Symmetries and Boundary Conditions with a Twist

Abstract

Interest in finite-size systems has risen in the last decades, due to the focus on nanotechnological applications and because they are convenient for numerical treatment that can subsequently be extrapolated to infinite lattices. Independently of the envisioned application, special attention must be given to boundary condition, which may or may not preserve the symmetry of the infinite lattice. Here, we present a detailed study of the compatibility between boundary conditions and conservation laws. The conflict between open boundary conditions and momentum conservation is well understood, but we examine other symmetries, as well: we discuss gauge invariance, inversion, spin, and particle-hole symmetry and their compatibility with open, periodic, and twisted boundary conditions. In the interest of clarity, we develop the reasoning in the framework of the one-dimensional half-filled Hubbard model, whose Hamiltonian displays a variety of symmetries. Our discussion includes analytical and numerical results. Our analytical survey shows that, as a rule, boundary conditions break one or more symmetries of the infinite-lattice Hamiltonian. The exception is twisted boundary condition with the special torsion Θ = πL/2, where L is the lattice size. Our numerical results for the ground-state energy at half-filling and the energy gap for L = 2–7 show how the breaking of symmetry affects the convergence to the L → ∞ limit. We compare the computed energies and gaps with the exact results for the infinite lattice drawn from the Bethe-Ansatz solution. The deviations are boundary-condition dependent. The special torsion yields more rapid convergence than open or periodic boundary conditions. For sizes as small as L = 7, the numerical results for twisted condition are very close to the L → ∞ limit. We also discuss the ground-state electronic density and magnetization at half filling under the three boundary conditions.

Appendices

Appendix A: Analytical Results for U = 0

1.1 A.1 Closed Boundary Conditions

For PBC or TBC, Bloch’s Theorem associates each single-particle eigenstate of H with a unique momentum k. Since the number of basis states $a_{\ell }^{\dagger }$ is L, we will have to define L distinct momenta. For now, however, we let the k’s be undetermined parameters.

Since H remains invariant under lattice translations, the model Hamiltonian commutes with the unit-translation operator T ₁, defined by the identity

$$\begin{array}{@{}rcl@{}} T_{1}a_{\ell}^{\dagger}\vert{\text{\o}}\rangle = a_{\ell+1}^{\dagger}\vert{\text{\o}}\rangle. \end{array} $$

(42)

with the operators a _ℓ defined by (27).

We seek eigenvectors of T ₁. Promising candidates are defined by the normalized Fermi operator

$$\begin{array}{@{}rcl@{}} b_{k}^{\dagger} = \frac{1}{\sqrt L}\sum\limits_{\ell=1}^{L}e^{-ik\ell}a_{\ell}^{\dagger}. \end{array} $$

(43)

To verify that the $b_{k}^{\dagger }$ diagonalize T ₁, we only have to compute $T_{1}b_{k}^{\dagger }\vert {\text {\o }}\rangle$. From (42) and (43) we can see that

$$\begin{array}{@{}rcl@{}} T_{1}b_{k}^{\dagger}\vert{\text{\o}}\rangle = \frac{1}{\sqrt{L}}\left( \sum\limits_{\ell=1}^{L-1}e^{-ik\ell}a_{\ell+1}^{\dagger}+ e^{-ikL}a_{1}^{\dagger}\right)\vert{\text{\o}}\rangle, \end{array} $$

(44)

where we have separated the last term from the sum on the right-hand side to emphasize that, under PBC or TBC, the translation displaces $a_{L}^{\dagger }$ to $a_{1}^{\dagger }$, as prescribed by (25b).

