# The clarity of Heisenberg uncertainty

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Narasoma
We know from basic quantum mechanics that the Heisenberg uncertainty states that position and momentum can not be measured simultaneously with arbitrary precision.

My question is, is this relation is due to the nature of the quantum system itself, or "merely" unbreachable limitation to our measurement apparatus?

Gold Member
My question is, is this relation is due to the nature of the quantum system itself, or "merely" unbreachable limitation to our measurement apparatus?
Commutation relation or Robertson's inequality belongs to quantum nature. AFAIK the idea of Heisenberg's uncertainty relation have been investigated in more detail on measurement error and natural deviation, and some progress including Ozawa’s relation was brought up. You might be interested in these anaysis.

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gentzen, Narasoma and topsquark
Gold Member
MHB
We know from basic quantum mechanics that the Heisenberg uncertainty states that position and momentum can not be measured simultaneously with arbitrary precision.

My question is, is this relation is due to the nature of the quantum system itself, or "merely" unbreachable limitation to our measurement apparatus?
It's due to the quantum nature of the system.

QM is a probabilistic Science: measurements of a system generate distributions. What you get when you measure a quantity is a data set that measures, not a single particle, but the same property of a system of particles. Each measurement will give a different eigenvalue so there will be a statistical variance in the measurements and it is this that causes the Uncertainty Principle.

Now, technology is certainly a factor. The UP predicts the best that we can do. Obviously, we need to be able to measure it that well.

-Dan

Narasoma
Gold Member
2022 Award
We know from basic quantum mechanics that the Heisenberg uncertainty states that position and momentum can not be measured simultaneously with arbitrary precision.
Although it is often formulated in this way even in physics textbooks, it's wrong. The correct statement is that in any state a particle can be prepared in the standard deviations of position-vector and momentum components in the same direction obey the uncertainty relation:
$$\Delta x \Delta p \geq \hbar/2.$$
That means if you prepare very often the particle always in the same quantum state and measure the position component very accurately such that you can measure the uncertainty ##\Delta x## and then again you prepare the particle very often in the same state but measure momentum accurately such that you can measure the uncertainty ##\Delta p##, these two uncertainties (standard deviations) obey this uncertainty relation.
My question is, is this relation is due to the nature of the quantum system itself, or "merely" unbreachable limitation to our measurement apparatus?
This is due to the nature. It's not a limitation of our ability to measure position or momentum accurately. Both quantities can be measured as accurately as you want.

gentzen, LittleSchwinger, malawi_glenn and 2 others
Gold Member
In case it's not intuitive that it's due to the nature of the system and not a measurement problem, nature has been kind enough to provide a real-world demonstration in the form of Bose-Einstein Condensates.

Take a bunch of atoms - normally very discrete objects - and cool them down arbitrarily close to absolute zero. Effectively, all motion stops, meaning their momentum approaches arbitrarily close to zero. Do you now know the exact position and momentum of these erstwhile discrete objects? Nope. Their normally discrete nature starts to smear out into a hazy cloud, until the concept of "position" no longer applies. The more you restrict their momentum, the bigger the smear gets.

Narasoma, topsquark, PeroK and 1 other person
Narasoma
Commutation relation or Robertson's inequality belongs to quantum nature. AFAIK the idea of Heisenberg's uncertainty relation have been investigated in more detail on measurement error and natural deviation, and some progress including Ozawa’s relation was brought up. You might be interested in these anaysis.
1. When I said, measurement, I also said "unbreachable." Unlike classical physics, in which measurement of any physical quantities can be made arbitrary precise, for example due to the technological advancement.

Let me put my question another way. If the UP is due to the nature of the system, then whether we measure it or not, it is fuzzy. If it is due to the unbreachable limitations of the measurement, then the UP comes from the interaction of the system with the apparatus and not necessarily from the quantum system itself.

Now, from the measurement only, we would have the same result, same fuzzyness. But, what I want to know is whether there is a way to confirm whether UP is due to the nature of the system or to the interaction with apparatus.

2. Ozawa relation is a new term for me. Let me check it first.

gentzen
Narasoma
It's due to the quantum nature of the system.

QM is a probabilistic Science: measurements of a system generate distributions. What you get when you measure a quantity is a data set that measures, not a single particle, but the same property of a system of particles. Each measurement will give a different eigenvalue so there will be a statistical variance in the measurements and it is this that causes the Uncertainty Principle.

Now, technology is certainly a factor. The UP predicts the best that we can do. Obviously, we need to be able to measure it that well.

-Dan
I don't undestand how to conclude that the UP is due to the nature of the system from statistical variance. Classical system has variance as well, right? But the UP does not emerge from it.

Narasoma
In case it's not intuitive that it's due to the nature of the system and not a measurement problem, nature has been kind enough to provide a real-world demonstration in the form of Bose-Einstein Condensates.

Take a bunch of atoms - normally very discrete objects - and cool them down arbitrarily close to absolute zero. Effectively, all motion stops, meaning their momentum approaches arbitrarily close to zero. Do you now know the exact position and momentum of these erstwhile discrete objects? Nope. Their normally discrete nature starts to smear out into a hazy cloud, until the concept of "position" no longer applies. The more you restrict their momentum, the bigger the smear gets.
Well. I think I understand the conclusion better this way. Perhaps, I should read solid state physics more. Thanks man.

Gold Member
MHB
I don't undestand how to conclude that the UP is due to the nature of the system from statistical variance. Classical system has variance as well, right? But the UP does not emerge from it.
When we take a measurement we expect that the measuring device will give us a set of values that are clustered around what we believe the actual value will be. That gives us what we call a variance in the data. This is strictly due to the measuring device.

