Why doesn't QM make particles zig-zag as they travel?

[vpoko]

Pardon my ignorance, I'm pretty sure this has an obvious answer, but it's something I've been wondering about.

In QM it's said that particles don't exist in a definite state between measurements, but rather exist as a wave function, with a probability of being found in a certain place when they're measured. Neutrinos are very weakly interacting, and travel through the partial vacuum of space. Since they're not being "measured" when they're not interacting with other particles, they should exist in their wave-function state during those times. How come, then, don't they zig-zag as the fly across the universe? My intuition tells me that between interactions (measurements), their position and momentum should be determined by a probability function, and there should be a possibility of them changing direction during those times. Obviously, this doesn't seem to happen (from what I've read), so I'm guessing my intuition is wrong. Could anyone shed some light on this for me? Thanks.

 

[tim_lou]

I think you are mixing up Newtonian physics with quantum mechanics.

In quantum mechanics, there is no particle traveling around in some direction. All you can speak of is a wave, which assign a complex number to the 3D real space. the wave can move and evolve according to the Schrödinger's equation. The wave function probably would not zig-zag around. (unless perhaps for a very weird potential). You can think of a strongly peaked wave function as a representation of a particle.

anyway, when you take a measurement of momentum, it could be in the opposite directions (depending on the wave function), the probability of getting a momentum measurement opposite to the general direction in which the wave travels is probably pretty low though. but if you calculate <p> (the mean value of momentum for an ensemble of identical system) , you will get a classically consistent answer and <p> would not zig-zag around.

 

[belliott4488]

To speak of a "zig-zag" path - or any other path, for that matter - means that you're thinking of the particle as having definite positions between observations, which is inherently non-quantum mechanical. On the other hand, you can think of the uncertainty in the particle's position, which grows the longer it propagates, as the effect of all the possible paths it could follow. This is the basis of the path integral formulation of QM. In that picture, the probability of finding a particle at a given position is given by the sum (i.e. the integral) of all possible paths leading to that point, including your zig-zag paths. It turns out that when you do this, you get the same answer as from the Schroedinger Eq. So, yeah, you have to consider the contribution of zig-zag paths, but the more outlandish they are, the less they contribute to the end result.

 

[vpoko]

So a particle's wave function describes the entire possible path the particle could take, not just the segments between measurements? I think I'm still far from clear on this, but you can only lead a horse to water, and I'm a pretty dumb horse :)

 

[nrqed]

So a particle's wave function describes the entire possible path the particle could take, not just the segments between measurements? I think I'm still far from clear on this, but you can only lead a horse to water, and I'm a pretty dumb horse :)

I think that the problem is that you are thinking about paths between measurements. You have to give up that notion. To talk about the path of a particle one must observe its position at different times. Therefore to even talk about a path there must be measurements. In between measurements, the concept of path is not even defined. You can't associate a path to a wavefunction because once a measurement is made, the wavefunction changes. So no measurement -> no concept of path.

 

[smallphi]

The trace of a charged particle in a bubble chamber detector can be considered a sequence of measurements of particle's positions and is usually a smooth curve. Why is that not a zig zag?

 

[strangerep]

The trace of a charged particle in a bubble chamber detector can be considered a sequence of measurements of particle's positions and is usually a smooth curve. Why is that not a zig zag?

In Schiff's QM book, there's a 2nd-order perturbation calculation of
this sort of thing. I.e: he works out the probability of two particles in
a cloud chamber interacting with an incoming charged test particle.
It turns out that the probability is negligible unless the two cloud
chamber particles lie on a line that corresponds to the test particle's
momentum direction.

So QM doesn't rule out zig-zag completely. But the probability of it
happening and being observed is vanishing small.

 

[Demystifier]

Pardon my ignorance, I'm pretty sure this has an obvious answer, but it's something I've been wondering about.

In QM it's said that particles don't exist in a definite state between measurements, but rather exist as a wave function, with a probability of being found in a certain place when they're measured. Neutrinos are very weakly interacting, and travel through the partial vacuum of space. Since they're not being "measured" when they're not interacting with other particles, they should exist in their wave-function state during those times. How come, then, don't they zig-zag as the fly across the universe? My intuition tells me that between interactions (measurements), their position and momentum should be determined by a probability function, and there should be a possibility of them changing direction during those times. Obviously, this doesn't seem to happen (from what I've read), so I'm guessing my intuition is wrong. Could anyone shed some light on this for me? Thanks.

