I On the Heisenberg uncertainty relation

[bhobba]

How are you defining "accuracy" of the measurements?

That would be my question as well.

The HUP applies to idealized measurements of exact accuracy and is a statement about the statistical spread of such.

We all know such idealized measurements do not exist. That's why a state of exact position is modeled by a Dirac Delta function which is a mathematical abstraction - nothing can have zero width and an infinite height. It took a long time for mathematicians to develop a rigorous theory of such involving such luminaries as the great Grothendieck. This I suspect was at least in part from scathing criticisms Von-Neumann made in his classic - Mathematical Foundations of QM. I was first told about it when it was introduced in my class on applied differential equations - I read it (its in the first few pages) and thought - now that's a criticism. My teacher never did tell me the solution - I asked - he just smiled and said - Bill you can look that one up yourself.

Thanks
Bill

 

[bhobba]

Hi Arnold, I found the section about the virtual statistical ensembles from my QM course I took part in Nov-Dec. 2003. It is all written in Romanian, so I need to translate it to German or English. Leave me some time, please. :) [to be continued in the dedicated thread].

Its also in Ballentine.

Thanks
Bill

 

[Mordred]

Wouldn't most of this be under the Von-Neumann measurement problem?

this article has a couple of good sections ie section 2.1
"The probability of detecting a probe position z as a function of time"

http://www.johnboccio.com/research/quantum/notes/13117649.pdf

looks like it has decent details on the Stern_Gerlach experiment and the width aspects under section 3.1
for the probability density of Q after interaction.

Not sure if this is what your after but if it is I saw a decent chapter on this in one of the QFT manuals I've been reading the past week

 

[rubi]

Let's approach it in a deductive way. We define the uncertainty of an observable $A$ in a state $\Psi$ as $\sigma_A^\Psi =\sqrt{\left<\Psi,A^2\Psi\right>-\left<\Psi,A\Psi\right>^2}$. Without loss of generality, let $\left<\Psi,A\Psi\right>=0$ (otherwise, use $A'=A-\left<\Psi,A\Psi\right>$). We are interested in bounds on the product $\sigma_A^\Psi \sigma_B^{U(t)\Psi}$, where $U(t)$ is the time evolution operator. We find:
$$\sigma_B^{U(t)\Psi}=\sqrt{\left<U(t)\Psi,B^2 U(t)\Psi\right>} = \sqrt{\left<\Psi,U^\dagger(t) B U(t) U^\dagger(t) B U(t) \Psi\right>} = \sqrt{\left<\Psi,(U^\dagger(t)BU(t))^2\Psi\right>}=\sigma_{U^\dagger(t) B U(t)}^\Psi=\sigma_{B(t)}^\Psi$$
We have defined the Heisenberg observable $B(t)=U^\dagger(t)B U(t)$. Now, by the usual uncertainty relation, we find
$$\sigma_A^\Psi \sigma_B^{U(t)\Psi}=\sigma_A^\Psi \sigma_{B(t)}^\Psi \geq \frac 1 2 \left|\left<\Psi,\left[A,B(t)\right]\Psi\right>\right|$$

Right now, this is just a mathematical bound on the product of uncertainties $\sigma_A^\Psi \sigma_B^{U(t)\Psi}$, but how does it relate to measurement?
If we decide to measure the operator $A$ at time $t=0$ and we have designed the measurement apparatus to measure at a certain accuracy $\epsilon_A$, which means that the statistical uncertainty in the measurement results will be $\epsilon_A$, then after a measurement, the state $\Psi$ will have collapsed onto a state $\frac{P_A(S)\Psi}{\lVert P_A(S)\Psi\rVert}$ with uncertainty $\sigma_A^{\frac{P_A(S)\Psi}{\lVert P_A(S)\Psi\rVert}} \approx \epsilon_A$. Thus, by our uncertainty relation at different times, we find
$$\sigma_A^{\frac{P_A(S)\Psi}{\lVert P_A(S)\Psi\rVert}}\sigma_B^{U(\Delta t)\frac{P_A(S)\Psi}{\lVert P_A(S)\Psi\rVert}}\geq \frac 1 2 \left|\left<\frac{P_A(S)\Psi}{\lVert P_A(S)\Psi\rVert},\left[A,B(\Delta t)\right]\frac{P_A(S)\Psi}{\lVert P_A(S)\Psi\rVert}\right>\right|=:c_\Psi^{\epsilon_A}$$
Thus, quantum mechanics predicts that if you have measured $A$ with accuracy $\epsilon_A$ at $t=0$, then you can measure $B$ with accuracy at most $\frac{c_\Psi^{\epsilon_A}}{\epsilon_A}$ at time $t=\Delta t$. If you perform the experiment a large number of times, then no matter how many digits your measurement apparatus can display, if you calculate the uncertainty $\sigma_B$ afterwards, you will get a number that exceeds $\frac{c_\Psi^{\epsilon_A}}{\epsilon_A}$. You can't build a device that can produce a smaller uncertainty. This is not because you are a bad engineer, but because quantum mechanics limits your ability to prepare a state in such a way that it produces a statistically robust definite outcome for $B$ at time $t=\Delta t$ (assuming you first measured $A$ with accuracy $\epsilon_A$). (See post #43 for more.)

