Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I have a spin system: enter image description here

As shown in the picture, there are two spins S1 and S2, and a pair of interactions between them. One is a ferromagnetic interaction and the other is anti ferromagnetic interaction. I am trying to calculate the Hamiltonian of this system.

The Hamiltonian of the system is:

$$ H = -J_F S1_z S2_z +J_{AF} S1_z S2_z $$

$S1_z$ is the spin matrix for Z direction for spin 1 and $S2_z$ is the spin matrix for Z direction for spin 2. If we allow two random values for $J_F$ and $J_{AF}$, -0.5 and 0.5 respectively the Hamiltonian of the system is as follows.

$$ H = 0.5 S1_z S2_z + 0.5 S1_z S2_z $$ $$ = S1_z S2_z $$ $$ = \begin{pmatrix} 1 & 0 \\ 0 & -1 \\ \end{pmatrix} \times \begin{pmatrix} 1 & 0 \\ 0 & -1 \\ \end{pmatrix} $$ $$ = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix} $$

Am I able to calculate the Hamiltonian correctly?

share|improve this question
add comment

1 Answer 1

up vote 2 down vote accepted

The Hamiltonian of this system lives in a 4-dimensional Hilbert space since you have two spin $1/2$. Therefore, you should represent the spin matrix in this four dimensional space like this:

$S_1^z=\begin{pmatrix} -0.5 & 0 &0 &0 \\ 0&-0.5 &0 &0 \\ 0 &0 &0.5 &0 \\ 0 &0 &0 &0.5 \end{pmatrix}$ , $S_2^z=\begin{pmatrix} -0.5 & 0 &0 &0 \\ 0&0.5 &0 &0 \\ 0 &0 &-0.5 &0 \\ 0 &0 &0 &0.5 \end{pmatrix}$

The order of the four states along the rows and columns is $|DD\rangle,|DU\rangle, |UD\rangle, |UU\rangle$ where $U$ stands for spin up and $D$ stands for spin down.

In this case $S_1^z.S_2^z=\begin{pmatrix} 0.25 & 0 &0 &0 \\ 0&-0.25 &0 &0 \\ 0 &0 &-0.25 &0 \\ 0 &0 &0 &0.25 \end{pmatrix}$

share|improve this answer
    
I am little bit confused here. Let me define a single bit $\frac{1}{2}$ spin as $ |Z\rangle = \begin{pmatrix} 1 \\ 0 \\ \end{pmatrix}$ or logic zero and a single bit $-\frac{1}{2}$ spin as $ |O\rangle = \begin{pmatrix} 0 \\ 1 \\ \end{pmatrix}$ or logic one. (contd...) –  Omar Shehab Mar 12 '13 at 2:13
    
So, for a two spin spin system, the state vectors are $|ZZ\rangle = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \\ \end{pmatrix}$, $|ZO\rangle = \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \\ \end{pmatrix}$, $|OZ\rangle = \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \\ \end{pmatrix}$ and $|OO\rangle = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \\ \end{pmatrix}$. To know the spins I operate your operators on the state vectors. –  Omar Shehab Mar 12 '13 at 2:14
    
When I operate $S_1^z$ on $|ZZ\rangle$ I get the following result. $$ S_1^z . |ZZ\rangle = \begin{pmatrix} -0.5 & 0 & 0 & 0 \\ 0 & -0.5 & 0 & 0 \\ 0 & 0 & 0.5 & 0 \\ 0 & 0 & 0 & 0.5 \\ \end{pmatrix} . \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \\ \end{pmatrix} = \begin{pmatrix} -0.5 \\ 0 \\ 0 \\ 0 \\ \end{pmatrix} $$ I don't understand what to make of it. But when I calculate the expected value, i.e., $\langle ZZ | S_1^z | ZZ \rangle$ (contd...) –  Omar Shehab Mar 12 '13 at 2:15
    
I get $\begin{pmatrix} -0.5 \\ \end{pmatrix}$. I am not sure if the expectation value I am getting is the $-\frac{1}{1}$-spin. If so, I think I almost understand you. But in page 7 of web.uconn.edu/~ch351vc/pdfs/spin1.pdf, the operator is different. It is $\Sigma_z = \begin{pmatrix} 2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -2 \\ \end{pmatrix}. $ Any clue? –  Omar Shehab Mar 12 '13 at 2:15
1  
$\Sigma_z=S_1^z+S_2^z$ The first matrix you wrote for $\Sigma_z$ is consistent with the definitions for$S_1^z$ and $S_2^z$. What is the problem in $S_1^z |ZZ\rangle=-0.5|ZZ\rangle$ ? Isn't this expected? –  Tarek Mar 13 '13 at 18:13
show 3 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.