Understanding the Uncertainty Principle and Its Effects on Atomic Particles

[Deepak Kapur]

According to uncertainty principle, it's not possible to measure the position and momentum of an atomic particle ( say electron) simultaneously.

Now, suppose a scientist grows so small in size that an electron is the size of a big ball (or planet) for him.

Is uncertainity principle applicable for the scientist also. Won't he be able to measure both these quantities simultaneously?

Does it mean that it's only the size difference that we construe as quantum phenomenon.

Just being curious!

 

[Rajini]

Hi,
I guess uncertainty would more in that case! and may be everything would be exaggerated.
Because the scientist will also in the same scale of atomic dimension.

 

[hisham.i]

To measure the position or the velocity in anything in life you have to send a signal to the measured body(light, sound...). Any signal you send is an energy and when the signal hits the body there will be an exchange of energy between the signal and the body..
In particle scale when you send an energy to the particle to know the position for example, this energy will change the energy the particle is moving with so it changes its speed, this what uncertainty means...
So in both cases where the scientist is big or small the principle is the energy and not the size. So i think that the scientist will not be able to measure both quantities.

 

[Fredrik]

According to uncertainty principle, it's not possible to measure the position and momentum of an atomic particle ( say electron) simultaneously.

According to quantum mechanics, it's worse than that. The particle doesn't have a well-defined position or a well-defined momentum at any time.

Now, suppose a scientist grows so small in size that an electron is the size of a big ball (or planet) for him.

I don't think the question makes sense. You're attributing classical properties like "size" to a quantum particle, and you're assuming that this "scientist" would also have classical properties even though we would expect QM to describe a physical system of that size.

Is uncertainity principle applicable for the scientist also. Won't he be able to measure both these quantities simultaneously?

The uncertainty theorem isn't a statement about practical difficulties associated with performing measurements. It's a statement about how the results will vary if we prepare a very large number of identical particles in the same state, and perform a position measurement on half of them and a position measurement on the other half.

Does it mean that it's only the size difference that we construe as quantum phenomenon.

It's not about size. It's about how strongly the system interacts with its environment. A particle in a vacuum doesn't interact significantly with its environment, but I'm interacting significantly with the air and the light around me. Even if you'd put me in a vaccum, the gravitational interaction between me and the rest of the universe is much stronger than the gravitational interaction between an electron and its environment.

 

[Deepak Kapur]

According to quantum mechanics, it's worse than that. The particle doesn't have a well-defined position or a well-defined momentum at any time.


I don't think the question makes sense. You're attributing classical properties like "size" to a quantum particle, and you're assuming that this "scientist" would also have classical properties even though we would expect QM to describe a physical system of that size.


The uncertainty theorem isn't a statement about practical difficulties associated with performing measurements. It's a statement about how the results will vary if we prepare a very large number of identical particles in the same state, and perform a position measurement on half of them and a position measurement on the other half.


It's not about size. It's about how strongly the system interacts with its environment. A particle in a vacuum doesn't interact significantly with its environment, but I'm interacting significantly with the air and the light around me. Even if you'd put me in a vaccum, the gravitational interaction between me and the rest of the universe is much stronger than the gravitational interaction between an electron and its environment.

You pop up everywhere!

I didn't mean to attribute classical properties to this hypothetical scientist.

1.To what extent is size a deciding factor in the situation that I have mentioned.

2. Suppose the scientist also is a matter-wave (very-2 small) as compared to an electron (an extremely big matter-wave in this case). Won't it be relatively easy for this small entity to determine the big entity.

 

[Deepak Kapur]

Hi,
I guess uncertainty would more in that case! and may be everything would be exaggerated.
Because the scientist will also in the same scale of atomic dimension.

Hello,

Plz elaborate.

Even in the quantum world, I think, there is a difference in small (extremely small )and big (extrememy large) entities (whatever these might be).

 

[Deepak Kapur]

Hi,
I guess uncertainty would more in that case! and may be everything would be exaggerated.
Because the scientist will also in the same scale of atomic dimension.

According to quantum mechanics, it's worse than that. The particle doesn't have a well-defined position or a well-defined momentum at any time.

I don't think the question makes sense. You're attributing classical properties like "size" to a quantum particle, and you're assuming that this "scientist" would also have classical properties even though we would expect QM to describe a physical system of that size.