There is an interpretation of QM that says that a sort of "zig zag" particle motion between experiments is exactly what happens:
http://en.wikipedia.org/wiki/Bohm_interpretation

 

[belliott4488]

Both the Bohm interpretation and Feynman's Path Integral formulation are ways of interpreting what happens between observations, but it's critical to remember that observations are the only things we can verify - not what happens between them. The predictions of the Schroedinger Eq. have been well verified, so any alternate interpretation must reproduce the same predictions, or it will likely be wrong. The alternate interpretation might help you to think about what's going on, or it might just be more comfortable, but in the end, the predicted observations are the same.

So ... think about zig-zag paths between observations if that works for you. They're definitely allowed in QM, but there will be infinite numbers of them, as there will be infinite numbers of smoother, straighter paths. When you work out the probabilities of all of those paths and predict what we'll actually see (expectation values), the most likely outcome will be a classical type path, i.e. smooth and straight.

There is a very tiny probability of a sequence of observations that zig and zag, but there's also a tiny probability that you'll tunnel through your chair in the next ten seconds and end up on the floor - but you don't worry about that, do you?

 

[Demystifier]

Both the Bohm interpretation and Feynman's Path Integral formulation are ways of interpreting what happens between observations,

I don't think that the path integral formulation says what happens between observations. If one would try to interpret it in this way, one would say that the particle passes all different paths at once, which would be logically meaningless.

 

[ueit]

Why doesn't QM make particles zig-zag as they travel?

Momentum conservation does not allow it.

 

[Demystifier]

Momentum conservation does not allow it.

QM is full of weird things when questions are asked on times BETWEEN measurements. So, perhaps momentum is not conserved between measurements? Indeed, when the wave function is such that the momentum is uncertain, Bohmian interpretation says that the momentum is certain but not conserved.

 

[belliott4488]

I don't think that the path integral formulation says what happens between observations. If one would try to interpret it in this way, one would say that the particle passes all different paths at once, which would be logically meaningless.

Well, no, of course it doesn't say that the particle actually follows all those paths. I just meant to point out the weirdness that occurs whenever we try to explain the observable results of quantum theory by using the natural language of classical physics, e.g. well-defined paths. Maybe I should just have said that it's not only correct to consider the possibility of a zig-zag path, but you must consider it - along with all other possible paths - to predict the actual outcome. Or maybe that's not any better ... it's hard to speak sensibly about these things.:uhh:

 

[ueit]

QM is full of weird things when questions are asked on times BETWEEN measurements. So, perhaps momentum is not conserved between measurements? Indeed, when the wave function is such that the momentum is uncertain, Bohmian interpretation says that the momentum is certain but not conserved.

Yeah, but when momentum is uncertain the wave function is constrained somehow (for example when the particle travels through a narrow slit). In this situation there is a momentum transfer between the particle and the slit, therefore the particle's momentum is not conserved. This is true even in classical physics. Doesn't BM predict straight trajectories for a particle departing from a point source?

 

[marlon]

The trace of a charged particle in a bubble chamber detector can be considered a sequence of measurements of particle's positions and is usually a smooth curve.

No, because you are not observing nor measuring the particle's position. What you observe is the emitted EM radiation generated by the interaction of the particle with the medium through which it passes.

Why is that not a zig zag?

Why would it be ?

Point is that one cannot speak about a particle's position, let alone its trajectory, unless you measure the position. But once you measure its position, all info on previous measurements is gone. So you certainly cannot build up the trajectory. In short : wavefunction collapse !

marlon

 

[vpoko]

No, because you are not observing nor measuring the particle's position. What you observe is the emitted EM radiation generated by the interaction of the particle with the medium through which it passes.

Why is that not considered a measurement of position? Don't we measure objects' positions constantly based on their interactions with photons? It seems to me that at least the classical concept of position is being imprinted on the emitted EM radiation. You could follow the emitted EM radiation back in time to the point of emission to determine the position of the particle that generated it. Of course I'm not sure if the the concept of position as piece of quantum information is different from what I've described.

 

[marlon]

Why is that not considered a measurement of position?

Because the electron does NOT interact with photons here but with the atoms that constitute the medium through which it passes. These atoms can get excited and emit EM radiation during the de-excitation process. You observe that radiation arising because of the atomic de-excitation. Interaction with phonons (ie crystal lattice vibrations) is also possible.

My point is that you observe radiation that is the CONSEQUENCE of the electron-medium interaction.

marlon

 

[lightarrow]

Because the electron does NOT interact with photons here but with the atoms that constitute the medium through which it passes. These atoms can get excited and emit EM radiation during the de-excitation process. You observe that radiation arising because of the atomic de-excitation. Interaction with phonons (ie crystal lattice vibrations) is also possible.

