Position & Momentum: Detectors, Measurements & Identifiers

[jaketodd]

If you have an array of momentum detectors, if there is such a thing, couldn't the detector that is hit in the array simply report its momentum measurement along with its identifier in the array, which would reveal the position?

Thanks,

Jake

 

[jtbell]

Sure. The uncertainty principle isn't about the limitations of detectors. You can measure the position and momentum of a single particle to any precision your detectors are capable of.

If you now repeat the experiment many times, using identically prepared particles, in general you get different results for the momentum and position each time. The momentum values have a mean (average) and a standard deviation \Delta p. The position values also have a mean and a standard deviation \Delta x. These standard deviations are what appear in the HUP:

\Delta x \Delta p \ge \hbar / 2

 

[jaketodd]

Why is there a minimum uncertainty as expressed in the equation you provided? Couldn't preparation get more exact and detectors more accurate to where the product of the standard deviations of momentum and position are less than hbar/2?

Thanks,

Jake

 

[elibj123]

Why is there a minimum uncertainty as expressed in the equation you provided? Couldn't preparation get more exact and detectors more accurate to where the product of the standard deviations of momentum and position are less than hbar/2?

Thanks,

Jake

No, and that is what the principle exactly states.

You can measure precisely one number, with the cost of an uncertain measurement of the second.

 

[DrChinese]

You may find it easier to think of this way: You have 2 non-commuting properties, p and q, and choose to measure them to unlimited precision in successive experiments. The question: does that mean it had those values simultaneously? No, the uncertainty principle expresses a different conclusion.

Now, this has nothing to do with the idea that the act of measurement imparts some specific change to the particle as a deterministic interaction (like billiard balls). If it did, you could consider the change to the measurement apparatus and deduce what the change was. But that won't work.

In fact, pairs of entangled particles respect the HUP. Measure p on Alice and q on Bob. What then happens when you measure q on Alice and p on Bob? You see the HUP in action. That would not be true if you could measure Alice and Bob without consideration of the HUP.

 

[dx]

It may be worth mentioning that when one says that "the position and momentum of a particle cannot be measured simultaneously", we are not imagining a picture of a classical particle which has an objective momentum and position in the classical sense which cannot be simultaneously "known", but are in fact saying that the ordinary classical picture has a "latitude" where in any given experimental situation the use of certian space-time relations in effect excludes the simultaneous application of energy-momentum relations, and therefore makes room for the existence of the non-classical quantum-mechanical regularities, which cannot be visualized in the customary manner without leading to contradictions.

 

[Count Iblis]

Although the quantum picture looks very counter intuitive, you can still see some problems with the classical picture from a purely intuitive point of view.

If there were no uncertainty relation then you could take a particle in a box and store an infinite amount of information in that system with the particle having only a finite amount of energy.

It can also be shown that the laws of classical mechanics allow you to build a computer whose clock cycle halves every cycle, so you can do an infinite amount of computations in a finite time. This together with the fact that you have an infinite amount of memory space available, makes it possible to build a computer that can settle whether or not certain mathematical theorems are true or false, even if they are unprovable, using brute force alone. It would even be able to check whether or not a proof exists.

 

[Frame Dragger]

Although the quantum picture looks very counter intuitive, you can still see some problems with the classical picture from a purely intuitive point of view.

If there were no uncertainty relation then you could take a particle in a box and store an infinite amount of information in that system with the particle having only a finite amount of energy.

It can also be shown that the laws of classical mechanics allow you to build a computer whose clock cycle halves every cycle, so you can do an infinite amount of computations in a finite time. This together with the fact that you have an infinite amount of memory space available, makes it possible to build a computer that can settle whether or not certain mathematical theorems are true or false, even if they are unprovable, using brute force alone. It would even be able to check whether or not a proof exists.

Was that the motivation for Beckenstein re the Beckenstein Bound?

 

[Count Iblis]

Was that the motivation for Beckenstein re the Beckenstein Bound?

The Bekenstein bound is obtained by going a step further. So, quantum mechanics implies that in some volume containing one particle, you can store only a limited amount of information using less than some given amount of energy. Also this amount of information depends on the mass of the particle you use. So, you can still store an arbitrary amount of information if only you can use arbitrary large amounts of energy. However, there is then still a limit due to General Relativity. Above a certain limit, the volume will collapse into a black hole.

