A problem with Quantum Mechanics?

Imo · Mar 19, 2005

Can anyone explain why this doesn't disprove QM?

By SR, inertial reference frames are equivalent, so the physics (and I'm assuming the Quantum Mechanical nature) must be equivalent. But then, if I'm traveling at a constant speed, the speed relative to myself is zero, and my position relative to myself is also zero. So Heisenburg's Uncertainty principle breaks down?

kleinwolf · Mar 20, 2005

Uncertainty principle is describing the uncertainty due to a "measurement" of a physical quantity. Since the measurement disturbs the system, I don't think u can use the uncertainty principle intrinsically, based on the hypothese: a particle does not measure itself.

jackle · Mar 20, 2005

You wouldn't notice the uncertainty principle when applied to yourself, the numbers are too small.

If you were an electron, you would have trouble traveling at a constant speed. You may well be traveling at different speeds all at the same time until someone measures you.

ZapperZ · Mar 20, 2005

You, like a bowling ball next to me, are not a quantum object. Various parts of your body are in contact with a gazillion degrees of freedom and you have "decohere". The same way you can classically determine exact position (and other observables) of a ball, this is in no way an indication of the breakdown of the HUP.

Look at the diffraction phenomenon from a single slit. THAT is the HUP staring at you right in your face (I have explained this elsewhere).

Zz.

Nacho · Mar 20, 2005

I think the answer lays in that when a "measurement" is made, it's always between two different objects. How can one object measure its speed or position relative to itself?

nickdanger · Mar 21, 2005

Heisenberg said that the uncertainty principle does not apply to the position and momentum determination of a location of where a real object or event has occurred, let's say the point on a scintillation screen where a photon has hit. We could fire photons of a not precisely identified frequency (here the uncertainty does apply), and then move the screen as a target, further and further away with each new photon that is fired, and so determine the path traveled and final position, which increasingly becomes as precise as we would like to make it. Instead the uncertainty applies only to the uncertainty of the future position and momentum of that photon had it not hit the screen.

So I would say you don't have to measure the coordinates on the screen where the scintillation occurred to know that it was indeed at a precise location. That location itself is classically definite and so are you in your location. So yes, you occupy a precise location. But the others on this thread are still correct. If you try to measure that location or even 'observe' it you necessarily interfer with the object sitting at its definite location and have uncertainty trying to make that determination.

Antiphon · Mar 21, 2005

Fundamentally, you simply don't have a unique position.

Even if you knew at some instant where you were, your wavefunction
(describing what might happen if you look at yourself after that)
spreads out and you yourself don't know where you are anymore.
(To make things worse, where you could be literally inerferes with
where you may be, hence the diffraction phenomenon of particles.)

DrChinese · Mar 21, 2005

If you actually performed an experiment to test this hypothesis, you would discover that the HUP does hold. And yet you would find no specific disagreement with SR either. The reason is that your hypothesis is invalid, i.e. that the deltas are zero. You can't prove that, and you would need to for your idea to make sense.

In other words, your statement works out to saying: "If I assume the HUP is wrong, then I therefore conclude that the HUP is wrong." Not a useful statement. Sorta like assuming SR is wrong by saying: "Assume I am traveling at the speed of light..."

reilly · Mar 21, 2005

Imo -- Clever question, better answers. You, however forget that QM is the most tested theory ever. It has yet to fail. This level of empirical success strongly indicates that verbal, nominally rational arguments will not put any dents in QM. If QM is found wanting, it will come through empirical evidence. Further non-mathematical arguments are at best metaphors. To convince physicists, you will need to provide careful mathematical analysis of your idea. If you do so, you'll find your verbal argument seriously wanting. Verbal cleverness is not good enough. Read some history, get the facts.

Regards,
Reilly Atkinson

kleinwolf · Mar 21, 2005

This is exactly the point with the EPR paradox, and particularly, the correlation of 2 spin-1/2 particles...(EPRB)

If you do only "verbal reasoning", without calculation, you get the correlation obtnained when measured along the same direction of the singlet state has to be perfect (if A gets -, then B gets +, vice versa).

However, a careful mathematical analysis proves this is wrong, and that on average, you get only 87.5% of correlation in that configuration.

Nacho · Mar 21, 2005

Yeah .. this basically boils down to the argument Einstein brought up early in his debates with Bohr, before EPR (completeness argument) .. "And in no way disturbing the 1st particle". Einstein gave up on that argument after awhile.

Computations of what a particle "should do" are not enough; as the Einstein-Bohr debates showed, you have to have the process of taking a measurement to realize the state of any particle.

I might add also, what I was trying to get at in my previous post, if you did know your own speed and position, then quite simply it wouldn't be termed "Relativity"; it would be termed "Absolute".

kleinwolf · Mar 21, 2005

Yes...I mean there is something weird that I suppose is still not clear in QM : intuitively, a non disturbing measurement is given by the identity operator : 1, since applied on any quantum state |S> : 1|S>=|S>...