In QM things are a bit different. Even if we had a perfect measuring device we would still get a variance. Say we are dealing with a system that has 4 possible energy states. When we take a measurement we will get a measurement that is one of those states. But we will not get the same measurement each time; we will get a distribution corresponding to how probable each state is likely to be. So we might get 10% in the lowest state, 25% in the next up, 45% in the next up and 20% in the highest. This generates a variance that we cannot get rid of, no matter how good our equipment is. This is what the UP is addressing.

-Dan

vanhees71
Mentor
is this relation is due to the nature of the quantum system itself
Yes.

"merely" unbreachable limitation to our measurement apparatus?
No.

topsquark
Mentor
If the UP is due to the nature of the system, then whether we measure it or not, it is fuzzy.
No, that's not correct. You can make a very precise measurement of one observable, reducing "fuzziness" to be as small as you like. The HUP doesn't put any limitations on that.

What the HUP says is that it is impossible for the system to be in a state that has sharp values for both of a pair of non-commuting observables, like position and momentum. There simply are no such states.

vanhees71 and topsquark
Homework Helper
Gold Member
2022 Award
I don't undestand how to conclude that the UP is due to the nature of the system from statistical variance. Classical system has variance as well, right? But the UP does not emerge from it.
The difference between QM and CM (Classical Mechanics) is more fundamental than the UP. This can be seen by looking at the spin on the electron. A classical object, like a spinning ball, has the following characteristics:

1) The ball has a definite axis of rotation (which can be experimentally detected).
2) The ball can, within reason, have any angular momentum: you can speed up or slow down its rotation.

The electron has the following characteristics:

1) It has no well-defined axis of rotation (this isn't due to "fuzziness").
2) Only a single value of angular momentum is possible: ##\pm \frac \hbar 2##.

The electron is fundamentally not a classical object and these characteristics are not due to limitations in measurement technology.

If you want to learn about QM it's a good idea to start with electron spin, as it is theoretically clear that quantum spin simply does not obey the principles of classical mechanics.

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topsquark
Gold Member
2022 Award
We know from basic quantum mechanics that the Heisenberg uncertainty states that position and momentum can not be measured simultaneously with arbitrary precision.
That's NOT what the Heisenberg uncertainty states. It's a misinterpretation, unfortunately found in most popular-science publications and even in some textbooks.

The quantum state operationally describes a preparation procedure. E.g., you use some trap to localize an electron in some region of space. The meaning of the quantum state is to provide the probablities for the outcome of measurements on the particle. You can measure any observable of the particle as accurately as you wish. There are no fundamental restrictions to the accuracy from quantum mechanics. E.g., you can measure the momentum or the position of the electron as precisely as you wish.

The Heisenberg-Robertson uncertainty principle then states that for any quantum state (preparation of the system) the standard deviations of two observables ##A## and ##B## represented by the corresponding self-adjoint operators ##\hat{A}## and ##\hat{B}## obey the uncertainty relation,
$$\Delta A \Delta B \geq \frac{1}{2} |\langle [\hat{A},\hat{B}] \rangle|.$$
For components of the position and momentum in the same ##x##-direction this implies
$$\Delta x \Delta p_x \geq \frac{\hbar}{2}.$$
As I said you can measure ##x## as precisely as you wish or ##p##.

To test the uncertainty relation you have to prepare many particles in the same state (e.g., putting it in a trap with a certain spatial extent) and measure very accurately its position and do the statistical analysis to get the standard deviation ##\Delta x##. To ensure that you measure the true uncertainty due to quantum fluctuations you must ensure that the resolution of your position measurement is much smaller than this quantum-mechanical standard deviation. Then you can again prepare many particles in the same state and do the same procedure with the momentum measurement.

Another question is the influence of the measurements on the system, i.e., the disturbance of the system due to measurements. This is a much more complicated issue than the Heisenberg-Robertson uncertainty relation, which can be derived after a few weeks in the QM1 introductory lecture. For more on the measurement-disturbance relations, see

P. Busch et al, Quantum Measurement, Springer (2016)
https://dx.doi.org/10.1007/978-3-319-43389-9

Also the issue of how to measure incompatible observables at the same time on a single particle. For this you have to extent the description of measurements from the standard idealized von Neumann (projective) measurements to socalled positive-operator-valued-measure measurements, which is also a quite complicated issue. For more on this, see

A. Peres, Quantum Theory: Concepts and Methods, Kluwer
Academic Publishers, New York, Boston, Dordrecht, London,
Moscow (2002).
My question is, is this relation is due to the nature of the quantum system itself, or "merely" unbreachable limitation to our measurement apparatus?
It's due to the nature of the quantum system itself, as detailed above.

andresB, topsquark, gentzen and 1 other person
Gold Member
We know from basic quantum mechanics that the Heisenberg uncertainty states that position and momentum can not be measured simultaneously with arbitrary precision.

My question is, is this relation is due to the nature of the quantum system itself, or "merely" unbreachable limitation to our measurement apparatus?
No, the Heisenberg uncertainty states that position and momentum can not be simultaneously predicted with arbitrary certainty. Due to the probabilistic nature of quantum mechanics, the quantum uncertainty of an observable tells us that we do not know in advance what the result of measurement will be.

Ultimately, the measuring apparatus must also be described by quantum theory, so the uncertainty is due to the nature of the total quantum system, including both the measured system and the apparatus. If you ask what's the nature of uncertainty in small simple systems without the apparatus, the answer is that we don't know. Different interpretations of quantum theory offer different answers to that question, and we don't know which interpretation, if any, is right. Physicists understand very well how to use quantum theory to make measurable predictions, but they don't fully understand what this means in measurement-independent terms.

topsquark, gentzen, PeroK and 1 other person