# Basic Heisenberg Uncertainty Principle

1. Mar 22, 2007

### the keck

1. The problem statement, all variables and given/known data

According the the Uncertainty Principle, an electron of mass m when placed on a 'flat' tabletop will actually 'float' above the table i.e. the vertical position of the electron will be spread out over some distance.

Use the uncertainty principle to find an expression for the equilibrium average height that the electron floats above the table's surface in terms of m,g etc...

2. Relevant equations

delta(p) x delta (y) >= h/(4*Pi)

3. The attempt at a solution

I used the fact that F=mg=dp/dt and hence delta(p)= mg x delta(t)

Hence delta (y) >= h/(4*Pi*mg*delta(t))

Thus in determining the average height, the delta(t) will cancel, leaving us with :

Average y = h/(4*Pi*mg)

This however seems to be too easy a solution even though unit analysis verifies that it gives us distance.

N.B. I have used both * and x to refer to multiplication

Regards,
The Keck

2. Mar 22, 2007

### dextercioby

Your analysis looks okay.

3. Mar 22, 2007

### nrqed

How did you get that?
This does not have the dimensions of a distance.

The steps before that last equation were ok, I don't see how you got this last expression.

PS: welcome to the forums!

4. Mar 22, 2007

### Dick

Yeah, I don't follow the logic at all. I'm used to problems like this being treated as energy minimization problems. Ie minimize kinetic plus potential energy:

p^2/(2m)+mgy

subject to the constraint py>=h/(4*pi). This doesn't seem to be the same thing...

5. Mar 23, 2007

### the keck

Yeah...Sorry about that guys, I had a look at the dimensional analysis again, and realised that it doesn't match.

So how do I actually do it? I'm not exactly sure how one determines the average equilibrium average position.

Thanks

Regards.
The Keck

6. Mar 23, 2007

### Dick

I already pointed this out, but minimize p^2/(2*m)+mgy subject to the constraint of the uncertainty principle (where p and y are 'average equilibrium' values and ought to be roughly the same as $$\Delta x$$ and $$\Delta p$$). Don't worry about being terribly exact about defining things like 'average equilibrium' - this is an estimate, not an exact calculation.

Last edited: Mar 23, 2007
7. Mar 28, 2007

### 12thee

who the hell are u guys?

r u guys university students or school students ?
im just worried . . im preparing for one of the toughest exams in the world and couldnt solve that one . . . https://www.physicsforums.com/images/smilies/surprised.gif [Broken]
:surprised

Last edited by a moderator: May 2, 2017
8. Mar 28, 2007

### Dick

I suspect everybody who posts here may give you a different answer.

Last edited by a moderator: May 2, 2017
9. Mar 28, 2007

### the keck

Indeed, there are a variety of ways to obtain the answer. The answer one gets will vary by 0.1-0.5 mm, but as long as the unit analysis is correct, then I doubt they can really mark you down...if you provide valid reasoning.

Dick's method is the one I used, and is quite valid. Thanks a lot!!!

Regards,
The Keck

10. Mar 29, 2007

### theshyamal

From the uncertainty principle you can write
(delta y)*(delta p)~hbar/2
You can also write (delta p)~squire root{2m(delta E)}
You can also put (delta E)~mg(delta y)
From these relations you can have

(delta y)~((hbar)^{2/3})/(2m^{2/3} g^{1/3})

Shyamal Biswas

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