 

[ShayanJ]

Thanks @rubi, incredible post!
I just have one question. It seems to me that there is no difference between $\epsilon$ and $\sigma$. Why do you use different names for them?
The reason I say this, is that we don't assume the quantum mechanical observables have any "real" value before measurement so it doesn't make sense to talk about the difference between the measured value and the real value. So there is nothing called the accuracy of a single measurement. All we can talk about is the uncertainty $\sigma$. And it seems to me that you're using them as the same quantity too. Because from $\sigma_A \sigma_B \geq c$, you get $\epsilon_B \geq \frac{c}{\epsilon_A}$.

EDIT: Unless $\epsilon$ is interpreted as some kind of a resolution, i.e. the ability of the measuring device to distinguish between two values of the observable. Then it makes sense but again, how do you get $\epsilon_B \geq \frac{c}{\epsilon_A}$ from $\sigma_A \sigma_B \geq c$? The resolution should be a property of our measurement device. How can it be related to any other measurement or whether we have a lot of systems or not?

 

[rubi]

It seems to me that there is no difference between $\epsilon$ and $\sigma$. Why do you use different names for them?

I wanted to distinguish between irreducible quantum uncertainty and experimental shortcomings such as systematic errors. I'm not an experimentalists, so I can't say much about it. $\epsilon$ is the precision of the measurement apparatus. In general, $\epsilon > \sigma$, but if the experimeter has built an ideal device, then they should be almost equal.

I said that after a measurement, the state will collapse onto $P_A(S)\Psi$ with associated uncertainty $\epsilon_A$, but of course, this is an idealization. In general, you will have to take measurement theory into account and put more work into analyzing what the state and its uncertainty will be after a measurement.

 

[Mordred]

nice post Rubi well detailed.

 

[Mentz114]

Let's approach it in a deductive way. We define the uncertainty of an observable $A$ in a state $\Psi$ as $\sigma_A^\Psi =\sqrt{\left<\Psi,A^2\Psi\right>-\left<\Psi,A\Psi\right>^2}$. Without loss of generality, let $\left<\Psi,A\Psi\right>=0$ (otherwise, use $A'=A-\left<\Psi,A\Psi\right>$). We are interested in bounds on the product $\sigma_A^\Psi \sigma_B^{U(t)\Psi}$, where $U(t)$ is the time evolution operator. We find:
[]
This is not because you are a bad engineer, but because quantum mechanics limits your ability to prepare a state in such a way that it produces a statistically robust definite outcome for $B$ at time $t=\Delta t$ (assuming you first measured $A$ with accuracy $\epsilon_A$).

That is much needed and answers a question that has troubled me for ages.

The statement ''non-commuting variables cannot be measured simultaneously' which is bandied about like gospel is actually incomplete and should read "non-commuting variables cannot be measured simultaneously with arbitrary precision". In fact, if the wave function is an not eigenvalue of the operator, then the operator will show some fluctuation. Bohm states this clearly in his book 'Quantum Theory' and leaves the proof as an exercise.

