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Fermions and the Jordan-Wigner transform IV: Diagonalizing Fermi Quadratic Hamiltonians

by Michael Nielsen on June 26, 2005

I’ve finally had a chance to get back to Fermions. Today’s post explains how to diagonalize a Hamiltonian which is quadratic in operators satisfying the Fermionic CCRs. Remarkably, we’ll do this using only the CCRs: the operators could arise in many different ways physically, but, as we shall see, it is only the CCRs that matter for determining the spectrum! This class of Hamiltonians arises in a lot of realistic physical systems, and we’ll see an explicit example later on, when we show that a particular spin model (the X-Y model) is equivalent to a Fermi quadratic Hamiltonian.

(Unfortunately, there seems to be a bug in WordPress that required me to strip most of the tags denoting emphasis (e.g. bold or italics) out of this post. Weird.)

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).

Diagonalizing a Fermi quadratic Hamiltonian

Suppose a_1,\ldots,a_n satisfy the Fermionic CCRs, and we have a system with Hamiltonian

   H_{\rm free} = \sum_j \alpha_j a_j^\dagger a_j,

where \alpha_j \geq 0 for each value of j. In physical terms, this is the Hamiltonian used to describe a system of free, i.e., non-interacting, Fermions.

Such Hamiltonians are used, for example, in the simplest possible quantum mechanical model of a metal, the Drude-Sommerfeld model, which treats the conduction electrons as free Fermions. Such a model may appear pretty simplistic (especially after we solve it, below), but actually there’s an amazing amount of physics one can get out of such simple models. I won’t dwell on these physical consequences here, but if you’re unfamiliar with the Drude-Sommerfeld theory, you could profitably spend a couple of hours looking at the first couple of chapters in a good book on condensed matter physics, like Ashcroft and Mermin’s “Solid State Physics”, which explains the Drude-Sommerfeld model and its consequences in detail. (Why such a simplistic model does such a great job of describing metals is another long story, which I may come back to in a future post.)

Returning to the abstract Hamiltonian H_{\rm free}, the positivity of the operators a_j^\dagger a_j implies that \langle \psi |H_{\rm   free} |\psi\rangle \geq 0 for any state |\psi\rangle, and thus the ground state energy of H_{\rm free} is non-negative. However, our earlier construction also shows that we can find at least one state |\mbox{vac}\rangle such that a_j^\dagger a_j|\mbox{vac}\rangle = 0 for all j, and thus H_{\rm   free}|\mbox{vac}\rangle = 0. It follows that the ground state energy of H_{\rm free} is exactly 0.

