# Uncertainty versus impossibility of measurement

• B
Gold Member

## Summary:

Can you really not measure the position and momentum of a particle simultaneously or is it just that there is uncertainty in what you will get?

## Main Question or Discussion Point

The uncertainty principle tells us that there is no state that a particle can be in such that we can predict with certainty both what the result of a position measurement will be and what the result of a momentum measurement will be. This statement is not the same as saying we can't measure the position and momentum of a particle at the same time. The latter statement is often repeated. Is it correct?

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Summary:: Can you really not measure the position and momentum of a particle simultaneously or is it just that there is uncertainty in what you will get?
The HUP says that there is a limit to the precision with which the values for certain pairs of physical quantities of a particle can be predicted from initial conditions. In short, this means that you can very much measure both, but only to some limited precision which is less than if you measured either one of them but not the other.
For instance, to get a high accuracy measurement of position of the particle, you need to interact with it using considerable energy. It might as well have been kicked by a mule; it goes skittering off somewhere, momentum completely unmeasured.

Gold Member
The HUP says that there is a limit to the precision with which the values for certain pairs of physical quantities of a particle can be predicted from initial conditions. In short, this means that you can very much measure both, but only to some limited precision which is less than if you measured either one of them but not the other.
I do not believe your second statement follows from the first. In fact, this is what my whole question is about. The statement "The outcome of the measurement cannot be predicted with certainty" is often taken to be synonymous with the statement "The measurement cannot be made precisely". But the latter does not follow from the former. As an analogy, consider a long-jumper in the Olympic games. The statement "No one can predict with certainty how far the long-jumper will jump" is not the same as "The distance the long-jumper jumped cannot be measured precisely." Now, if a particle is in some eigenstate of momentum, we would be able to predict with certainty the outcome of a measurement of momentum but not the outcome of a position measurement. My question is, can both of these things be measured at the same time even if only one of them has an outcome which we can predict with certainty?

PeroK
PeterDonis
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The uncertainty principle tells us that there is no state that a particle can be in such that we can predict with certainty both what the result of a position measurement will be and what the result of a momentum measurement will be.
More precisely, that there is no state which is an eigenstate of both the position operator and the momentum operator. There cannot be any such state because the two operators do not commute.

This statement is not the same as saying we can't measure the position and momentum of a particle at the same time. The latter statement is often repeated. Is it correct?
Yes, but not for any reason to do with the uncertainty principle. It's more fundamental than that.

What would it mean to measure both position and momentum (or any two different observables) "at the same time"? Different observables correspond to different operators. You can only apply one operator in any given measurement. So you can't measure two different observables "at the same time" because you can't make two different measurements "at the same time". You can only make one.

atyy and SamRoss
Demystifier
My question is, can both of these things be measured at the same time even if only one of them has an outcome which we can predict with certainty?
No, we cannot. More precisely, if we turn on two apparatuses (say one for measuring the momentum and the other for measuring the position) at the same time, both apparatuses will produce definite outcomes, but we will not be able to predict with certainty those outcomes for neither of them.

A. Neumaier
2019 Award
if a particle is in some eigenstate of momentum
A particle cannot be in a momentum eigenstate since a particle's wave function must be normalizable.

The problem with your initial question is that the meaning of ''simultaneous measurement of position and momentum'' is quite fuzzy when the required precisions are high. Within a higher precision than given by the uncertainty principle, the meaning cannot be unambiguously related to theory, and hence is physically meaningless.

vanhees71
PeroK
Homework Helper
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Summary:: Can you really not measure the position and momentum of a particle simultaneously or is it just that there is uncertainty in what you will get?

The uncertainty principle tells us that there is no state that a particle can be in such that we can predict with certainty both what the result of a position measurement will be and what the result of a momentum measurement will be. This statement is not the same as saying we can't measure the position and momentum of a particle at the same time. The latter statement is often repeated. Is it correct?
The central point you are making is valid. The HUP does not, per se, put any constraints on what you can choose to measure. The theoretical problem with measuring position and momentum simultaneously is the state of the particle after that measurement. There are a number of examples where you can try to outwit the UP in some way - and try to force the simultaneous measurement of incompatible observables. In each case, the actual physical constraints that prevent this may be different. Einstein and Bohr famously debated this for many years. There's also an interesting discussion of it here (section 4.3) and also (in a later section I think) on attempts to measure spin in two directions simultaneously:

https://physics.mq.edu.au/~jcresser/Phys304/Handouts/QuantumPhysicsNotes.pdf

SamRoss
PeroK
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A particle cannot be in a momentum eigenstate since a particle's wave function must be normalizable.

