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Fermions and the Jordan-Wigner transform V: Diagonalization Continued

by Michael Nielsen on June 29, 2005

In today’s post we continue with the train of thought begun in the last post, learning how to find the energy spectrum of any Hamiltonian quadratic in Fermi operators. Although we don’t dwell in this post on the connection to specific physical models, this class of Hamiltonians covers an enormous number of models of relevance in condensed matter physics. In later posts we’ll apply these results (together with the Jordan-Wigner transform) to understand the energy spectrum of some models of interacting spins in one dimension, which are prototypes for an understanding of quantum magnets and many, many other phenomena, including many systems which undergo quantum phase transitions.

Update: Thanks to Steve Mayer for fixing a bug in the way matrices are displayed.

Note: This post is one in a series describing fermi algebras, and a powerful tool known as the Jordan-Wigner transform, which allows one to move back and forth between describing a system as a collection of qubits, and as a collection of fermions. The posts assume familiarity with elementary quantum mechanics, comfort with elementary linear algebra (but not advanced techniques), and a little familiarity with the basic nomenclature of quantum information science (qubits, the Pauli matrices).

The Hamiltonian H = \sum_{jk} \alpha_{jk} a_j^\dagger a_k we diagonalized in the last post can be generalized to any Hamiltonian which is quadratic in Fermi operators, by which we mean it may contain terms of the form a^\dagger_j a_k, a_j a_k^\dagger, a_j a_k and a_j^\dagger a_k. We will not allow linear terms like a_j+a_j^\dagger. Additive constant terms \gamma I are easily incorporated, since they simply displace all elements of the spectrum by an amount \gamma. There are several ways one can write such a Hamiltonian, but the following form turns out to be especially convenient for our purposes:

   H = \sum_{jk} \left( \alpha_{jk} a_j^\dagger a_k -\alpha^*_{jk} a_j a_k^\dagger +   \beta_{jk} a_j a_k – \beta^*_{jk} a_j^\dagger a_k^\dagger \right).

The reader should spend a little time convincing themselves that for the class of Hamiltonians we have described, it is always possible to write the Hamiltonian in this form, up to an additive constant \gamma I, and with \alpha hermitian and \beta antisymmetric.

This class of Hamiltonian appears to have first been diagonalized in an appendix to a famous Annals of Physics paper by Lieb, Schultz and Mattis, dating to 1961 (volume 16, pages 407-466), and the procedure we follow is inspired by theirs. We begin by defining operators b_1,\ldots,b_n:

   b_j \equiv \sum_k \left( \gamma_{jk} a_k + \mu_{jk} a_k^\dagger \right).

We will try to choose the complex numbers \gamma_{jk} and \mu_{jk} to ensure that: (1) the operators b_j satisfy Fermionic CCRs; and (2) when expressed in terms of the b_j, H has the same form as H_{\rm free}, and so can be diagonalized.

A calculation shows that the condition \{ b_j, b_k^\dagger \} = \delta_{jk} I is equivalent to the condition

   \gamma \gamma^\dagger + \mu \mu^\dagger = I,

while the condition \{ b_j, b_k \} = 0 is equivalent to the condition

   \gamma \mu^T+\mu \gamma^T = 0.

These are straightforward enough equations, but their meaning is perhaps a little mysterious. More insight into their structure is obtained by rewriting the connection between the a_js and the b_js in an equivalent form using vectors whose individual entries are not numbers, but rather are operators such as a_j and b_j, and using a block matrix with blocks \gamma, \mu, \mu^* and \gamma^*:

   \left[ \begin{array}{c} b_1 \\ \vdots \\ b_n \\     b_1^\dagger \\ \vdots \\ b_n^\dagger \end{array} \right]   = \left[ \begin{array}{cc} \gamma & \mu \\ \mu^* & \gamma^*       \end{array} \right]    \left[ \begin{array}{c} a_1 \\ \vdots \\ a_n \\     a_1^\dagger \\ \vdots \\ a_n^\dagger \end{array} \right].

The conditions derived above for the b_js to satisfy the CCRs are equivalent to the condition that the transformation matrix

   T \equiv \left[ \begin{array}{cc} \gamma & \mu \\ \mu^* & \gamma^*       \end{array} \right]

is unitary, which is perhaps a somewhat less mysterious condition than the earlier equations involving \gamma and \mu. One advantage of this representation is that it makes it easy to find an expression for the a_j in terms of the b_j, simply by inverting this unitary transformation, to obtain:

   \left[ \begin{array}{c} a_1 \\ \vdots \\ a_n \\     a_1^\dagger \\ \vdots \\ a_n^\dagger \end{array} \right]   = T^\dagger   \left[ \begin{array}{c} b_1 \\ \vdots \\ b_n \\     b_1^\dagger \\ \vdots \\ b_n^\dagger \end{array} \right].

The next step is to rewrite the Hamiltonian in terms of the b_j operators. To do this, observe that:

   H = [ a_1^\dagger \ldots a_n^\dagger a_1 \ldots a_n ]   \left[ \begin{array}{cc} \alpha & -\beta^* \ \beta & -\alpha^*       \end{array} \right]     \left[ \begin{array}{c} a_1 \\ \vdots \\ a_n \\ a_1^\dagger \\         \vdots \\ a_n^\dagger \end{array} \right].

