Is it possible to measure both position and momentum simultaneously?

[Clueless123]

A simultaneous measurement of both a particle's position and momentum may be successfully accomplished if more than one photon were utilized for the measurement. A non-demolishing measurement is possible if the emitters were aligned such that each would offset the other’s recoil of the target through the simultaneous emission of a photon. By using multiple photons for the measurement, the recoils of the target would cancel each other out, and allow for a simultaneous measurement of both position and momentum. So, is it actually possible to measure both position and momentum simultaneously if the recoil is offset by using more than one photon?

 

[PeroK]

There is also uncertainty in the position and momentum of a photon. More than you might imagine. In fact, a photon does not have a position observable at all.

 

[DrChinese]

Apparently you are aware of the Heisenberg Uncertainty Principle (HUP), which has been around for nearly 100 years. Yet you are imagining that some simple experiment will demonstrate a violation of it. So there are several issues here.

1. The HUP does not itself prevent a simultaneous measurement of non-commuting observables. It prevents such measurement(s) from being performed with unlimited precision. If you measure position quite accurately, you will not get a precise value for momentum (and vice versa).

2. And you wouldn't notice that immediately. It would take repeated measurements of identically prepared systems to notice that you are getting a spread of values.

3. And how would you prepare a series of identical systems in the first place? I.e. with known position and momentum? That ends up being a circular problem (since the HUP is standing in your way).

I could go on, but the bottom line answer to your idea is: you end up with the limits of the HUP in every scenario.

 

[vanhees71]

Apparently you are aware of the Heisenberg Uncertainty Principle (HUP), which has been around for nearly 100 years. Yet you are imagining that some simple experiment will demonstrate a violation of it. So there are several issues here.

1. The HUP does not itself prevent a simultaneous measurement of non-commuting observables. It prevents such measurement(s) from being performed with unlimited precision. If you measure position quite accurately, you will not get a precise value for momentum (and vice versa).

That's also not entirely true.

The usual Heisenberg uncertainty relation, proved within the first few lectures in QM 1, is about properties of a particle, i.e., it says that it is impossible to prepare a particle such that both position and momentum are arbitrarily accurately determined. The general theorem says that
$$\Delta A \Delta B \geq \frac{1}{2} |\langle \mathrm{i}[\hat{A},\hat{B}] \rangle|,$$
where the expectation value can be taken wrt. an arbitrary state $\hat{\rho}$,
$$\langle \hat{A} \rangle=\mathrm{Tr}(\hat{\rho} \hat{A}).$$
Only compatible observables can be made simultaneously well determined for all possible values they can take, because this is possible only if there's a common complete set of eigenvectors of the corresponding self-adjoint operators.

There are exceptions in the sense that it can be that there are states, for which two incompatible observables take a determined value. An example is angular momentum. If you are in a state with $J=0$, then all components of the angular momentum have eigenvalue 0. In general, of course you can only determine one component of angular momenum, for which usually one chooses the 3-component.

Any observable can, at least in principle, be measured at any precision you are able to achieve practically, independent of the state the system is prepared in. The Heisenberg uncertainty relation does not tell you anything about the disturbance of the system due to the measurement. That's a much more complicated question, which also depends on the details of the measurement apparatus used.

2. And you wouldn't notice that immediately. It would take repeated measurements of identically prepared systems to notice that you are getting a spread of values.

That's a very important point. At least within the minimal statistical interpretation of QT the only thing we have are the probabilistic laws provided by the formalism, i.e., to, e.g., verify the uncertainty relation for position and momentum of a particle, you have to prepare the particle very often in the same state and measure the position at high accuracy to obtain the statistics of measurement outcomes at the level of significance you want. Then you get an expectation value and the standard deviation. The same you do with momentum for another ensemble of equally prepared particles. Then the QT prediction is that $\Delta x \Delta p_x \geq \hbar/2$, and this you can verify in the above described way.

Note that we haven't measured position and momentum simultaneously. The possibility to measure incompatible observables on a single quantum system in an arbitrary state is a much more complicated question, and can be described adequately only with a generalized more modern framework of measurement theory, described by socalled positive operator-valued measures.

3. And how would you prepare a series of identical systems in the first place? I.e. with known position and momentum? That ends up being a circular problem (since the HUP is standing in your way).

That's the point! You cannot prepare (!!!) the particle in such a way that position and momentum are better determined than the Heisenberg uncertainty relation allows.

I could go on, but the bottom line answer to your idea is: you end up with the limits of the HUP in every scenario.