This result is easily generalized to the case where the \alpha_j have any sign, with the result that the ground state energy is \sum_j \min(0,\alpha_j), and the ground state |\psi\rangle is obtained from |\mbox{vac}\rangle by applying the raising operator a_j for all j with \alpha_j < 0. More generally, the allowed energies of the excited states of this system correspond to sums over subsets of the \alpha_j. Exercise: Express the excited states of the system in terms of |\mbox{vac}\rangle. Just by the way, readers with an interest in computational complexity theory may find it interesting to note a connection between the spectrum of H_{\rm free} and the Subset-Sum problem from computer science. The Subset-Sum problem is this: given a set of integers x_1,\ldots,x_n, with repetition allowed, is there a subset of those integers which adds up to a desired target, t? Obviously, the problem of determining whether H_{\rm free} has a particular energy is equivalent to the Subset-Sum problem, at least in the case where the \alpha_j are integers. What is interesting is that the Subset-Sum problem is known to be NP-Complete, in the language of computational complexity theory, and thus is regarded as computationally intractable. As a consequence, we deduce that the problem of determining whether a particular value for energy is in the spectrum of H_{\rm free} is in general NP-Hard, i.e., at least as difficult as the NP-Complete problems. Similar results hold for the more general Fermi Hamiltonians considered below. Furthermore, this observation suggests the possibility of an interesting link between the physical problem of estimating the density of states, and classes of problems in computational complexity theory, such as the counting classes (e.g., #P), and also to approximation problems. Let's generalize our results about the spectrum of H_{\rm free}. Suppose now that we have the Hamiltonian    H = \sum_{jk} \alpha_{jk} a_j^\dagger a_k.  Taking the adjoint of this equation we see that in order for H to be hermitian, we must have \alpha_{jk}^* = \alpha_{kj}, i.e., the matrix \alpha whose entries are the \alpha_{jk} is itself hermitian. Suppose we introduce new operators b_1,\ldots,b_n defined by    b_j \equiv \sum_{k=1}^n \beta_{jk} a_k, where \beta_{jk} are complex numbers. We are going to try to choose the \beta_{jk} so that (1) the operators b_j satisfy the Fermionic CCRs, and (2) when expressed in terms of the b_j, the Hamiltonian H takes on the same form as H_{\rm free}, and thus can be diagonalized. We begin by looking for conditions on the complex numbers \beta_{jk} such that the b_j operators satisfy Fermionic CCRs. Computing anticommutators we find    \{ b_j, b_k^\dagger \} = \sum_{lm} \beta_{jl} \beta_{km}^* \{ a_l,a_m^\dagger \}. Substituting the CCR \{ a_l,a_m^\dagger \} = \delta_{lm} I and writing \beta_{km}^* = \beta_{mk}^\dagger gives    \{ b_j, b_k^\dagger \} = \sum_{lm} \beta_{jl} \delta_{lm} \beta_{mk}^\dagger I= (\beta \beta^\dagger)_{jk} I where \beta \beta^\dagger denotes the matrix product of the matrix \beta with entries \beta_{jl} and its adjoint \beta^\dagger. To compute \{b_j,b_k\} we use the linearity of the anticommutator bracket in each term to express \{b_j,b_k\} as a sum over terms of the form \{ a_l,a_m \}, each of which is 0, by the CCRs. As a result, we have:    \{ b_j,b_k \} = 0. It follows that provided \beta \beta^\dagger = I, i.e., provided \beta is unitary, the operators b_j satisfy the Fermionic CCRs. Let's assume that \beta is unitary, and change our notation, writing u_{jk} \equiv \beta_{jk} in order to emphasize the unitarity of this matrix. We now have    b_j = \sum_k u_{jk} a_k. Using the unitarity of u we can invert this equation to obtain    a_j = \sum_k u^\dagger_{jk} b_k. Substituting this expression and its adjoint into H and doing some simplification gives us    H = \sum_{lm} (u \alpha u^\dagger)_{lm} b_l^\dagger b_m. Since \alpha is hermitian, we can choose u so that u \alpha u^\dagger is diagonal, with entries \lambda_j, the eigenvalues of \alpha, giving us    H = \sum_j \lambda_j b_j^\dagger b_j. This is of the same form as H_{\rm free}, and thus the ground state energy and excitation energies may be computed in the same way as we described earlier. What about the ground state of H? Assuming that all the \lambda_j are non-negative, it turns out that a state |\psi\rangle satisfies a_j^\dagger a_j |\psi\rangle = 0 for all j if and only if b_j^\dagger b_j|\psi\rangle = 0 for all j, and so the ground state for the two sets of Fermi operators is the same. This follows from a more general observation, namely, that a_j^\dagger a_j |\psi\rangle = 0 if and only if a_j|\psi\rangle = 0. In one direction, this is trivial: just multiply a_j|\psi\rangle = 0 on the left by a_j^\dagger. In the other direction, we multiply a_j^\dagger a_j |\psi\rangle = 0 on the left by a_j to obtain a_j a_j^\dagger a_j |\psi\rangle = 0. Substituting the CCR a_j a_j^\dagger = -a_j^\dagger a_j + I, we obtain    (-a_j^\dagger a_j^2+a_j)|\psi\rangle = 0. But a_j^2 = 0, so this simplifies to a_j|\psi\rangle = 0, as desired. Returning to the question of determining the ground state, supposing a_j^\dagger a_j|\psi\rangle = 0 for all j, we immediately have a_j|\psi\rangle = 0 for all j, and thus b_j|\psi\rangle = 0 for all j, since the b_j are linear functions of the a_j, and thus b_j^\dagger b_j|\psi\rangle = 0 for all j. This shows that the ground state for the two sets of Fermi operators, a_j and b_j, is in fact the same. The excitations for H may be obtained by applying raising operators b_j^\dagger to the ground state. Exercise: Suppose some of the \lambda_j are negative. Express the ground state of H in terms of the simultaneous eigenstates of the a_j^\dagger a_j. Okay, that's enough for one day! We've learnt how to diagonalize a fairly general class of Hamiltonians quadratic in Fermi operators. In the next post we'll go further, learning how to cope with additional terms like a_j a_j and a_j^\dagger a_k^\dagger.

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