Ground State of Quantum System: l=0?

lion8172
Messages
27
Reaction score
0
ground states

Homework Statement



Is it generally true that the ground state of a given quantum system corresponds to the lowest quantum numbers? For instance, is it generally true that the ground state of a system governed by a radial potential always corresponds to l=0? If not, how do we know, in particular, that the ground state of the finite, 3-D potential well corresponds to l=0? Griffiths seems to implicitly assume this fact in one of his problems (4.9): "Find the ground state (of the 3-D finite spherical well) by solving the radial equation with l=0."

Homework Equations





The Attempt at a Solution

 
Last edited:
Physics news on Phys.org
Here is a hint, l = 0 corresponds to a spherical configuration, which is usually the lowest eneergy state.
 
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