We then change the summation index to ℓ ^′ = ℓ + 1 in the sum on the right-hand side of (44), which shows that

$$\begin{array}{@{}rcl@{}} T_{1}b_{k}^{\dagger}\vert{\text{\o}}\rangle = \frac{1}{\sqrt{L}}\left( \sum\limits_{\ell^{\prime}=2}^{L}e^{-ik(\ell^{\prime}-1)} a_{\ell^{\prime}}^{\dagger}+e^{-ikL}a_{1}^{\dagger}\right)\vert{\text{\o}}\rangle. \end{array} $$

(45)

To include the last term within parentheses in the sum on the right-hand side, we now impose the condition

$$\begin{array}{@{}rcl@{}} e^{ikL}=1, \end{array} $$

(46)

so that (45) reduces to the compact expression

$$\begin{array}{@{}rcl@{}} T_{1}b_{k}^{\dagger}\vert{\text{\o}}\rangle = \frac{1}{\sqrt{L}}\sum\limits_{\ell^{\prime}=1}^{L}e^{-ik(\ell^{\prime}-1)} a_{\ell^{\prime}}^{\dagger}\vert{\text{\o}}\rangle, \end{array} $$

(47)

which shows that

$$\begin{array}{@{}rcl@{}} T_{1}b_{k}^{\dagger}\vert{\text{\o}}\rangle = e^{ik}b_{k}^{\dagger}\vert{\text{\o}}\rangle. \end{array} $$

(48)

From (48), we can see that, for momenta satisfying (46), the $b_{k}^{\dagger }$ are eigenstates of the translation operator. Equation (46) is equivalent to the expression

$$\begin{array}{@{}rcl@{}} k = \frac{2n\pi}{L}, \end{array} $$

(49)

where n is an integer.

To generate L distinct eigenstates, we could let n run from unity to L on the right-hand side of (49). It is nonetheless customary to choose the integers so that the momenta lie in the first Brillouin Zone, i.e., for − π < k ≤ π. The following sequences are therefore defined:

$$\begin{array}{@{}rcl@{}} n= \left\{\begin{array}{ll} -\frac{L}{2}+ 1, \ldots,\frac{L}{2}\qquad&(L=\text{even})\\ -\frac{L-1}{2}, \ldots,\frac{L-1}{2}\qquad&(L=\text{odd}). \end{array}\right. \end{array} $$

(50)

Equations (43), (49) and (50) define a set of L non-degenerate eigenstates of the translation operator T ₁. Since the latter commutes with the Hubbard Hamiltonian H under PBC or TBC, we can see that the $b_{k}^{\dagger }$ also diagonalize H.

To complete the diagonalization, we have to find the eigenvalues associated with the $b_{k}^{\dagger }$. On the basis of the latter, the Hubbard Hamiltonian takes the form

$$\begin{array}{@{}rcl@{}} \mathbf{H} =\sum\limits_{k}(\epsilon_{k}-\mu)b_{k}^{\dagger}b_{k}, \end{array} $$

(51)

from which we have that

$$\begin{array}{@{}rcl@{}} \left[\mathbf{H},b_{k}^{\dagger}\right] = (\epsilon_{k}-\mu)b_{k}^{\dagger}. \end{array} $$

(52)

To identify the eigenvalues 𝜖 _k , we therefore need to compute the commutator on the left-hand side of (52) and start out by computing the commutator $\left [\mathbf {H},a_{\ell }^{\dagger }\right ]$ between the Hamiltonian and a local operator $a_{\ell }^{\dagger }$ (ℓ = 1,…, L). From (29), with U = 0, we have that

$$\begin{array}{@{}rcl@{}} \left[\mathbf{H},a_{\ell}^{\dagger}\right]= -ta_{\ell+1}^{\dagger}-t^{*}a_{\ell-1}^{\dagger}-\mu a_{\ell}^{\dagger}\quad(\ell=1,\ldots,L). \end{array} $$

(53)

Reference to (43) now shows that

$$\begin{array}{@{}rcl@{}} \sqrt{L}[\mathbf{H},b_{k}^{\dagger}] &=& -t\sum\limits_{\ell=1}^{L}\left( e^{-ik\ell}a_{\ell+1}^{\dagger}\right) -t^{*}\sum\limits_{\ell=1}^{L}\left( e^{-ik\ell}a_{\ell-1}^{\dagger}\right)\\ &&-\mu -\sum\limits_{\ell=1}^{L}e^{-ik\ell}a_{\ell}^{\dagger}. \end{array} $$

(54)

