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In summary, the uncertainty principle contradicts the idea that an object can be at rest relative to an observer.

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They will have uncertain positions, too.

However, I would expect thermal energy to introduce more error into this experiment than uncertainty.

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If you assume something to be true, then it is true.

The uncertainty principle is a restriction on what can be measured, not what can be assumed.

The uncertainty principle is a restriction on what can be measured, not what can be assumed.

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Not so. Certain things can be assumed to be true in a thought experiment, but if such assumptions contradict the laws of nature then the thought experiment achieves nothing. It is like assuming speed of light is greater than c to deduce a fact. The fact will be irrelevant to physics, as in physics the speed of light is c. The uncertainty principle is more generally that one cannot know exact position and momentum rather than one cannot measure exact position and momentum. This is because in order to know something about an observable, you need to measure it.

Basically, for the conclusion to be correct, the premises of the argument must also be correct. I am wondering if the assumption is an incorrect premise.

Basically, for the conclusion to be correct, the premises of the argument must also be correct. I am wondering if the assumption is an incorrect premise.

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I don't see why not. It can have these properties, but good luck measuring both at once.

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Well, one reason why not is because relative to the observer, he would observe a momentum of 0, and therefore he would know the exact momentum of the object. Uncertainty principle is error in position times error in momentum is greater than hbar/2. However if he knows exact momentum, the left hand side of the equation will be 0 violating the law

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If you had asked "Can an object be **measured** at rest relative to an observer and not violate the uncertainty principle?" so the answer is still yes.

Now, if you asked, "Would an electron ever be**measured** at rest relative to an observer?" then the answer is no. You'd get an electron cloud if you measured an electron that's part of an atom. If you confined an electron with a very focused laser beam, you could get it to stay within a very tiny area. With good enough equipment, you could still notice it move when it's position is measured optically.

Now, if you asked, "Would an electron ever be

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That is good news. But i still don't see how it is possible looking at the mathematics.

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chrisphd said:That is good news. But i still don't see how it is possible looking at the mathematics.

Holy mother of..., you've never seen anything stay still before??

All sorts of objects stay still relative to an observer when measured.

I'm looking around my apartment, and most of the objects I see appear still.

Sure, there is large error in my measuring tool (bad eyesight).

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The bottom line is, as Fredrik said, we're not talking about a

We're talking about a wave function.

I think that's what was initially misleading.

Say we capture an electron with an electric field.

We measure the position of the electron a bunch of times.

The probability to find the electron at any location is the square of the wave function.

The wave function has a maximum, and that maximum may be stationary.

This is not to say that the electron is stationary.

Hope this is helpful.

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That question is not unique.

If the object is small enough so that uncertainty is significant, then uncertainty effects the results of experiment.

If the object is large, then you can measure it to within the accuracy of your instruments.

You could measure an electron in the same exact place twice, but it would be a coincidence. The next time it may be somewhere else, because the photon you used to measure it last time sent it off on on a new trajectory.

Furthermore, the more accurate you want your measurements to be, the shorter wavelength of light you need to use. Shorter wavelength means higher energy, means higher momentum delivered to the poor electron being so cruelly observed.

If the object is small enough so that uncertainty is significant, then uncertainty effects the results of experiment.

If the object is large, then you can measure it to within the accuracy of your instruments.

You could measure an electron in the same exact place twice, but it would be a coincidence. The next time it may be somewhere else, because the photon you used to measure it last time sent it off on on a new trajectory.

Furthermore, the more accurate you want your measurements to be, the shorter wavelength of light you need to use. Shorter wavelength means higher energy, means higher momentum delivered to the poor electron being so cruelly observed.

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It isn't possible to measure something to be stationary, period. This would make the wave function zero everywhere (and turn the uncertainty principle into [itex]\infty\cdot 0\geq\frac\hbar 2[/itex])chrisphd said:

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Why would the wavefuntion be 0?

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What about my bookshelf? It's stationary, in a classical sense.Fredrik said:It isn't possible to measure something to be stationary, period...

I think what chrisphd is struggling with is that our classical concept of stationary is as unreal in quantum mechanics as absolute space is in general relativity.

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chrisphd said:Why would the wavefuntion be 0?

Actually the wave function would be zero everywhere but one point, where it would be 1.

You don't see particles behave like this in nature, though.

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The wave function is a mathematical function which, when squared, produces the probability distribution for finding a particle over all space.

Integrating the probability distribution from negative infinity to infinity must result in one, meaning the probability of finding the particle somewhere in all of space is one.

