that we should always measure in the coordinate system you are defining? To be honest I would like you to connect coordinate transformation to my question because you are stating a fact that is irrelevant to my question. – user11937 Oct 17 '21 at 23:41

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It would be far simpler to just directly measure $x$ of your first beam and $p_x$ of the second beam.

Both of which, $x$ and $p_x$ can be measured to arbitrary precision thus violating the uncertainty principle.

There is no violation of the uncertainty principle. If you have an unlimited supply of identically prepared systems you can measure to arbitrary precision (in principle). The uncertainty principle limits what you can simultaneously measure on a single system.

Mark Mitchison
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  • If I do as you say I will localize a particle in one beam and lose all info about its momenta and vice versa in the second beam. Which doesn’t help me to find the momentum belonging to an electron at a specified region in space. Whereas in my experiment I am measuring both the region the electron occupies and it’s momentum within the region of space. – user11937 Oct 16 '21 at 08:41
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    @user11937 As per the comment I just left on the original question, your proposal differs from Mark's merely by a trivial $\pi/4$ rotation of your reference axes (and scaling by $\sqrt{2}$) – J. Murray Oct 16 '21 at 19:54
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    @user11937 The same happens if you measure $x+y$: the particle's position in the $x+y$ direction becomes localized so you lose all information about $p_x+p_y$. – Carmeister Oct 18 '21 at 01:24
  • @Carmeister we are measuring p_x - p_y which commutes with x+y and so measuring both observables to arbitrary precision is not a problem. – user11937 Oct 18 '21 at 01:30
  • @user11937 Yes but then you are measuring $p_x+p_y$ of the second electron beam and claiming that it tells you about the value of $p_x+p_y$ for the first beam. But when you measured $x+y$ of the first beam you changed the value of $p_x+p_y$ for that beam. – Carmeister Oct 18 '21 at 01:33
  • @Carmeister I measure the sum of the momenta after the particles pass through the x-y filter. I don’t see how the measurement of the sum coordinates is in the first beam affects the measurement of sum of momenta in the second beam since they are divided and there is no entanglement of any sort. – user11937 Oct 18 '21 at 01:47
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    @user11937 “I don’t see how the measurement of […] the first beam affects the measurement of […] the second beam” that is the point. The two measurements are not correlated! So adding them tells us nothing about individual measurements. – Superfast Jellyfish Oct 18 '21 at 03:15
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You are assuming that the two beams of electrons are two different systems in identical quantum states. The uncertainty principle limits measurement of two non-commuting observables on one system, but says nothing about measurements on separate systems. If I had two identical systems in identical states, I could just measure $x$ in one system and $p_x$ in the other system which would give me an accurate measurement of $x$ and $p_x$ at the same time. There is no need to go through the complicated process you have described.

Prahar
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    The no-cloning theorem deals with an unknown state... Further, why would the measurement of $p_x$ in $A_1$ even violate the HUP? If you'd repeat the measurements many times, then you would find (for a 'generic' state) $\sigma (x) \sigma(p) \geq \hbar$. – Tobias Fünke Oct 16 '21 at 10:56
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    @Jakob It would not violate HUP. That's my point. Measuring one variable to arbitrary high accuracy does not violate HUP. OPs idea is to take two identical systems, measure $x$ in one system to high accuracy and then measure $p_x$ in the second identical copy to high accuracy. In doing this, you would have measured both $x$ and $p_x$ to very high accuracy and not have violated HUP. The point I am making is that this sort of measurement is NOT a violation of HUP at all. – Prahar Oct 16 '21 at 10:59
  • @Jakob The problem here is the requirement of having two identical systems in the first place. – Prahar Oct 16 '21 at 11:00
  • You say: Then measure $x$ in $A$ and $p_x$ in $A_1$ and that way I would have violated the uncertainty principle - and I asked: why? I don't see a violation of the HUP: If you create an ensemble of identical states (which, in principle, is possible if you know the state), then the corresponding variances of the measurement outcomes cannot become arbitrarily small. I don't see the any violation of the HUP here. – Tobias Fünke Oct 16 '21 at 11:18
  • @Jakob - you misunderstood that sentence. I will change it so it doesn't cause confusion. PS - I was describing what a counter-argument to my answer might look like but clearly I didn't get that point across. – Prahar Oct 16 '21 at 11:20
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    While the bulk of your answer is correct, the no-cloning part of your answer is wrong. The theorem restricts us from copying/cloning arbitrary states. It does not restrict preparation of identical states. We can always prepare identical states of our choice (by the means of an operator whose eigenstate is our state of choice). – Superfast Jellyfish Oct 16 '21 at 18:54
  • @SuperfastJellyfish removed it. – Prahar Oct 16 '21 at 18:55
  • @Prahar please read my comment to Mark’s answer above because your answer is very similar to his. – user11937 Oct 16 '21 at 23:36
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Then, adding the results of the measurements we have $(x+y)+(x−y)=2x$ and then $(p_x−p_y)+(p_x−p_y)=2p_x$.

Not quite, there are two problems with this. The first problem is that, since there are two electrons, there need to be two sets of position and momentum operators, so the $x$ in the first term is different from the $x$ in the second term.

The second problem is that you're being sloppy about the distinction between the operators and eigenvalues. For instance, in the first case, you don't actually have $x$ and $y$ eigenvalues, since you didn't measure $\hat x$ and $\hat y$; you only measured $\hat x+\hat y$ (I use hats to denote operators). As such, you can't actually break up $(x+y)$ into $x+y$. For this reason, it makes more sense define new operators, based on the rotation and scaling mentioned by J. Murray, $\hat u_i=\hat x_i+\hat y_i,\hat v_i=\hat x_i-\hat y_i$, and similarly for momentum (the subscripts denote the two sets of operators for the two particles).