However, the final state can also by anything, since every vector is eigenvector of 1. Hence you get an infinite number of possible end states by measuring without disturbing a state...which seems quite dumb.

seratend · Mar 22, 2005

Simple question and a lot of answers!
In fact, your question simply deals with the interpretation problem of the position and momentum states in QM.
As we suppose that classical paths do not exist in the QM world (e.g. the which path nightmare in a double slit experiment), we must take care when we want to associate a classical position and a classical path to a quantum state.
In QM, the position and momentum are conjugate variables. This is a basic postulate of QM (i.e. [Q,P]=ihbar, from where we deduce the HUP). Now, each time you use the QM position and momentum states, never forget that these individual states are not "classical fixed positions" or "classical constant motions" (interpretation problem).

In fact, you do not have to go to a so complicated example (change of a reference frame, use of SR or galilean frames) to understand the interpretation problem of QM position and momentum states. Just take one free particle in the state of a null momentum (|p=0>). Can we say that this particle has a defined position? (This is the reformulation of your problem in a simpler form, where we are trying to interpret the null momentum state as the path of a particle with a null speed).

The "standard" QM answer is no. If you do the experimental position measurement on a set of particles with a state |p=0>, you will obtain a different position "x" result for each trial, where x belongs to the |R line. That is the formal result of standard QM: a null momentum state does not say that you have a defined position (as shown by the experimental trials you can realize).
Therefore, each time you confuse the QM state with classical representations (such as the path of a classical particle), you will encounter logical incoherencies.

However if you still need to "see"(interpret) a classical path behind a QM momentum state, just have a look at the de brooglie-bohm formulation of QM. With this reformulation of QM, you have a logically coherent description of particles based on "almost classical paths" (at least in the non-relativistic approximation).

I hope I have answered to your question.

Seratend.

Imo · Mar 22, 2005

Thanks for everyone's answers. Near the end, there seemed to be a small discussion dealing with interpretations of QM, which I'm not all that familiar with (since we're simply taught the Bohr interpretation).

I think I understand the answer to the problem. I stated that the momentum and position are simply zero "by definition" so to speak. In fact, it's been pointed out, I can't say that.

Some people tried to explain why position and speed aren't zero, but I'm still not so sure why it's not so (other than the fact that QM is just intrinsically weird). Would anyone mind a go at a (really) simplistic reason for this? thanks

oh, one final thing. This was by no means a thread to say that the Bohr interpretation was wrong. I put that title to catch people attention, but the question mark was there because I knew there had to be a reason why it didn't discard the interperation (I know I'm not that smart as to be able to disprove Bohr overnight). Sorry if it put some people on edge.

Imo

bw · Mar 22, 2005

I think there are different ways to answer your question(as the formulation is not very precise).

There are actually two uncertainty principles, which are conceptually very different, though they are often confused as the same thing in most physics textbooks.

One version pertains to a SINGLE measurement on the SAME object, as demonstrated by Hesienberg's famous microscope. The deltas roughly correspond to the resolution of the quantities being measured.

I don't know how valid this is, because as far as I know it is not derivable from the QM formalism. The actual content of this statement is kind of unclear. For example, Heisenberg's picture conveys the wrong impression that the object did have well defined momentum and position prior to the measurement. Only that the act of measuring "disturbes" the system so violently that in gaining one piece of information(position, say), we inevitible destroy the other piece(momentum).

But according to the standard formalism of QM, the object existed in some "superpostion" state of limbo prior to the measurement, it didn't have any clearly defined momentum and position (assumption that it did lead to contradictions with experiments,as all quantim interferences would disappear). Hence it is not correct to think that uncertainty is the result of these quantities being "disturbed" by of the act of measurement. It is also unclear how this sort of limitations come about except in specific examples.

On the other hand, a technically precise version of the UP is a STATISTICAL statement.

It says that if you carry out REPEATED measurements of two "non commuting" observables(say, position and momentum, or two different spin components of an electron) on an ENSEMBLE of identically prepared systems(a stream of electrons described by the same wave function,say), the product of the STANDARD DEVIATIONS of the measured quantities(the delta's) cannot be made arbitrarily small(typically bounded below by a number of the magnitude of Plank's constant).

This version of the UP is precise. It is an easy consequence of the axioms of QM (or the Fourier transform, in the case where the observables happen to be momentum and position of a "point particle").

This version, while precise, seems a bit limited becuase there are times we DO want to talk about single measurement type of situations. Moreover, I should point out that there is still no consensus as to what the wave function actually means. Problems can arise if we use the notion too freely. Howvever, it is too restrictive if we insist that the wave function must be used in connection to an ensemble(so what is the wave function of the universe a la Hawking?)

Now if we use the second version of UP, your original question is clearly meaningless as obviously there isn't an ensemble of you all prepared to the same quantum state.

Saratend converts your question to an ensemble of paricles with definite momentum and explains why the position is "uncertian". By performing REPEATED measurements of the particles' position we find particles all over the places, the standard deviation is infinite.

