The 1st excitad state in 1st order correction

  • Thread starter Thread starter eman2009
  • Start date Start date
  • Tags Tags
    Correction State
eman2009
Messages
32
Reaction score
0

Homework Statement


find the 1st order correction for the 1st excited state with spin when the
H'=(eB0/2mc).Lz+2Sz z the direction
this proplem for the charge particleon sphere of radius R with magnatic field B


Homework Equations



l=1 m=0,1,-1
3. ]The attempt at a solution[/b
work out for the matrix 2x3 for

<Y10lH'lY10> SAME WITH SPIN DOWEN (-)
<Y11lH'lY11> ,,,,,,,,,
<Y1-1lH'lY1-1> ,,,,,,,,,,

And then find the digonal matrix solve it then find E01
 
Physics news on Phys.org
can i know why nobody answer me.
 
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