# Heisenberg and quantum mechanics

1. Jul 22, 2006

### superweirdo

Heisenberg showed that, even in theory with a hypothetical infinitely precise instrument, no measurement could be made to arbitrary accuracy of both the position and the momentum of a physical object.

What I want to know is how? How did he show it coz that conters my belief of predictability. I believed that w/ infinite knowledge and tools, we could see the future coz we would have the ability to predict everything.

2. Jul 22, 2006

Staff Emeritus

Actually the way Heisenberg did it is this: In quantum mechanics each of position and momentum come out as probabilities when you perform a measurement. And the quantum math says that these two probability distributions are Fourier Transforms of each other. Well there is an old theorem about fourier transforms of each other that says the product of their standard deviations is constant. So the sharper one is, the more diffuse the other one is, in a seesaw relationship. Hence the smaller a variance in position your measurement produces, the bigger a variance in momentum it entails, and vice versa.

3. Jul 23, 2006

### Locrian

The HUP is hardly the biggest hurdle in predicting everything! In fact, the HUP is unimportant in predicting almost anything.

4. Jul 23, 2006

### ZapperZ

Staff Emeritus
There is a again a misconception here. Let's see if you might be surprised by this:

The HUP says nothing about how precise a measurement you can make in a single measurement. I can make as precise a measurement as I want of a particle's position, and after that, I can make as precise a measurement of its momentum as I want, limited only to my technical limitation. There's nothing in the measurement accuracy in here that is limited by the HUP.

Look again at the HUP and look at the statistical nature of the observable that is being measured to arrive the the $\Delta(x)$ value, for example.

Zz.

5. Jul 23, 2006

### MeJennifer

In your example you are making two measurements not one.

6. Jul 23, 2006

### ZapperZ

Staff Emeritus
Which is what is contained in the HUP.. the relationship between two pairs of non-commuting operators. If you make a measurement of A, then your knowledge of what B would be IF you were to measure it NEXT would depend on how well you know about A. It doesn't say you make just ONE measurement.

Zz.

7. Jul 23, 2006

Staff Emeritus
Kind of missing the point aren't you MeJennifer? The point is that you can know the position accurately as long as you don't ask about momentum, or momentum accurately as long as you don't ask about position, and if you like you can do these two measurements successively on the same particle, and this is all in accord with the uncertainty principle. If you don't like seeing both measurements discussed, just read one.

8. Jul 23, 2006

### moving finger

To be fair to superweirdo, what he should have said is “we cannot simultaneously know to arbitrary precision both the position and the momentum”

Best Regards

9. Jul 23, 2006

### -Job-

Can we get arbitrarily close to the values of position and momentum of a particle at a time T by measuring these quantities separately & successively with an arbitrarily small interval between measurements?

Or maybe by measuring momentum at time T1, and measuring position at time T3 and using this information to estimate the quantities at time T2, then again measuring momentum at time T5 and position at time T7, using these values to estimate position and momentum at time T6, then using T2 and T6 to estimate the quantities at time T4 and so on. Through a process of eliminating error little by little can't we catch up to the actual values for momentum and position at a given T?

Last edited: Jul 23, 2006
10. Jul 23, 2006

### moving finger

No, because if you measure position to absolute precision (ie with zero uncertainty), that automatically makes momentum completely unknown. If you then subesequently measure the momentum to absolute precision, your earlier position measurement becomes invalid for making any predictions - measure the position a second time (after the momentum measurement) and it will be different to what you thought it was going to be based on your earlier measurment.

There is an analogous (but perhaps easier to visualise) problem with "vertical spin" versus "horizontal spin". If you measure the vertical spin of a spin-1/2 quantum object (eg an electron) then you will find that it is either spin-up or spin-down. But having done this, the horizontal spin is now completely unknown. If you subsequently measure the horizontal spin then you will find that it is either spin-left or spin-right, but having done this the vertical spin is now completely unknown! You cannot "know" both the horizontal and vertical spin directions of a spin-1/2 object at the same time, because the question has no meaning. In the same way, you cannot know the position and momentum (with arbitrary position) at the same time, because the question has no meaning.