We then let ℓ → ℓ − 1 in the first sum on the right-hand side and ℓ → ℓ + 1 in the second sum. The limits of the first and second sums will change. Nonetheless, thanks to boundary condition, which makes ℓ = 0 (ℓ = N + 1) equivalent to ℓ = N (ℓ = 1), the sums will still cover all lattice sites, ℓ = 1,…, L. It therefore follows that

$$\begin{array}{@{}rcl@{}} \left[\mathbf{H},b_{k}^{\dagger}\right] = -te^{ik}b_{k}^{\dagger}-t^{*}e^{-ik}b_{k}^{\dagger}-\mu b_{k}^{\dagger}. \end{array} $$

(55)

We next recall that t ≡ t ₀ e ^i𝜃, and compare with (52) to see that

$$\begin{array}{@{}rcl@{}} \epsilon_{k} = -2t_{0}\cos(k+\theta). \end{array} $$

(56)

In particular, under PBC (𝜃 = 0) (56) reduces to the equality

$$\begin{array}{@{}rcl@{}} \epsilon_{k} = -2t_{0}\cos(k), \end{array} $$

(57)

and under TBC with the special torsion Θ = (π/2)L, to the equality

$$\begin{array}{@{}rcl@{}} \epsilon_{k} = 2t_{0}\sin(k). \end{array} $$

(58)

1.2 A.2 Open Boundary Condition

Open boundary condition invalidates Bloch’s Theorem. Instead of a running wave, we may visualize a wave-function that vanishes at ℓ = 0 and ℓ = L + 1, an image that associates the following single-particle operator with the single-particle eigenvectors:

$$\begin{array}{@{}rcl@{}} d_{k}^{\dagger} = \sqrt{\frac{2}{L+1}}\sum\limits_{\ell=1}^{L}\sin(k\ell)c_{\ell}^{\dagger}, \end{array} $$

(59)

subject to the condition that sin(kℓ) vanish for ℓ = L + 1, i.e., for momenta given by the equality

$$\begin{array}{@{}rcl@{}} k = \frac{\ell\pi}{L+1}\qquad(\ell=1,\ldots,L). \end{array} $$

(60)

To show that the $d_{k}^{\dagger }$ in (59) diagonalize (1), we again compute the commutator $[\mathbf {H},c_{\ell }^{\dagger }]$. Under OBC we find that

$$\begin{array}{@{}rcl@{}} &&\left[\mathbf{H},c_{\ell}^{\dagger}\right]= -\mu c_{\ell}^{\dagger} -t_{0}c_{2}^{\dagger}\qquad\qquad\qquad\qquad\,(\ell=1);\\ &&\left[\mathbf{H},c_{\ell}^{\dagger}\right]= -\mu c_{\ell}^{\dagger} -t_{0}c_{\ell+1}^{\dagger}\,-\,t_{0}c_{\ell-1}^{\dagger} \qquad(1<\ell\!<L);\\ &&\left[\mathbf{H},c_{\ell}^{\dagger}\right]= -\mu c_{\ell}^{\dagger} -t_{0}c_{L-1}^{\dagger}\qquad\qquad\qquad\quad(\ell=L). \end{array} $$

(61)

From (59) and (62), we then have that

$$\begin{array}{@{}rcl@{}} [\mathbf{H},d_{k}^{\dagger}] =& -&\sqrt{\frac{2}{L+1}}t_{0} \left( \sum\limits_{\ell=2}^{L} \sin\left( k(\ell-1)\right)c_{\ell}^{\dagger} \right.\\ &+&\left.\sum\limits_{\ell=1}^{L-1} \sin\left( k(\ell+1)\right) c_{\ell}^{\dagger}\right)-\mu d_{k}^{\dagger}.\\ \end{array} $$

(62)

Since sin(kℓ) vanishes for ℓ = 0, we can let the summation index in the first sum on the right-hand side of (62) run from ℓ = 1 to ℓ = L. Likewise, given that sin[k(L + 1)] = 0 [see (60)], we can extend the second sum to ℓ = L, to obtain the expression

$$\begin{array}{@{}rcl@{}} \left[\mathbf{H},d_{k}^{\dagger}\right] &=& -\sqrt{\frac{2}{L+1}}t_{0}\sum\limits_{\ell=1}^{L}\left( \sin\left( k(\ell-1)\right) \right.\\ &&\left.+\sin\left( k(\ell+1)\right)\right)c_{\ell}^{\dagger}-\mu d_{k}^{\dagger}. \end{array} $$

