Why Is My Calculation of the Particle Energy in a Rectangular Box Different?

cgw
Messages
42
Reaction score
0
I am missing something.

The question is to find the 6 lowest energy states of a particle of mass m in a box with edge lengths of L_{1}=L, L_{2}=2L, L_{3}=2L.
The answer gives E_{0}=\frac{\pi^2\hbar^2}{8mL^2}.

I would have said E_{0}=\frac{3\pi^2\hbar^2}{4mL^2} .

What am I missing?
(the answer given for the actual question is 6, 9, 9, 12, 14, 14)
 
Last edited:
Physics news on Phys.org
Finally figured out the latex.
 
Last edited:
Well done with the latex. Why would you say what you say ?

I don't know what you mean by "6, 9, 9, 12, 14, 14".
 
The question asks for the energy of the six lowest states \frac{E}{E_{0}}. The textbook answer gives E_{0}=\frac{\pi^2\hbar^2}{8mL^2}

The way I see it is:

E = E_0 at n1=1, n2=1, n3=1


E_{0}=\frac{\pi^2\hbar^2}{2m}\left{\left(\frac{n_{1}}{L}\right)^2+\left(\frac{n_{2}}{2L}\right)^2+\left(\frac{n_{3}}{2L}\right)^2\right}

which should be E_{0}=\frac{3\pi^2\hbar^2}{4mL^2}
 
Last edited:
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