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Talk:Uncertainty Principle

Are you quite sure it's h/4π? I've always seen the Uncertainty Principle written as Δx Δp ≥ h-bar, which is equal to h/2π. -- Xaonon

I'm not sure, and neither are the experts:

I just checked Encyclopedia Britannica, and they give both 2π and 4π in different articles. A case for m:Making fun of Britannica :-)

I think we should leave it at 4π -- at least we are on the safe side. --AxelBoldt

I think it's 4π, though it's a while since I did an y QM. Doesn't it come from:

(ΔA)(ΔB) >= (1/2) |[A,B]|

and

[x^,p^] = h_bar / i ,

(Δx^)(Δp^) >= h_bar / 2 .

-- DrBob

Ok, I will then happily make fun of Britannica now... --AxelBoldt

I think to remember that DrBob is right. The thing is that when you use it to estimate a value, sometimes you say

(Δx^)(Δp^) ~ h_bar

that is compatible with the previous--AN

Sadly, the source is a bit old, but in Feynman's lectures, Feynman defines the measurements Δx and Δp to be the width of a gaussian distribution, which may be the source of the confusion. He does, however, state precisely:

ΔxΔp ≥ h/2π

It was originally published 1963-1965, so I guess it may have changed since then.

BlackGriffen

Ah, that must be it. They never clearly say what they mean with "uncertainty". Feynman takes thh "width", which is probably twice the standard deviation. We take the standard deviation itself, that's why we get half his number. So we are fine, but EB is still screwed :-) --AxelBoldt

Not quite Axel. Check the math. If Δx' and Δp' are the standard deviations, equal to half the width, then:

(2Δx')(2Δp') ≥ h/2π

=> Δx'Δp' ≥ h/8π

That might be the source of the confusion if the competitors were:

ΔxΔp ≥ h_bar

ΔxΔp ≥ h_bar/4

I'm not familiar enough with statistical data analysis to say much more, however. Does the width of a gaussian divided by √2 mean anything? --BlackGriffen

Yup, √2 times standard deviation is the width of the range where 50% of the values will be. But it only works for a gaussian distribution, and there is no reason to assume that all observables are normally distributed, in fact they're most definitely not, so our use of the standard deviation is much cleaner. --AxelBoldt

So, the width over √2 corresponds to the 50% of observations. Doesn't that mean that the formula:

ΔxΔp ≥ h/2

is the incorrect one? the correct ones being:

ΔxΔp ≥ h

σ_xσ_p ≥ h/4

lower case sigma is the standard character for a standard deviation, right?--BlackGriffen

In stats, they use sigma, but it seems that physisists use Δx both for standard deviation and for the 50% range, and call both "uncertainty". If we used Δx for the 50% range and σ_x for the standard deviation, then the correct formulas would be

ΔxΔp ≥ h

σ_xσ_p ≥ h/2

But the first of those is really pointless since it makes the unjustified assumption that the variables are normally distributed. --AxelBoldt

If there are two differing definitions of Δx and Δp we should note this, and that the uncertainity principle takes different forms depending on what definition is chosen. -- SJK

I removed the reference to the energy--time uncertainty principle, since it is not really correct. While energy surely is an operator in quantum mechanics, time is not, it's a parameter. One derives uncertainty inequalities from commutators of operators (i.e. if the commutator between two operators is not zero then there is an uncertainty relationship between the standard deviations). Since time is not an operator, it commutes with everything. I would suggest that the primary author of this article read the relvent sections of L. Ballentine's _Quantum_Mechanics_A_Modern_Development_ for a very clear discussion of the HUP.

While I'm ranting, I would suggest dropping the stuff at the end about Einstein. First off, he never denied the empirical validity of the HUP, and second even if he did his opinion on the the structure of QM is best left to another page.

It's actually messier than that. In standard non-relativistic QM, position is an operator and time isn't. If you extend this to relativistic QM, things get very messy, since there isn't even a quantity that stands up and says that "I'm the position operator".

