Estimating with uncertainty principle

dsdsuster
Messages
30
Reaction score
0
http://quantummechanics.ucsd.edu/ph130a/130_notes/node98.html

Can someone help me understand what's going on here?

He says "The idea is that the radius must be larger than the spread in position, and the momentum must be larger than the spread in momentum." which I suppose must be true in order to have a well-defined position and momentum? And then he uses the uncertainty of the momentum as a substitute for the actual momentum. This part in particular I don't really understand.

Thanks!
 
Physics news on Phys.org
I think it's because the average momentum is 0 (Closed orbit), so the time-average of the momentum squared (Which is what actually enters the equation), is exactly the variance in the expectation value of the momentum.
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top