My point is that you observe radiation that is the CONSEQUENCE of the electron-medium interaction.

marlon

Do you mean that those atoms moves, so if a pulse of EM radiation emitted from one of them in a specific point, can actually corrispond to a particle that has passed 1 metre away?
I don't think it would be quite useful, in physics!
I really cannot understand why you don't consider that interaction as a measure of position. You consider "measure of position" only in the case of interaction with photons? So, how would you perform a photon's measure of position, considered there isn't any photon-photon scattering (at least at common energies)? Maybe with a photographic film? Isn't the photon or particle's position measured through the interaction with atoms in that case?

 

[marlon]

I really cannot understand why you don't consider that interaction as a measure of position.

Well, you tell me, what are we actually observing ? An electron that makes a certain trajectory or EM radiation that is emitted ? I guess you agree with me that an electron is not the same as EM radiation, no ?

I don't quite understand what your problem is here. We KNOW there is an electron through the observed EM radiation, we know how it behaves and we can check that behaviour with respect to theoretical models. Theory and experiment give the same results for the same initial parameters and THAT IS IT !

marlon

 

[Demystifier]

Doesn't BM predict straight trajectories for a particle departing from a point source?

Not necessarily. Only when the wave function is an eigenstate of the radial-momentum operator.

 

[lightarrow]

Well, you tell me, what are we actually observing ? An electron that makes a certain trajectory or EM radiation that is emitted ?

Indirect measure, as many measures we perform in physics, I don't see the problem.

 

[marlon]

Indirect measure, as many measures we perform in physics, I don't see the problem.

But are we "seeing" the electron or the emitted radiation ?

My point is that the "lines and circles" we observe in the Bubble chamber spectra are emitted radiation, not electrons that are moving along these trajectories.

marlon

 

[K.J.Healey]

But then we don't "see" anything do we? We only see the radiation emitted or reflected from a source.
I'd look to a more realistic definition of measurement where you can say you measure where a wall is by looking at it by using light to reflect off the surface. Or you can choose to argue all day that you're not measuring the wall but the LIGHT that came off it. Then you can argue that as a human you're not measuring the light but rather the effect the light had on the chemicals in the back of your eye, and so on and so forth.

 

[marlon]

But then we don't "see" anything do we? We only see the radiation emitted or reflected from a source.

EXACTLY ! The difference between classical and quantum physics is the influence of that interaction between matter and photons. This is essentially the crux of the uncertainty principle and that is why people need to distinguish between seeing a car and seeing an electron.

marlon

 

[lightarrow]

EXACTLY ! The difference between classical and quantum physics is the influence of that interaction between matter and photons. This is essentially the crux of the uncertainty principle and that is why people need to distinguish between seeing a car and seeing an electron.

marlon

Where is the difference between seeing a car crossing a certain area and seeing an (high energy) electron crossing a bubble chamber? Yes, you see light emitted from atoms of the bubble, in the last case, but would you see a different electron's trajectory if it traveled in the void and you would observe it through light scattered off it? To measure a particle's position you exploit an interaction between that particle and something else, be it atoms of matter, light or else.

 

[Demystifier]

Where is the difference between seeing a car crossing a certain area and seeing an (high energy) electron crossing a bubble chamber?

It depends on the interpretation.
For example, in the Bohmian classical-like interpretation of quantum mechanics there is no difference at all.
But also in the quantum-like interpretation of classical mechanics
http://xxx.lanl.gov/abs/quant-ph/0505143
http://xxx.lanl.gov/abs/0707.2319
there is no difference too.

 

[K.J.Healey]

I agree that there's no difference. As long as we keep in mind that whatever we're measuring is that particles state only when we measured it, and it is forever changed after we have measured it.

As for the electron bubbles experiment, is there any way of knowing what effect the bubbles have on the electrons trajectory? Such that being in a medium that absorbs the radiation. Is there any effect that is causing the radiation?
Or is it more like watching a car that has its headlights on, where regardless of whether or not we're choosing to measure it we can "see" it. We don't have a state where the car has its headlights off (electron not emitting) to compare it to.

 

[gptejms]

It depends on the interpretation.
For example, in the Bohmian classical-like interpretation of quantum mechanics there is no difference at all.
But also in the quantum-like interpretation of classical mechanics
http://xxx.lanl.gov/abs/quant-ph/0505143
http://xxx.lanl.gov/abs/0707.2319
there is no difference too.

When the particle no. is not fixed as, in rel. quantum mechanics,what will become of Q(quantum potential)?Will it keep changing depending on the no. of particles?
Or treating the wavefunction as a functional of the field and using the Schrodinger equation,should one write Q--in this case Q would also be a functional of the field.

 

[ueit]

Not necessarily. Only when the wave function is an eigenstate of the radial-momentum operator.

Can you give me some examples of how to prepare such a state?

 

[Demystifier]

Can you give me some examples of how to prepare such a state?

