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Consider a quantum particle in a box (everything one dimensional, for simplicity). The box is isolated from everything else. We could not measure simultaneously the particle's position and momentum, because they are complementary variables,

$$[x,p] = i\hbar$$.

Since the system particle+box is isolated from everything else, we could say that it has a well defined total momentum $p+P$, where $P$ is the momentum of the box.

Despite the absolute position $x$ being incompatible with the total momentum $p+P$, if we consider the box coordinates as quantum variables, $[X,P] =i\hbar$, the relative position $x-X$ commutes with $p+P$, i.e.,

$$ [x-X,p+P] = 0. $$

If we consider the quantities $x-X$ and $p+P$ are the actual particle properties, defined up to reference frame variables. They are compatible so they could be simultaneously measured. It does not hurt the complementary principle, since they are well defined up to values $(X,P)$ that are complementary. We can't get rid of complementarity, but it seems that it got swept away by the reference frame coordinates.

My question is: Can we think about the complementarity of particle properties as a consequence of assuming that we have a Classical reference frame?

If we assume that $Q,P$ are classical variables and $Q=P=0$, the quantities $x-X$ and $p+P$ reduces to $x,p$, which are complementary. What if the problem of complementary is to assume a classical reference frame in first place?

If the answer is yes, I have a related question:

The box is arbitrary stuff. It can be the whole universe. In this sense, everything inside the universe would have well defined properties, up to universal variables $(X,P)$ that are the only complementary variables. Could we sweep away complementarity to universe, to recover everything else as well defined relative quantities?

EDIT:

After the Confuse-ray30 answer and arguments, I decided to clarify some points:

  • The idea of a box is no central to my argument. I thought about it being the laboratory or something like that, but what is central is the idea of a reference frame. It is not really constraining the particle, or interacting with it. I think this point also could clarify some controversy that emerged in Agnius Vasiliauskas answer, about $x$ being or not a function of $X$.
  • As liangre pointed out, the variables $x-X$ and $p+P$ lose properties of conjugated variables, like $p+P$ not being the generator of position translations. But in my point of view it was expected, since they commute, so they have simultaneous values. The point is that they could be regarded as particle properties not being subjected to complementarity.
  • As I said in the comments, I'm not arguing against the uncertainty principle. I'm just curious if this could be regarded as a property of the reference frame instead of the particle being measured. It is not a new interpretation of QT either. I thought that both "yes" or "no" could be interesting answers for this question, which turns it worth to be asked.
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    $\begingroup$ I would rather think about it as "it matters in which order you measure position/momentum because the operators don't commute, so you can't really know what either one is "supposed" to be if you also know the other". Is that not sufficient given that it comes straight from the operator formalism? $\endgroup$
    – controlgroup
    Commented 23 hours ago
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    $\begingroup$ From the perspective of an observer outside of the box-particle system, the relative position and total momentum are a pretty odd set of variables. It would be much more natural to look at the center of mass position (or for simplicity just call it the average position) and total momentum. These don't commute. $\endgroup$
    – Andrew
    Commented 23 hours ago
  • $\begingroup$ @controlgroup it is true that I loose information about the particle and the box individually, but the idea is to consider just the relative properties as being particle properties, and the variables of the box alone in the end are not relevant (fundamentally, for defining what is inside) $\endgroup$
    – Ruffolo
    Commented 15 hours ago
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    $\begingroup$ I don't know how to respond to that. You want to model the whole universe as a single quantum variable in non-relativistic quantum mechanics without gravity? It doesn't seem like a very good model. Furthermore, an observer in the Universe (like us) would be in the box, and therefore interested in $x$ on it's own, the position of the particle, which doesn't commute with $p$. I don't even know what it would mean to measure $X$, the "position of the Universe". $\endgroup$
    – Andrew
    Commented 9 hours ago
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    $\begingroup$ I'm not trying to go away with uncertainty, I just noted that the question of reference frames in quantum theory is strange it seems that the uncertainty could be swept to the reference frame variables. In the end, I'm not trying to reinvent the wheel, but I was curious about what kind of arguments we could have to prohibit such scheme $\endgroup$
    – Ruffolo
    Commented 8 hours ago

4 Answers 4

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The answer might be "No"

Every times we consider the variables as X of bos or x of particle could not be the generator of spatial translations.The reason why is that each observable physical quantity is only the properties of each physical object not for the space.Even more the generator of spatial translations is only $e^{-\frac{i}{\hbar}lp}$,for the reason as $e^{-\frac{i}{\hbar}lp}|x\rangle=|x+l\rangle$.

