I The Hamiltonian elements in Anderson dimer

  • I
  • Thread starter Thread starter hokhani
  • Start date Start date
  • Tags Tags
    Hamiltonian
hokhani
Messages
561
Reaction score
18
TL;DR Summary
I don't know the occur of - sign in some elements of the Hamiltonian
In a system with two orbitals ##c## and ##d## (each with two spin degrees of freedom), consider the Hamiltonian ##H=V(d^{\dagger}_{\uparrow} c_{\uparrow} + c^{\dagger}_{\uparrow}d_{\uparrow}+d^{\dagger}_{\downarrow} c_{\downarrow} + c^{\dagger}_{\downarrow}d_{\downarrow})##. Also suppose that the many body states are indexed as ##|n_{d_{\uparrow}}, n_{d_{\downarrow}}, n_{c_{\uparrow}} , n_{c_{\downarrow}}\rangle## where ##n## shows the occupation number of each spin-orbital.
As mentioned in the literature, ##\langle1001|H|1100\rangle=V## and ##\langle0110|H|1100\rangle=-V##. However, according to my calculations, ##H|1100\rangle=V(|0110\rangle+|1001\rangle)## which gives ##V## for the two matrix elements ##\langle1001|H|1100\rangle## and ##\langle 0110|H|1100\rangle##. I would like to know where my calculations goes wrong! Any help is appreciated.
 
Last edited:
Physics news on Phys.org
Can you provide a citation to an example of the literature you refer to?
 
Haborix said:
Can you provide a citation to an example of the literature you refer to?
Of course. In the following virtual lecture on youtube at the minute 52 (the blue text):

The Hamiltonian is written in the subspace ##Q=2, S_z=0## and I am only interested in the difference between the arrays ##H_{1,3}## and ##H_{2,3}## so I neglected the other terms of the Hamiltonian in my main post.
 
I think you've probably just gotten the order of the basis vectors mixed up. I get ##H|1100\rangle=V(-|0110\rangle+|1001\rangle)+(2\epsilon+U)|1100\rangle##. In the notation of the video I can represent ##|1100\rangle## as $$\begin{pmatrix}0 \\ 0 \\ 1 \\ 0\end{pmatrix}.$$

EDIT: I think I may have misunderstood your question. Are you essentially asking how to get the matrix representation of the Hamiltonian?
 
Last edited:
Haborix said:
I think you've probably just gotten the order of the basis vectors mixed up. I get ##H|1100\rangle=V(-|0110\rangle+|1001\rangle)+(2\epsilon+U)|1100\rangle##. In the notation of the video I can represent ##|1100\rangle## as $$\begin{pmatrix}0 \\ 0 \\ 1 \\ 0\end{pmatrix}.$$

EDIT: I think I may have misunderstood your question. Are you essentially asking how to get the matrix representation of the Hamiltonian?
Thank you so much for your attention. Could you please explain how did you obtain ##c^{\dagger}_{\uparrow} d_{\uparrow} |1100 \rangle =- |0110 \rangle##? In other words, my question is about the appearance of the minus sign.
EDIT: Your answer is quite to the point and I am lookig for the reason for the minus sign.
 
Last edited:
I am not sure if this falls under classical physics or quantum physics or somewhere else (so feel free to put it in the right section), but is there any micro state of the universe one can think of which if evolved under the current laws of nature, inevitably results in outcomes such as a table levitating? That example is just a random one I decided to choose but I'm really asking about any event that would seem like a "miracle" to the ordinary person (i.e. any event that doesn't seem to...
Not an expert in QM. AFAIK, Schrödinger's equation is quite different from the classical wave equation. The former is an equation for the dynamics of the state of a (quantum?) system, the latter is an equation for the dynamics of a (classical) degree of freedom. As a matter of fact, Schrödinger's equation is first order in time derivatives, while the classical wave equation is second order. But, AFAIK, Schrödinger's equation is a wave equation; only its interpretation makes it non-classical...

Similar threads

Replies
12
Views
1K
Replies
1
Views
1K
Replies
1
Views
1K
Replies
0
Views
1K
Replies
12
Views
2K
Replies
1
Views
1K
Replies
6
Views
2K
Back
Top