It turns out that the maximum amount of information you can store in a volume if the amount of energy you can use is unrestricted, is proportional to the surface area bounding the volume. This is the Bekenstein bound.

 

[Frame Dragger]

The Bekenstein bound is obtained by going a step further. So, quantum mechanics implies that in some volume containing one particle, you can store only a limited amount of information using less than some given amount of energy. Also this amount of information depends on the mass of the particle you use. So, you can still store an arbitrary amount of information if only you can use arbitrary large amounts of energy. However, there is then still a limit due to General Relativity. Above a certain limit, the volume will collapse into a black hole.

It turns out that the maximum amount of information you can store in a volume if the amount of energy you can use is unrestricted, is proportional to the surface area bounding the volume. This is the Bekenstein bound.

Ahhhh, and this is why you can infer that a BH doesn't have a "bound breaking" remnant, based on the area of its event horizon?

 

[thoughtgaze]

Sure. The uncertainty principle isn't about the limitations of detectors. You can measure the position and momentum of a single particle to any precision your detectors are capable of.

If you now repeat the experiment many times, using identically prepared particles, in general you get different results for the momentum and position each time. The momentum values have a mean (average) and a standard deviation \Delta p. The position values also have a mean and a standard deviation \Delta x. These standard deviations are what appear in the HUP:

\Delta x \Delta p \ge \hbar / 2

I hope you don't mean to say you can measure the position and momentum of a single particle to any precision simultaneously.

 

[Matterwave]

Are simultaneous measurements even allowed in QM? Can you in fact measure 2 quantities at the exact same time? I always thought in QM you had to measure 1 first and then measure the other one - hence the operators representing measurements kind of thing.

 

[ansgar]

I hope you don't mean to say you can measure the position and momentum of a single particle to any precision simultaneously.

yes you can in Principle, the HUP is about statistics, i.e many measurements.

Just have a look at the derivation of HUP and the definition of standard deviation which is what the delta x and delta p are referring to.

 

[ansgar]

Are simultaneous measurements even allowed in QM? Can you in fact measure 2 quantities at the exact same time? I always thought in QM you had to measure 1 first and then measure the other one - hence the operators representing measurements kind of thing.

You should also just go through the derivation of HUP and the definition of standard deviation again.

 

[thoughtgaze]

yes you can in Principle, the HUP is about statistics, i.e many measurements.

Just have a look at the derivation of HUP and the definition of standard deviation which is what the delta x and delta p are referring to.

I see what your saying I think. But explain this to me...

If you make both measurements on position and momentum, at the same time... one of the postulates of quantum mechanics says that measurement leaves the particle in an eigenstate it was observed to be in, but the position and momentum eigenstates are not the same. Are you telling me the particle can be observed to be in both a position eigenstate and a momentum eigenstate at the same time? What exactly are you saying?

 

[Matterwave]

You should also just go through the derivation of HUP and the definition of standard deviation again.

My question is less to do with the HUP than the simple question of whether one can in fact make simultaneous measurements of 2 observables. I suppose one technically could, since measuring x will allow the measurement of x^2 or some such...but I'm thinking more in terms of "independent" (at least explicitly, perhaps not implicitly) operators.

 

[ansgar]

I see what your saying I think. But explain this to me...

If you make both measurements on position and momentum, at the same time... one of the postulates of quantum mechanics says that measurement leaves the particle in an eigenstate it was observed to be in, but the position and momentum eigenstates are not the same. Are you telling me the particle can be observed to be in both a position eigenstate and a momentum eigenstate at the same time? What exactly are you saying?

But as been told, it is possible to have wavefunctions which have such behaviour under such application of operators.

Can you in classical dynamics measure two things at the same time? the position and the momentum? No you need two position measurments to determine the momentum.

 

[Count Iblis]

To see the difference between exp[i k x] and exp[i (k + delta k) x] requires one to consider a range of length 1/delta k. So, to get more and more precise momentum measurements, you need to use larger and larger measurement devices.

