Measurements and uncertainty principle

[daniel444]

TL;DR Summary

Which state takes a quantum object after a simultaneous measurement of coordinate and momentum?

Dear friends

please help me, for I am completely confused and can not understand the logical connection between two postulates of quantum mechanics.

One postulate states, that if some observable is being measured, for instance coordinate, then the superposition of many possible states, which correspond possible coordinate values, collapses into one eigenstate of this operator (in our case coordinate)

The other postulate states, that the product of the uncertainties of coordinate and momentum can not be smaller then h/2pi.

So

Me measure coordinate and momentum simultaneously. That means me must get eigenstates of coordinate and momentum operators according to the 1. Postulate. But it is impossible according to the 2. Postulate, because that would mean, that the uncertainties of both coordinate and momentum are equal zero!

Tell me where I am wrong

 

[PeterDonis]

Me measure coordinate and momentum simultaneously.

You can't. That is one consequence of the uncertainty principle.

More precisely, you can't make a measurement that simultaneously gives exact results for both position and momentum. You can make a measurement that gives a result that restricts both position and momentum to a certain range, as long as the product of the ranges satisfies the uncertainty principle.

 

[daniel444]

You can't. That is one consequence of the uncertainty principle.

More precisely, you can't make a measurement that simultaneously gives exact results for both position and momentum. You can make a measurement that gives a result that restricts both position and momentum to a certain range, as long as the product of the ranges satisfies the uncertainty principle.

Yes, exactly this I mean. I can make a me a measurement that gives a result that restricts both position and momentum to a certain range. So I´ve made such sort of a measurement. It is still a measurement. Which state takes the quantum object after this measurement? Will it be a superposition?

 

[PeroK]

Yes, exactly this I mean. I can make a me a measurement that gives a result that restricts both position and momentum to a certain range. So I´ve made such sort of a measurement. It is still a measurement. Which state takes the quantum object after this measurement? Will it be a superposition?

All states are a superposition of other states. In this case, the state after measurement will not be an eigenstate of position or momentum. Although, technically those are not physically viable states in any case.

 

[daniel444]

All states are a superposition of other states. In this case, the state after measurement will not be an eigenstate of position or momentum. Although, technically those are not physically viable states in any case.

so that means, the 1. postulate is wrong? Because in all quantum mechanics books is written, that after a measurement the wave function collapses in an eigenstate of the operator, which corresponds to the observable, which is being measured

 

[PeroK]

so that means, the 1. postulate is wrong? Because in all quantum mechanics books is written, that after a measurement the wave function collapses in an eigenstate of the operator, which corresponds to the observable, which is being measured

That sounds like your interpretation of one of the postulates. Observables that have a continuous spectrum, such as position and momentum, do not obey that postulate precisely as you have stated it. The eigenstates of position and momentum are not physically realisable. The result of a measurement of either may result in a wave packet with a range of energies, momentum and position.

 

[daniel444]

That sounds like your interpretation of one of the postulates. Observables that have a continuous spectrum, such as position and momentum, do not obey that postulate precisely as you have stated it. The eigenstates of position and momentum are not physically realisable. The result of a measurement of either may result in a wave packet with a range of energies, momentum and position.

so that means, no matters which type of measurement me make, we will always have a superposition of coordinate/momentum eigenstates (wave packet) after the measurement?

 

[PeroK]

so that means, no matters which type of measurement me make, we will always have a superposition of coordinate/momentum eigenstates (wave packet) after the measurement?

Yes.

 

[PeterDonis]

Which state takes the quantum object after this measurement?

A state which has some finite uncertainty of both position and momentum, where the product of the uncertainties satisfies the uncertainty principle. For example, it could be a Gaussian.

 

[PeterDonis]

in all quantum mechanics books is written, that after a measurement the wave function collapses in an eigenstate of the operator

No, "all quantum mechanics books" do not state this the way you have stated it.