 

[sbrothy]

This article suggests that you could if you could reach absolute zero which is of course impossible. So academic it probably makes no practical sense.

 

[Clueless123]

Well, if the problem is due to the measurement apparatus, then is it possible to reduce that margin of error by conducting a nano-experiment near absolute zero (to increase precision)? Not easy, but with multiple trials with minimal displacement, not impossible to do. Correct me if I'm wrong, but the original reason why the Heisenberg Uncertainty Principle was formulated was due to the recoil effect of the target. So, why not just use another photon to counteract the recoil? Not a big deal. No reason to go off on a tangent saying the universe is uncertain due to an experimental error with the apparatus or the lack of countering the recoil. Just a thought. May not be easy to accomplish, but I wouldn't say it is impossible to duplicate with multiple trials.

 

[Lord Jestocost]

The uncertainty principle is not really about measurements. It would be better to call it indeterminacy principle. It is inherent – so to speak – to “objects”. They simply do not have exact momenta, positions and energies in general. This has basically nothing to do with measurements or other setups. We cannot know the future because we cannot know the present for sure.

 

[Clueless123]

That's circular logic in relying upon the uncertainty principle while at the same time, trying to prove it.

How about this. Pretend you're back in time before the uncertainty principle even existed, and the experiment showed that the recoil of the target was causing an issue. So, you add another photon into the mix to counteract that recoil. Then precision became a problem, so you try to minimize that by conducting a nano-experiment near absolute zero to minimize the misalignment of the measurement apparatus.

Upon multiple trial and error runs, the experiment should be able to accurately measure both position and momentum simultaneously. May not be easy to do, but not impossible.

The uncertainty principle was derived by an experimental error, not indicative of the universe's behavior. Or, so it seems to me...

 

[Hyperfine]

The uncertainty principle was derived by an experimental error, not indicative of the universe's behavior. Or, so it seems to me...

What seems to you is both irrelevant and incorrect.

The previous respondents have made good-faith efforts to provide you with accurate information. Take advantage of it.

 

[Nugatory]

Well, if the problem is due to the measurement apparatus....

It isn't. This is one of the most stubborn pop-sci myths of all time.

What the uncertainty principle does say is something along the lines of...
The state in which a position measurement will give us an exact infinite-precision result X is also a state in which we cannot exactly predict the result of an exact infinite-precision measurement of the momentum P.

We test this proposition by repeating the same experiment: initialize the system in the state such that if we measure the position we will get X; and then measure the momentum instead. Do this over and over again and we will find that even though the initial state is the same every time, and even though we are doing an exact measurement of P, we get different results for P each time. The uncertainty principle describes the statistical spread of these results.
(You should be aware that I have cut some corners, as well as ignoring the mathematical complications that come from position and momentum being continuous variables)

If you have any interest in actually understanding quantum mechanics, the best thing you can do is to try to forget everything you think you know and start over with a real college-level textbook. If that's not practical for you (the big problem will be the amount of math required), Giancarlo Ghirardi's book "Sneaking a look at God's cards" is much more layman-friendly and at least unwinds the most pernicious pop-sci myths.

 

[DrChinese]

The uncertainty principle was derived by an experimental error, not indicative of the universe's behavior. Or, so it seems to me...

Again, you are missing what everyone is telling you. The HUP has absolutely nothing to do with experimental error.

1. That should be obvious from the actual formula for the HUP, which can be expressed a number of ways. See @vanhees71 post #4 for example. If there were only experimental limits, why would there be a relationship between measurement precision of non-commuting observables? And yet there is no such limitation on the measurement precision of commuting observables? 2. And in fact, so you can see that experimental precision is NOT an issue, consider an electron (rather than a photon per your example). Electron position and momentum do NOT commute on the same axis - X, Y or Z. But they DO commute on different axes. So X momentum and Y position commute, and there is no limitation on the precision of simultaneously knowing both of these. See for example from (9-5) and after in the following:

https://faculty.washington.edu/seattle/physics227/reading/reading-24-25.pdf3. Furthermore, your idea about non-commuting limits (HUP) being solely due to the experimental issues tacitly assumes that both momentum and position have well-defined simultaneous values at all times. They don't. Essentially, that idea was rejected about 50 years ago once Bell's Theorem appeared (along with other no-go theorems) and experimental results confirmed the predictions of QM. We now know that the outcomes of measurements depends on context - i.e. how the observer chooses to measure. It would be off-topic to explain in detail Bell and the predecessor paper EPR here, but it is a compelling argument against your assumption.

