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

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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