Best Regards

11. Jul 23, 2006

### Rach3

Problem is, there really does not exist such a thing at given time T! Think of a wave packet; a very narrow wave packet involves plane waves of a wide spread of momentum (and v.v.), in its Fourier representation. Is there a wave packet well-localized in both space and momentum? It's quite absurd. In QM this is essential to measurements; if you make a precise position measurement, the wavefunction is very narrow in position space in consistency with that measurment, so the momentumspace wavefunction is very wide and has little information. You could measure momentum here, but that would collapse the wavefunction into something very wide in position space, thus erasing information about your earlier position measurement.

12. Jul 23, 2006

### Rach3

The essential feature here is that the "spin up" and "spin left" states are not orthogonal as you'd expect, but at a 45deg angle. The basis state for a spin-1/2 object is quite non-intuitive; it's a two-dimensional space, spanned by {spin-up, spin-down} basis states, or equally well by {spin-left, spin-right} basis states, at 45deg and 135deg angles to the previous two.

Thus, "spin up" is equally well described as a superposition of "spin left" and "spin right".

13. Jul 23, 2006

### -Job-

In other words, is it as if any process measuring the momentum at a given time can change the position and vice-versa?
In macroscopic objects this problem doesn't seem to exist. At what point in the scale of matter does accurate knowledge of both position and momentum become impossible to gain? Can we at least use the values obtained at the smallest "allowed" scale to bound the values at the "prohibited" scale?

Is it also impossible to purposedly give a particle exact values for position and momentum at the same time?

Last edited: Jul 24, 2006
14. Jul 24, 2006

### Rach3

At the scale where $$\Delta x \Delta p$$ are on the order of Planck's constant. As far as we're dealing with coherent systems, there's no abrupt boundrary, but a gradual transition - large systems involve large X's and P's, and so the HUP becomes much less important, till it becomes negligble at macroscopic scales. The intermediate region is known as the "mesoscopic" region, ('meso' for 'middle'), and quantum effects at this scale are an active area of research. (by which I mean - there's much we don't know!)

15. Jul 24, 2006

### moving finger

when one makes the measurement of horizontal spin (for a spin-1/2 object) the physical configuration of the system is orthogonal to the configuration one uses for vertical spin - see http://www.upscale.utoronto.ca/GeneralInterest/Harrison/SternGerlach/SternGerlach.html

This is another way of saying that if we measure vertical spin, horizontal spin is unknown.

Best Regards

Last edited: Jul 24, 2006
16. Jul 24, 2006

### Rach3

The physical system yes, but the physical quantum states, in the Hilbert space, are not orthogonal. That's exactly the common source of confusion; orthogonal states are emphatically not the ones corresponding to measurements orthogonal in physical position.

Apologies for any confusion.

17. Jul 24, 2006

### moving finger

It doesn't "seem" to exist because the uncertainty is normally so very small in the macroscopic world - so we can usually ignore it.

The process of measuring momentum doesn't exactly change the position - but there is no easy way of understanding what is going on.

Explaining exactly what is going on in the apparently simple Stern-Gerlach case of spin-measurement is not easy - it does not matter which "direction" you measure the spin of a spin-1/2 particle to be, you will always find the spin is quantised in one of two ways - +1/2 or -1/2 relative to the orientation of the measurement system.

Best Regards

18. Jul 24, 2006

### moving finger

OK, agreed. To avoid confusion one obviously needs to be careful to specify whether one is talking about Hilbert space or physical space when referring to orthogonality in measurements.

Best Regards

19. Jul 24, 2006

### lalbatros

That's right the uncertainty principle doesn't allow you much predictions.
It is merely the root of quantum mechanics.
Further, it explains the most fundamental predictions.

Like the size of the atoms.
Quantum tunneling, nuclear decay.
Vacuum quantum noise.
Electron diffraction.
Line width.
Resolution of electron microscope.
Quantum cryptography.

.... any more ?

Let's start a big list ...

Michel

Last edited: Jul 24, 2006
20. Jul 24, 2006

### ZapperZ

Staff Emeritus
Yes we can!

The precision of a SINGLE MEASUREMENT of position and momentum is limited only via our technology. Furthermore, the more precise I measure momentum is INDEPENDENT of how precise I can measure momentum. This does not follow the HUP! For example, in my avatar, would you be surprise to know that the horizontal axis is in fact, a measure of the momentum of a bunch of photoelectrons, and that how precise I can measure it depends only on the pixel size of my CCD? Yet, I haven't told you anything about what happened before, i.e. did I pass it through a slit?