 

[rubi]

Ok, let's say we measured the eigenvalue of the observable $A$ to be $A_i$ using a device with a measurement accuracy $\epsilon_A$. Then (if we neglect measurement theory) the state $\Psi$ will collapse to the state $\Psi_{(A_i,\epsilon_A)}=\frac{P_A(S(A_i,\epsilon_A))\Psi}{\lVert P_A(S(A_i,\epsilon_A))\Psi\rVert}$, where $P_A(S(A_i,\epsilon_A))$ is a projector of $A$ and $S(A_i,\epsilon_A)$ is some Borel set in the spectrum of $A$. For example, it could be that $S=(A_i-\epsilon_A,A_i+\epsilon_A)\cap\mathrm{spec}(A)$. In any case, we get $\sigma_A^{\frac{P_A(S(A_i,\epsilon_A))\Psi}{\lVert P_A(S(A_i,\epsilon_A))\Psi\rVert}}\approx\epsilon_A$. By our uncertaincy relation, we have:
$$\sigma_A^{\Psi_{(A_i,\epsilon_A)}}\sigma_B^{U(\Delta t)\Psi_{(A_i,\epsilon_A)}} \geq \frac 1 2 \left|\left<\Psi_{(A_i,\epsilon_A)},\left[A,B(\Delta t)\right]\Psi_{(A_i,\epsilon_A)}\right>\right|$$
Thus, independent of the outcome $A_i$, we have:
$$\sigma_B^{U(\Delta t)\Psi_{(A_i,\epsilon_A)}}\geq \sigma_B^{\Psi,\Delta t,\epsilon_A} :=\inf_{A_i \in \mathrm{spec}(A)} \sigma_B^{U(\Delta t)\Psi_{(A_i,\epsilon_A)}}=\inf_{A_i \in \mathrm{spec}(A)} \frac{1}{2\epsilon_A}\left|\left<\Psi_{(A_i,\epsilon_A)},\left[A,B(\Delta t)\right]\Psi_{(A_i,\epsilon_A)}\right>\right|$$
In general, $\sigma_B^{\Psi,\Delta t,\epsilon_A}\neq 0$. For example in our case, we have $\sigma_p^{\Psi,\Delta t,\epsilon_x} = \frac{\hbar}{2\epsilon_x}+O(\Delta t^2)$.

Now, if we prepare a quantum system in the state $\Psi$, measure $A$ at time $t=0$ to accuracy $\epsilon_A$ and $B$ at time $t=\Delta t$ to accuracy $\epsilon_B$ and we repeat this procedure $N$ times, we will get a list of $N$ pairs $(A_i,B_i)_i$. We can now take an estimator for the variance and calculate for example $\mathrm{Var}_N((B_i)_i)=\frac 1 N \sum_{i=1}^N (B_i-\bar B)^2$. Then the prediction of quantum mechanics is:
$$\lim_{N\rightarrow\infty} \mathrm{Var}_N((B_i)_i) \geq (\sigma_B^{\Psi,\Delta t,\epsilon_A})^2$$
This has nothing to do with the accuracy $\epsilon_B$ at which you have measured the $(B_i)_i$. You can measure them to any desired accuracy, but statistically, it will be impossible for you to get the variance down beyond the uncertainty limits of quantum mechanics.

 

[Teclis]

Inherent fuzziness as the state of a particle isn't distinct as defined by the no go theorem.

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 vice versa.

In the double slit experiment, when a electron both leaves the source and impacts the screen wouldn't you know both its momentum and position?

 

[DrChinese]

In the double slit experiment, when a electron both leaves the source and impacts the screen wouldn't you know both its momentum and position?

Do you realize that your question reduces to: why can't you know the simultaneous position and momentum of a free electron? So you would want to go back to square one in the Uncertainty principle and go from there.

 

[PeterDonis]

when a electron both leaves the source and impacts the screen

There is no "when" at which both of these events take place. They take place at different times.

wouldn't you know both its momentum and position?

No. See above.

 

[Teclis]

Do you realize that your question reduces to: why can't you know the simultaneous position and momentum of a free electron? So you would want to go back to square one in the Uncertainty principle and go from there.

What if the electron where traveling between two parallel infinite sheets of uniform charge? Then the net force on the electron would be zero but the electron would exert a force on the sheets of charge that would not be zero and would indicate both its position and velocity.

 

[PeterDonis]

What if the electron where traveling between two parallel infinite sheets of uniform charge?