 

[thoughtgaze]

ansgar,

Ah yes, I think I see now. Let's say the position of a particle is measured to high accuracy. This leaves the particle in an eigenstate such that the value obtained in a successive measurement of the momentum is uncertain to a degree in accord with HUP. However, one can still measure to arbitrary accuracy what this momentum is. In an ensemble the values measured of position and momentum will have a distribution with a standard deviation given by the HUP.

And just to be clear for myself, if the operators of two observables, p and q, commute , then they share common eigenstates and it is possible to measure p and know with complete confidence what successive measurement of q will be.

Tell me if this is the correct way of looking at it. Thanks.

 

[LostConjugate]

Can you in classical dynamics measure two things at the same time? the position and the momentum? No you need two position measurments to determine the momentum.

This is a great point. If momentum is m(dx/dt) and the object's position is measured to an accuracy so small that even the distance dx goes to 0 the term momentum no longer has any meaning.

This is also a good quote:

"In much the same way that the existence of perpetual motion
machines is ruled out by the Second Law of Thermodynamics, the existence of an
apparatus which would give a precise determination of position and momentum is in
conflict with the nature of physical states in quantum mechanics."

I would also like to point out that this was always a consequence in wave mechanics equations far before QM came about.

 

[DrChinese]

And just to be clear for myself, if the operators of two observables, p and q, commute , then they share common eigenstates and it is possible to measure p and know with complete confidence what successive measurement of q will be.

It is possible to know 2 COMMUTING observables simultaneously, but not 2 NON-commutung observables.

I guess a good example would be photon polarization and frequency, which commute. You can know these to unlimited precision simultaneously.

 

[LostConjugate]

It is possible to know 2 COMMUTING observables simultaneously, but not 2 NON-commutung observables.

I guess a good example would be photon polarization and frequency, which commute. You can know these to unlimited precision simultaneously.

Is that simply because they share eigenvectors?

 

[thoughtgaze]

on the question of simultaneous measurements...

on the American Institute of Physics webpage... http://www.aip.org/history/heisenberg/p08a.htm ,

they say ...

"The uncertainty relations may be expressed in words as follows.

The simultaneous measurement of two conjugate variables (such as the momentum and position or the energy and time for a moving particle) entails a limitation on the precision (standard deviation) of each measurement. Namely: the more precise the measurement of position, the more imprecise the measurement of momentum, and vice versa. In the most extreme case, absolute precision of one variable would entail absolute imprecision regarding the other."

i'm thinking the statistical interpretation is just another way of looking at HUP

Is that simply because they share eigenvectors?

Basically..but there is a subtlety. Even if they commute, the particle must be in an eigenstate of both observables if you are to measure simultaneously the values to arbitrary precision. If they don't commute, then this can never be the case.

 

[eaglelake]

I see what your saying I think. But explain this to me...

If you make both measurements on position and momentum, at the same time... one of the postulates of quantum mechanics says that measurement leaves the particle in an eigenstate it was observed to be in, but the position and momentum eigenstates are not the same. Are you telling me the particle can be observed to be in both a position eigenstate and a momentum eigenstate at the same time? What exactly are you saying?

We are still confused about the difference between a single measurement and repeated measurements. A single measurement has nothing to do with uncertainty! There is no way to determine the uncertainty when we have only a single measurement.

Many repeated measurements give many different results. If we repeat the experiment enough times, then we obtain the entire eigenvalue spectrum of the observable being measured. The probability distribution of all those values can be used to determine the uncertainty in the observable being measured.

Position and momentum can be measured simultaneously with any precision allowed by the measuring apparatus. The uncertainty principle does not say that p_y y \ge \frac{\hbar }{2}, where p_y is the measured value of momentum and y is the measured value of position Rather, it does say that \Delta p_y \Delta y \ge \frac{\hbar }{2}, where \Delta p_y is the uncertainty in momentum and \Delta y is the uncertainty in the position.

Consider a slit experiment where a photon passes through slits and is detected on a distant screen. We see a dot on the screen at the position of the photon. Thus, we know the photon's position at the instant it hit the screen. But, we also know the momentum p_y = p\sin \theta by measuring the scattering angle \theta. But these single values of position and momentum have nothing to do with uncertainty or the uncertainty principle.