It shows that the reference frame transformation does not base on the physical obeject ,just only base on the real space.

The next:transformaton of "each object" is the transformation of each space properties based on those objects.

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    $\begingroup$ It makes sense that $x-X$ and $p+P$ don't have the same properties of $x$ and $p$, since they are not canonically conjugated. But now, since they commute, they could have well defined values simultaneously and can be thought as classical variables. $\endgroup$
    – Ruffolo
    Commented 14 hours ago
  • $\begingroup$ yeah, it is the thing that[x-X,p+P]equal to 0,but i meant that x-X is not simply to space transformation. $\endgroup$
    – liangre
    Commented 12 hours ago
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No, relative commutator of $[x-X,p+P]$ does not commute.

To see this you need to compute it, namely by expressing full operators $$ \tag 1 \left[ x-X,~ -i\hbar\frac{\partial}{\partial x}-i\hbar \frac{\partial}{\partial X} \right] $$

This get's you to computing,

$$ \tag 2 \left(x-X \right) \cdot \left(-i\hbar\frac{\partial}{\partial x}-i\hbar \frac{\partial}{\partial X} \right) - \left(-i\hbar\frac{\partial}{\partial x}-i\hbar \frac{\partial}{\partial X} \right) \cdot \left( x-X \right) $$

Applying first multiplication terms to $\psi$ and expanding, get's you to :

$$ \tag 3 -i\hbar x \frac{\partial \psi}{\partial x} + i\hbar X \frac{\partial \psi}{\partial x} -i\hbar x \frac{\partial \psi}{\partial X} +i\hbar X \frac{\partial \psi}{\partial X}$$

Doing same with second multiplication of (2), we get : $$ \tag 4 -i\hbar \frac{\partial (x \psi)}{\partial x} +i\hbar \frac{\partial (X\psi)}{\partial x} -i\hbar \frac{\partial (x\psi)}{\partial X} +i\hbar \frac{\partial (X\psi)}{\partial X}$$

Now applying differentials multiplication rule to the (4) and then subtracting (4) from (3) we arrive at final commutator,

$$ \tag 5 [x-X,p+P] = i\hbar \frac{\partial x}{\partial X} .$$

I.e. simple commutator of $[x,p]$ is adjusted by a term/differential- how much particle coordinate changes when you move box around. So to say in general due to box movement uncertainty of particle wave-function increases.

P.S. And just thinking intuitively, it must be clear that uncertainties of a particle in a box just can't disappear just because we shuffle box system around, that would be crazy.

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    $\begingroup$ I think $\frac{\partial x}{\partial X}=0$. Also, in the abstract, $[x-X,p+P]=[x,p]+[x,P]-[X,p]-[X,P]=0$ $\endgroup$
    – Ryder Rude
    Commented 12 hours ago
  • $\begingroup$ @RyderRude So you think, that moving box from $X_1$ to new the $X_2$ location,- particle coordinate $x$ will be still located in a region bounded by old box position $X_1$ ?? How it can be so ? BTW, keep in mind that moving box by $\Delta X$ is equivalent to simply changing box reference frame from $O(X)$ to $O(X-\Delta X)$. I don't see how follows that $\frac{\partial x}{\partial X}=0$,- can you base your this assumption on something ? $\endgroup$
    – Agnius Vasiliauskas
    Commented 12 hours ago
  • $\begingroup$ I think that if we consider the particle and box as different quantum systems, the state space is defined as the tensor product $\mathcal H_p\otimes \mathcal H_b$, and the operators $x$ and $X$ could be simultaneously diagonalized in the product basis $|x\rangle|X\rangle$. At least it was my reasoning when writing the question $\endgroup$
    – Ruffolo
    Commented 11 hours ago
  • $\begingroup$ I mean that the joint "box CoM + particle" wavefunction is of the form $\psi (x,y)$, where $y$ is the CoM eigenvalue of the box. Now, $x$ and $y$ are independent variables. Neither is a function of the other, so $\frac{\partial x}{\partial y}=0$. I would impose the boundary conditions on the particle wavefunction using delta potentials. Perhaps we can use $V=\delta (y-a)+\delta(y+a)$ as the potential. This potential should reproduce the physics of a free particle inside a box that can move around. $\endgroup$
    – Ryder Rude
    Commented 11 hours ago
  • $\begingroup$ @RyderRude I don't see how $x$ and $X$ are independent variables. Again by definition of OP, $x$ is absolute coordinate position of particle, and so does box com position $X$. And so $x=f(X)$, because when you move box,- particle moves together with it. This proves that $x$ and $X$ are dependent variables. $\endgroup$
    – Agnius Vasiliauskas
    Commented 11 hours ago
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Partial answer:

Your assumptions confuse me and might be the problem. The box has momentum, i.e. it is part of the Hamiltonian. Yet, if it is and can constrain the particle, it must interact with the particle. Thus, we have a position dependent interaction term between box and particle. But then the total momentum does not commute with the Hamiltonian, neither do the two separate momenta in general. (Then again, maybe there is an example of a interaction which localizes the wave function and commutes with the derivative, I dont think so though)

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  • $\begingroup$ I don't think it should interact with the particle. I'm not actually imposing that it should constraint the particle, I'm just using it as a reference frame. The central potential could localize the wave function and conserve total momentum $\endgroup$
    – Ruffolo
    Commented 5 hours ago
  • $\begingroup$ Yeah, I noticed you dont do much with the box. It doesnt even need to be around the particle. I guess my answer is wrong then. @Ruffolo $\endgroup$
    – Confuse-ray30
    Commented 1 hour ago
  • $\begingroup$ Maybe I should edit it, changing the box for a point or anything else $\endgroup$
    – Ruffolo
    Commented 1 hour ago
  • $\begingroup$ @Ruffolo If I replace the box by a new particle species instead that does not interact, then we have the same argument. Now Im wondering if we can understand this better under this light... You commented thaz while I was typing away. I think you should, yeah $\endgroup$
    – Confuse-ray30
    Commented 1 hour ago
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No. You have introduced ancilla observables $X$ and $P$ in your scheme in addition to your system variables $x$ and $p$. Your system variables $x$ and $p$ are still complementary and in fact the effect of introducing the ancilla has added noise to the system. You have simply defined new variables which you may be able to measure jointly, but you would need to go back to the system variables (your original observables of interest) and the effect of the ancilla is non-trivial.

The best known discussion of this is due to

Arthurs, E., and J. L. Kelly. "BSTJ briefs: On the simultaneous measurement of a pair of conjugate observables." The Bell System Technical Journal 44.4 (1965): 725-729.

The idea was "rediscovered" in

Stenholm, Stig. "Simultaneous measurement of conjugate variables." Annals of Physics 218.2 (1992): 233-254.

and a different approach was proposed by

She, C. Y., and H. Heffner. "Simultaneous measurement of noncommuting observables." Physical Review 152.4 (1966): 1103.

Basically the ancilla functions as a pointer but has some noise, and IIRC Stenholm shows that the optimal state preparation and optimal ancilla preparation are Gaussian with the same width (basically harmonic oscillator coherent states). This type of approach has spawned a mini-industry on optimal joint measurements of pairs of observables, and the general technique is called a Naimark extension. See for instance

Levine, Robert Y., and Robert R. Tucci. "On the simultaneous measurement of spin components using spin-1/2 meters: Naimark embedding and projections." Foundations of physics 19.2 (1989): 175-187.

Chiribella, Giulio, G. M. D'Ariano, and M. F. Sacchi. "Joint estimation of real squeezing and displacement." Journal of Physics A: Mathematical and General 39.9 (2006): 2127.

or

Mišta Jr, Ladislav, et al. "Angle and angular momentum: Uncertainty relations, simultaneous measurement, and phase-space representation." Physical Review A 106.2 (2022): 022204.

as other examples where this kind of construction is considered.

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