(63)

Expansion of the sines in the summand on the right-hand side reduces (63) to the form

$$\begin{array}{@{}rcl@{}} \left[\mathbf{H},d_{k}^{\dagger}\right] = -2t_{0}\cos(k)\sqrt{\frac{2}{L+1}}\sum\limits_{\ell=1}^{L}\sin(k\ell)c_{\ell}^{\dagger}-\mu d_{k}^{\dagger}. \end{array} $$

(64)

Comparison with (59) then shows that

$$\begin{array}{@{}rcl@{}} \left[\mathbf{H},d_{k}^{\dagger}\right] =\left( -2t_{0}\cos(k)-\mu\right)d_{k}^{\dagger}, \end{array} $$

(65)

which allows us to write the OBC Hamiltonian in a diagonal form akin to (51):

$$\begin{array}{@{}rcl@{}} \mathbf{H} =\sum\limits_{k}(\epsilon_{k}-\mu)d_{k}^{\dagger}d_{k}, \end{array} $$

(66)

with the 𝜖 _k from (57).

Equation (57) describes the dispersion relations for both OBC and PBC. Nonetheless, the single-particle energies 𝜖 _k for OBC are distinct from the 𝜖 _k for PBC, because the allowed momenta are boundary-condition dependent. For OBC, the k are given by (60); for PBC, they are determined by (49) and (50).

1.3 A.3 Dispersion Relations

Figure 12 compares the dispersion relations for PBC, TBC, and OBC. As an illustration, the single-particle levels for L = 4 are depicted for each condition. The single-particle levels for PBC or OBC are given by (57) with k defined by (49) or (60), respectively. Under TBC the levels are given by (56), with k defined by (49). With μ = 0, which corresponds to ground-state occupation N = 4, the levels with 𝜖 _k < 0 are doubly occupied in the ground state, while the levels at 𝜖 _k = 0 have single occupation.

Fig. 12

Dispersion relations for a periodic boundary condition, b twisted boundary condition with torsion Θ = (π/2)L (local torsion 𝜃 = π/2), c twisted boundary condition with torsion Θ = (π/6)L (local torsion 𝜃 = π/6), and d open boundary condition. In each plot bold blue dashes indicate the single-particle levels for L = 4. At half filling, the chemical potential is μ = 0, the negative-energy levels are doubly occupied, the zero-energy levels are singly occupied, and the positive-energy levels are vacant, as indicated by the vertical arrows. The dispersion relation is an even function of k for periodic boundary condition, and an odd function for twisted boundary condition with the special torsion Θ = (π/2)L. By contrast, for Θ = (π/6)L the dispersion relation is asymmetric

With L = 4, the special torsion in (22) is Θ = 2π, equivalent to Θ = 0. The energy levels for PBC and for TBC must therefore be identical. Comparison between panels (a) and (b) in the figure shows how two distinct sets of allowed moment can yield the same single-particle energies. Under TBC with Θ≠2π [panel (c) in Fig. 12] or OBC [panel (d)] the single-particle energies are different; there is no zero-energy level, for instance.

The single-particle spectra in panels (a), (b), and (d) of Fig. 12 are particle-hole symmetric. The bold dashes occur in pairs with energies ± 𝜖, even though their momenta are changed under particle-hole transformation: for positive k, for instance, k → π − k in panels (a), (b), and (d). The dispersion relation in panel (b), TBC with torsion Θ = (π/2)L, is an odd function of k, a symmetry that, for all L, introduces a zero-energy level, at k = 0 in the single-particle energy spectrum. The special torsion Θ = (π/2)L therefore reproduces the feature of the infinite-lattice U = 0 model responsible for the vanishing energy gap at half-filling. No such zero-energy level is found in panel (c) of Fig. 12, which is particle-hole asymmetric, like all spectra for TBC with Θ≠(π/2)L.