The uncertainty theorem isn't a statement about practical difficulties associated with performing measurements. It's a statement about how the results will vary if we prepare a very large number of identical particles in the same state, and perform a position measurement on half of them and a position measurement on the other half.


It's not about size. It's about how strongly the system interacts with its environment. A particle in a vacuum doesn't interact significantly with its environment, but I'm interacting significantly with the air and the light around me. Even if you'd put me in a vaccum, the gravitational interaction between me and the rest of the universe is much stronger than the gravitational interaction between an electron and its environment.

If we keep on bombarding an electron with a wave of extremely small wavelength, we can have a fair idea of it's position ( if it's our choice) for all the time to come (theoretically).

So, I think it's only one thing that is uncertain not both.

Moreover, the presence of an observer that has only wavelengts at his disposal is a big hinderance in this case.

It may be possible (to be frank I doubt what I am stating) that by some psychic means (telepathy etc.) we are able to judge postion etc. without bombarding things with waves, bullets etc.

You may consider it a joke also. Some fun is necessary even in a Physics Forum.

But replies are warranted on jokes also.

 

[Rajini]

According to uncertainty principle, it's not possible to measure the position and momentum of an atomic particle ( say electron) simultaneously.

Now, suppose a scientist grows so small in size that an electron is the size of a big ball (or planet) for him.

Is uncertainity principle applicable for the scientist also. Won't he be able to measure both these quantities simultaneously?

Does it mean that it's only the size difference that we construe as quantum phenomenon.

Just being curious!

Werners principle applies to particles..So if you thing scientist as a particle..then it will be applied to that scientist also..
What ever be the situation..you cannot measure the position and the product of mass and velocity at the same time..if position is accurate then the product of mass and vel. won't be accurate.
Remember: uncertainty is not due to difficulties in measurement or due to experimental limits.. Uncertainty is something like intrinsic behavior/property of the quantum system.

 

[Rajini]

Yes exactly:
You cannot measure position and product of vel. and mass at the same time.
That also means if your position is accurate then the product of vel. and mass won't be accurate..this is the principle.
see the Werner relation:
$$\Delta x\; \Delta p\geq\hbar$$

 

[Fredrik]

You pop up everywhere!

I tend to do that. Especially in the relativity and quantum physics forums.

I didn't mean to attribute classical properties to this hypothetical scientist.

The thing is, if it doesn't have a lot of classical properties, it's not a scientist.

1.To what extent is size a deciding factor in the situation that I have mentioned.

The bigger the object, the more difficult it is to isolate it from its environment. It's the degree of interaction with the environment that matters.

2. Suppose the scientist also is a matter-wave (very-2 small) as compared to an electron (an extremely big matter-wave in this case). Won't it be relatively easy for this small entity to determine the big entity.

I suppose, but that would spread out the electron's momentum over a larger range. The uncertainty relation is a theorem in QM, so there's no way around that.

This post might be useful: https://www.physicsforums.com/showthread.php?p=2340718

 

[Lojzek]

According to quantum mechanics, it's worse than that. The particle doesn't have a well-defined position or a well-defined momentum at any time.

That is not completely true. In theory a particle can have a well defined position (if wave function is delta) or momentum (straight wave), just not both at the same time.

 

[Fredrik]

That is not completely true. In theory a particle can have a well defined position (if wave function is delta) or momentum (straight wave), just not both at the same time.

There are some problems with that claim, both mathematical and physical. States are usually defined as the unit rays of a Hilbert space, but a plane wave isn't a member of the Hilbert space and therefore doesn't define a ray, and delta isn't even a function. Also, we shouldn't think of the whole Hilbert space as defining "physical" states. We should at least require that $$\hat p^n\psi$$ is square integrable for all positive integers n, and probably that $$\hat x^n\psi$$ is square integrable too. The justification for the first condition is that if you describe the system as being in the state $$\psi$$, an observer a distance L to your left would describe the system as being in the state

$$x\mapsto\psi(x-L)=\sum_{n=0}^\infty\frac{(-L)^n}{n!}\frac{d^n}{dx^n}\psi(x)=\sum_{n=0}^\infty\frac{(-i\hat pL)^n}{n!}\psi(x)=\exp(-i\hat pL)\psi(x)$$

So the condition ensures that translation operators exist.