It is actually this expression for H which motivated the original special form which we chose for H. The expression is convenient, for it allows us to easily transform back and forth between H expressed in terms of the a_j and H in terms of the b_j. We already have an expression in terms of the b_j operators for the column vector containing the a and a^\dagger terms. With a little algebra this gives rise to a corresponding expression for the row vector containing the a^\dagger and a terms:

   [a_1^\dagger \ldots a_n^\dagger a_1 \ldots a_n]   = [b_1^\dagger \ldots b_n^\dagger b_1 \ldots b_n] T.

This allows us to rewrite the Hamiltonian as

   H =    [b^\dagger b]   T M T^\dagger   \left[ \begin{array}{c} b \\     b^\dagger \end{array} \right],

where we have used the shorthand [b^\dagger b] to denote the vector with entries b_1^\dagger, \ldots, b_n^\dagger,b_1,\ldots,b_n, and

   M = \left[ \begin{array}{cc} \alpha & -\beta^* \\ \beta & -\alpha^*       \end{array} \right].

Supposing we can choose T such that TMT^\dagger is diagonal, we see that the Hamiltonian can be expressed in the form of H_{\rm   free}, and the energy spectrum found, following our earlier methods.

Since \alpha is hermitian and \beta antisymmetric it follows that M also is hermitian, and so can be diagonalized for some choice of unitary T. However, the fact that the b_js must satisfy the CCRs constrains the class of Ts available to us. We need to show that such a T can be used to do the diagonalization.

We will give a heuristic and somewhat incomplete proof that this is possible, before making some brief remarks about what is required for a rigorous proof. I’ve omitted the rigorous proof, since the way I understand it is uses a result from linear algebra that, while beautiful, I don’t want to explain in full detail here.

Suppose T is any unitary such that

   T M T^\dagger = \left[ \begin{array}{cc} d & 0 \\ 0 & -d \end{array} \right],

where d is diagonal, and we used the special form of M to deduce that the eigenvalues are real and appear in matched pairs \pm \lambda. We’d like to show that T can be chosen to be of the desired special form. To see that this is plausible, consider the map X \rightarrow S X^* S^\dagger, where S is a block matrix:

   S = \left[ \begin{array}{cc} 0 & I \\ I & 0 \end{array} \right].

Applying this map to both sides of the earlier equation we obtain

   ST^* M^* T^T S^\dagger =  \left[ \begin{array}{cc} -d & 0 \\ 0 & d \end{array} \right] = -TMT^\dagger.

But M^* = -S^\dagger M S, and so we obtain:

   -ST^* S^\dagger M S T^T S^\dagger = -TMT^\dagger.

It is at least plausible that we can choose T such that ST^*S^\dagger = T, which would imply that T has the required form. What this actually shows is, of course, somewhat weaker, namely that T^\dagger ST^* S^\dagger commutes with M.

One way of obtaining a rigorous proof is to find a T satisfying

   T M T^\dagger = \left[ \begin{array}{cc} d & 0 \\ 0 & -d \end{array} \right],

and then to apply the cosine-sine (or CS) decomposition from linear algebra, which provides a beautiful way of representing block unitary matrices, and which, in this instance, allows us to obtain a T of the desired form with just a little more work. The CS decomposition may be found, for example, as Theorem VII.1.6 on page 196 of Bhatia’s book “Matrix Analysis” (Springer-Verlag, New York, 1997).

Problem: Can we extend these results to allow terms in the Hamiltonian which are linear in the Fermi operators?

In this post we’ve seen how to diagonalize a general Fermi quadratic Hamiltonian. We’ve treated this as a purely mathematical problem, although most physicists will probably have little trouble believing that these techniques are useful in a wide range of physical problems. In the next post we’ll explain a surprising connection between these ideas and one-dimensional spin systems: a tool known as the Jordan-Wigner transform can be used to establish an equivalence between a large class of one-dimensional spin systems and the type of Fermi systems we have been considering. This is interesting because included in this class of one-dimensional spin systems are important models such as the transverse Ising model, which serve as a general prototype for quantum magnetism, are a good basis for understanding some naturally occurring physical systems, and which also serve as prototypes for the understanding of quantum phase transitions.

From → General

3 Comments
  1. Those 038; are probably coming because &038; is the ampersand symbol in html!

    Anway, I’m a bit confused as to what you might mean for the problem: “Can we extend these results to allow terms in the Hamiltonian which are linear in the Fermi operators?” What would the goal of such a transform be? In the quadratic case you are turning this into an expression in terms of the number operators. What would you be doing with the linear terms?

    As always nice article. A cool application, which I’m sure you know, is Terhal and DiVincenzos application of this to showing that certain gate sets are not universal for quantum computation.

  2. You’re probably right about the ampersands. This is a pity – it means the standard latex way of rendering matrices doesn’t work in wordpress + latexrender. I’ll have to figure out another way of doing this.

    As regards your question, I’m not exactly sure what I’d aim to get out. If I was, I probably wouldn’t have stated it as a problem. What I had in mind was something like the way we complete the square to solve quadratic equations, but I can’t quite see how to make it work. At a guess, the way to approach this is to start by diagonalizing a Hamiltonian which involves only number operators and linear terms.

    The Terhal-Divncenzo-Knill-Valiant result is very cool. I must admit, this and other results (Gottesman-Knill, Vidal et al, and so on) sometimes make me wonder about the dogma that quantum systems are intrinsically hard to simulate on a classical computer. There’s so many which aren’t.

  3. One other thing. If my suggestion does work with new matrices, you will need to get rid of any old gifs with 038 in them which are lying about in the cache. Then WordPress will regenerate the correct gifs.

    That means the boring task of identifying the gif file names in the browser and deleting them, which if there’s a lot of them, is a boring task!

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