Depending on boundary condition, the U = 0 infinite-lattice Hamiltonian can have any of the dispersion relations represented by red solid lines in Fig. 12. With L → ∞, all momenta in the range − π < k ≤ π are allowed, and at least one of them will satisfy 𝜖 _k = 0. Under OBC, for example, the single-particle energy vanishes at k = π/2. If N = L, at zero temperature all levels below (above) 𝜖 _k = 0 will be filled (vacant), and the zero-energy level guarantees that it will cost zero energy to add or to remove an electron from the ground state. There is no energy gap.

1.4 A.4 Ground-State Energy

In the ground state, all levels below the Fermi level are filled. If we introduce the notation k = occ to denote the momenta of the occupied levels, the expression for the ground-state energy under PBC or OBC reads

$$\begin{array}{@{}rcl@{}} E_{\Omega}=-4t_{0} \sum\limits_{k=\text{occ}}\cos(k), \end{array} $$

(67)

where the momenta are specified by (49) or (60), respectively, and the single-particle energies from (57) have been doubled to account for spin degeneracy.

Under TBC, the momenta are again given by (49), and the single-particle energies, from (56), which yields

$$\begin{array}{@{}rcl@{}} E_{\Omega}=-4t_{0} \sum\limits_{k=\text{occ}}\cos(k+\theta), \end{array} $$

(68)

which for the special torsion Θ ≡ L𝜃 = (π/2)L reduces to the form

$$\begin{array}{@{}rcl@{}} E_{\Omega}=4t_{0} \sum\limits_{k=\text{occ}}\sin(k). \end{array} $$

(69)

For all L, the ground-state energy can always be analytically computed, but the computation depends on boundary condition and L parity. For OBC and even L, for instance, (67) reads

$$\begin{array}{@{}rcl@{}} E_{\Omega}=-4t_{0} \sum\limits_{\ell=1}^{L/2}\cos\left( \frac{\pi\ell}{L+1}\right). \end{array} $$

(70)

It proves convenient to rewrite the right-hand side of (70) as the real part of a complex number:

$$\begin{array}{@{}rcl@{}} E_{\Omega}=-4t_{0} \Re\sum\limits_{\ell=1}^{L/2}\exp\left( \frac{i\pi\ell}{L+1}\right), \end{array} $$

(71)

because the summand then defines a geometric progression, which can be easily summed. The following expression results:

$$\begin{array}{@{}rcl@{}} E_{\Omega}=-4t_{0}\Re\frac{i\exp\left( \frac{i\pi}{2(L+1)}\right)-\exp\left( \frac{i\pi}{L+1}\right)}{\exp\left( \frac{i\pi}{L+1}\right) -1}. \end{array} $$

(72)

We then multiply the fraction on the right-hand side of (72) by the complex conjugate of the denominator to show that

$$\begin{array}{@{}rcl@{}} E_{\Omega}=-2t_{0}\frac{2\sin\left( \frac{\pi}{2(L+1)}\right)-1+\cos\left( \frac{\pi}{L+1}\right)}{1-\cos\left( \frac{\pi}{L+1}\right)}. \end{array} $$

(73)

which immediately leads to the expression

$$\begin{array}{@{}rcl@{}} E_{\Omega}=2t_{0}\left( 1-\frac{1}{\sin\left( \frac{\pi}{2(L+1)}\right)}\right). \end{array} $$

(74)

Similar analyses yield the other expressions in Table 2, which compares the ground-state energies for OBC, PBC, and TBC. In the L → ∞ limit, the right-hand sides of (67) or (68) can be more easily computed. For PBC, for instance, we find that

$$\begin{array}{@{}rcl@{}} E_{\Omega}= \frac{L}{\pi} {\int}_{-\pi/2}^{\pi/2}\epsilon_{k}\,\mathrm{d} k. \end{array} $$

(75)

Here, the prefactor of the integral on the right-hand side is the density L/(2π) of allowed k levels in momentum space multiplied by the spin degeneracy, and the energies 𝜖 _k are given by (57). The integral on the right-hand side of (75) computed, we find that

$$\begin{array}{@{}rcl@{}} E_{\Omega} = - \frac{4L}{\pi}t_{0}, \end{array} $$

(76)

which amounts to the per-particle energy in (30).