The physical difficulties are of course that even though a position measurement "squeezes" the wavefunction into a sharply peaked shape, there's no physical interaction that can make that peak infinitely sharp, because potentials in nature are smooth except at the location of the point particle that causes it (and of course, the wavefunction of that particle isn't going to be infinitely sharp either). Consider e.g. the Coulomb potential, which is proportional to 1/r.

 

[Sillyboy]

To measure the position or the velocity in anything in life you have to send a signal to the measured body(light, sound...). Any signal you send is an energy and when the signal hits the body there will be an exchange of energy between the signal and the body..
In particle scale when you send an energy to the particle to know the position for example, this energy will change the energy the particle is moving with so it changes its speed, this what uncertainty means...
So in both cases where the scientist is big or small the principle is the energy and not the size. So i think that the scientist will not be able to measure both quantities.

Really, I think you explanation is weird. At least I cannot accept it. I think HUP is a principle which has nothing to do with measruement. It tells us that the world is uncertain. Maybe my answer is wrong. If so, please point them out!

 

[ZapperZ]

Really, I think you explanation is weird. At least I cannot accept it. I think HUP is a principle which has nothing to do with measruement. It tells us that the world is uncertain. Maybe my answer is wrong. If so, please point them out!

Your answer is correct. It has nothing to do with the instrumentation. It also has nothing to do with a single measurement. I can measure, to arbitrary accuracy of my technology at hand, the position and then momentum of a particle. Each of the measurement uncertainty associated with those two measurements are NOT the HUP. They can be independently improved with no direct effect on each other.

I've described the misunderstanding of the HUP several times on here, but it appears that it should be reiterated again since there's a severe misunderstanding in the OP.

The most direct way to illustrate this is using a single-slit measurement. Let's say you have a source of classical particle that emits this particle one at a time on demand, and emits it with a constant velocity and kinetic energy. At some distance from this source is a single slit. For clarity sake let's say the slit is alligned along the x-direction, so that the width of the slit is along the y-direction. The orientation of the x and y coordinate axes is in such a way that (using the right-handed coordinate system) the z-axis is along the direction of propagation of the particles. So the direction of the z-axis is from the source to the slit, and beyond.

Beyond the slit is a screen of detectors (could be a photographic plate, a CCD, etc.) This detector records where the particle hits after it passes through the slit. Let's say this screen is a distance L after the slit.

Now, let's get some basics out of the way:

1.If a particle gets through the slit, then I can say that my knowledge of the position of the particle at the slit has an uncertainty equal to the width of the slit. Thus, the width of the slit $\Delta y$ is the uncertainty in the position of the particle when it passes through the slit.

2. The y-component of the momentum of the particle can be found by looking at how far the particle drifts along the y-direction when it hits the screen. This makes the explicit assumption that no external forces acts on the particle at and after it passes through the slit, so that it's momentum remains constant from the slit to the screen (which is a reasonable assumption). Let's say the particle drifts from the center, straight-through line and hits the screen at a distance Y. If it takes the particle a time T to reach the screen (which we can assume to be a constant if screen distance from the slit is much larger than the width of the slit (i.e. L >> $\Delta y$), then the y-component of the momentum is $p_y \propto Y/T$. Now, there is a measurement uncertainty here in determining where exactly the particle hits the detector. This measurement uncertainty depends on the resolution of the detector, how fine is the "mesh", etc. But this is NOT the "uncertainty" that is meant in the HUP. We haven't gotten to the uncertainty of the momentum YET. All we have is a measurement of the y-component of the momentum of the particle.

Now, let's do the experiment with the classical particles. You shoot the particles one at a time and record where it hits on the screen. Ideally, what you will end up on the screen is only one spot where the particles that pass through the slit hit. However, closer to reality is that you end up with a gaussian distribution at the slit, where the peak lies directly along the straight-through direction that has zero y-component of momentum. The uncertainty in the momentum then corresponds to the width of the gaussian distribution (full width at half maximum). Now THIS is $\Delta p_x$ as referred to in the HUP!