Granted, I was thinking of the non-relativistic theory. Perhaps a paragraph about the relativisitic theory is in order? --matthew

--- I'll need to read Ballentine to see what he says, but this seems wrong. The variable t does not commute with the Hamiltonian operator, and mathematically, a wave function of finite time does not have a defined energy and the mathematical relationship between energy and time appears to be the same as momentum and position. -- Chenyu

The more I think about it, the more I think that Ballentine is wrong if he is asserting that there is no energy-time uncertainty relationship

That's not what I (or Ballentine) said. I merely recommended Ballentine's book as a very clear discussion of the HUP. --Matthew Nobes

or that time commutes with energy.

Huh? Time commutes with *EVERYTHING* in non-relativistic QM. It's a parameter, there is no time operator. --Matthew

I've found these links on the web

[[:http://www.aip.org/history/heisenberg/p08a.htm|http://www.aip.org/history/heisenberg/p08a.htm]] [[:http://press.web.cern.ch/pdg/cpep/unc_vir|http://press.web.cern.ch/pdg/cpep/unc_vir]]

Granted, these are popular pages, but one presumes that AIP and CERN had someone proofread the pages. There is also the discussion in

[[:http://math.ucr.edu/home/baez/uncertainty|http://math.ucr.edu/home/baez/uncertainty]]

which I think we should fold into the article -- Chenyu

Umm the link to Dr. Baez's page renforces what I said. Notice the derved relationship is between the total energy and the time derivative of some operator. That's why I deleted the energy--time reference, because this type of UP requires more careful thought --Matthew

I don't think that time does commute with energy in NR QM. To calculate the time expectation value of wave function phi, you use the expression <phi|t|phi>. To calculate the energy function you use an operation which has a time derivative in it. t and d/dt do not commute. It's formally exactly the same relationship as x and d/dx. Yes you can make a distinction between t the parameter and t the operator, but you can do exactly the same thing between x the operator and x the parameter.

Okay, I think you realize the error below, but just in case let me reiterate, there is no time operator in non relativistic QM. Such a thing *does*not*exist* in the theory. Dr. Baez's webpage, which you cite, gives a plausible way of constructing something that might look like a time operator, but it won't be time itself. And things get worse in the relativistic theory, since there is not good position operator either. Your expression above shows this as well, say you set

<t> = <phi|t|phi>.

Now t is not a operator, so this is

<t> = t <phi|phi> = t (<phi|phi>=1).

This is *not* the same thing as x=<phi|X|phi> since there is no particular reason to assume that |phi> is a position eigenstate. -- Matthew

I really don't see any reason why t should be treated differently in NR QM than position. -- Chenyu

Becuase time is not an operator. X and P~d/dx are operators. -- Matthew

Never mind - I think I just saw it.

On second thought I don't see it --- Chenyu

AARRGRRHHHHH!!!!! My mind is frying.....

Anyway, I don't have any objection to the article as it stands. I misunderstood the original comment to say that there wasn't an Energy-Time relationship rather than to say that its derviation is tricky. Maybe we need another article just for the energy-time relationship. --- Chenyu

I suppose `one copy of THE system ... and another, identical one' is meant to refer to two real life systems with the same specs. But the formulation sounds much like ensembles of systems from conventional statistics, i.e. just thought experiments. I feel the statement would be stronger if the fact that the systems are real is stated more explicitely. That is, if I'm correct about that. RitaBijlsma

If you mean

Disturbance plays no part since the principle even applies if position is measured in one copy of the system and momentum is measured in another, identical one.

then I agree that the statement needs to be clarified. Check out the no cloning theorem. -- CYD

Recently, an addition was made to the article claiming that "Conciousness interpretation" of QM have been proven wrong. I seem to remember an experiment where the two slits of a double slit experiment are equipped with detectors. If the detectors record which slit an electron took, and somebody looks at the results, then no interference pattern occurs. But if the detectors record the information, and subsequently the information is destroyed before anyone has a chance to see it, then the interference pattern does occur. How are these results being interpreted nowadays? --AxelBoldt