Appendix B: Analytical Results for U → ∞

1.1 B.1 Open Boundary Condition

Under OBC, the energy levels are given by (57), with k defined by (60). At half filling, with N = L, each level is singly occupied in the ground state. Since the distribution of energy levels is particle-hole symmetric, the contribution of the kinetic energy to the ground-state energy vanishes, so that

$$\begin{array}{@{}rcl@{}} E_{\Omega}^{N=L}=-\mu L. \end{array} $$

(77)

By contrast, in the N = L − 1 ground state, the topmost level, with single-particle level

$$\begin{array}{@{}rcl@{}} \epsilon_{k_{max}} = 2t_{0}\cos\left( \frac{\pi L}{L+1}\right), \end{array} $$

(78)

is vacant, and the corresponding many-body eigenvalue will include the negative of $\epsilon _{k_{max}}$, that is

$$\begin{array}{@{}rcl@{}} E^{L-1}_{\Omega}=-2t_{0}\cos\left( \frac{\pi}{L+1}\right)-\mu(L-1). \end{array} $$

(79)

1.2 B.2 Closed Boundary Conditions

The exact results under PBC [34, 39, 43] support the attractive image of individual levels labeled by momenta. The same image holds under TBC. Either under PBC or TBC, however, only for U = 0 are the allowed k given by (49). Without Coulomb interaction, the momentum states are decoupled from each other, and the allowed k are solely determined by boundary condition. For U ≠ 0, by contrast, the k states are interdependent, and the allowed momenta depend on the spin degrees of freedom. Even in the U → ∞ limit, which is relatively simple under OBC, as discussed in Section B.1, the conditions determining the allowed momenta under PBC or TBC depend on the total spin S and its component, S _z .

As an illustration, consider the Hamiltonian (1) with L = 4 under TBC with the special torsion Θ = (π/2)L, which is equivalent to PBC, and let U → ∞. The conservation laws divide the Fock space into sectors labeled by the charge N, total spin S, total spin component S _z and momentum k. We choose the chemical potential μ so that the ground state lies in one of the sectors with N = 3.

Coulomb repulsion forces the three electrons to occupy three distinct sites. For definiteness, let us assume that the unoccupied state is at site ℓ = 4. The total spin S is the sum of three spin- 1/2 variables. Each variable can have S _z = ↑ or S _z = ↓. The three spins can therefore be found in 2³ = 8 configurations. The maximum spin resulting from addition of the three variables is S = 3/2. The minimum is S = 1/2. With S = 3/2, S _z can take four distinct values—a quadruplet. Out of the eight possible configurations, four states must therefore constitute two doublets, with S = 1/2.

Quadruplet

The S _z = S = 3/2 member of the quadruplet, known as the fully-stretched state because the three spin components are aligned, is given by the expression

$$\begin{array}{@{}rcl@{}} \vert\frac{3}{2},\frac{3}{2};\ell=4\rangle = c_{1\uparrow}^{\dagger}c_{2\uparrow}^{\dagger} c_{3\uparrow}^{\dagger}\vert{\text{\o}}\rangle, \end{array} $$

(80)

where the label ℓ = 4 on the left-hand side reminds us that the fourth site is vacant.

Cyclic permutation of both sides of (80) yields the spin eigenstates |3/2, 3/2, ℓ〉 (ℓ = 1, 2, 3). In analogy with (43), we can then construct four eigenstates of the translation operator T ₁:

$$\begin{array}{@{}rcl@{}} \vert\frac{3}{2},\frac{3}{2},k\rangle = \frac{1}{2}\sum\limits_{\ell=1}^{4}e^{-ik\ell}\vert\frac{3}{2}, \frac{3}{2};\ell\rangle. \end{array} $$

(81)

