Particle of mass m in a box of length L Quantum Mechanics

FloridaGators
Messages
11
Reaction score
0
1. http://img10.imageshack.us/img10/8602/232cw.jpg /b][/URL]

Homework Equations


http://img199.imageshack.us/img199/484/232ag.jpg

sorry about the insert pictures, i don't know how to type up the eqns easily

The Attempt at a Solution


Now I know that outside the box the schrodinger equation should give 0 right. I'm just not sure how to go about (a) and (b). (c) The expectation value is pretty straightforward in that you just evalutate <psi|omega|psi>/<psi|psi> with the boundaries from 0 to L and we know the denominator here is 1 since it's normalized.
 
Last edited by a moderator:
Physics news on Phys.org
I think for part(a) .. you have to solve the schroedinger`s equation to obtain the energy and you know that the potential inside the box is zero , so go from there ..
 
One fundamental postulate of QM is that if you make an observation, then you collapse the state of the particle into an eigenstate of the observable. Eigenstates of the Hamiltonian (the energy observable) are exactly the solutions to the Schroedinger equation. From this information, can you see what are the possibilities when you measure the energy of the system?
 
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