As to why your position and momnetum are not "really zero". One easy answer is that all measurements have errors. In determining the momentum and position of a massive object such as yourself, any small error in the meausrement, though too small to be detected in instruments usually used in our "macroscopic" scale, is huge comapring to Plank's constant. You are a massive object, your de Broglie wave length would be so tiny that quantum effects would be undetectable( I neglect the complication that you're actually a composite object made up of a large number of partcles, which somebody has already mentioned)

A key point is of course, what is really a measurement? By that it usually means some action that creates a "record" in the "macroscopic" world and therefore it is tied to irreversibility. But what is "macroscopic" and how irreversibility arises? These questions go to the very foundation of QM. The last word has certainly not been spoken yet.

bw · Mar 22, 2005

There are a lot of weird things and mystries in QM. But this is not one of them.What you discribed is not a measurement.

A measurement, mathematically speaking is projecting a state vector onto a certain subspace of the Hilbert space(assumming finite dimesnion for simplicity) in question. If you try to measure a quantity represented by a self adjoint operator A(assumming the spectrum to be non degenerate for simplicity), say, this would mean projecting the state vector |S>, onto an eigen space of A, the corresponding eigen value a would be the outcome of the measurement.

Immediatly after the measurement, the state |S> "jumps" to an eigen state |a> corresponding to a(not the projection, as projections don't preseve length of the state vector).This happens with probability |<a|S>|^2 .It then continues to evolve according to the Schrodinger equation.

Note that the measurement DOES NOT convert the state vector |S> to the "end state" A|S>! This won't make sense because self adjoint operators in general don't preserve norm,--hence probability,--you don't "apply" a self adjoint operator to a state vector to get another state vector.

So by applying the identity I repeatedly you're not making any measurement.You're just saying the state has not changed.So the identity basically means do nothing, as expected.

But here is a non trivial way to use the identity operator.

Given an orthonormal basis which are eigenstates of an observable A as above. We can write I = ∑|a><a|, the sum is over all the eigen vectors of A in our basis. Now suppose I have another observable B. If I perform a measurement of B on the same state |S>, the probability amplitude of getting the value b would be

<b|S>=<b|IS>=<b|∑|a><a|S>=∑<b|a><a|S>

The right hand side expresses the probablity amplitude for each possibility if we measure A(but we don't, the identity means "do nothing"!). When computing the square modulus of this equation to get the probability the sum on the left hand side gives rise to "cross terms", which correspond to quantum interference, which would be absent if we simply add the probabilities instead of the amplititude, as in classical physics.

Consider the two slit experement. B in this case would be the particle's position on the screen. A would represent "which hole the particle goes through"(this is not very precise because it is not clear what the eigen values of A would be, as the "measurment" here results in yes/no, rather than a numerical answer, but you get the idea), the eigen states of A would be "top hole" and "bottom hole".

The equation above says that when computing the amplitude of detecting the particle at b, we must sum over the aplitudes of all possible channels through which the particle may have explored.

So the identity operator represents "opening all possible channels" . Suppose there are several alternative ways through which an event might have occured. If no "irreversible record" was left (i.e. no "measurement" was made)to indicate which alternative was followed(all channels was opened), we must add all the amplitudes for all the channels (or "paths")before we square to get the probability.

This is the mathematical reason of quantum interference.

Properly refined this idea leads to Feynman's path integral.

Imo · Mar 22, 2005

In response to bw on his response to my question:
So what your saying is my question is fundamentally flawed (which i thought was the case but I wanted to make sure I was right). Even if I were on the quantum level, I would still have an "intrinsic unknown" in my measurements for both speed and position simply based on my wave function and that this would compensate for the HUP.

Basically, my assumption that the speed and position relative to yourself is zero is just not true because of the statistical wave function.

Thanks again everyone

seratend · Mar 23, 2005

I think you still not understand the problem. As long as you keep thinking that the operators position and momentum of QM (and their associated eigenvalues) are the classical position and classical speed (in case of no em field), you are in trouble.
In classical mechanics, you have mdq/dt=p (no em field), i.e. you have a simple relation between the position and the momentum. You have no more this case in QM. You just have a weak equivalence for the mean values of the position and momentum operators: m.d/dt<Q>=<P> that's all (no em field).
If you prefer, we could have renamed the results of a QM position measurement, "square" and the results of a QM momentum measurement, "circle".
Now, why should we say that the QM "square" is the classical position and the QM "circle" is the classical momentum? ("Some people tried to explain why [QM] position and speed aren't zero")
It is not because 2 different theories use the same label for different objects that the objects are the same. These theories use the same label, because we have the same results under some restricted conditions/limits (i.e. the classical limit hbar -->0).

Seratend.

kleinwolf · Mar 26, 2005

BW :

This is clear (a measurement is an isometry, since initial and final states are normalized). The complete measurement process is however more complicated, since every rotation corresponding to a possible outcome, should be represented with given probabilities...

For what I said : since every vector is eigenvector of 1, then, 1¦S>=¦S> means :

"the final state is ¦S> with probability 1"...

However, there are other possible endstates that happens with nonvanishing probabilities...does it make sense, since there existe one particular eigenstate, namely ¦S> (the initial state), that happens with probability (not probability density) 1 ?

marlon · Mar 26, 2005

Check out my latest entry :
https://www.physicsforums.com/journal.php?s=&journalid=13790&action=view

regards
marlon

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