Inconsistancy in QM?

1. Nov 20, 2008

michael879

a friend of mine gave me this thought experiment which appears to show an inconsistency in QM. I realize there must be some false assumption being made, but I don't see it. This thought experiment takes place in a 1D world completely described by QM (no QFT, QCD, GR, SR, string theory etc). As far as I can tell, it does not deviate from the QM model but manages to violate the uncertainty principle. Heres the experiment:

At t=0, an electron exists at x=0 with dx = 0 and dp=inf. You are at x=1 and send a photon towards the origin.

At t=t', you detect that the photon has returned. you can immediately calculate that the position of the photon-electron collision happened at x' = ct'/2 and that the electron's velocity at the point of collision was x'/t'.

Therefore the error in both the electron's position and momentum at the time of collision is 0. Isn't this a violation of the uncertainty principle?

I realize that setting dx=0 at t=0 is not physically possible. However, if dx is sufficiently low, than dx*dp will be less than h-bar/2 at the point of collision.

2. Nov 20, 2008

ZapperZ

Staff Emeritus
Let me illustrate this a different way. Let's say you have a slit of a slit that is the exact diameter of an electron (hey, if you can have an electron sitting still at a particular location, I can also have an electron having a diameter and a slit having that exact diameter).

At t=0, an electron is detected having passed through the slit and hitting a screen with infinite resolution (we are continuing our unrealistic scenario here). Because of diffraction effects, that single electron has a slight momentum in the x direction, and so, it hits the screen off-center. But since we know where it hits, and how far off center it hits, we can calculate its p_x momentum.

So what did we know? That electron, when it passed through that slit, had an EXACT position (since the width is exactly the width of the slit) equivalent to the center of the slit, and then it has an EXACT momentum p_x, since our detector has an infinite resolution.

Same scenario as what you've described.

Nowhere in here was the HUP invoked. The HUP is not about the accuracy of ONE measurement of the position, and ONE measurement of momentum. The HUP involved the statistical spread of A NUMBER OF MEASUREMENTS. In other words, if you repeat the same experiment with the slit, even when the exact value of the position is still the same (same slit), the value of the NEXT p_x can be very different than the one obtained previously. If you do this MANY times, the spread in the value of p_x will be close to infinite, because your slit has uncertainty in position of zero! Now THIS is the HUP.

This is a common misconception of the HUP by many people. They often mistaken the instrument/measurement uncertainty as the HUP. Often, people associate the act of measurement as using higher photon energy onto something like an electron, which relegates the HUP as simply a matter of technique rather than an intrinsic property.

Zz.

3. Nov 20, 2008

cesiumfrog

Consider three parallel screens, each with a pinhole. If a flash of light is shone on the first screen, some light will go through the first pinhole to the second screen. A few photons will go through the second pinhole, and for these photons you claim to simultaneously know both their position and momentum. Nonetheless, each such photon has a chance of also getting through third pinhole, regardless of whether this last trajectory corresponds with an inconsistent direction of momentum. So can you really say you knew the photon's position and momentum, if you aren't able to predict where it is going?

edit: too slow

4. Nov 20, 2008

michael879

ah yea I realized something along those lines after I posted this. Thanks for the quick reply.

5. Nov 21, 2008

buffordboy23

Here is another way to look at it. The position and momentum operators do not commute, so they are incompatible observables (they don't share a complete set of common eigenfunctions) and therefore must obey the uncertainty principle. This is true for any two observables whose operators fail to commute.

6. Nov 21, 2008

ueit

Sure I can. The most important thing one should understand about QM is that particles are not billiard balls. They interact with other particles through fields. A photon, when passing through a slit, has a good chance of being absorbed by the electron cloud at that place. So, the fact that the photon passed through two slits does not allow you to tell if it will also pass through a third slit, not because you didn't know its position and momentum but because you didn't know the exact configuration of the electron cloud there. Also, you didn't know if the photon has been absorbed and re emitted at the second slit. So, there is nothing "inconsistent" about the direction of momentum. The momentum has to be exactly conserved, being transferred from the photon to the particles situated at those slits where a momentum change occured.

7. Nov 21, 2008

cesiumfrog

Maybe the particles of the standard model are not billiard balls, but the theory of QM would give the exact same result for ideal billiard balls. (Consider whether the Hamiltonian requires the long range interaction term.) I think Zz has already addressed your argument:

As for conservation, do we really expect it for each particle or just the ensemble? It would appear in this case that the uncollapsed wave conserves momentum, but your point reminds me of tunnelling, might you say that some particles borrowed momentum?

8. Nov 22, 2008

ueit

Unlike billiard balls, quantum particles interact through fields that can be infinite in range (electromagnetism) or short range (weak force). I'm not quite sure about the color force.

Ideal billiard balls interact only through direct collisions. For the assumed size of fundamental particles (zero) there will be no interaction at all.

The electrostatic, long range force, appears in the Hamiltonian of every atom or molecule.

My understanding of HUP is that it is a limitation of what can be predicted about a system. I am not sure what you mean by " intrinsic property". Property of what?

As far as I know momentum conservation holds for each interaction between particles, not only on a statistical level.

No, this momentum "borrowing" comes from another bad analogy between quantum and classical experiments. In tunneling, we don't know for example the exact value of the potential at the moment the particle passes through the barrier at that specific place. We only know the average value of the potential.

9. Nov 22, 2008

ZapperZ

Staff Emeritus
Intrinsic property of our universe, NOT of the measurement technique.

Zz.