The problem with your initial question is that the meaning of ''simultaneous measurement of position and momentum'' is quite fuzzy when the required precisions are high. Within a higher precision than given by the uncertainty principle, the meaning cannot be unambiguously related to theory, and hence is physically meaningless.
The point is that there is nothing in the theory that stops you trying; and, it's not immediately obvious from the theory what stops you succeeding (in a simultaneous measurement of two incompatible observables).

A. Neumaier
2019 Award
The point is that there is nothing in the theory that stops you trying; and, it's not immediately obvious from the theory what stops you succeeding (in a simultaneous measurement of two incompatible observables).
My point was that the notion of simultaneous measurement at extremely high precision is theoretically ill-defined. Hence though nothing prevents one to try, there is no way to check whether one succeeded!

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DrChinese
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...My question is, can both of these things be measured at the same time even if only one of them has an outcome which we can predict with certainty?
Actually, a variation of this can be performed using spin-entangled pairs of particles.

1. You can measure A's polarization at angle X. A subsequent measurement of A at angle X will confirm the outcome of the first measurement. So will a measurement pf particle B at angle X. You can confirm this arrangement at any spin value for X. Of course, and actual measurement of B at angle X is redundant since they are spin entangled.)

2. Next you measure particle B at some non-commuting angle Y (relative to X). This after measuring particle A at angle X as per 1.

The question is: do you now know particle A's value for spin at angle Y, as well as particle B's value at angle X? Because if you did, you would now "beat the HUP". After all, A and B are entangled so a measurement on one yields information about the other. What do you think?

PeterDonis
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if we turn on two apparatuses (say one for measuring the momentum and the other for measuring the position) at the same time
First, what kind of apparatuses are you thinking of, and how could they both be turned on at the same time for the same particle?

Second, supposing this could be done, what operator would such a measurement realize? As I said in post #4, you can only apply one operator per measurement.

PeterDonis
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the meaning of ''simultaneous measurement of position and momentum'' is quite fuzzy when the required precisions are high
Isn't it also "fuzzy" (as in, not well defined) when the precisions aren't high? How can you measure two different operators simultaneously?

PeterDonis
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there is nothing in the theory that stops you trying
How would you describe "trying" using the math of the theory?

Gold Member
The question is: do you now know particle A's value for spin at angle Y, as well as particle B's value at angle X? Because if you did, you would now "beat the HUP". After all, A and B are entangled so a measurement on one yields information about the other. What do you think?
Having measured A’s polarization at angle X and B’s polarization at angle Y, it would seem like we have enough information to predict the outcome of an experiment measuring either A at X or A (same particle) at Y. However, this should not be possible though I’m not well-versed enough to state where exactly the argument breaks down.

Demystifier
vanhees71
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2019 Award
The Heisenberg-Robertson uncertainty relation for any position and momentum component in the same direction says that for any (pure or mixed) state fulfills $\Delta x \Delta p_x \geq \hbar/2$. Here $\Delta x$ and $\Delta p_x$ are defined as the standard devitions of the corresponding probability distributions. This means that it is impossible to prepare a particle in any state such that this product is smaller than $\hbar/2$. This means the better you determine the particle's position in $x$ direction the less determined is the particle's momentum component in the $x$ direction and vice versa.

It doesn't say anything about, how precisely you can measure $x$ or $p_x$ nor how precisely you can meausure $x$ and $p_x$ "simultaneously". That's a much more complicated question, which needs a lot of careful analysis for any given concrete measurement device that you use to do the one or the other measurement of position and/or momentum.

Often the HUP is errorneously mixed up with the other issue of disturbance of the system by a measurement, which is due to Heisenberg's first paper on the subject, which was immediately criticized by Bohr precisely of this misunderstanding: Again: The HUP as derived in standard textbooks thus says something about the impossibility of state preparation such $x$ and $p_x$ have standard deviations not fulfilling the uncertainty relation, but it doesn't say anything about what's measureable in principle and how much a precise measurement of the one observable disturbs the determination of the other observable.

