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Confused with notation

by LagrangeEuler
Tags: confused, notation
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LagrangeEuler
#1
Aug15-13, 02:38 AM
P: 316
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}_2-J_{2,3}\vec{S}_2 \cdot \vec{S}_3-J_{3,4}\vec{S}_3 \cdot \vec{S}_4-J_{4,1}\vec{S}_4 \cdot \vec{S}_1 ##
I can't see that this is the same.
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Hypersphere
#2
Aug15-13, 11:09 AM
P: 182
So what do you get if you write out the first sum?

Hint: Your second Hamiltonian only has nearest neighbors.
Jolb
#3
Aug15-13, 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 nearest-neighbor 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 4-element 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 double-count a single interaction. But this only works for symmetric interactions, i.e. Jij = Jji, 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 nearest-neighbors" is often a more specific description than physicist notation like Ʃ<i,j>.

LagrangeEuler
#4
Aug15-13, 03:12 PM
P: 316
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##?
Hypersphere
#5
Aug15-13, 05:07 PM
P: 182
Quote Quote by LagrangeEuler View Post
Yes just that. Strange. And how to write ##\sum_{i<j}## if you have vectors ##\vec{i},\vec{j}##?
Yeah, it doesn't make sense if you think of i,j as vectors. The upshot is that you don't have to - the indices are just supposed to label all sites. As long as your system has finite length, width, height (and so on, if you want to) - all you need is one number. It doesn't need to have anything to do with the structure of the lattice or your intuition. See this example for a rectangular grid:
1 2 3 4
5 6 7 8
Every site is characterized by the number [itex]i=x+4*(y-1)[/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.

Quote Quote by LagrangeEuler View Post
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##?
Are you confused about that factor of 1/2? It only changes the definition of [itex]J_{ij}[/itex], so it's not that important physically. You can just pick either convention at the start of your calculations.

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.
LagrangeEuler
#6
Aug16-13, 03:46 PM
P: 316
Are you confused about that factor of 1/2? It only changes the definition of [itex]J_{ij}[/itex], so it's not that important physically. You can just pick either convention at the start of your calculations.

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.
Yes but you know authors very often didn't write what they mean by
[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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