- #1
Given a Hamiltonian, find eigenvalues and eigenvector
- Thread starter mudyos
- Start date
- #2
malawi_glenn
Science Advisor
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what have you done and though so far? If you don't tell us and show us, you'll get no help-
- #3
malawi_glenn
Science Advisor
- 6,735
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mudyos: This forum works as this: You post question and work and thougts done. If not, then no one will help you. The RULES of this forum is those, didn't you read the rules before posting here?
I give you some questions for you to answer, then maybe you can do something.
What is a matrix representation of an operator?
What is an hermitian operator and matrix?
What is an eigenvector and eigenvalue?
What is a commutator?
What is a projector?
I give you some questions for you to answer, then maybe you can do something.
What is a matrix representation of an operator?
What is an hermitian operator and matrix?
What is an eigenvector and eigenvalue?
What is a commutator?
What is a projector?
- #4
christianjb
- 529
- 1
I'm not about to download a .doc file.
- #5
- #6
malawi_glenn
Science Advisor
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It doesn't matter christianjb and mudyos. Follow the rules and show work.
mudyos: why can't you answer my questions?
mudyos: why can't you answer my questions?
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...
Boundary conditions:
The derived Dirichlet and Neumann boundary conditions
In terms of the magnetic scalar potential:
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...
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