I Generic Form of Hermitian Matrix

  • Thread starter Thread starter thatboi
  • Start date Start date
  • Tags Tags
    Quantum physics
Click For Summary
The discussion revolves around the confusion regarding the generic form of a Hermitian matrix, specifically equation (2.5) in a referenced thesis. The participant questions why the bottom right element is denoted as ##\bar{A}## and how this relates to the commutation relation stated in equation (2.4). They express uncertainty about the connection between Hermiticity and the commutation relations, indicating a potential oversight in their understanding. An example is provided with a specific matrix, demonstrating that it can yield a Hermitian Hamiltonian. The conversation highlights the need for clarity on the interplay between matrix elements and commutation relations in Hermitian matrices.
thatboi
Messages
130
Reaction score
20
I am reading the following thesis: https://www.kip.uni-heidelberg.de/Veroeffentlichungen/download/6387/pdf-6387.pdf
Specifically, I am confused about equation (2.5), where they give the generic form of the matrix ##\mathcal{M}## due to the Hermiticity of ##\mathcal{H}## and the commutation relation (2.4). I am confused about why the bottom right element is ##\bar{A}##. I'm sure this is related to the commutation relation but I'm confused as to how they enter into the picture. ##\mathcal{H}^{\dagger} = (\mathcal{M}a)^{\dagger}(a^{\dagger})^{\dagger} = a^{\dagger}\mathcal{M}^{\dagger}a## so where do the commutation relations come from or what step did I skip?
 
Physics news on Phys.org
Yes, something is missing, because I can take ##N=1## and ##M=\begin{pmatrix} 0&1\\1&2 \end{pmatrix}##, which make ##H=a^\dagger a^\dagger + aa+2aa^\dagger##, which is hermitian.
 
For the quantum state ##|l,m\rangle= |2,0\rangle## the z-component of angular momentum is zero and ##|L^2|=6 \hbar^2##. According to uncertainty it is impossible to determine the values of ##L_x, L_y, L_z## simultaneously. However, we know that ##L_x## and ## L_y##, like ##L_z##, get the values ##(-2,-1,0,1,2) \hbar##. In other words, for the state ##|2,0\rangle## we have ##\vec{L}=(L_x, L_y,0)## with ##L_x## and ## L_y## one of the values ##(-2,-1,0,1,2) \hbar##. But none of these...

Similar threads

Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 77 ·
3
Replies
77
Views
8K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 24 ·
Replies
24
Views
3K
Replies
10
Views
2K
  • · Replies 27 ·
Replies
27
Views
3K
  • · Replies 4 ·
Replies
4
Views
4K