Let's make the width of the slit smaller. This means $\Delta y$ is smaller. You are now letting a smaller possible angle of incidence of the particle from the source to get through the slit. This means that there will be a smaller spread that is detected on the screen. The gaussian distribution will be thinner. So classically, what we expect is that as $\Delta y$ gets smaller, $\Delta p_y$ also correspondingly becomes smaller.

This is what we expect in classical mechanics. If all the initial conditions remans identical (I have the same source), then the more I know where the object is at any given instant, the more I can predict its subsequent properties. I can say what its y-momentum will be with increasing accuracy as I increase my certainty of its position by decreasing the width of the slit. I can easily predict where the next particle is going to hit since I will know what its momentum is going to be after it passes through a very narrow slit. My ability to predict such things increases with decreasing slit width.

Fine, but what happens with a quantum particle such as a photon, electron, neutron, etc.?

We need to consider two different cases. If the slit width is considerably larger than the deBroglie wavelength (or in the case of a photon, its wavelength) of the particle, then what you have is simply the image of the slit itself. The ideal situation would give you simply a "square" or gaussian distribution at the screen of the intensity of particles hitting the detector. This is no different than the classical case.

It gets interesting as you decrease the slit. By the time the width of the slit is comparable to the deBroglie wavelength, something strange happens. On the screen, the spread of the particles being detected start expanding! In fact, the smaller you make the slit width, the larger the range of values for Y that you detect. The "gaussian spread" now is becoming fatter and fatter. This is the single-slit diffraction pattern that everyone is familiar with.

Now THIS is the uncertainty principle at work. The slit width, and thus $\Delta y$ is getting smaller. This implies that $\Delta p_y$ is getting larger. Take note that the measurement uncertainty in a single is still the same as in the classical case. If I shoot the particle one at a time, I still see a distinct, accurate "dot" on the screen to tell me that this is where the particle hits the detector. However, unlike the classical case, my ability to predict where the NEXT one is going to hit becomes worse as I make the slit smaller. As the slit and $\Delta y$ becomes smaller and smaller, I know less and less where the particle is going to hit the screen. Thus, my knowledge of its y-component of the momentum correspondingly becomes more uncertain.

The moral of the story is that the HUP applies either (i) on a statistical measurement of many positions and momentum, or (ii) on our ability to make the prediction of the NEXT outcome of a measurement. It is not the instrumentation accuracy, nor in the lack of ability to measure both position and momentum with equal or arbitrary accuracy.

Zz.

 

[thoughtgaze]

According to quantum mechanics, it's worse than that. The particle doesn't have a well-defined position or a well-defined momentum at any time.

Usually this is the case but, according to quantum mechanics, one could measure the momentum to arbitrary accuracy and the particle would be left in that state with a well-defined momentum.

 

[Fredrik]

Usually this is the case but, according to quantum mechanics, one could measure the momentum to arbitrary accuracy and the particle would be left in that state with a well-defined momentum.

This would make the state more well-defined than before, or less ill-defined if you prefer, but not completely well-defined. See #12 for comments about "states" with perfectly well-defined values of momentum or position.

 

[thoughtgaze]

This would make the state more well-defined than before, or less ill-defined if you prefer, but not completely well-defined. See #12 for comments about "states" with perfectly well-defined values of momentum or position.

Hey, I think I see what you mean. For physical reasons you want your wavefunction to be square integrable. And by requiring the wavefunction to be square integrable for an observer displaced a distance L, further requires that the nth derivative is square integrable as well, which is not the case for a perfectly well-defined momentum or position. However, I have 2 questions:

1) Why shouldn't we think of the whole Hilbert space as defining "physical" states? We've already emitted the momentum eigenstates and delta functions to be given membership into the Hilbert space? I guess I just don't see what you mean here.

2) Where did you get that expression for the translation in terms of an infinite sum?

 

[yoda jedi]

You pop up everywhere!

Very funny ! ...laughs...


like a quantum fluctuation...

 

[eaglelake]

Much of the discussion in the "position and momentum" thread is pertinent to this topic.

 

[Fakorede O.]

According to uncertainty principle, it's not possible to measure the position and momentum of an atomic particle ( say electron) simultaneously.

Now, suppose a scientist grows so small in size that an electron is the size of a big ball (or planet) for him.

Is uncertainity principle applicable for the scientist also. Won't he be able to measure both these quantities simultaneously?

