Probability for a particle in a box

rmjmu507
Messages
34
Reaction score
0

Homework Statement


A particle in a box of width 2a is in a state \psi=\frac{1}{\sqrt{2a}} for |x| less than or equal to a and 0 for |x| greater than a. What is the probability of finding the particle in [-b, b] inside the box? What is the probability of finding the particle with momentum p inside the box?

Homework Equations


None

The Attempt at a Solution


So, for the probability of finding the particle in [-b,b], I simply integrated the function from -b to b and found that the probability is b/a.

I'm not sure how to approach the probability calculation for finding the particle with momentum p though...can someone please provide some guidance?
 
Physics news on Phys.org
What measurement values of the momentum operator are possible?
 
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