Uncertainty Principle logic help

[chrisphd]

Does the uncertainty principle contradict the idea that an object can be at rest relative to an observer? For example, in physics class we examined an electron in a box, and we assumed the walls of the box are stationary relative to the observer. But the uncertainty principle says delta p times delta x > hbar/2. So the walls of the box would have a definite position and momentum relative to the observer, violating the principle. Where is my error in logic.

 

[gendou2]

The atoms in the walls of the box behave according to quantum mechanics.
They will have uncertain positions, too.
However, I would expect thermal energy to introduce more error into this experiment than uncertainty.

 

[chrisphd]

Your missing the point here. I know that the particles in the box exhibit uncertainty. But for the thought experiment, it was assumed that the walls of the box where stationary. However, i am asking, doesn't assuming another object is stationary to an observer contradict the uncertainty principle, as we are assuming we are able to know the exact position and momentum of that object.

 

[gendou2]

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.

 

[chrisphd]

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.

 

[Fredrik]

When you calculate the energy eigenstates and the corresponding eigenvalues (i.e. the energy levels) of a "particle in a box" you're just solving the Schrödinger equation for a potential with a certain shape (zero in a certain interval and infinity everywhere else). The uncertainty principle says something about the shape of the wave function, not about the shape of the potential, so there's no disagreement there.

 

[chrisphd]

My question in simplest terms is, can an object be at rest relative to an observer and not violate the uncertainty principle?

 

[gendou2]

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

 

[chrisphd]

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

 

[gendou2]

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.

 

[chrisphd]

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

 

[gendou2]

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).
The uncertainty in the position of my bookshelf is well under the threshold of detection.

 

[chrisphd]

LOL. One could argue that the things are you are looking at only appear to be stationary. They are still moving at very low speeds, but you have failed to measure their speed to high enough degree of accuracy. However, I'm not saying that its not possible to measure the speed of an object to be 0 relative to an observer. I'm just wondering how it doesn't violate the uncertainty principle.

 

[gendou2]

I'm sorry for not explaining myself well.

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

 

[chrisphd]

Ok. But the question still remains, is it possible to measure something to be stationary and not violate the uncertainty principle?

 

[gendou2]

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.

 

[Fredrik]

Ok. But the question still remains, is it possible to measure something to be stationary and not violate the uncertainty principle?

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 \infty\cdot 0\geq\frac\hbar 2)

 

[chrisphd]

Why would the wavefuntion be 0?

 

[gendou2]

It isn't possible to measure something to be stationary, period...

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.

 

[gendou2]

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.

 

[chrisphd]

So the wavefunction is more like a kronika delta (however you spell it) function, which would actually be a point rather than a wave?

 

[gendou2]

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.

 

[Fredrik]

Why would the wavefuntion be 0?

The wave function (at a fixed time) can be written as

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

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 will actually be infinite at 0 in this limit, but it's not enough. g becomes a Dirac delta function, so

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

Now try to normalize this:

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

 

[Fredrik]

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

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 be infinite).

 

[Fredrik]

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.

Yeah, I know. We've all been there.

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

 

[gendou2]

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 be infinite).

Is that so?

 

[gendou2]

Yeah, I know. We've all been there.

In quantum mechanics, the answer to the question "where is the particle now?" 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 t₁ and t₂. It's possible. Not likely with an accurate detector and a small particle like an electron.

 

[Fredrik]

A measurement is never exact. If the precision of the measurement is dx, then you have only measured the position to be in the interval (x-dx,x+dx). The result of the measurement is a number, x. The position of the particle is (still) represented by a wave function. The only thing that has changed is that the new wave function is zero everywhere outside that interval.

 

[gendou2]

A measurement is never exact. If the precision of the measurement is dx, then you have only measured the position to be in the interval (x-dx,x+dx). The result of the measurement is a number, x. The position of the particle is (still) represented by a wave function. The only thing that has changed is that the new wave function is zero everywhere outside that interval.