Now, if we repeat the same slit experiment many times we get many different values for p_y. There is a statistical distribution of those different values and the standard deviation of that statistical distribution is what we call the uncertainty in momentum for this experiment. Likewise for the position.

Many reject the projection postulate. It is more of an interpretation than a postulate. There is no need to know the state of the photon after it is detected. Usually the photon is lost in the detection material after detection and we have no idea what its quantum state is.

What we should say is, "there is no experiment with a state function that is an eigenfunction of both momentum and position". There is no preparation procedure for which repeated measurements always yield the same momentum value and the same position value. For such a case \Delta p_y = 0, \Delta y = 0, and the Heisenberg uncertainty principle would be violated. We believe this to be impossible and, up till now, all experiments agree.

Finally, a specific answer to your question:
No, we cannot observe a particle to be in an eigenstate of momentum and in an eigenstate of position at the same time.
(But, then you must reject the projection postulate.)

 

[LostConjugate]

The measurement process must leave the system in one of the eigenstates of the observable operator. States with no uncertainty are eigenstates of Hermition operators.

If you measured two observables simultanously the state would be left with zero uncertainty in both observables. The physical state must then be in an eigenstate of both observables simulatanously following the measurement.

In the case of position and momentum, there are no physical states that are eigenstates for both x and p.

 

[DrChinese]

The measurement process must leave the system in one of the eigenstates of the observable operator. States with no uncertainty are eigenstates of Hermition operators.

If you measured two observables simultanously the state would be left with zero uncertainty in both observables. The physical state must then be in an eigenstate of both observables simulatanously following the measurement.

In the case of position and momentum, there are no physical states that are eigenstates for both x and p.

This does not contradict anything you say above: It would be possible to prepare a photon in eigenstates of polarization and frequency, and verify those with repeated measurements of unlimited precision (in principle).

 

[LostConjugate]

This does not contradict anything you say above: It would be possible to prepare a photon in eigenstates of polarization and frequency, and verify those with repeated measurements of unlimited precision (in principle).

As long as the state exists it should be possible. I think polarization commutes with the hamiltonian.

 

[dextercioby]

\Delta p_y \Delta y \ge \frac{\hbar}{2} (1)

is (as stated above in the post #2 of the thread) a matter of statistics of infinitely many and not of single measurements. The common and wrong interpretation of (1) in terms of the accuracy of a simultaneous measurement stems from the blind acceptance of von Neumann's projection postulate.

But this postulate can be rejected thus eliminating the justification for an interpretation of (1) based on simultaneous measurements.

 

[LostConjugate]

\Delta p_y \Delta y \ge \frac{\hbar}{2} (1)

is (as stated above in the post #2 of the thread) a matter of statistics of infinitely many and not of single measurements.

I disagree, the possibility of a single simultaneous measurement does not exist. It is proven this way in that the state of the particle after the measurement could not exist.

 

[Frame Dragger]

Does polarization commute with the Hamiltonian?... and... I'm genuinely confused now.

 

[LostConjugate]

Does polarization commute with the Hamiltonian?... and... I'm genuinely confused now.

You should be able to find simultaneous eigenstates of either position and polarization, or momentum and polarization. Though I don't know much about the "photon polarization operator".

 

[thoughtgaze]

\Delta p_y \Delta y \ge \frac{\hbar}{2} (1)

is (as stated above in the post #2 of the thread) a matter of statistics of infinitely many and not of single measurements. The common and wrong interpretation of (1) in terms of the accuracy of a simultaneous measurement stems from the blind acceptance of von Neumann's projection postulate.

But this postulate can be rejected thus eliminating the justification for an interpretation of (1) based on simultaneous measurements.

Has it been verified that one interpretation is true and the other is not? Doing a quick wiki one sees that Heisenberg himself interpreted it as a single measurement.

In fact, the wiki page gives two forms of the uncertainty relations... one as given by Heisenberg and one as refined by Kennard. They say the following...