This is a classical model, not a quantum model, so it's pointless to use it to try to answer a quantum question.

 

[Teclis]

This is a classical model, not a quantum model, so it's pointless to use it to try to answer a quantum question.

Ok sorry, but what if you performed the double slit experiment with protons or something really large like a buckyball . Even if there is no detector to cause the wave function to collapse the proton or buckyball will still have a gravitational field and the field will be stronger at the slit through which it passes regardless of whether there is detector to fire a photon at it and cause the wave function to collapse via the Compton effect so wouldn't that tell you which slit the proton or bucckyball went through without collapsing the wave function?

 

[Ilja]

(ii) a constraint on the measured uncertainty of both position and momentum in a simultaneous measurement. I am interested in the (ii)-part.

Then start with standard theory about such simultaneous measurements.

It should define, for every density operator, a probability distribution on the (p,q) plane. The map should be linear, for not contradicting the meaning of density operators, and have some correspondence between the actions of p and q shifts for the density operator and the corresponding one on the (p,q) plane. This gives a positive operator-valued measure on the plane, and the symmetry properties reduce this to a quite simple simultaneous measurement: Introduce a second test particle in some fixed state, approximately $p_1\sim 0, q_1\sim 0$ and measure $q-q_1$ and $p+p_1$, which commute.

The accuracy of this measurement you can evaluate independently, by measuring whatever you like of the prepared state of the second test particle. The natural most accurate measurements are those using harmonic oscillator ground states.

 

[Quandry]

there seems to be no restriction on the precision with which $q$ at time $t$ and $p$ at a slightly later time $t'$ can be measured.

I have no concern about whether or not measurements can be done, or if they are done, what effect they have, nor am I concerned about large ensembles of particles.
I am concerned about basic principles, i.e can position and momentum of a single particle be known at the same time?
Momentum is a vector quantity. So assuming that we already know the mass because we know what type of particle we are measuring, how do we know the direction at a single point in time (assertion alert) - we can't.
But that is not what the question is. The question relates to the precision of measurements taken at two different times, one for q and one for p. To determine the direction for p its position must be 'knowable' at two times to determine the vector value. And assuming we can 'know' where it is twice, regardless of how small Δt is, can we know with certainty what happened in the universe which may have impacted on the particle during Δt.

I find the proposal that we can know momentum of a single particle at a specific time more mystifying than QM.

 

[A. Neumaier]

I am trying to draw some conclusion from the discussion so far, based on my question, my clarifying remarks in posts #15, #16, #20, #28, and the contributions cited below.

If you're picturing a situation where the measurements take time, then once $\Delta t$ is lower than that time your results will become junk. The specific way in which they become junk depends on how you're doing the measurement. There are many ways to model this, creating many varieties of junk. [...] The main points I wanted to make with it are:

1. You can make explicit models of the "measurements happening so close that they overlap" situation.
2. Those models fail to distinguish between upward and rightward with certainty.

[...] If your measurements aren't arbitrarily fast, then they will overlap and generally break each other's effects. The way they break depends on how the measurements are implemented.

From its general sound I like this answer, but could you please give the criterion by which you decide under which circumstances a spin measurement actually measures spin, and under which conditions it must be considered junk. This cannot only depend on the timing because even if there is only a single measurement done, what precisely qualifies it as a measurement of spin up, say?

I am not even sure what it means to have measured position, let alone how to specify the accuracy of the measurement. If someone knows (with authoritative references) I'd like to be enlightened.

This request still holds. If I have a detector that measures something, what is the criterion that allows me to say that I have measured the position of a particle?

Taking rubi's explanation ''we have just measured the position. The quantum state will be sharply peaked on some definite position'' for what it implies to have done an accurate position measurement as a temporary definition, and demand the same for momentum, an accurate momentum measurement should imply that the state is projected to a quantum state sharply peaked around the measured momentum. The measurement is presumably the more accurate the more narrow the peak.

rubi seems to suggest that the criterion for being a position measurement is whether the posterior state is sharply peaked on some definite position. But if I take the analogous criterion for measuring the presence of a particle in a beam than a valid photodetection should put any initial state, a superposition of the no-photon state and the 1-photon state into a definite 1-photon state, while actually the posterior state is always the 0-photon state. Thus this cannot be the general criterion.