People forget that the HUP is a STATISTICS of measurement, not just ONE single measurement. The average of a measurement makes no sense when you have just one measurement to play with. No, what the HUP says here is that if you have a pair of non-commuting observables A and B, then if you make a measurement of A, then your knowlege of the outcome of a measurement on B depends on how well you knew A. This is contradictory to classical physics, where the more you know A, then you will have a better prediction of B for identically-prepared system.

Zz.

21. Jul 24, 2006

### moving finger

Do you literally mean measuring position AND momentum simultaneously? If so, this would seem to be in conflict with the HUP. Can you provide any published information to support this claim?

Is this a typo? Did you mean to say that momentum measurement precision is independent of momentum measurement precision? If yes, what on earth does it mean?

OK - but how does measuring momentum alone have any bearing on the question of measuring momentum and position simultaneously?

That's just what I said. Position and momentum are non-commuting variables. This means that we cannot know both simultaneously to arbitrary precision - the precision to which we can know both simultaneously is limited by the HUP.

Best Regards

Last edited: Jul 24, 2006
22. Jul 24, 2006

### ZapperZ

Staff Emeritus
Again, let's go back to the single slit example that I've used numerous times.

As I make the slit narrower and narrower (let's say the width is along x), then the uncertainty in position of the photon or electron that pased through the slit $\Delta(x)$ is getting smaller.

After it passed through the slit, I want to know its x-component momentum. What do I do? I let it hit a detector behind the slit. How do I measure $p_x$? I measure how far it has deviated from the center line, because this tells me how much momentum in the x-direction it has picked up after it passed through the slit. Now, how well can I measure this momentum? It depends on how FINE of a detector I have! If I have a very fine CCD, I can measure where the particle hit the detector to very high accuracy! This accuracy has nothing to do with how fine I measure $\Delta (x)$! This is why I said that my ability to finely measure this momentum is INDEPENDENT of how well I determine the position. In a single measurement, the HUP has no role!

However, and this is where the HUP kicks in, if I were to perform the IDENTICAL experiment again, even if $\Delta (x)$ is the same as before, the value that I would obtain for $p_x$ may NOT be the same. In fact, the smaller $\Delta (x)$ is, the MORE VARIED $p_x$ can get as I do this experiment many times. I will see the spread in momentum that I measure getting larger and larger as I know more precisely where it passed by making the slit smaller.

What does this mean? The smaller the slit and the better I know where the particle was when it passed through the slit, the poorer is my knowledge of what momentum value the particle will have after it passed through the slit. This is a reflection of the STATISTICAL spread of the momentum values, NOT from the accuracy from a single measurement!

This is the one aspect of the HUP that many people do not get.

Zz.

23. Jul 24, 2006

Staff Emeritus
What I hear you saying is that the Born rule associating a probability with the squared amplitude has only a frequentist interpretation at the phenomenal level. You can't really see the probabilistic aspect of one particle; you have to use the collective statistics of many particles.

Is that a correct interpretation of what you wrote?

24. Jul 24, 2006

### ZapperZ

Staff Emeritus
Probably. :)

The "statistics" here isn't any different than classical statistics. You make a measurement of something repeatedly, and only after a "sufficient" number of sampling are you able to deduce what the "actual" value is supposed to be. Typically, you get a Lorentzian or Gaussian distribution centered at the "actual" value. But how is one to know that when one only made ONE measurement. The sampling domain is not sufficient to convey any degree of confidence.

What makes a QM system different than a classical system in this case is that the "widths" of such distributions (more commonly known as the standard deviation in statistics) between a pair of non-commuting observables are related to each other via the HUP. This is where the HUP kicks in, in the spread of values of each of the observable, but NOT in the value of each individual measurement itself. The latter is determined by the ability of the instrumentation.

Zz.

25. Jul 24, 2006

### moving finger

You're not measuring position and momentum simultaneously in your experiment, you are measuring them sequentially, via two separate measurements (the "slit" measures position, and a certain time later your CCD detector measures another position, from which you infer a momentum). But you can only correctly infer momentum from this second measurement if you assume that the electron has behaved like a classical macroscopic object between the two measurements. The measurements are separated in time, thus not simultaneous. Check any good text on QM, they all say the same thing - one cannot measure position and momentum simultaneously to arbitrary precision.

Best Regards

Last edited: Jul 24, 2006