I am one of those hard-core theoreticians who typically do not know how to prepare states in practice. :tongue2:

 

[Demystifier]

When the particle no. is not fixed as, in rel. quantum mechanics,what will become of Q(quantum potential)?Will it keep changing depending on the no. of particles?
Or treating the wavefunction as a functional of the field and using the Schrodinger equation,should one write Q--in this case Q would also be a functional of the field.

To include particle creation/destruction, Bohmian mechanics needs to be modified.
There are several ideas how to do that. Most of them, in one way or another, require introduction of fields. In my opinion, the most elegant way to introduce particle creation/destruction, which is the only way to do it without fields, is by string theory. See, e.g., Sec. IV.B of
http://xxx.lanl.gov/abs/hep-th/0702060
In particular, Fig. 1 summarizes various approaches to particle creation/destruction.
For an additional argument that one has to introduce strings in order to make Bohmian mechanics consistent with particle creation/destruction see
http://xxx.lanl.gov/abs/0705.3542

For those who do not want to read these papers, let me briefly explain the main idea. Kinematically, there is no difference between one string and many strings, because many strings can be viewed as one string splitted in many pieces. The process of particle creation is a continuous process of string splitting (it is continuous in spacetime, not in space). Therefore, all you need is a quantum potential for one string. No fields, no separate quantum potentials for states with different numbers of particles/strings.

 

[marlon]

Where is the difference between seeing a car crossing a certain area and seeing an (high energy) electron crossing a bubble chamber?

The difference is in the fundamentally distinct nature between classical physics and QM. Seeing a car allows you to derive an eauqtion of its movement in terms of time or position, ie you can rewrite the equation of the trajectory. In QM, you cannot do this for an electron because of the HUP and because of the fact that you are seeing emitted EM radiation.


marlon

 

[lightarrow]

The difference is in the fundamentally distinct nature between classical physics and QM. Seeing a car allows you to derive an eauqtion of its movement in terms of time or position, ie you can rewrite the equation of the trajectory. In QM, you cannot do this for an electron because of the HUP and because of the fact that you are seeing emitted EM radiation.
marlon

If you remember I talked about high energy electrons. I don't think a 1 TeV electron has "problems" of HUP. About radiation, if the problem is light emitted from the atoms that have interacted with the electron, we use another way: we simply enlighten it. We will see the electron's trajectory as well.

 

[marlon]

If you remember I talked about high energy electrons. I don't think a 1 TeV electron has "problems" of HUP.

Well, i am talking about ANY electron. But that's irrelevant to this discussion.

About radiation, if the problem is light emitted from the atoms that have interacted with the electron, we use another way: we simply enlighten it. We will see the electron's trajectory as well.

"we simply emlighten it", what does that mean ? How does one achieve that experimentally.

I really feel that we are turning in circles here. I have said a thousand times that knowing an electrons orbit or trajectory VIOLATES the uncertainty principle. I would like to ask you how you can unify those two "arguing" concepts ?

Check out how an ordinary cathode ray tube operates : http://en.wikipedia.org/wiki/Cathode_ray_tube
An early version of this device was used in the first electron discovery by Thomson. Now, you tell me, when we use such a device, WHAT DO WE OBSERVE ?


Besides, the light emission process you describe there is not accurate. But, ok, that is not relevant now.

marlon

 

[lightarrow]

Well, i am talking about ANY electron. But that's irrelevant to this discussion.
"we simply emlighten it", what does that mean ? How does one achieve that experimentally.

Compton scattering.

I really feel that we are turning in circles here. I have said a thousand times that knowing an electrons orbit or trajectory VIOLATES the uncertainty principle.

The same for a car. The difference clearly is in the values. Let's say we want to know an electron's trajectory with 0.01 mm precision in its width. Since

$$\Delta p_x \Delta x & \ge\frac{\hbar}{2}$$

we have:

$$\Delta p_x & \ge\frac{\hbar}{2\Delta x} \simeq \ 10^{-34}/ 2*10^{-5} = 5*10^{-30}\ kg\ m\ s^{-1}$$

which corresponds, for an electron's mass $$\simeq\ 10^{-30}\ kg$$ to a

$$\Delta v_x\ge\frac{5*10^{-30}}{10^{-30}}\ = \ 5\ m\ s^{-1}$$

A 1 TeV electron has a speed v $$\simeq\ (1 - 10^{-11})\ c\simeq\ c$$

and so travels 1 m in $$\frac{1}{300,000,000}$$ s

The corrisponding transversal space is:

$$\Delta s\simeq\frac{5}{300,000,000}\ <\ 2*10^{-8}\ m$$

which is negligible with respect to 0.01 mm.