 

[Drakkith]

Perhaps it's best to first look at a non-quantum (classical) example of the uncertainty principle. See this video by 3Blue1Brown:


The uncertainty principle in quantum physics is simply a specific example of this general principle and happens to result in strange, non-intuitive things that we aren't used to seeing in the real world.

 

[vanhees71]

We are very much used to see this in the real world since the real world is best described in terms of field theory, classical and quantum. The uncertainty principle of position and momentum always occurs as soon as wave equations are involved, and indeed it can be understood in terms of Fourier transforms. If you have a wave function, representing a pure quantum state, you can look at it in the position or momentum representation. These are in fact just different representations of the same ket $|\psi \rangle$. The position representation is just $\psi(x)=\langle x|\psi \rangle$ and the momentum representation is $\tilde{\psi}(p)=\langle p|\psi \rangle$. The physical meaning is that $P_{\psi}(x)=|\langle x|\psi \rangle|^2$ is the position-probability-density distribution and $\tilde{P}_{\psi}(p)=|\langle p|\psi \rangle|^2$.

Operationally, i.e., in the real world such a state ket represents the preparation procedure the particle has undergone before it's measured. As stressed already above, you can prepare the particle many times in this state and either measure position or momentum as accurately as you like and use the results to test the statistical predictions of QT.

Now the relation between position and momentum representation is given by the Fourier transformations,
$$\psi(x)=\int_{\mathbb{R}} \mathrm{d} p \frac{1}{\sqrt{2 \pi \hbar}} \exp(\mathrm{i} x p) \tilde{\psi}(p)$$
and
$$\tilde{\psi}(x)=\int_{\mathbb{R}} \mathrm{d} x \frac{1}{\sqrt{2 \pi \hbar}} \exp(-\mathrm{i} x p) \psi(x).$$
You can convince yourself, using e.g., Gaussian wave packets as an example that, if $P_{\psi}(x)$ is sharply peaked around some position $x_0$, then $\tilde{P}_{\psi}(p)$ has a pretty wide spread, i.e., the more localized your particle is the less well known is its momentum and vice versa. That's another point of view on the Heisenberg uncertainty relation, but it says the same thing: It's about the impossibility to prepare a particle such that both position and momentum are very precisely determined. Rather the standard deviations of these particles always fulfill $\Delta x \Delta p \geq \hbar/2$, and this uncertainty relation is a property of the possitiblity of the particle to be prepared in any state you like. It doesn't say anything about the possibility or impossibility to measure position and momentum accurately.

It's clear that there cannot be a general law of physics concerning measurements, because it depends on the specific measurement device, how the quantity of interest is measured and how accurately that can be done with the given device, i.e., you have to analyze the specific device to understand how precise it can measure the quantity under consideration and what the systematical and statistical uncertainties are. Often the latter is the much more challenging task of experimental physics than to construct a measurement device per se!

 

[Fra]

All the explanations given already are good, and effectively follows from the definition of momentum in quantum mechanics... which leads to a subtle point which may explain some confusion in the philosophical or ontological transition from classical mechanics to "information mechanics":

That's circular logic in relying upon the uncertainty principle while at the same time, trying to prove it.

How about this. Pretend you're back in time before the uncertainty principle even existed, and the experiment showed that the recoil of the target was causing an issue.

Think about how do you define momentum in back in the days of "classical mechanics"? It was essentially defined as p=mv.

And compare with how momentum is defined in quantum mechanics. Here momentum is defined by means of an operator acting on an "information state", which is probabilistic in nature.

One can argue that using the second definition, already implies the HUP, as it follows from the definition via the conjugate relation of a fourier transform. So how we "explained something" by changing the definition, which admittedly may appear silly.

But the reason for this is still that the empirical foundation of QM is more solid. You can construct this from essentially considering a distinguishable event labeled by x, with a time stamp. From the information about this (or the distributions, and adopting the wavefunction) can can define conjugate variables.

To construct the "classical mechanics" defintion in QM relating only to empirical "samples", we would need to make up a new distinguishable events p or v. But then these have an undefined relation, and we don't know how to combined information about them in a single "state of information". (QM solves it be allowing one state to encode simultaenously information of non-commutative measurements; how would you do that in classical mechanics?)

So the only think one need to accept is, why do we "redefine" things in QM? It's because the empirical justification of the old definitions, can't be rooted in the same way in theory which puts more emphasis on detection events. We can't define things from mental pictures of bullets flying, we need to define them in terms of empirically accessible information. Ie. in principle detector counts; and derivetives thereof.

/Fredrik