Solving the 1D Ising ModelToday (Wed Week 2) we went through the solution to the 1D Ising model in detail. Outline of this lectureBig pictureWhat are we trying to do?Our end goal is to find various thermodynamic properties of the 1D Ising model. Remember that thermodynamics means that
and we're trying to find various properties of these thermodynamic systems. What sorts of properties?
How do we calculate properties?Prof. Kivelson outlined the procedure for us on Monday.
Here the sum runs over all the states of the system , means ‘‘the energy of the system when it's in state s’’, and is the inverse temperature of our system. We spend most of our effort trying to figure out how to compute the partition function, which begs the question… Why is the partition function useful?Once we've found the partition function , we can calculate pretty much everything else. For instance:
To explain the notation: I'm summing over all states (this time I call the states rather than ). Inside the sum I am multiplying the spin of the 'th site (which is ) by the Boltzmann weight . The number is the energy of the system when it's in the state , and we find this by plugging in each of the spins into the Hamiltonian.
Hopefully, this serves as a sort of useful roadmap for where we're going. Solving the 1D Ising ModelGame Plan
These quantities are best calculated in the eigenbasis of the transfer matrix. So our next steps are to
Afterwards, we will diagonalize the transfer matrix and explicitly calculate these quantities. Throughout these steps, we'll appeal to Pauli matrices and our intuition about the quantum mechanics of spinhalf to help us calculate things.
Writing the Hamiltonian as a sum over bondsOur first step is to rewrite the Hamiltonian as a single sum over bonds, rather than two separate sums. Later on, we'll see that this form helps us neatly separate the partition function in a nice way. The Hamiltonian was defined as where the first term is the energy of each of the bonds between neighboring sites, and the second is the energy of each of the sites. Since we want to rewrite the Hamiltonian as a sum over bonds, we reassign the energy of 'th site to its two neighboring bonds (bond to , and to ). (We also introduce a factor onehalf to compensate for the fact that each site will now be counted twice). Now the Hamiltonian is sum over bonds where is the energy of the bond between sites and . (Note that takes on four possible values, since there's four combinations of what the spins on sites and : ++, +, +, and .) Defining the transfer matrixWith the Hamiltonian written in this form, we can calculate the partition function more easily. Remember that the partition function is the sum over all states of the Boltzmann weight . Since the Hamiltonian can be written as a sum, the Boltzmann weight can be written as a product At this point we introduce the transfer matrix as a notational trick to make the expression much look nicer. If we rename each of the factors in the product by defining then looks like ie, there's just a factor of for each of the bonds in the Hamiltonian. Remark
As another notational trick (remember this whole business with transfer matrices is sort of a notational trick anyways!), we can also rewrite the matrix entries of in a quantummechanicsesque manner as . In this picture, we interpret the transfer matrix as an operator in the Hilbert space spanned by and . Since we're well versed with manipulating bras and kets from quantum mechanics, this alternative notation might be helpful for building our intution. Using this braket notation, our expression for now looks like This notation can be more enlightening, or more confusing, depending on your tastes! Personally, I like this braket notation because my eyes are pretty bad and I find it annoying to squint at subscripts. Interpreting the transfer matrixThe transfer matrix answers the question, if there's a bond between a spin and another spin , then what is the value of for that bond? Since there's two possibilities for the first spin (+1 or 1) and two possibilities for the second spin (+1 or 1), that means that is a 2x2 matrix with four entries. (It's a funny sort of matrix where the indices take on values of .) To be explicit, we can write the components of as Rewriting the partition function as a traceNow that we have a nice expression for , we turn back to our task of finding the partition function , which means we need to sum over all the possible configurations of the system. How do we do this? We need to sum over all possible configurations of the spins; each spin can either be up or down, so there's a total of configurations. Symbolically, we write this sum as that is, the first spin takes on values , the second spin takes on values , etc., and we're performing this sum for each of the spins. In class, Prof. Kivelson wrote a confusing expression with a product sign; personally, I'm not a fan. I think that expicitly writing out each of the summations is much more understandable. Anyways, to find the partition function, we need to calculate When we expand out the sum, we realize the key trick: the transfer matrices are matrixmultiplied with each other, because you sum over the repeated index. Rememeber that matrix multiplication is defined as . If we zoom in on the multiplication between the 12 transfer matrix and the 23 transfer matrix, we see that indeed, the transfer matrices are being multiplied by each other when we sum over their shared index : So when we sum over spin #2, those two transfer matrices ‘‘collapse’’ together and we're left with a squared transfer matrix between spin #1 and spin #3. If we repeat this process of ‘‘collapsing’’ all the transfer matrices together, we end up with which we recognize as the formula for the trace of , Remark
