# Apparent violation of the px uncertainty principle in bubble chamber measurements

1. Nov 26, 2009

### LAncienne

In thinking about QM, I was wondering if anyone could comment on this? Back in prehistoric times, I did research using bubble chambers. In a bubble chamber, you see a number of curved (because of the magnetic field the chamber is imbedded in) tracks corresponding to different particles resulting from well defined collisions. After kinematic and dynamic reconstruction, the tracks can be identified with specific particles. NOW, in looking at these tracks, first, I know the position (those little bubbles ;-{) ) very well, and from the curvature of the track and the identified mass, I know the momentum quite well. So, I have both , with I suspect, less uncertainty than implied by the uncertainty principle. What's wrong with this analysis? Can this, even though dealing with individual relativistic particles, still be considered the macro (as opposed to QM) realm?

2. Nov 26, 2009

### cesiumfrog

How about doing the math on that?

3. Nov 26, 2009

### hamster143

For an electron, track width of, say, 10 micrometers, corresponds to Heisenberg transverse velocity uncertainty of 10 m/s. Particles observed in cloud chambers have velocities of thousands of km/s.

4. Nov 27, 2009

### DrChinese

Suppose you did have P at T1 and Q at T2? Would that violate the HUP? There is nothing that stops you from measuring them at different points in time, that wouldn't necessarily be indicative of their values at any single point in time. That would become very clear were you to re-measure P and Q, in which case one or the other would be quite different.

Also, as ZapperZ often points out: there is actually nothing that prevents you from measuring 2 non-commuting observables at the same time anyway - to any level of precision. But yet again, if you measure them at a later time, their values will have changed in accordance with the HUP.

For example: such measurements can be performed on entangled particles Alice and Bob. Measure Alice's X-spin and Bob's Y-spin, then measure them again later. One will be the same as before, the other will be purely random in accordance with the HUP. So what you thought were simultaneous observable values don't any longer appear to violate the HUP.

5. Nov 27, 2009

### LAncienne

Interesting! I really need to get my mind around how time t enters into the px HUP and how E does as well. p and x are the three space vector parts of covariant 4 vectors, while E,t are the rest (the timelike parts) so the two HUP aspects are sort of fundamentally different. I'm wondering, since I effectively measured the momentum over some small but definitely detectable span of time, whether this is sort of illegally finessing the HUP, or smearing it out. What I have done is measure a lot of locations (bubbles), and then use this information to get a really very good handle on the momentum p.

Anyway, a usual physics back-of-the-envelope estimate gives dx = .5x10-2 cm (per bubble) and dp = 2x10-2p. I need to work out the units

Off topic, the invariants , scalars are mass (units where c = 1) for E,p and proper time for t,x. I sort of remember that another quasi-HUP (or maybe just UP) (tied more directly into the Fourier transform way of looking at things) was mass m of resonances and how long they apparently lived (much more statistical). i.e bumps in an invariant mass distribution.

6. Nov 27, 2009

### LAncienne

Stepping back a bit (sorry to use this to think on-line) the actual measurement that must be dealt with, vis a vis HUP, is a photograph (actually three for reconstruction purposes), since the whole event is 3 dimensional, not planar. The photograph(s) capture, in 3 dimensions, the entire observable lifetime of the particle (one of three or four) in the bubble chamber, progressing from creation to leaving the scene of the crime (or decaying). The photographs are the measurement, capturing all the information (kinematics anyway) available. Classical apparatus, augmented by classical additional parameters (like magnetic field). The scnitt may occur, in capturing the light (or absence) in the silver iodide particles (pretty small) in the emulsion of the photographic film. I think that may be going too far, however, but, who knows?

7. May 6, 2011

### jcsubmit

The popular presentation of the HUP is incorrect. The HUP is about state preparation, not measurement. The bubble chamber is a good example of measuring on a continuous basis the position and momentum of a particle. What you can't do is prepare a state where particles in that state have both precise position and momentum values. Eg passing particles through a slit gives a spread of momenta in the plane of the slit.

8. May 6, 2011

### DrChinese

Welcome to PhysicsForums, jcsubmit!

A comment: Not sure what popular presentation you are objecting to, and it really doesn't matter. However, I would like to point out that 2 entangled particles can be prepared and their identical spins/polarizations can be measured to any degree of certainty in non-commuting bases (i.e. one on Alice and the other on Bob). But you still will not have measured anything that violates the HUP.

9. May 7, 2011

### maverick_starstrider

Admittedly I basically have no idea what a bubble chamber is but h-bar is exceptionally small. However, I find it hard to believe that any contraption could have the necessary positional and temporal resolution for it to become a real issue. Keep in mind that most, though not all, means of experimentally determining velocity (and thus momentum) really just make two position measurement separated by some finite time. Thus whether or not the commutator of the exact observables you are actually measuring (averaged over some sort of gaussian error in temporal resolution) is non-zero to begin with, even if you did have the required precision to concern yourself with positions on the order of h-bar, is questionable. Keep in mind that Robertson's Inequality really only describes the commutator of certain operators, the specific case of HUP arises from the commutator of position and momentum AT THE SAME TIME.

10. May 7, 2011

### DrChinese

The thing to remember is that entangled particles obey the HUP. If experimental precision were an issue, then this would not be true. That is because essentially, entangled particles are clones of each other (when suitable adjustments are made).

11. May 8, 2011

### Fredrik

Staff Emeritus
When two observables don't commute, it's often (always?) because the corresponding measuring devices would interfere. Consider e.g. an attempt to simultaneously measure Sx and Sy of a silver atom. You would measure one of those observables with a Stern-Gerlach apparatus, which involves a magnet and some kind of particle detector. How would you measure both? If you put two of those magnets next to each other, the fields would add up, and you would end up measuring neither of the two observables you were interested in. (I'm guessing that you would end up measuring some multiple of Sx+Sy).

I think the answer to this is just that the interactions with the liquid that produce the bubbles don't determine the positions accurately enough to make the wave function spread out much in momentum space. Perhaps you can tell us how accurate these momentum measurements are? This would tell us how accurately the position needs to be determined to be in apparent conflict with the uncertainty relation.

Last edited: May 8, 2011