To determine the allowed momenta k in (81), we translate both sides by one lattice parameter, that is,

$$\begin{array}{@{}rcl@{}} T_{1} \vert\frac{3}{2},\frac{3}{2},k\rangle = \frac{1}{2}\sum\limits_{\ell=1}^{4}e^{-ik\ell}\vert\frac{3}{2}, \frac{3}{2};\ell+1\rangle, \end{array} $$

(82)

or if we let ℓ → ℓ − 1 in the sum on the right-hand side,

$$\begin{array}{@{}rcl@{}} T_{1} \vert\frac{3}{2},\frac{3}{2},k\rangle = e^{ik}\frac{1}{2}\sum\limits_{\ell=0}^{3}e^{-ik\ell}\vert \frac{3}{2},\frac{3}{2};\ell\rangle. \end{array} $$

(83)

Under closed boundary condition, ℓ = 0 is equivalent to ℓ = L ≡ 4, and it follows that

$$\begin{array}{@{}rcl@{}} T_{1} \vert\frac{3}{2},\frac{3}{2},k\rangle = e^{ik}\vert\frac{3}{2},\frac{3}{2},k\rangle, \end{array} $$

(84)

provided that e ^ikL = 1, which condition determines the allowed momenta:

$$\begin{array}{@{}rcl@{}} k= -\frac{\pi}{2},0, \frac{\pi}{2},\pi. \end{array} $$

(85)

According to (84) and (85), the $\vert \frac {3}{2},\frac {3}{2},k\rangle$ are non-degenerate eigenstates of the translation operator T ₁, which commutes with the model Hamiltonian. It follows that the momentum eigenvectors $\vert \frac {3}{2},\frac {3}{2},k=n\pi /2\rangle$ (n = −1,…, 2) are eigenstates of H. In fact, straightforward algebra shows that

$$ \mathbf{H}\vert\frac{3}{2},\frac{3}{2},k\rangle = (2t_{0}\sin{k}-3\mu)\vert\frac{3}{2},\frac{3}{2},k\rangle\qquad \left( k=-\frac{\pi}{2},0,\frac{\pi}{2},\pi\right). $$

(86)

The momentum k = −π/2 yields the lowest eigenvalue,

$$\begin{array}{@{}rcl@{}} E_{S=3/2}=-2t_{0}-3\mu. \end{array} $$

(87)

Equation (86) has simple physical interpretation. The vacancy—a hole—at site ℓ in the state |3/2, 3/2; ℓ〉 (ℓ = 1, 2, 3, 4) can hop to either neighboring site, ℓ − 1 or ℓ + 1, just as the electron at site ℓ in (43) can hop to the neighboring sites. The spectrum of the model Hamiltonian in the S = S _z = 3/2 sector therefore define single-particle energies forming a band analogous to the ones in Fig. 12b, with single-particle energies given by (58).

Doublets

The quadruplet (80) is unique, but the two doublets are not. Two doublets are the symmetric combinations

$$ \vert\frac{1}{2},\frac{1}{2},g;\ell=4\rangle = \frac{1}{\sqrt6} \left( c_{1\uparrow}^{\dagger}c_{2\uparrow}^{\dagger} c_{3\downarrow}^{\dagger} -2c_{1\uparrow}^{\dagger}c_{2\downarrow}^{\dagger}c_{3\uparrow}^{\dagger}+ c_{1\downarrow}^{\dagger} c_{2\uparrow}^{\dagger}c_{3\uparrow}^{\dagger}\right), $$

(88)

which is even (g) under left-right inversion of the lattice segment ℓ = 1, 2, 3, and

$$\begin{array}{@{}rcl@{}} \vert\frac{1}{2},\frac{1}{2},u;\ell=4\rangle = \frac{1}{\sqrt2} \left( c_{1\uparrow}^{\dagger}c_{2\uparrow}^{\dagger} c_{3\downarrow}^{\dagger}-c_{1\downarrow}^{\dagger}c_{2\uparrow}^{\dagger}c_{3\uparrow}^{\dagger}\right), \end{array} $$

(89)

which is odd (u). To verify that the right-hand sides are doublets, we only have to check that $S_{+}\vert \frac {1}{2},\frac {1}{2},p;\ell =4\rangle =0$ (p = g, u), where the raising operator is $S_{+} \equiv {\sum }_{\ell }c_{\ell \uparrow }^{\dagger }c_{\ell \downarrow }$.