It's also possible to do this manipulation in the braket notation. (Personally, I find this notation a bit more enlightening.) Remember that a matrix element in braket notation is written as . If we use this notation, the partition function looks like If we suggestively scooch the sum signs next to their respective ketbras, it looks like and we realize that there's just a copy of the identity lying between neighboring transfer matrix operators. (Remember the resolution of the identity .) So simplifies to as advertised. Also, I can't help but to notice how familiar the expression for looks! It's reminiscent of the propogator in the path integral formulation of quantum mechanics, where we split up the time evolution operator into chunks, and then sandwich a resolution of the identity between each of the chunks ;) Rewriting average spin as a traceNow that we know how the partition function looks like, let's go on to compute the ensemble average of the spin, . I'll be using the braket notation since I find it hard to squint at all the puny subscripts. As with all expectation values, we can calculate the expectation of by summing up the value of for each microstate, weighted by its probability. This has the form and if we plug in the Boltzmann weights for the probabilities , our expression looks like The sum looks quite similar to our expression for earlier, with the exception of an extra factor of . To see the effect of what that factor does, let's zoom in on everything that involves inside the expression: How do we make sense of this expression? If there wasn't the factor of , this would just be the resolution of the identity, , but it's not immediately obvious what the factor of does. Let me try explicitly writing out the sum: This is a matrix with +1 in the (+1,+1) entry and a 1 in the (1,1) entry. In other words, it's the Pauli z matrix defined by Now when we write out the transfer matrices in our sum, all the ketbras resolve to the identity expect for the 'th one, which instead becomes a Pauli z matrix. So we end up with a trapped between a bunch of 's in our expression. Rewriting the spinspin correlation as a traceWhen we find the spinspin correlation function , a very similar trick happens; you can work out the details for yourself. The end result is that Calculating these tracesWe still need to calculate these traces in terms of some basis. Remember from linear algebra that we can calculate traces in terms of any basis we want. In our case, the best basis to use is the eigenbasis of the transfer matrix . Since the transfer matrix is a real symmetric 2x2 matrix, it has two orthogonal eigenvectors that span its whole Hilbert space. We'll write these two eigenvectors and their corresponding eigenvalues as We won't actually explicitly find these eigenvalues and eigenvectors just yet; I'll leave that for the next section because there's some neat tricks you can do with Pauli matrices by drawing an analogy to a spinhalf particle. For now I want to pretend we know the eigenstates already, and just focus on the algebraic manipulations. (Throughout this section I'll slip into quantum mechanics lingo, and say ‘‘operator’’ in place of ‘‘matrix’’ or ‘‘state’’ in place of ‘‘vector’’) Simplifying the partition functionWhat's so nice about the eigenstates of is that they make it much easier to find . When you apply an operator multiple times to an eigenstate, you just pull out an eigenvalue each time: (Remember that is an operator, but are just numbers.) So if we use this basis to evaluate the partition function, it simplifies very nicely: Since eventually we'll be taking the thermodynamic limit , it's nice to rewrite this expression a bit to understand what happens when we take that limit. We find that In the expression above, is less than 1, so when we raise it to a huge power like , it goes to 0. Simplifying the average spinWe can do something quite similar for finding the average spin which can be written explicitly in the eigenbasis of as I'll describe in words what happens when we sandwich our eigenstates around the operator. The copies of to the left of the can act backwards on the braeigenstate to pull out powers of the eigenvalue (no complex conjugate since is Hermitian with real eigenvalues!); the other copies of to the right of act forwards on the keteigenstate to pull out another powers of the eigenvalue. We're left with the trapped inside the sandwich: Again, when we take , only the first term survives, because the ratio of the terms goes to 0. So we find that Diagonalizing the Transfer MatrixCheck out my answer on piazza, where I explain how to find the eigenvalues of by decomposing it in terms of the Pauli matrices as . Check out the picture here for a graphical explanation of why . A picture is worth a thousand words! Work in progress, check back soon… Our first problem setTo be honest, we pretty much did the problem set during class today. On the problem set, Prof. Kivelson first asks us to calculate the magnetization density , which we did in class (!). Later he asks us to express the transfer matrix in terms of Pauli matrices (which we also did in class (!?)) and to discuss the correspondence between the 1D Ising Model and a spinhalf quantum system (again, we also did this in class!!!). It looks like the main purpose of the problem set is
Anyways, I'm not sure how much I can discuss here without violating the honor code. But I did have an interesting remark on the definition of magnetization density. Remark
tl;dr: is a subtle statement. On the problem set, the magnetization density is defined as which seems sort of bizarre since we add up terms and then divide by again. What's going on here? The answer has to do with intensive vs extensive quantities, and experimental observables vs mathematical expressions. To start off, notice that the average of any particular microscopic spin is impossible to measure! (Spin would have a puny magnetic field, and besides, there's a whole bunch of other spins nearby that would mess up the measurement.) Our actual experimental observable is not the magnetization of one particular spin, but rather the magnetization of the whole magnet, which is given by That is, when we perform an experiment, we measure the total magnetization of the magnet, which we get by adding up the contributions from each of the spins. (Notice this is capital , not lowercase !) The problem with total magnetization is that it's an extensive rather than intensive property – that is, it scales with the system size . If we doubled the size of the magnet, we would also double the total magnetiation . Now, we don't want to use an extensive property that grows with , since it'll blow up when we take the thermodynamic limit . Rather, we want to use an intensive property independent of system size, a density rather than a total quantity. To solve this conundrum, just divide the total magnetization by . We define the magnetization density as and voila, no longer blows up as . So we've achieved our goal of finding an experimental quantity that's intensive. Finally, just to make this whole affair even more ridiculous, it turns out that is actually the same value for all ! (Remember, the Ising model Hamiltonian is translationally invariant.) So when we find , we're just adding up the same number times, and then dividing by …and at the end of the day, the magnetization density is just the same as the average spin . Leave a Comment Below!Comment Box is loading comments...
