╔(σ_σ)╝ said:
you can not know the momentum and position of an electron at a given instant but, you can only know one.
It's not just that you can't
know both at the same time. It doesn't
have both at the same time. Any state that has a well-defined position doesn't have a well-defined momentum, and vice versa. In classical mechanics, the answer to the question "where is the particle now?" is always an ordered triple (x,y,z) of real numbers. In QM, the best answer is...the wavefunction, i.e. a function from \mathbb R^3 to \mathbb C (when the time coordinate is held fixed), and if the wavefunction is sharply peaked (i.e. if the position is well-defined), then its Fourier transform is smeared out over a large region, which makes the momentum ill-defined.
I recommend that you read the first few pages of
https://www.amazon.com/dp/0131118927/?tag=pfamazon01-20 (use the "look inside" feature), if you haven't already, to make sure that you understand wave functions and their interpretation.
There is an argument, which I believe is originally due to Heisenberg, that basically says that the only way to measure the position of a physical system is to have it interact with other particles, which then "nudges" the system, making our knowledge of the momentum less precise. Lots of books include a slightly more detailed version of this argument.
Here's one I saw recently. Read the paragraph on page 40 that starts with "As argued by Heisenberg" (and don't try to read the rest of the book until you've had at least one more QM class and 3-4 more math classes, if you want to stay sane).
This argument gets a lot of negative comments in these forums, because people who have studied QM know that the version of the uncertainty principle that actually appears in QM has nothing to do with "nudges" or "knowledge" about the position and momentum. It's much more profound than that. So some of those negative comments are valid, because the argument
gives you the wrong idea about what QM actually says. However, I think a lot of people (including me, until recently) have failed to understand the significance of this argument. It plays the same role in the development of QM as Einstein's postulates in the development of SR. They are both somewhat sloppy, but insightful statements that can help you
guess what mathematical model to use in the new theory you're trying to find. Einstein's postulates eventually gave us Minkowski space, and Heisenberg's argument eventually gave us Hilbert space.
Let me clarify an important detail. Heisenberg's argument leads to an approximate inequality called the uncertainty principle. The theory that was eventually found using Heisenberg's insights is quantum mechanics. QM includes an exact inequality which is
also called the uncertainty principle. That one is what people in this forum have in mind when they talk about the uncertainty principle. That's why Heisenberg's argument gets more negative comments than it deserves. People just see an argument that doesn't use any of the concepts that are defined by QM and a result that has a very different interpretation than what
they call the uncertainty principle, so to them it looks like complete nonsense.
This is a statement and proof of the modern version of the uncertainty principle (i.e. the one that actually appears in the theory of QM):
If A=A^\dagger and B=B^\dagger, then
\Delta A\Delta B\geq\frac{1}{2}|\langle[A,B]\rangle|
where the uncertainty \Delta X of an observable X is defined as
\Delta X=\sqrt{\langle(X-\langle X\rangle)^2\rangle}
Proof:
\frac{1}{2}|\langle[A,B]\rangle| =\frac{1}{2}|(\psi,[A,B]\psi)| =\frac{1}{2}|(B\psi,A\psi)-(A\psi,B\psi)| =|\mbox{Im}(B\psi,A\psi)|
\leq |(B\psi,A\psi)| \leq \|B\psi\|\|A\psi\| =\sqrt{(B\psi,B\psi)} \sqrt{(A\psi,A\psi)}=\sqrt{ \langle B^2\rangle}\sqrt{\langle A^2\rangle}
The second inequality is the
Cauchy-Schwarz inequality Now replace A and B with A-\langle A\rangle and B-\langle B\rangle respectively. This has no effect on the left-hand side since the expectation values commute with everything, so we get
\frac{1}{2}|\langle[A,B]\rangle| \leq \sqrt{\langle (B-\langle B\rangle)^2\rangle}\sqrt{\langle (A- \langle A\rangle)^2\rangle}=\Delta B\Delta A
That's how I wrote it down in my personal notes, but I should clarify a few things. I'm writing scalar products as (x,y) here. The expectation value \langle X\rangle is defined as (\psi,X\psi), or in
bra-ket notation (|\psi\rangle,X|\psi\rangle)=\langle\psi|X|\psi \rangle, so it would be more appropriate to use the notation \langle X\rangle_\psi for the expectation value, and \Delta_\psi X for the uncertainty. We should really be talking about the uncertainty of X
in the state \psi.
By the way, you have an awesome name.
