Uncertainty versus impossibility of measurement

[SamRoss]

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.

I read the section you recommended, 4.3, which had to do with the double-slit experiment and a quantum eraser. It was an interesting discussion about how knowledge of a system results in the loss of interference but I'm not sure how it ties into this discussion. Did I read the wrong part?

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.

The generalised UP doesn't seem to me to provide any new insight into this discussion than the HUP. In fact, many of the posts so far have already implied the GUP by discussing spin operators rather than position and momentum. I agree that disturbing a system is often a red herring when the goal is an understanding of the HUP or GUP but isn't it relevant in a discussion about simultaneous measurements?

 

[PeroK]

I read the section you recommended, 4.3, which had to do with the double-slit experiment and a quantum eraser. It was an interesting discussion about how knowledge of a system results in the loss of interference but I'm not sure how it ties into this discussion. Did I read the wrong part?
The generalised UP doesn't seem to me to provide any new insight into this discussion than the HUP. In fact, many of the posts so far have already implied the GUP by discussing spin operators rather than position and momentum. I agree that disturbing a system is often a red herring when the goal is an understanding of the HUP or GUP but isn't it relevant in a discussion about simultaneous measurements?

What precisely is the outstanding question?

 

[SamRoss]

What precisely is the outstanding question?

As far as I'm concerned, my original question has been answered satisfactorily by you and others. The HUP does not tell us that we can't measure the position and momentum at the same time but there are other reasons why we cannot do so. I do not have any outstanding questions at the moment; I was simply responding to your post. In particular, I was a bit confused by the suggestion that I should focus on the generalized uncertainty principle which I thought we already established is not helpful in understanding why position and momentum cannot be measured simultaneously. As vanhees71 suggested and A. Neumaier confirmed, there is a separate disturbance inequality which is more useful here. In any event, I will go back and read the analysis of the HUP vs. the "old" HUP as you suggest. It sounds interesting. Thanks everybody for your help!

 

[SamRoss]

The HUP does not tell us that we can't measure the position and momentum at the same time but there are other reasons why we cannot do so.

Have I spoken too soon? Here's a quote from pg. 19 of Cresser's Quantum Physics Notes: "However, the uncertainty relation does not say that we cannot measure the position and the momentum at the same time. We certainly can, but we have to live with the fact that each time we repeat this simultaneous measurement..." So just to be clear, Cresser is wrong here, right? It has been the conclusion of this thread that position and momentum cannot be measured simultaneously albeit for reasons other than the uncertainty principle.

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

 

[PeterDonis]

It has been the conclusion of this thread that position and momentum cannot be measured simultaneously albeit for reasons other than the uncertainty principle.

It depends on what you mean by "measure position and momentum simultaneously".

It is not possible to apply two different measurement operators in the same measurement. So if by "position" and "momentum" we mean the two different measurement operators that usually are called by those names in QM textbooks, we can't measure them both simultaneously because we can't apply both of them in the same measurement; we have to pick one. (Strictly speaking, what I have just said only applies to operators that do not commute, but this is the case for the QM textbook position and momentum operators, and for spin operators in different directions.)

However, it is possible to find single measurement operators that, when applied in a measurement, give information about both position and momentum. For example, we can find operators of the kind @A. Neumaier has described, which, as I said in post #30, give results that say things like: "the position is within the range x1 to x2, and the momentum is within the range p1 to p2". This is not an exact measurement of either position or momentum, but it gives information about both position and momentum.

Probably the quote in the text you refer to means the second of the two options above.

 

[A. Neumaier]

You can only measure one operator with a single measurement, but the choice of possible operators to measure is much wider than just "position" and "momentum" as those operators are described in introductory QM textbooks. (And pretty much all of the operators described in introductory QM textbooks are highly idealized ones that can never be exactly realized in practice.) In particular, it is possible to choose an operator that gives results that say something like: "the position is within some range x1 to x2, and the momentum is within some range p1 to p2", where different operators of this type will have different widths of the ranges for position and momentum, and the product of the two ranges can be no smaller than a finite limit.

More precisely, the product in question is the product of the widths of the ranges.

For example, consider the operator $\Delta:=(Q-x)^2/x_0^2+(P-p)^2/p_0^2$, where $x_0$ and $p_0$ are suitable units of position and momentum. One can measure (in principle) one of its eigenvalues $\delta$ and interpret the result as verification of the joint values $x\pm\sqrt{\delta}x_0$ of $Q$ and $p\pm\sqrt{\delta}p_0$ of $P$. But the spectrum of the operator $\Delta$ is bounded from below by a constant depending on $x_0$ and $p_0$ in a way such that the resulting accuracy is never better than what the uncertainty relation says.

