
#1
Aug1513, 02:38 AM

P: 277

Notation for Heisenberg Hamiltonian
## H_{exch}=\sum_{i<j}2J_{i,j}\vec{S}_i \cdot \vec{S}_j ## where the sumation extends over all spins pair of the crystal lattice. Suppose we have chain with four sites and periodic boundary condition ## H_{exch}=J_{1,2}\vec{S}_1 \cdot \vec{S}_2J_{2,3}\vec{S}_2 \cdot \vec{S}_3J_{3,4}\vec{S}_3 \cdot \vec{S}_4J_{4,1}\vec{S}_4 \cdot \vec{S}_1 ## I can't see that this is the same. 



#2
Aug1513, 11:09 AM

P: 178

So what do you get if you write out the first sum?
Hint: Your second Hamiltonian only has nearest neighbors. 



#3
Aug1513, 02:18 PM

P: 419

Is this all the information you have? It seems like if this is all given to you, there is an obvious language issue and a potential notational issue.
First "the sumation extends over all spins pair of the crystal lattice." is not correct English. Does "spins pair" mean all nearest neighbors? If it were a nearestneighbor Hamiltonian (which is what you would expect for an Ising or Heisenberg model), then I would write it as: [tex]H=\sum_{\left \langle i,j \right \rangle}J_{ij}\vec{S}_i\cdot\vec{S}_j [/tex] where the notation [itex]\left \langle i,j \right \rangle [/itex] generically means "nearest neighbors." The notation is useful because the indexing of lattice sites might be tricky when you deal with higher dimension. [Note that the way you have it written above would include an interaction between i=1 and j=3, which would not be nearest neighbors in a circular chain. The fact that i=4 and j=1 appears is because in a 4element circular chain, 4 would be a nearest neighbor with 1.] Factors of 2 or 1/2 are usually included in Hamiltonians with symmetric interactions so that you can make sure to remember that if spin i interacts with spin j, then both i and j contribute their interaction to the hamiltonian, or conversely, so that you do not doublecount a single interaction. But this only works for symmetric interactions, i.e. J_{ij} = J_{ji}, which is typically but not always the case. Condensed matter hamiltonians are usually written with intuitive notation that can be confusing if you try and write it down like a mathematician with no idea of what the actual "neighbor" structure is. You'll find that the explanation in words, like "The Hamiltonian consists of pairwise interactions between nearestneighbors" is often a more specific description than physicist notation like Ʃ_{<i,j>}. 



#4
Aug1513, 03:12 PM

P: 277

Confused with notation
Yes just that. Strange. And how to write ##\sum_{i<j}## if you have vectors ##\vec{i},\vec{j}##? Notation ##\sum_{\langle i,j \rangle}## looks also confusing for me sometimes. Because I'm never sure
##\sum_{\langle i,j \rangle}J_{i,j}\vec{S}_i \cdot \vec{S}_j## or ##\frac{1}{2}\sum_{\langle i,j \rangle}J_{i,j}\vec{S}_i \cdot \vec{S}_j##? 



#5
Aug1513, 05:07 PM

P: 178

1 2 3 4 5 6 7 8 Every site is characterized by the number [itex]i=x+4*(y1)[/itex], where x=1,2,3,4 and y=1,2,... Then, if the actual positions are interesting, we can form the vector [itex]\mathbf{r}_i = x \hat{x} + y \hat{y}[/itex] using modulus operations and the basis vectors of the lattice. Actually, the position vectors are often not that interesting for spin chain systems as the positions are usually fixed, i.e. not dynamical. However, it can be important if you want to relate two sums. The factor of 1/2 is then used to avoid double counting. I.e. [tex]\sum_{ij} \mathbf{S}_i \cdot \mathbf{S}_j = 2 \sum_{i<j} \mathbf{S}_i \cdot \mathbf{S}_j[/tex] since the first notation contains both [itex]\mathbf{S}_1 \cdot \mathbf{S}_2[/itex] and [itex]\mathbf{S}_2 \cdot \mathbf{S}_1=\mathbf{S}_1 \cdot \mathbf{S}_2[/itex] while the second sum only contains one instance of [itex]\mathbf{S}_1 \cdot \mathbf{S}_2[/itex]. When using the [itex]\sum_{\langle i,j\rangle}[/itex] notation that Jolb described, it generically contains both the ij and the ji terms, so you are in fact counting all symmetric terms twice. In that case using the Hamiltonian [tex]H = \frac{1}{2} \sum_{\langle i,j \rangle} J_{ij} \mathbf{S}_i \cdot \mathbf{S}_j [/tex] makes sense, but again, it's only a scale factor in the end. 



#6
Aug1613, 03:46 PM

P: 277

[tex]H = \sum_{\langle i,j \rangle} J_{ij} \mathbf{S}_i \cdot \mathbf{S}_j [/tex] for me this is correct way to write it [tex]H = \frac{1}{2} \sum_{\langle i,j \rangle} J_{ij} \mathbf{S}_i \cdot \mathbf{S}_j [/tex] know its all explicit. Because if they calculate critical temperature or any other parameter I wish to know what is exact form of Hamiltonian and what is ##J## in Hamiltonian. Because ##J## you can measure. 


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