# B Possible Loophole in the uncertainty principle

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1. Nov 2, 2017

### RevolutionaryThinker

The uncertainty principle states that any one given person who knows a lot about a given particles velocity will know less about the location, but if then they try to observe the location, the less they will know about the velocity. What if two scientists observed one particle though. One observed its speed and on observed its location. Then after observing they would communicate their results to each other. Due to their being two observers, would this nullify the principles previously mentioned stipulations?

Last edited: Nov 2, 2017
2. Nov 2, 2017

### phinds

No. The HUP doesn't say what you think it says. What it really says is not that you can't measure both the location and velocity simultaneously to any degree of accuracy on a single particle one time, what it says is that if you create IDENTICAL starting conditions over and over the results of making such measurements will have a probability distribution, not be the same every time. In classical mechanics, that is insanity, but in quantum mechanics it's the way the world really works at the micro level.

3. Nov 2, 2017

### StevieTNZ

The number of observers involved, when making a position and/or momentum measurement, does not matter.

4. Nov 2, 2017

Eh... ??

5. Nov 2, 2017

### NFuller

The simple answer is that the first scientist's measurement will be changed by the second scientist's measurement. So only the second measurement will be valid after this.

The HUP is a manifestation of the non-commutative behavior of the momentum and position operators in QM. Non-commutative means that if you apply the operators in different orders, you get different answers. Since this is a B-level thread, I don't want to start throwing a bunch of math at you, but I would recommend that you look into what is meant by these operators and how they relate to the HUP. Perhaps try to get a few intro to QM books, and try to trudge through them.

I know telling you to just look up the math is a dodgy answer, but you really need to have a firm understanding of the mathematical formalism of QM to have any hope of getting a satisfactory answer to such questions.

6. Nov 2, 2017

### Staff: Mentor

This doesn't seem like a good way to describe what's going on. The first measurement's result was what it was; the second measurement doesn't change the fact that a particular result was observed on the first measurement.

What the second measurement does change is the state of the particle that you have to use to predict the results of future measurements. After the first measurement, but before the second, you use the state indicated by the result of the first measurement; after the second measurement, you use the state indicated by the result of the second measurement.

For example, if the first measurement observed momentum and the second observed position, the state indicated by the second measurement's result would have a much larger uncertainty in momentum than the state indicated by the first measurement's result; so even though the first measurement might have given a very precise result for the particle's momentum, the state indicated by that result can no longer be used to make predictions after the second measurement is made, and its precision no longer does you any good in making predictions.

7. Nov 3, 2017

### NFuller

Yes, this is what I'm trying to say. I apologize if it wasn't clear. It should be obvious that making future measurements will not change a value recorded in the past.

8. Nov 3, 2017

### Demystifier

The point is that observation of a system involves a physical interaction with the measuring apparatus, which affects the system you want to observe. If there are two measuring apparatuses, one measuring position and the other momentum, the result is an interaction that affects the system in such a way that at the end you don't know very well any of the two quantities.

9. Nov 3, 2017

### phinds

Yes, but I think you are complicating things for a novice because you give the impression that the HUP is a measurement problem, which it is not.

10. Nov 6, 2017

### Demystifier

You have a point, but I think your answer to his question does not say what happens if position and momentum are attempted to be measured simultaneously, which is how I interpreted his question.

Last edited: Nov 6, 2017
11. Nov 6, 2017

### vanhees71

Well, the HUP you learn in the first few weeks of the QM 1 lecture (one can call it the Heisenberg-Robertson uncertainty relation) independently of the interpretation refers to the implication of the meaning of the quantum state as a probability (distribution) for the outcome of measurements.

In general form it says that for two observables $A$ and $B$, represented by self-adjoint operators $\hat{A}$ and $\hat{B}$ for any state, represented by the Statistical Operator $\hat{\rho}$ for the standard deviations for the outcomes of measurements of these observables you have
$$\Delta A \Delta B \geq \frac{1}{2} \left |\langle [\hat{A},\hat{B}] \rangle_{\rho} \right |,$$
where the expectation value is given by
$$\langle O \rangle_{\rho}=\mathrm{Tr} (\hat{\rho} \hat{O}),$$
and the standard deviation by
$$\Delta O=\langle \hat{O}^2 \rangle_{\rho} - \langle \hat{O} \rangle^2.$$
The question, in which sense quantitatively a measurement of observable $A$ influences the outcome of subsequent measurements of another observable $B$ is much less easy to answer and indeed still a matter of debate. For sure it depends on the details of each concrete measurement. There's a recent textbook on the subject:

P. Busch et al, Quantum Measurement, Springer 2016

12. Nov 9, 2017

### Zafa Pi

As @phinds said, you have not stated the HUP correctly., but I find him a bit vague. And as usual what @vanhees71 said is precise and correct, but hardly level B.
I'll give it a try.
Suppose you have a zillion particles in the same state ( prepared in an identical fashion). Now suppose you measure the position of 1/2 zillion of them (the measurements can be as accurate as you like) and measure the velocity of the remainder. Next you calculate the standard deviation of the position measurements, call it Δposition, and do the same with the velocity measurements and call it Δvelocity.
Then the HUP states that there is a constant k > 0 which is independent of the initial state such that Δposition⋅Δvelocity ≥ k.
I.e., you cannot make both standard deviations (for a given state) as small as you like.

Last edited: Nov 9, 2017
13. Nov 9, 2017

### DrChinese

I'd say that is pretty good.

14. Nov 9, 2017

### jerromyjon

I like all of your replies, but I'd like to go one step deeper and say that any position measurement you make to any degree of certainty means that the particle you measured inherently has less certain momentum. The uncertainty is not in the measurements or the means used to measure, it is in the quantum state which has been proven to inherently contain probabilistic uncertainty.

15. Nov 9, 2017

### Zafa Pi

If this is directed at me, thank you. However, assuming I understand you correctly, I beg to differ. A quantum state is a fixed unit vector in a Hilbert space (old school) and is not inherently probabilistic. The stochastic aspect of QM lies with the measurement, which is a random variable.
I will provide the simplest concrete example that I know.

Let |θ⟩ = [cosθ,sinθ] which is a unit vector in the Hilbert space R². This is meant to represent the state of a polarized photon whose axis of polarization is θ degrees (counterclockwise) from the horizontal. There is nothing stochastic about this state.
Let Pauli Z and X be two observables/measurement operators. These represent measuring with polarization analyzers set at 0º and 45º. (This can be done with just polarized lenses, but is harder.)

Measuring |θ⟩ with Z yields 1 with probability cos²(θ-0), and -1 with probability sin²(θ-0).
Measuring |θ⟩ with X yields 1 with probability cos²(θ-45) and -1 with probability sin²(θ-45).
Notice that variance of measuring by Z + variance of measuring by X = 1 independent of θ, which is analogous to the HUP, in that we can't make both variances small.