# Uncertainty principle using 2 detectors

1. Apr 26, 2013

This may be a dumb question, but I am still new the the concept of the quantum world... If we can't measure position and momentum precisely at the same time, why can't we use 2 detectors? One to measure position, and one to measure momentum. Have them take the measurement at the same time, and you have a "Work around". I fully understand it won't work, but I don't understand WHY! Could someone explain why this doesn't work?

Thanks to all in advance, and I appreciate everyone here as you have provided answers to many questions I have had in the past.

2. Apr 27, 2013

### vanhees71

Quantum theory does not say that you cannot measure position and momentum precisely but it says a particle cannot be prepared in a state where both position and momentum are determined arbitrarily precisely. This is a misconception of the principle going back to Heisenberg's very first paper on it. It has been corrected by Bohr shortly thereafter.

I have discussed a modern example of experiments providing evidence for this view in this forum, unfortunately with no response, here:

3. Apr 27, 2013

### Simon Bridge

That's intreguing vanhees71
For the two detectors proposed to measure the position and momentum simultaneously would require them to be in the same place at the same time ... i.e. the same detector. The detector, even if perfect, would just measure the uncertainty in the prepared states. It's nice - I like it. Though it does suggest that an individual particle could be measured to have a precise momentum and position at the same time, without disturbing it's state - the stats belong to the ensemble? (Should I take this to the other thread?)

The usual response would just be that the act of measuring one disturbs the other.
For the simultaneous measurement the experimenter has to come up with a way to hit the particle very hard (to measure position precisely) and softly (to get the momentum) at the same time - a contradiction.

4. Apr 27, 2013

### dipole

I really hate the explanation that "measuring position disturbs the momentum" or vice versa. That is not what the uncertainty principle states, even if it is a fact of reality...

You can't measure position and momentum at the same time because if position is well defined, then the momentum is fundamentally ill-defined. That is, if the particle is in a state that has well-defined position, then it doesn't even make sense to talk about it having some well-defined momentum!

This has nothing to do with measurement, and it's a simple consequence of position and momentum being Fourier transforms of one another. Any electrical engineer could explain this to you without ever needing to know a thing about quantum mechanics.

5. Apr 27, 2013

### vanhees71

That's to point. To make it very clear again: The particle cannot be prepared in a state that violates the Heisenberg-Robertson uncertainty relation $\Delta x_j \Delta p_j \geq \hbar/2$, but you can measure both position and momentum such that the disturbance of the other quantity is much less than this. This is the content of the other thread I pointed to. There, as incompatible observables not position and momentum have been considered but the spin states of neutrons, which is simpler to analyze (because you don't have to deal with the subtleties of unbound operators with continuous spectra) and also very precisely to measure.

6. Apr 27, 2013

### HomogenousCow

Quantum mechanics is very different from classical mechanics and it is where all the confusion and strangeness come from.
In classical mechanics we always talk about what result you would get if you measured the system, and the theory dictates that a perfect measurement does not change the system.
In quantum mechanics however, more often than not we are talking about a system that is not being measured, and what happens to it when it is measured, when the system is measured (according to the most mainstream interpretation) the state "collapses" onto one eigenstate, the eigenstate corresponding to the result of your measurement.
The crux of this is that the state, when not being measured can be in a mixture of many states, it is only when it is measured that the state collapses to one of them.
The uncertainty principle provides a lower bound for the product of the standard deviation for the position uncertainty and the momentum uncertainty. Because the two must multiply to a certain value (h-bar divided by 2), if one is very very small (for example if the state is very localized in position) the other number must be very very big to compensate.

7. Apr 27, 2013

### hankaaron

Isn't that how quantum computers work?

8. Apr 27, 2013

### Staff: Mentor

I don't think so. Can you point to some reference somewhere that says that's how they work?

9. Apr 27, 2013

### Simon Bridge

... not to my knowledge, but I'm prepared to be proved wrong.

However, I suspect that you are not asking because you want to find out how quantum computers work :), instead you hope to make some other point - perhaps?
Whatever that point is, I can only guess.

I shall guess, off the context, that you mean to point out that it is possible to have the one detectors which measures both momentum and position.

It certainly is!
Well done.

10. Apr 28, 2013

### dlgoff

11. Apr 28, 2013

### Sonderval

I'll give a simple classical analogy: Consider a wave, let's say a water wave, like a ripple on a pond. You can determine where it is, but not too precisely because the ripple has some extension. You can also determine the wave length of such a ripple by measuring the distance between wave crestst. This will work the better, the more wave crests you can measure. (That's basically what the so-called Fourier transform tells you - it's the same with sound; a single bang has no clear frequency because it is too short). The wave length is analoguous to the particle's momentum. So the longer the ripple is, the easier it is to determine its wavelength, but the less you know about the position (it spreads over a wide range). If you have a sharply defined wave (like a single tsunami wave), you can say where it is, but you cannot say much about its wave length.

For an electron, things are similar - momentum takes the role of the wave length. But things are also different because when you actually measure position, you will find a definite value. That's due to the fact that the electron wave is a probability wave (and that's the reason why the behavior of the water wave is only analoguos) and that a measurement actually realises a new state of the electron (at the point where you measured it), but this is really a different issue from uncertainty.

12. Apr 28, 2013

### DrewD

[Mentor's note: I edited out a reply to a post that's been deleted]

vanhees71
I haven't had a chance to finish your other post nor to read the paper, but wouldn't measuring a state more precisely than HUP would allow then prepare the state in a similar manner? This seems too obvious, so I assume there is a subtlety that I am missing that will be addressed in your post/the paper

Last edited by a moderator: Apr 29, 2013
13. Apr 28, 2013

### TheNutyProfesr

It all boils down to the math, plus its interpretation.

Our interpretation of the math (the Copenhagen interpretation) is that no matter how one tries to measure the position and the speed+direction of a particle we cannot be more precise than some tiny number (planks constant (h)). Now this constant is so small that when we look at large atom (ie U-253) or molecules where the size is larger compared to h, we can pretty much measure position and momentum rather accurately.

(there are infinite interpretations, each just as valid as the other)

If I got any of this a lil off, please correct me.

14. Apr 29, 2013

### vanhees71

To stress it again: This is a misinterpretation of the Heisenberg-Robertson uncertainty relation, which makes a statement about the observables position and momentum of a particle in any possible state according to quantum mechanics. It has nothing to do with the measurement-disturbance relation, i.e., the fact that the measurement of one observable affects the system's state in such a way that another observable becomes indetermined. There is another relation describing this measurement-disturbance relation, which is very different from the Heisenberg-Robertson uncertainty relation. This has been demonstrated recently in an experiment with photons:

L. A. Rozema et al, PRL 109, 100404 (2012)

Also one should be very careful with the Copenhagen interpretation. The assumption of a collapse when measuring an observable is by no means necessary for the interpretation of quantum theory and it causes more trouble than it helps in interpreting the formalism. I'm a follower of the Minimal Statistical interpretion (ensemble representation) a la Ballentine et al, and I never had any trouble with interpretion since I learnt about it anymore. The only thing is that you have to accept that the description of nature due to quantum theory is inherently probablistic, and for me the violation of Bell's inequality shows that it is very likely to be the correct description.

Maybe there is some non-local deterministic theory, we are not aware of, which can account for all phenomena, but as long as nobody has found one, I stick to quantum theory in the minimal interpretation.

15. Apr 29, 2013

### Fredrik

Staff Emeritus
I have removed some incorrect claims that were made in an authoritative way. I also had to remove the rest of the discussion about those claims.

Last edited: Apr 29, 2013