# Uncertainity Principle

1. Apr 11, 2010

### 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!

2. Apr 11, 2010

### 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.

3. Apr 11, 2010

### 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.

4. Apr 11, 2010

### Fredrik

Staff Emeritus
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.

5. Apr 11, 2010

### Deepak Kapur

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.

6. Apr 11, 2010

### Deepak Kapur

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).

7. Apr 11, 2010

### Deepak Kapur

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.

Last edited: Apr 11, 2010
8. Apr 11, 2010

### Rajini

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. wont 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.

9. Apr 11, 2010

### 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 wont be accurate..this is the principle.
see the Werner relation:
$$\Delta x\; \Delta p\geq\hbar$$

10. Apr 11, 2010

### Fredrik

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

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

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.

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

11. Apr 12, 2010

### Lojzek

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.

12. Apr 12, 2010

### Fredrik

Staff Emeritus
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.

13. Apr 13, 2010

### Sillyboy

Really, I think you explaination 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!

14. Apr 13, 2010

### ZapperZ

Staff Emeritus
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 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.

15. Apr 13, 2010

### thoughtgaze

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.

16. Apr 14, 2010

### Fredrik

Staff Emeritus
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.

17. Apr 14, 2010

### thoughtgaze

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?

18. Apr 16, 2010

### yoda jedi

Very funny !!!! ...laughs.....

like a quantum fluctuation...........

19. Apr 16, 2010

### eaglelake

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

20. Apr 19, 2010

### Fakorede O.

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.

21. Apr 19, 2010

### ZapperZ

Staff Emeritus
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.

22. Apr 19, 2010

### Fredrik

Staff Emeritus
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).

Last edited: Apr 19, 2010
23. Apr 19, 2010

### thoughtgaze

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

24. Apr 21, 2010

### yoda jedi

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