# Uncertainty principle and single slit experiment

1. Jan 11, 2012

### Gavroy

hi

my concern is, that there is a huge amount of equations that are somehow related to the uncertainty principle and it is not straightforwardly obvious to me, when i have to use which of them.

let me put it this way: as far as i see, there is this overall applicable equation that says
ΔxΔp ≥ h/(4π)

but for instance, my schoolbook proposes the relation ΔxΔp=h/2, when they talk about the uncertainty of a single slit diffraction pattern that is made with electrons, where they regard Δx as half the slit-width and Δp as the x-deflection of the electron in units of momentum, that it needs to reach the first minimum. it appears to me, that this approach is extremely arbitrary and now there are lots of excercises, where they e.g. say that by using light, Δx is half the coherence length of light and use the same relation.

and now i found an excercise, where they assume that the uncertainty of an electron in a hydrogen atom is about 10^(-10) m, and ask for the uncertainty in momentum.
the solution to this excercise is easily derived from the assumption that ΔxΔp≈h<---but where does this come from? how do i get this?

so my question is, how do i find the appropriate relation to the given problem?

Last edited: Jan 11, 2012
2. Jan 11, 2012

### Naty1

english is great!!!

Some background here:
Your schoolbok equation appears as #2 (except with "h bar" instead of h):

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

The overall Wikipedia discussion is, I think, a good one.

If you care enough to see several in this forum argue check out this long discussion:(maybe look at the first page and last page for conclusions)

what is it about position and momentum that forbids knowing both quantities at once?

3. Jan 11, 2012

### eaglelake

I agree that your English is fine!

In order to calculate the uncertainty exactly you must know the corresponding wavefunction. Not knowing the wavefunction for a given experiment, we make an educated guess by assuming that the uncertainty in position will be comparable to the size of the slit, or of the atom. That is what your textbook, like many others, does.

The true expression is $$\Delta x\Delta p \ge \hbar /2.$$.

Best wishes

4. Jan 11, 2012

### Gavroy

okay, thanks to you two!

so, deciding whether one has to use an equation like ΔxΔp=h/2 or ΔxΔp=h, which are probably in most cases also inappropriate, is impossible, if the actual wavefunction of the underlying experiment is not known. thus, maybe the guys who wrote this book calculated the actual uncertainty by using this more elaborated structure of quantum mechanics and gave us these equations, so that we could get an impression of the dimension of uncertainty in these experiments? ergo, one would have to calculate the uncertainty by using the mathematical apparatus of quantum mechanics in order to get it right? is this what you meant?

Last edited: Jan 11, 2012
5. Jan 13, 2012

### eaglelake

I meant that the textbook author was most likely trying to demonstrate the use of the uncertainty principle in order to give you a “feel” for it without doing any advanced math. He is giving you an introduction to the heart of quantum mechanics without all the hairy details, which will come later. I do not think that is inappropriate. He wanted you to see how an uncertainty in position is accompanied by an uncertainty in momentum. This is almost always the case in quantum mechanics. Either of the equations you mention will give the correct order of magnitude, so they are appropriate as a starting position.

What we call the quantum uncertainty is called the standard deviation in classical statistics. If we repeatedly measure the position in a given experiment we get a statistical distribution of all possible results. That statistical distribution of those results is given by $$\left| {\psi (x)} \right|^2$$ and the position uncertainty is defined as
$$\Delta x = \sqrt {\left\langle \psi \right|\left. {\left. {x^2 } \right|\psi } \right\rangle - \left| {\left\langle \psi \right|\left. {\left. x \right|\psi } \right\rangle } \right|^2 }$$. Thus, we must know the wavefunction in order to make these calculations.

Best wishes

6. Jan 13, 2012

### Gavroy

thank you, that was exactly what i wanted to know.