Does it mean that it's only the size difference that we construe as quantum phenomenon.

Just being curious!

I really agree with this principle. Because it is very impossible to measure the momentum and position of a moving substance simultaneously e.g aeroplane.

 

[ZapperZ]

I really agree with this principle. Because it is very impossible to measure the momentum and position of a moving substance simultaneously e.g aeroplane.

Very impossible? Really?

I shoot photons at an incoming airplane. I get reflected signal that is Doppler shifted. That tells me how fast the plane is moving, and thus, tells me the momentum. I also know how long it took for the light signal to go out and come back. That tells me how far away the airplane is from me.

I've just obtained both position and momentum simultaneously, in one measurement. Why is this "very impossible"? How do you think air traffic controller could know where you are and how fast your plane is moving?

Zz.

 

[Fredrik]

Hey, I think I see what you mean. For physical reasons you want your wavefunction to be square integrable. And by requiring the wavefunction to be square integrable for an observer displaced a distance L, further requires that the nth derivative is square integrable as well, which is not the case for a perfectly well-defined momentum or position. However, I have 2 questions:

1) Why shouldn't we think of the whole Hilbert space as defining "physical" states? We've already emitted the momentum eigenstates and delta functions to be given membership into the Hilbert space? I guess I just don't see what you mean here.

2) Where did you get that expression for the translation in terms of an infinite sum?

Not sure what you mean by "emitted" (Admitted? Omitted?), or if you're saying that they're members of the Hilbert space or that they're not. (They're not).

Looks like you didn't quite get the argument I used. I'm saying that if there's a function $$\psi$$ that you can use as a mathematical representation of the state of a particle, then should also exist a function $$\psi_L$$ that a translated observer would use as a representation of the same state. The relationship between them should be $$\psi_L(x)=\psi(x-L)$$.

And we know that $$\psi(x-L)=\exp(-i\hat p L)\psi(x)$$. If $$\psi$$ is infinitely differentiable, it's just a consequence of Taylors formula (and the definitions of $$\hat p$$ and the exponential). It can also be thought of as a consequence of the very natural requirement that translation operators must satisfy T(a+b)=T(a)T(b), because Stone's theorem ensures that there exists a self-adjoint operator $$\hat p$$ such that $$T(L)=\exp(-i\hat p L)$$ for all L. These two facts imply that we have

$$\hat p=-i\frac{d}{dx}$$

if and only if $$\psi$$ is infinitely differentiable. (The "if" part is very easy. The "only if" part is more difficult).

 

[thoughtgaze]

Not sure what you mean by "emitted" (Admitted? Omitted?), or if you're saying that they're members of the Hilbert space or that they're not. (They're not).

Looks like you didn't quite get the argument I used. I'm saying that if there's a function $$\psi$$ that you can use as a mathematical representation of the state of a particle, then should also exist a function $$\psi_L$$ that a translated observer would use as a representation of the same state. The relationship between them should be $$\psi_L(x)=\psi(x-L)$$.

And we know that $$\psi(x-L)=\exp(-i\hat p L)\psi(x)$$. If $$\psi$$ is infinitely differentiable, it's just a consequence of Taylors formula (and the definitions of $$\hat p$$ and the exponential). It can also be thought of as a consequence of the very natural requirement that translation operators must satisfy T(a+b)=T(a)T(b), because Stone's theorem ensures that there exists a self-adjoint operator $$\hat p$$ such that $$T(L)=\exp(-i\hat p L)$$ for all L. These two facts imply that we have

$$\hat p=-i\frac{d}{dx}$$

if and only if $$\psi$$ is infinitely differentiable. (The "if" part is very easy. The "only if" part is more difficult).

I meant "omitted" and I'm pretty sure I understood the argument just fine, thanks anyway.

 

[yoda jedi]

According to uncertainty principle, it's not possible to measure the position and momentum of an atomic particle ( say electron) simultaneously.

Now, suppose a scientist grows so small in size that an electron is the size of a big ball (or planet) for him.

Is uncertainity principle applicable for the scientist also. Won't he be able to measure both these quantities simultaneously?

Does it mean that it's only the size difference that we construe as quantum phenomenon.

Just being curious!

you said it "MEASURE". but has them...