As far as I know, what you described should not happen: in both circumstances there should be an interference pattern. Can you provide a link? We need an article quantum decoherence. I will get around to writing that one of these days, if no one else does it. -- CYD

Huh? This whole thing confuses me. Here's what you might be thinking of, if you put dectectors at the slits, and they function at 100% then there will be no interference. It doesn't matter wether somebody reads the output of the detectors or not. If the dectectors worked then the interference is destroyed. There is an interesting effect if the detectors are not 100% efficient. Then the interference pattern get's ``washed out. There is a brief attempt at an explanation on my home page see

http://www.sfu.ca/~manobes/posts/twoslit

for some details. Also I think what is needed is one long complete article on quantum mechanics, not a great mass of micro articles on various features. Actually even better would be two long articles, on for laymen and one more advanced. --Matthew Nobes

I think the current approach is fine. We already have what you suggest, in the articles quantum mechanics and mathematical formulation of quantum mechanics. These lay down the framework of the theory. However, it is necessary to have "micro articles" such as this, simply because the secondary requirements and implications of quantum mechanics are so numerous. The uncertainty principle is not a postulate of the theory.

That's true, it's a theorem about operators. This is my point though, for a layman's type article the HUP requires a couple of sentences, there is no need for a micro article. From a mathmatical standpoint it is an easy to prove theorem, again a couple of lines in a long article. However, since I don't have the time to write any long articles right now I'll quit critiquing the approach others take, and stick to physics issues. -- Matthew

In my understanding, position and momentum don't have to be noncommuting observables; the fact that they are is a result of our choice of Hamiltonian and state space, which is motivated by experimental results (phenomenology).

That's not how I'd put it. Momentum is the generator of translations, as such it will not commute with position. This is true even in classical mechanics where x and p~d/dx have a non-zero Poisson bracket. -- M.

The Pauli exclusion principle is another example of a principle often associated with, but not strictly required by, quantum mechanics. --CYD

That's not really true. The PEP is a theorem in the relativistic formulation. You cannot construct a consistent theory of fermions without it (at least in three space + one time dimension) look up the ``Spin-Statistics theorem. *Historically* it was a phenomenological principle, but from a logical perspective it is contained in the theory. -- M.

I put in the various statements about the consciousness interpretations

of QM. As far as I am aware of, few if any physicists have ever believed that consciousness has a special role in wave function collapse. The problem is that many popularizations of QM have made it seem that there is a connection, and the fact that a lot of people seem to be influenced by that is why I keep emphasizing the point that you can demonstrate that it isn't. I'm not aware of the experiment that you referred to.

I searched around and couldn't find it either anymore. The closest I found was an experiment involving polarization and a double slit, which might be interpreted as "the interference pattern reappears if the polarization information is destroyed by another polarization filter", but that hardly qualifies as an introduction of conciousness into QM. --AxelBoldt

In the sound wave example, it is possible to calculate the exact frequency of a sound wave at a given time. The mathematics for this calculation are exact, notwithstanding the measurement error introduced by the uncertainty principle.

Given a sound signal, you cannot talk about the exact frequencies contained in the signal at a specified point in time. In order to do Fourier analysis, you always need to look at the function over a whole interval; a single point is not enough. And if you make the interval smaller, the uncertainty in the frequencies increases proportionally. It's very much analogous to the uncertainty principle, and in fact, the same theorem underlies both effects. --AxelBoldt

That's not true. It is not the same theorem. Standard QM uncertaintaty relations are derived from a theorem about operators on a hilbert space. These are not things that occur in classical wave mechanics. This idea about frequency and intervals is what underscores the time-energy relations, which is precisly why I cautioned against talking about them in the same language as standard QM relations. -- Matthew

There is a theorem relating the "uncertainty" in a function and the uncertainty in its Fourier transform. In standard wave mechanics, the Fourier transform is a Hilbert space automorphism which translates between the position observable x and the momentum observable -i d/dx. So the Fourier theorem can actually be used to prove the space-momentum uncertainty relation. --AxelBoldt

I'd like to put this sentence back in the main article. It's a clear layman-accessible explanation of what the uncertainty principle is all about.

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