Therefore, if the particle is "still", it is described by a wave function that has the value one at some point, and zero elsewhere.

Like I said, this does not describe any particles of nature.

Integrating the probability distribution from negative infinity to infinity must result in one, meaning the probability of finding the particle somewhere in all of space is one.

Therefore, if the particle is "still", it is described by a wave function that has the value one at some point, and zero elsewhere.

Like I said, this does not describe any particles of nature.

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The wave function (at a fixed time) can be written aschrisphd said:Why would the wavefuntion be 0?

[tex]\psi(x)=K\int_{-\infty}^\infty g(p)e^{-ipx/\hbar}dp=[/tex]

where K is a normalization factor. (This is the most general solution of the time-independent Shrödinger equation). To measure the momentum to be between a and b is to change the state so that g is 0 everywhere except in the interval (a,b). To measure the momentum to be exactly 0 is to change g so that it's 0 everywhere except at 0. So the integral is definitely 0 unless g is infinite at 0. g

[tex]\psi(x)=K\int_{-\infty}^\infty\delta(p)e^{-ipx/\hbar}dp=K[/tex]

Now try to normalize this:

[tex]1=\int_{-\infty}^\infty|\psi(x)|^2dx=K^2\int_{-\infty}^\infty dx=K^2\cdot\infty \implies K=0[/itex]

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The wave function would be 0 everywhere, but its Fourier transform (i.e. the momentum-space wave function) would be 0 everywhere except at one point where it would begendou2 said:Actually the wave function would be zero everywhere but one point, where it would be 1.

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Yeah, I know. We've all been there.gendou2 said:What about my bookshelf? It's stationary, in a classical sense.

I think what chrisphd is struggling with is that our classical concept of stationary is as unreal in quantum mechanics as absolute space is in general relativity.

In quantum mechanics, the answer to the question "where is the particle

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Fredrik said:The wave function would be 0 everywhere, but its Fourier transform (i.e. the momentum-space wave function) would be 0 everywhere except at one point where it would beinfinite).

Is that so?

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Fredrik said:Yeah, I know. We've all been there.

In quantum mechanics, the answer to the question "where is the particlenow?" isn't a number (or 3 numbers), it's a complex-valued function. That takes a while to get used to.

A detector can measure the position of a particle as a number, right?

Q: Where is a particle going to be? A: Function.

Q: Where is that particle I measured? A: Number.

So...

Q: Is it possible to measure something to be stationary and not violate the uncertainty principle?

A: Sure. Detector reads position X at time t

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Fredrik said:

Fair enough.

I concede that there is no meaning to absolute position or absolute rest in quantum mechanics.

I maintain that in my electron experiment, I have a working definition of measured position and measured rest.

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Ok, but theoretically, what is the limit to how small we can make dx?

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There is no lower value for dx. Contrary to what Fredrik said it is quite possible for measurements to be exact. Being exact has nothing to do with the Uncertainty principle. In the uncertainty principle the value [itex]\Delta x[/itex] is not an imprecision in the measurement in position, it is thechrisphd said:Ok, but theoretically, what is the limit to how small we can make dx?

[itex]\Delta x[/itex] represents the indetermancy of the

Pete

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Not when we're measuring position, which is what we were talking about. But yeah, a measurement of e.g. the spin component in the z direction of an electron can of course be exact.pmb_phy said:Contrary to what Fredrik said it is quite possible for measurements to be exact.

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What in the world would give you that idea?Fredrik said:Not when we're measuring position, which is what we were talking about. But yeah, a measurement of e.g. the spin component in the z direction of an electron can of course be exact.

When measuring any physical observable one can theoretically obtain exact values. There is nothing in quantum mechanics which would indicate otherwise. Physical observables are the eigenvalues of operators.

I think you're confusing imprecision in single measurements with the uncertainty in physical observables. They are

For this reason dx could equal zero while the uncertainty in x, labeled [itex]\Delta x[/itex] would be non-zero. One could have a state for which [itex]\Delta x[/itex] is zero or as small as you'd like but then [itex]\Delta p[/itex] would be a infinite or very large according to the uncertainty principle. However both x and p themselves are eigenvalues and each can be measured exactly.

Let me ask you this - Do you know what it means to have precise measurements of x and p but have a finite uncertainty in x and p?

Pete

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It's absurd that you assume that I'm confusing the thing I called dx with the [itex]\Delta x[/itex] in the uncertainty relation.

This should be good... prove it.pmb_phy said:dx could equal zero while the uncertainty in x, labeled [itex]\Delta x[/itex] would be non-zero.

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