Given this, the values you have measured are $u_1$, $p_{v1}$, $v_2$ and $p_{u2}$. Since all four of the corresponding operators commute, there is no inconsistency here.

Sandejo
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  • I find it odd that a probabilistic description of the position and momentum operators is based on a single observation of a single electron. In other words, how would you measure standard deviation of momentum and position of a single electron in single observation. Therefore, in order to have a valid measurement of the standard deviation of each observable you would need to conduct a measurement on a collection of identically prepared particles - which is the case in my question. Also, for coordinate transformation read my last comment above. – user11937 Oct 17 '21 at 22:37
  • @user11937 It doesn't really make sense to talk about the standard deviation of a single observation, unless you're talking about the measurement uncertainty, which is something else. The standard deviation of an operator is defined by $\sigma_{\hat A}=\sqrt{\langle\hat A\rangle^2-\langle A\rangle^2}$. Since this expression contains expectation values, it makes sense that measuring it requires making measurements on ensembles of particles in the same state,—because the standard deviation is defined for the state, not the particle. – Sandejo Oct 18 '21 at 00:17
  • @user11937 Regarding the topic of coordinate transformation, I have not introduced any coordinate transformation—all I have done is define a new set of operators in terms of the old ones out of convenience. The $\hat u$ and $\hat p_u$ operators could still be represented in the $x,y$ basis by $\langle\psi\rvert\hat u\lvert x,y\rangle=(x+y)\langle\psi|x,y\rangle$ and $\langle\psi\rvert\hat p_u\lvert x,y\rangle=-i\hbar\left(\frac{\partial}{\partial x}+\frac{\partial}{\partial y}\right)\langle\psi|x,y\rangle$. – Sandejo Oct 18 '21 at 00:25
  • And how do you make a measurement of a state without a physical particle in that abstract state? To your second comment: so you are renaming the operator – user11937 Oct 18 '21 at 00:34
  • @user11937 You do still need a physical particle in order to make the measurement. I'm just saying that the standard deviation of an operator is a property of the state, rather than the particle. This is relevant because, when you make the measurement, you change the state of the particle. – Sandejo Oct 18 '21 at 00:59
  • First, the standard deviation refers solely to the outcomes of the operator (look up any textbook such as Sakurai) and hence a property of the operator itself. Second,if the two operators commute you are not destroying the state such as measuring the operators x+y and p_x - p_y in that order. – user11937 Oct 18 '21 at 01:12
  • @user11937 Of course, the standard deviation depends on the operator—that should go without saying. When you measure $\hat x+\hat y$ the state changes to an eigenstate of that operator; when you measure $\hat p_x-\hat p_y$, since that commutes with $\hat x+\hat y$, the state changes to a joint eigenstate of both operators. – Sandejo Oct 18 '21 at 01:18
  • If it bothers you to think that the system is divided into two i.e. the beam is split into two, you can think of it this way: we measure x+y and then p_x - p_y. Then we rotate the filter 90 degrees to x- y and measure p_x + p_y on the ensemble of particles. – user11937 Oct 18 '21 at 01:20
  • If the $\hat x+\hat y$ and $\hat x-\hat y$ measurements are made on different particles, then it is necessary to think of the system as divided in two. If the measurements are made on the same particle, that would make you get different results, since the measurements would not be made on the same state. – Sandejo Oct 18 '21 at 02:45
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Let me try to rephrase what all the other answers are trying to say.

Consider that the initial state is such that each (x,y) is equally likely within a square of unit length. What this means is that within that square each point is equally likely and outside it there is zero probability of detection. You can choose any state that you like, it will not affect the arguments that follow.

Now let us consider the two measurements we make; one of $\hat x + \hat y$ and another one of $\hat x - \hat y$. After you measure the first, your state localises to a point (p1) within the unit square. But here’s the thing, when you measure the second, the state localises to another point (p2) which more often than not is not the same as p1. enter image description here

Now do you see why it makes no sense to add the two measurements and call it a single measurement? The second measurement is independent of the first, so combining them is meaningless. It is more apparent when we label our measurement operators properly. $\hat x_1 + \hat y_1$ and $\hat x_2 + \hat y_2$.

If you repeat this set of measurements infinite times, you’ll fill up the square (complete information of initial state). But each subsequent measurement has no correlation with the previous one.

Superfast Jellyfish
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Then, adding the results of the measurements we have $(x+y)+(x−y)=2x$ and then $(px−py)+(px−py)=2px$.

This is the problem. You've measured $x+y$ on the first beam and $x-y$ on the second beam. These two quantities are not really related. You could write

$$ (x_1+y_1)-(x_2-y_2) $$

But we don't have $x_1=x_2$ or $y_1=y_2$, so you can't conclude anything interesting about the position or momenta of the individual beams.

That said, this idea has been explored in the cavity optodynamics literature: Møller et. al. "Quantum back action evading measurement of motion in a negative mass reference frame" (2017). Mytakeaway is that yes, when you have more than 1 system with position and momentum you can find linear combinations of their position and momenta which commute allowing for simultaneous measurement of those linear combinations. However, you are not getting full information about any particular subsystem in this case, so the usual constraints imposed by quantum mechanics are still valid.

Jagerber48
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