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.

 

[chrisphd]

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

 

[pmb_phy]

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

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 \Delta x is not an imprecision in the measurement in position, it is the uncertainty in position. \Delta x, also known as the stadard deviation, is a function only of the wavefunction and has nothing to do with a single measurement of a physical observable. Given the same initial state then subsequent measurements of, say, position will yield different values, each of which can be measured with arbitrary precision.

\Delta x represents the indetermancy of the state itself. It can be calculated without knowing anything about how it is measured. A large value of \Delta x means that there will be a wide spread in the values actually measured. While \Delta x is completely fixed and determined by the state itself the imprecision of the measurement itself depends only on the apparatus doing the measurement.

Pete

 

[Fredrik]

Contrary to what Fredrik said it is quite possible for measurements to be exact.

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.

 

[gendou2]

Not sure but I would guess the Planck length.

http://en.wikipedia.org/wiki/Planck_length

 

[pmb_phy]

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.

What in the world would give you that idea?

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 very different things. The quantity you described above and labeled dx is not an uncertainty and as such is not required to obey the uncertainty principle.

For this reason dx could equal zero while the uncertainty in x, labeled \Delta x would be non-zero. One could have a state for which \Delta x is zero or as small as you'd like but then \Delta p 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

 

[Fredrik]

This is at least the third time in a short time that you have claimed that something trivial and obviously correct in one of my posts is wrong, and then started talking to me as if you think I just started to learn this stuff yesterday. For some reason you don't seem to do a lot of thinking before you reply to one of my posts.

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

dx could equal zero while the uncertainty in x, labeled \Delta x would be non-zero.

This should be good... prove it.

 

[pmb_phy]

This is at least the third time in a short time that you have claimed that something trivial and obviously correct in one of my posts is wrong, and then started talking to me as if you think I just started to learn this stuff yesterday.

Its unfortunate that you feel the need to read things into my posts which are not there. However in this case you do demonstrate a lack of understanding of the uncertainty principle. This is very common even for people who have studied quantum mechanics, even at the college level.

For some reason you don't seem to do a lot of thinking before you reply to one of my posts.

I see no reason to start making accusations such as this. In case you're unaware of it this kind of response is insulting. I know very well what I'm saying. I can only infer what you mean by what you say. This material was already discussed recently in another thread of the same title. Its for these reasons I've asked you certain questions, i.e. so that from your responses I can determine exactly what you are are trying to say rather than trying to deduce it from what you posts. However if you find that you're unable to respond without this poor attitude then I will not respond to your posts and will place you on my igore list. I don't have the time nor the patience for this kind of nonsense. Your post will be reported to the powers that be in this case. The moderators do not allow posts which are problematic (harassment, fighting, or rude) and unappropriated.

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

Then please explain the difference between them as you understand them. Why do you think I asked you where you got that idea?? It sure can't be found in any text in quantum mechanics. It does not appear as if you understand the difference from what you've written in your posts, hence my question.

This should be good... prove it.

Your posts have a very combative attitude. This forum is moderated so that this kind of thing doesn't happen.

Answer - Simple. First off there is no reason to assume that one can't measure something like position exactly. So in what follows it will be assumed that x is exact. As an example consider an electron moving in the x-direction and for which the wave function initially has a Gaussian shape. Nowe measure the position. Repeat this process and for each value of position record how many times the electron was measured to be at this location. Repeat this process. In theory this will establish a probability density. In practice one cannot carry out infinite amounts of measurements. Note that in each case you start with the same state, i.e. the one described by the Gaussian function. After you have this data then you can determine the "spread" of the values of x. The standard deviation (sd) is the measure of this "spread" of values. Now calculate the standard deviation. The value of the sd is precisely identical to \Delta x. I.e. they are two different words which describe the exact same thing.