For a detailed study on the latter question and in which sense Heisenberg's old "micrsoscope agument" can be stated more correctly, see

https://arxiv.org/abs/1306.1565

SamRoss and PeroK
DrChinese
Gold Member
Having measured A’s polarization at angle X and B’s polarization at angle Y, it would seem like we have enough information to predict the outcome of an experiment measuring either A at X or A (same particle) at Y. However, this should not be possible though I’m not well-versed enough to state where exactly the argument breaks down.
The simple answer is that you don't know the "extra" information because it would defy the HUP.

This was in fact the essence of the famous EPR paper of 1935, written long before entanglement could be tested. EPR thought you could learn (using the described technique for momentum and position) more information than QM (the HUP) would allow. But now we know that the HUP applies here too. Therefore, it is NOT an issue of being unable to measure both simultaneously - because this setup allows that. And it clearly shows that the HUP is a valid limit to what any particle has for non-commuting observables.

SamRoss
Gold Member
Second, supposing this could be done, what operator would such a measurement realize? As I said in post #4, you can only apply one operator per measurement.
1. If I understand your posts correctly, the problem with measuring position and momentum simultaneously does not have to do with the HUP. Instead, simultaneously applying the techniques normally used to measure position with the techniques normally used to measure momentum does not result in knowledge of position and momentum but rather knowledge of a completely new operator. Is that right?
2. This is related to something I've always been curious about - is there any way to look at the equipment in a laboratory and say, "Well, this setup is clearly designed to measure the such-and-such operator"?
3. Further, while the HUP may not say that non-commuting operators cannot be measured simultaneously, is there still a proof of this statement?

Gold Member
The HUP as derived in standard textbooks thus says something about the impossibility of state preparation such $x$ and $p_x$ have standard deviations not fulfilling the uncertainty relation, but it doesn't say anything about what's measureable in principle and how much a precise measurement of the one observable disturbs the determination of the other observable.
So there is a "disturbance" inequality which is separate from the standard HUP?

PeterDonis
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2019 Award
simultaneously applying the techniques normally used to measure position with the techniques normally used to measure momentum does not result in knowledge of position and momentum but rather knowledge of a completely new operator. Is that right?
What I was saying is that you can only apply one operator at a given measurement, and position and momentum are two different operators, so you have to pick one. I'm not sure what it would even mean to simultaneously apply a position measurement technique and a momentum measurement technique, let alone what "completely new operator" such an application would correspond to. IMO, the first thing anyone who claims such a thing can be done should do is to explicitly describe how they would do it.

PeterDonis
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Both behind paywalls, unfortunately, so I can't read more than the abstracts, which aren't very informative. And these papers are from the mid-1960s. Why aren't they in the public domain by now?

PeroK
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So there is a "disturbance" inequality which is separate from the standard HUP?
If you read those notes by JD Cresser that I linked to, there is some excellent analysis of the HUP and the original "old" HUP.

In any case, QT has the HUP and the generalised UP (for any non-commuting operators). I'd focus on those. The concept of measurements "disturbing" a system turns out to be a red herring, IMO.

A. Neumaier
2019 Award
Isn't it also "fuzzy" (as in, not well defined) when the precisions aren't high? How can you measure two different operators simultaneously?
There is no problem to measure both to macroscopic accuracy. The fuzziness involved is far below the accuracy of the measurements, hence irrelevant. For high precision measurements, this is no longer the case and one needs careful investigations.

PeterDonis
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2019 Award
There is no problem to measure both to macroscopic accuracy.
In the sense that we can just use classical mechanics as an approximation for this case, yes. But classical mechanics doesn't model measurements with operators.

What I'm asking is, for the quantum case, where we can't use classical mechanics as an approximation, and we are modeling measurements with operators, how would we model "measure position and momentum simultaneously"? Or, in the case of spin, something like "measure spin-z and spin-x simultaneously"? Let alone physically realize such a measurement?

A. Neumaier
2019 Award
So there is a "disturbance" inequality which is separate from the standard HUP?
Yes. See the discussion in the references given here and here.
There is no problem to measure both to macroscopic accuracy.
What I'm asking is, for the quantum case, where we can't use classical mechanics as an approximation, and we are modeling measurements with operators, how would we model "measure position and momentum simultaneously"?
I gave here a very concrete example of a joint position and momentum measurement of quantum particles.

Note that all complex measurements make use of classical mechanics for the auxiliary part of the measurement setting. Generally, only the part sensitive to the quantum degrees of freedom is handles by quantum mechanics. Otherwise it would be impossible to measure anything quantum at all.

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SamRoss