 

[PeterDonis]

More precisely, the product in question is the product of the widths of the ranges.

Yes, agreed.

 

[SamRoss]

For example, we can find operators of the kind @A. Neumaier has described, which, as I said in post #30, give results that say things like: "the position is within the range x1 to x2, and the momentum is within the range p1 to p2".

But the spectrum of the operator $\Delta$ is bounded from below by a constant depending on $x_0$ and $p_0$ in a way such that the resulting accuracy is never better than what the uncertainty relation says.

I didn't know this. Thank you. I had been assuming that if it were possible somehow to make a measurement of position and momentum then both of those measurements would be exact even if they were unpredictable beforehand.

 

[PeroK]

I didn't know this. Thank you. I had been assuming that if it were possible somehow to make a measurement of position and momentum then both of those measurements would be exact even if they were unpredictable beforehand.

The eigenstates of position and momentum are physically not realisable, so in principle these measurements are never exact.

 

[Demystifier]

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?

If you have the book "The Quantum Theory of Motion" by Holland, there is a section in this book that reviews those results.

 

[Interested_observer]

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.

I would agree with the latter statement if the word "physically" is replaced with the word "theoretically". In my opinion this embodies the whole "problem" with the interpretation of quantum theory...trying to make the physical reality "match" the theory. Bass ackwards.

 

[vanhees71]

There's nothing problemantic with the fact that "momentum eigenstates" (plane waves) are not elements of the Hilbert space but of the dual of the dense subspace where position and momentum operators are defined. Mathematically the somewhat sloppy language of physicists is formulated as the socalled "rigged Hilbert space", which makes the sloppy language rigorous. As far as the "1st quantization formulation" of non-relativistic quantum theory is concerned, there are no problems with this theory whatsoever. There are also no contradictions from experiment of quantum theory, when extended to relativistic quantum field theory, which however has some mathematical problems left if you want to go beyond standard perturbation theory of effective field theories. There's nothing left to match with "reality" (which for a physicist is just what can be measured).

The problem with the question about "simultaneous measurements" indeed is that you need to give a precisely defined real-world measurement prescription. That holds for any concrete measurement, also for single-observable measurements or the simultaneous measurement of compatible observables. Only then you can give a specific answer to the question, in which sense the measurement disturbs the measured system. For this you need a model for the interaction between measurement devices and the system, as is given by the two older papers mentioned earlier in the thread. For a modern state-of-the art treatment see the papers/books by Busch (also cited above in this thread).

 

[atyy]

Have I spoken too soon? Here's a quote from pg. 19 of Cresser's Quantum Physics Notes: "However, the uncertainty relation does not say that we cannot measure the position and the momentum at the same time. We certainly can, but we have to live with the fact that each time we repeat this simultaneous measurement..." So just to be clear, Cresser is wrong here, right? It has been the conclusion of this thread that position and momentum cannot be measured simultaneously albeit for reasons other than the uncertainty principle.

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

I'd say he's wrong for the reason that PeterDonis gives in post #4.

PeterDonis charitably gives an option in post #35 that makes Cresser right, but basically those measurements of position and momentum are not accurate. An accurate measurement of position is one whose outcomes are distributed according to $|\psi(x)|^2$. So if you want accurate measurements of position and momentum, they cannot be simultaneously performed because the position and momentum operators don't commute. If you want inaccurate measurements of position and momentum, then they can be simultaneously performed.

https://arxiv.org/abs/quant-ph/0609185
"The uncertainty principle is usually described, rather vaguely, as comprising one or more of the following no-go statements, each of which will be made precise below:
(A) It is impossible to prepare states in which position and momentum are simultaneously arbitrarily well localized.
(B) It is impossible to measure simultaneously position and momentum.
(C) It is impossible to measure position without disturbing momentum, and viceversa."

[Obviously they use "uncertainty principle" loosely because they are experts. The usual textbook uncertainty principle refers to (A). ]

 

[atyy]

The eigenstates of position and momentum are physically not realisable, so in principle these measurements are never exact.

In principle position measurements can be exact. It doesn't matter if position eigenstates are not physically realizable, because the exact measurement of position doesn't require or result in the system being in a position eigenstate.

 

[PeterDonis]

the exact measurement of position doesn't require or result in the system being in a position eigenstate

What do you mean by "the exact measurement of position"?

 

[atyy]

What do you mean by "the exact measurement of position"?

