Position & Momentum: Detectors, Measurements & Identifiers

[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