"However, it should be noted that σx and Δx are not the same quantities. σx and σp as defined in Kennard, are obtained by making repeated measurements of position on an ensemble of systems and by making repeated measurements of momentum on an ensemble of systems and calculating the standard deviation of those measurements. The Kennard expression, therefore says nothing about the simultaneous measurement of position and momentum."

I think one can see the uncertainty relation in both interpretations.

 

[LostConjugate]

I think it all boils down to trying to compare velocity, a difference in position, with a specific position. You can't simultaneously say a particle has this difference in position and this exact position.

 

[eaglelake]

The measurement process must leave the system in one of the eigenstates of the observable operator. States with no uncertainty are eigenstates of Hermition operators.

If you measured two observables simultanously the state would be left with zero uncertainty in both observables. The physical state must then be in an eigenstate of both observables simulatanously following the measurement.

In the case of position and momentum, there are no physical states that are eigenstates for both x and p.

Why must the measurement process leave the system in an eigenstate of the observable being measured?

Assuming you are correct, then what do you do with that eigenstate? What is the system being described after particle detection? How long does that eigenstate last? Is there no end to the experiment? Assume a photon is detected via the photoelectric effect so that it no longer exists. Does that mean that a non-existent photon can be in an eigenstate of position, say?

I think you can see the mess we get into with this line of thought.

A measurement requires the detection of the particle which ends the experiment. Every experiment requires a measurement result that gives it closure (Bohr). Once the particle has been detected the experiment is finished. There is no "after" for us to fret over.

A system can be prepared in an eigenstate. We can make measurements on this system. Particle detection is not a preparation procedure. In order to verify that the system is projected into an eigenstate by the measurement process, we must immediately repeat the measurement. But, it is impossible to again detect the particle when it is hidden in the detector material; we cannot experimentally verify the projection postulate.

Given a particular experiment, assume the preparation procedure gives the state vector \left| \psi \right\rangle. If we decide to measure the position, then we write the state vector in the position representation and \left| {\left\langle {y}<br /> \mathrel{\left | {\vphantom {y \psi }}<br /> \right. \kern-\nulldelimiterspace}<br /> {\psi } \right\rangle } \right|^2 is the probability for finding the particle at position y when we measure position. If, on the other hand, we decide to measure the momentum, then we write the state vector in the momentum representation and
\left| {\left\langle {{p_y }}<br /> \mathrel{\left | {\vphantom {{p_y } \psi }}<br /> \right. \kern-\nulldelimiterspace}<br /> {\psi } \right\rangle } \right|^2 is the probability for finding the particle with momentum p_y when we measure momentum. The theory does not suggest that we cannot measure position and momentum simultaneously, although the experimental configuration might prevent it. If we do measure position and momentum simultaneously, we can calculate the probability for each measurement. Generally, neither observable will be in an eigenstate; there will be an uncertainty in each observable determined by the state vector.

For an interesting example, see arXiv:quant-ph/0703126 where we measure the momentum for a system known to be in a position state.

Actually, post #2 above said it all! But we repeat for emphasis:

A single measurement tells us nothing about uncertainty! A single measurement does not mean the system is an eigenstate! Only when repeated measurements all give the same value is the system in an eigenstate!

 

[eaglelake]

Can you in classical dynamics measure two things at the same time? the position and the momentum? No you need two position measurments to determine the momentum.

I remember an experiment where a ball is fired into a pendulum bob and held there. The pendulum recoiled and swung up some vertical distance allowing us to calculate the momentum at the instant of collision. The position of the bob is the ball's position at the instant of collision. Thus, we have measured both position and momentum simultaneously.

In classical physics a particle has a trajectory, meaning that it has both position and momentum at every instant. There are classical laws that allow us to calculate x(t) and p(t). Of course, we can verify these predictions experimentally. Classically, there is no prohibition against measuring position and momentum simultaneously.

(Unless, you want to discuss Zeno's paradox. But, that is another matter!)

 

[thoughtgaze]

the uncertainty principle can be seen with a single measurement interpretation or more accurately with an ensemble statistical interpretation.

scroll down to "Trying to beat the uncertainty principle" in the following website

http://galileo.phys.virginia.edu/classes/252/uncertainty_principle.html