Were you aware that the \Delta x is identical to the standard deviation? Were you unaware that the value of \Delta x is determined from the wave function and has nothing to do with how position is measured?

One last question so that I know we are talking about the same thing - Please post the exact mathematical definition of \Delta x and why you believe that it has anything to do with how precise a measurement of something like position is made. Thanks.

Pete

 

[Fredrik]

Its unfortunate that you feel the need to read things into my posts which are not there.

I'm not sure that I have, but you have read a lot of things into my posts that certainly aren't there. Your replies are often very far off. It's like you're quoting my posts but answering someone else's. I don't usually have a "combative attitude", but your posts have been increasingly annoying. You keep claiming incorrectly that the most obviously correct things I say are actually incorrect, and you keep asking me questions that are far below both of our levels of understanding of the subject (very far) for no good reason. I don't know if you're trying to be condescending, but you are.

I don't think you have a poor understanding of these subjects. I think the problem is that you have your own way of interpreting certain statements, and that you don't realize that no one else interprets them that way. For example in the other thread with the same title, you told me and other people that we were wrong to say that you can't measure two things at the same time. Then it turned out that you didn't actually disagree with what I said. You just had a unique definition of what it would mean to measure two things at the same time: First you talked about how you could do the measurements on two members of an ensemble of identical systems in the same state. Then you talked about how, if there's only one system instead of an ensemble, you have the option to measure either the first observable or the second. No one else would define "measure two things at the same time" that way.

I'm mentioning that because it's a very good example of how you twisted a trivial and completely correct statement around and claimed that it was false.

However in this case you do demonstrate a lack of understanding of the uncertainty principle.

I did not, but it's fascinating that you still think I did. This discussion is not the result of anyone's poor understanding of physics. The problem is that you keep misunderstanding things you read (in my posts) and never go back to check if that's what you did.

Answer - Simple. First off there is no reason to assume that one can't measure something like position exactly. So in what follows it will be assumed that x is exact. As an example consider an electron moving in the x-direction and for which the wave function initially has a Gaussian shape. Nowe measure the position. Repeat this process and for each value of position record how many times the electron was measured to be at this location. Repeat this process. In theory this will establish a probability density. In practice one cannot carry out infinite amounts of measurements. Note that in each case you start with the same state, i.e. the one described by the Gaussian function. After you have this data then you can determine the "spread" of the values of x. The standard deviation (sd) is the measure of this "spread" of values. Now calculate the standard deviation. The value of the sd is precisely identical to \Delta x. I.e. they are two different words which describe the exact same thing.

You claimed that the \Delta x=\sqrt{\langle(x-\langle x\rangle)^2\rangle}=\sqrt{\langle x^2\rangle-\langle x\rangle^2} of a particle in a state described by a wave function that's zero everywhere except at a single point is non-zero. I asked you to prove that, and this is your answer?

(Yes, that is what you said, but you probably don't even realize it. Go back and read what you wrote and the text you replied to).

Seriously, you're talking about determining the probability density associated with a particular quantum state by making a measurement on each member of an infinitely large ensemble of identical systems in the same state. There is absolutely nothing about what you said that proves your claim, or disproves mine, which is (still) that no single measurement on one particle (not an ensemble) will produce a wave function that's zero everywhere except a single point!

I don't have the time nor the patience for this kind of nonsense.

I don't either.

 

[Fredrik]

Please post the exact mathematical definition of \Delta x and why you believe that it has anything to do with how precise a measurement of something like position is made.

I didn't even notice this before. You are falsely claiming that I believe something ridiculous. Something that only a person who doesn't know what the uncertainty principle says could believe. I have definitely not given you any reason to believe that I believe anything like that.

Why did you ask me this? Do you actually believe that I believe this, or are you just trying to insult me in a way that you think isn't going to be obvious to the moderators?

 

[ZapperZ]

This thread is going nowhere, and it's time we all go find a lollipop.

Zz.