For an example of a more careful statement, see the following (which is a sticky at the top of the QM forum):

https://www.physicsforums.com/threads/the-7-basic-rules-of-quantum-mechanics.971724/

 

[PeroK]

There's an analysis of the minimal uncertainty wave packet here. See section 5.

https://www.lancaster.ac.uk/staff/schomeru/lecturenotes/Quantum Mechanics/S7.html

 

[PeterDonis]

The eigenstates of position and momentum are not physically realisable.

Nor are the operators corresponding to them, which describe exact (i.e,. infinite precision) measurements of position or momentum. Infinite precision is not physically realizable. The actual observables we can realize are things like "measure position to within an accuracy of dx" or "measure momentum to within an accuracy of dp" (or more complicated things like "make a measurement that gives position to within some accuracy and momentum to within some accuracy"). Textbooks don't normally try to write down the operators corresponding to these observables because it would be tedious; but the resulting states are normally assumed to be Gaussians.

 

[vanhees71]

so that means, the 1. postulate is wrong? Because in all quantum mechanics books is written, that after a measurement the wave function collapses in an eigenstate of the operator, which corresponds to the observable, which is being measured

In many but not all books. The collapse postulate is the most confusing of all the Copenhagen like interpretation schemes. Fortunately it's unnecessary to use QT to the real world, and it's almost never realized even approximately: What state the measured system is in after the measurment depends on the measurement you made on it, i.e., it depends on what the interaction of the measured system with the measurement apparatus does to it. E.g., if you detect a photon, usually it's absorbed, and after the measurement thus there's only the photodetector left but no photon, i.e., it doesn't make any sense to say the measured photon's state is then an eigenstate of the measured observable.

It's also important to remember that states, as far as the single system under consideration is concerned, describe the preparation of the system and not the measurement. The uncertainty relation for position and momentum $\Delta x \Delta p_x \geq \hbar/2$ says you cannot at the same time accurately localize a particle and make its momentum also very well defined. If $\Delta x$ is "small", then $\Delta p$ is necessarily large and vice versa.

It doesn't say anything about how accurate you can measure the one or the other observable, which entirely depends on the measurement setup. It's also another question in how far a measurement disturbs the measured system. This is also not what this simple usual uncertainty relations for incompatible observables say, i.e.,
$$\Delta A \Delta B \geq \frac{1}{2} |\langle [\hat{A},\hat{B}] \rangle|.$$
This refers to limitations in "preparability" of states concerning the standard deviations/uncertainties of the observables under consideration.

There is of course also a limit in the ability to measure an observable accurately without disturbing the system. E.g., if you want to measure the location of a charged particle accurately you need to, e.g., scatter light with a small wave length at it, and for this you need to interact with at least one single photon of energy $\hbar \omega$ with it. The smaller the wave-length of the photon the larger is the frequency and thus the more you'll disturb your particle by measuring its position. But that's not described by the uncertainty relation. You just have to calculate how much momentum you'll transfer by scattering the photon in a certain direction to estimate how much the particle is kicked around by measuring it by interacting with the photon.

Concerning incompatible observables there's usually also a tension between measuring one observable very accurately and also simultaneously the other, but that's also interrelated with the preparability. A famous example is Weizsäcker's analysis of the Heisenberg microscope, which is a gedanken experiment with the famous double-slit setup. I.e., you let a single photon run through a double slit and you use a lense to either measure its momentum, i.e., you put the photoplate in the focal plane. Then each point on the photo plate refers one-to-one to a measured momentum of the photon. Then the picture when repeating this with many equally prepared photons running through the double slit is the double-slit interference pattern, and it's impossible to know through which slit each single photon came. If you put the screen in the image plane, you'll resolve from which slit each photon came, but their spread in momenta is large. You can also make a compromise and put the screen somewhere else but neither in the focal nor the image plane of the lense. Then you get restricted accuracy for both observables "position" (through which slit came the photon) or momentum.