Let's approach it in a deductive way. We define the uncertainty of an observable $A$ in a state $\Psi$ as $\sigma_A^\Psi =\sqrt{\left<\Psi,A^2\Psi\right>-\left<\Psi,A\Psi\right>^2}$. [...]
Thus, quantum mechanics predicts that if you have measured $A$ with accuracy $\epsilon_A$ at $t=0$, then you can measure $B$ with accuracy at most $\frac{c_\Psi^{\epsilon_A}}{\epsilon_A}$ at time $t=\Delta t$. [For:] If you perform the experiment a large number of times, then no matter how many digits your measurement apparatus can display, if you calculate the uncertainty $\sigma_B$ afterwards, you will get a number that exceeds $\frac{c_\Psi^{\epsilon_A}}{\epsilon_A}$. You can't build a device that can produce a smaller uncertainty. This is not because you are a bad engineer, but because quantum mechanics limits your ability to prepare a state in such a way that it produces a statistically robust definite outcome for $B$ at time $t=\Delta t$ (assuming you first measured $A$ with accuracy $\epsilon_A$).

This seems to me like arguing that one cannot measure with high accuracy the number of eyes on an individual classical die (cast from a thoroughly mixing source) because upon performing the experiment a large number of times, you cannot get the standard deviation below $\sqrt{35/12}$.

In general, you will have to take measurement theory into account and put more work into analyzing what the state and its uncertainty will be after a measurement.

My question is not about the final posterior state but about the accuracy of a single pair of measurements $q(t)$ and $p(t+\Delta t)$ on a single particle. The underlying difficulty is to find out how to make the measurement concept clear enough that this can be distinguished from determining a probability distribution and its spread. Because identifying the two means making a measurement on the ensemble, not on the single case.

You can measure them to any desired accuracy, but statistically, it will be impossible for you to get the variance down beyond the uncertainty limits of quantum mechanics.

One has already the same situation classically with a randomly prepared die. But nobody disputes in this case that one can measure the eyes of a single die to higher accuracy than the square root of the variance.

(See post #43 for more.)

The reference is no longer correct; please edit it!

 

[Mentz114]

This request still holds. If I have a detector that measures something, what is the criterion that allows me to say that I have measured the position of a particle?

This is the closest thing to localizing an atom I ever came across -

arXiv:quant-ph/0512006v2 9 Mar 2006

Time-resolved and state-selective detection of single freely falling atoms

Torsten Bondo, Markus Hennrich, Thomas Legero, Gerhard Rempe and Axel Kuhn

We report on the detection of single, slowly moving Rubidium atoms using laser-induced fluorescence. The
atoms move at 3 m/s while they are detected with a time resolution of 60μs. The detection scheme employs a
near-resonant laser beam that drives a cycling atomic transition, and a highly efficient mirror setup to focus a
large fraction of the fluorescence photons to a photomultiplier tube. It counts on average 20 photons per atom.

 

[Strilanc]

From its general sound I like this answer, but could you please give the criterion by which you decide under which circumstances a spin measurement actually measures spin, and under which conditions it must be considered junk. This cannot only depend on the timing because even if there is only a single measurement done, what precisely qualifies it as a measurement of spin up, say?

The specific criteria I have in mind for "is a process a Z axis measurement?" is whether that process distinguishes between the up/down basis states. The process must return "Off!" for a qubit in state $|0\rangle$ and "On!" for a qubit in state $|1\rangle$. That's basically it. (I guess I also want it to leave the target qubit in the claimed state [no extra rotations], and to not mess with other qubits I care about.)

There is a huge variety of such processes. The effects of overlapping those processes with an X axis measurement are varied. The results won't be literally useless junk, in the sense of conveying no information about the system. But they will be junk in the sense of not doing what you expect, of not achieving the intended purpose.

I gave one example of a measurement process (continuous CNOT onto ancilla, measure ancilla at leisure, no three-qubit interaction despite overlapping the two CNOTs) that does one weird thing when measurements are overlapped. With the criteria of "normally it distinguishes between Up and Down", you can easily create other situations that do different weird things. The trick "think about measurement as controlled operations onto a fresh ancilla" is very helpful for understanding these situations.