# Uncertainty principle and electrons

1. Aug 19, 2011

### ralqs

I've been exposed to two different interpretations of the uncertainty principle.

1) If an electron is in a certain state, a measurement of its position will yield a definite result. However, if after the measurement the electron could be returned to the same state, then a repeated measurement of its position will yield a different answer. Same holds for measurements of momentum. However, the standard deviation of the distribution for positions * the standard deviation for the distribution for momentum will always be greater than a certain constant.

2) Measurements are fuzzy. Measurements of, say, position will never yield a definite result, unless the momentum becomes completely unknown.

Which one is right?

2. Aug 19, 2011

### DrChinese

The first answer is not always correct, it depends. If you prepare particles in an identical pure state, sometimes you will always get the same (or similar) answer for all of them. Ditto for repeated measurements on the same particle, i.e. it depends.

The second is true for non-commuting observables, following the Heisenberg Uncertainty Principle.

3. Aug 20, 2011

That's why I like working with Bose-Einstein Condensates: unless you've added energy to the system to stimulate radiation emission (and why do physicists ALWAYS do that?), you know precisely where every electron is: in its lowest possible energy state consistent with its parent atom's status as an independent atom or as a member of a molecule!.

Ain't that cute?

4. Aug 20, 2011

Where are the electrons in a plasma?

They're totally GONE!

All you have are nuclei.

Makes for very easy work.

But, am I cheating?

5. Aug 20, 2011

### Ken G

To clarify further, one must be careful to distinguish the concept of a "pure state" (or a "certain state" from the OP) from the concept of an "eigenstate." An eigenstate is always a pure state, but not the other way around. An eigenstate is only defined with respect to some observable, so a state could be an eigenstate of one observable but not another (if it does not "commute"). In that eigenstate, that observable is definite, and will always come out the same, but other observables won't. So in a pure state that is not an eigenstate of that observable, the observable is not definite, and will come out spread over a statistical range that can be called an "uncertainty." Thus #1 in the OP is only true if the "certain state" is not an eigenstate of position (i.e., not a delta function), but becomes a delta function after the position measurement.

For the case of complementary observables, an eigenstate of one implies infinite uncertainty in the other. That's not strictly physically possible, except as an idealization, so more typically, we have a pure state that is not an eigenstate of either of the complementary observables, but the product of the uncertainties is above some Planck limit.

Since delta functions don't really happen in practice, I might find fault with the wording of #1. A position measurement does not really give a definite position, it only says that the object is within some small bin. The more energy used, the smaller you can make the bin, but it's never a delta function. Thus #2 sounds generally better to me, and absent the "unless..." part at that.

Last edited: Aug 20, 2011
6. Aug 20, 2011

### ralqs

This is news to me; I was under the impression that a wavefunction automatically collapses into a delta function upon measurement. Are you saying this is wrong? How then would one determine the shape of a wavefunction after measurement?