By exact measurement of position (more usually called sharp or precise measurement), I mean a measurement that yields outcomes distributed according to $P(x) = |\psi(x)|^2$. The part in the usual "elementary rule of thumb QM" that is problematic is that after a position measurement, the system collapses into a position eigenstate. That is problematic because position eigenstates are not physical since they are not square integrable. Technically, we can apply the Born rule without specifying any post-measurement state, ie. apply the Born rule without collapse, if no measurement is made after the position measurement. We can also fix that by having it collapse into a square integrable state instead of a position eigenstate. There is a less handwavy discussion in section 2.3.2 of https://arxiv.org/abs/0706.3526.

 

[PeterDonis]

There is a less handwavy discussion in section 2.3.2 of https://arxiv.org/abs/0706.3526.

I'll take a look.

 

[martinbn]

By exact measurement of position (more usually called sharp or precise measurement), I mean a measurement that yields outcomes distributed according to $P(x) = |\psi(x)|^2$. The part in the usual "elementary rule of thumb QM" that is problematic is that after a position measurement, the system collapses into a position eigenstate. That is problematic because position eigenstates are not physical since they are not square integrable. Technically, we can apply the Born rule without specifying any post-measurement state, ie. apply the Born rule without collapse, if no measurement is made after the position measurement. We can also fix that by having it collapse into a square integrable state instead of a position eigenstate. There is a less handwavy discussion in section 2.3.2 of https://arxiv.org/abs/0706.3526.

Why not use rigged Hilbert spaces? That's what they are for.

 

[vanhees71]

To put it in more qualitative physical form: Of course, in a strict sense there's no perfectly exact measurement of position in the sense that you determine it to a geometrical point, i.e., you can measure the position of a particle only with a finite resolution, but there is no fundamental principle that sets a limit to this resolution. So given a particle prepared in some quantum state $\hat{\rho}$ (which may be a pure or mixed state) to measure the $\Delta x$ of the uncertainty relation you have to measure the position of the particle at a resolution much higher than $\Delta x$, and you have to repeat this sufficiently often on an ensemble of particles prepared in the specific state $\hat{\rho}$ such that you get the standard deviation $\Delta x$ at sufficient statistical significance (what's "sufficient" is defined by the experimentalist of course).

The same holds true for a single-momentum measurement, and what's meant by the textbook Heisenberg uncertainty relation $\Delta x \Delta p \geq 1/2$ refers to such single-position and single-momentum measurements each with a resolution much better than $\Delta x$ and $\Delta p$, respectively. It's not referring to a "simultaneous measurement of postion and momentum" nor to its theoretical possibility or impossibility.

To make statements about such a simultaneous measurement and how accurate you can make it in principle, you have to specify concrete measurement devices to do so.

One can, however, also look at it at from the point of view of information theory. Then the formalism gives you a theoretical limitation for any such simultaneous measurements. The argument is not that difficult. You suppose that you have measured somehow position and momentum on an ensemble such that you know that the average position and momentum are $\langle x \rangle$ and $\langle p \rangle$ and the standard deviations are $\Delta x^2$ and $\Delta p^2$. From the Shannon-Jaynes principle of maximum entropy this gives a statistical operator of minimal prejudice (maximum entropy),
$$\hat{\rho}=\frac{1}{Z} \exp[-\lambda_1 (\hat{x}-x_0)^2-\lambda_2 (\hat{p}-p_0)^2 ],$$
where $\lambda_1$, $\lambda_2$, and $x_0$ and $p_0$ are convenient choices of Lagrange parameters to fulfill the conditions that the expectation values and standard deviations of position and momentum take the given values. You find the details of the further calculation in

https://itp.uni-frankfurt.de/~hees/publ/stat.pdf

It turns out that you have to set $x_0=\langle x \rangle$ and $p_0=\langle p \rangle$. The partion sum is
$$Z=\frac{1}{2 \sinh(\sqrt{\lambda_1 \lambda_2}/2},$$
and the standard deviations simply
$$\Delta x^2=-2 \partial_{\lambda_1} \ln Z=\sqrt{\frac{\lambda_1}{4 \lambda_2}} \coth \left(\frac{\sqrt{\lambda_1 \lambda_2}}{2} \right),$$
$$\Delta p^2=-2 \partial_{\lambda_2} \ln Z=\sqrt{\frac{\lambda_2}{4 \lambda_1}} \coth \left(\frac{\sqrt{\lambda_1 \lambda_2}}{2} \right).$$
This implies that
$$\Delta x \Delta p = \frac{1}{2}\coth \left(\frac{\sqrt{\lambda_1 \lambda_2}}{2} \right) \geq \frac{1}{2},$$
i.e., no matter how accurately you measure simultaneously position and momentum, their standard deviations fulfill the Heisenberg uncertainty relation.