Is the uncertainty principle a consequence of classical reference frame?

Question

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?

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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.

Answer 1

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 (presumably the ancilla and the system have interacted but this is not strictly necessary) 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 infinite well is unnecessary to your example; best known discussion of this kind of setup is in

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 and ancilla preparations for the joint measurement are Gaussians 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.

Answer 2

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.

Answer 3

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)

Answer 4

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.

EDIT

Resolving partial derivative. Consider such schematics :

So we can rewrite (5) RHS as $$ \tag 6 i\hbar \frac{\partial \left(X+x'\right)}{\partial X}= i\hbar \left(1+ \frac{\partial x'}{\partial X} \right) .$$

Now since relative $x'$ is independent of box COM position (at least in a simple model), so we can assume that ${\partial x'}/{\partial X} = 0$ and this leaves us with standard commutator $i\hbar$. The conclusion from that is that $[\hat x,\hat p]$ commutator is invariant to box translation operations. Unless in more complex model we should account for a particle wave-function change when it hits box wall at the moment box starts to move (wall moves with acceleration). In this case probably we can't assume $x',X$ to be independent variables anymore.