The choices defined by (88) and (89) are not unique, because any linear combination between their right-hand sides will also have spin 1/2. One can easily verify that they are normalized and mutually orthogonal. Cyclic permutation of (88) and (89) yields three other pairs with vacancies at sites ℓ = 1, 2, and 3, from which eight eigenstates of the translation operator T ₁ can be constructed, as in (81). The allowed momenta are once more given by (85). For each sector with S = S _z = 1/2 and given k, two states |1/2, 1/2, p, k〉 (p = g, u) result. Projection of the model Hamiltonian upon the orthonormal basis formed by these two states yields a 2 × 2 matrix:

$$\begin{array}{@{}rcl@{}} \mathcal{H}_{S=1/2,k}=-t_{0} \left[\begin{array}{cc} \sin(k)&-\sqrt3i\cos(k)\\ \sqrt3i\cos(k)&\sin(k) \end{array}\right]-3\mu. \end{array} $$

(90)

Diagonalization yields the two eigenvalues of the Hamiltonian in the S = S _z = 1/2, k sector:

$$\begin{array}{@{}rcl@{}} E_{1/2,k}^{\pm}=-t_{0}\left( \sin(k)\pm\sqrt3\cos(k)\right)-3\mu. \end{array} $$

(91)

The lowest eigenvalues among the four S = S _z = 1/2, k (k = −π/2, 0, π/2, π) sectors lie in the k = 0 and k = π sectors:

$$\begin{array}{@{}rcl@{}} E_{1/2,0}^{+} = E_{1/2,\pi}^{-}=-\sqrt3t_{0}-3\mu. \end{array} $$

(92)

Bethe-Ansatz Approach

Unfortunately, the same analysis cannot be extended to longer lattices, because the number of basis states grows exponentially with L. The alternative is the Bethe-Ansatz solution [38, 39].

The Bethe-Ansatz solution covers any lattice size L, under OBC, PBC, or TBC. Instead of a closed expression for the eigenvalues of the Hamiltonian, it yields a set of coupled nonlinear equations, known as the Lieb-Wu equations. For most choices of the model parameters, the Lieb-Wu equations are notoriously difficult to solve, even numerically. The exceptions are the U = 0 limit, discussed in Section 4.1, the infinite system, L → ∞, to be discussed in Section B.3, and the U → ∞ limit, to which we now turn.

The notation we have adopted, in which N denotes the number of electrons and M, the number of ↓-spin electrons, follows Lieb and Wu [38]. The Bethe Ansatz approach seeks N-electron eigenstates described by real-space eigenfunctions Ψ(x ₁, x ₂,…, x _N ; σ ₁,…, σ _N ), dependent on the particle positions x _j and spin components σ _j (j = 1,…, N).

The eigenfunctions are parametrized by two sets of quantum numbers: k _n (n = 1,…, N) and λ _m (m = 1,…, M), associated with the charge and spin degrees of freedom, respectively. To determine the k _n and λ _m , a system of N + M non-linear coupled algebraic equations must be solved.

The eigenvalues of the Hamiltonian depend only on the k _n , which can be formally identified with momenta. If U = 0, the k _n coincide with the single-particle momenta k in Section 4.1. With U ≠ 0 they are no longer given by (49) or by (60) and have to be determined from the Lieb-Wu equations.

Once the k _n are found, the eigenvalues of the Hamiltonian under OBC or PBC can be computed from a sum analogous to (67) [34, 38, 39]:

$$\begin{array}{@{}rcl@{}} E= -2t_{0}\sum\limits_{n=1}^{N}\cos(k_{n}) -\mu N, \end{array} $$

(93)

where the sum runs over the N occupied k _n .

For TBC, the sum is analogous to the right-hand side of (68)

$$\begin{array}{@{}rcl@{}} E= -2t_{0}\sum\limits_{n=1}^{N}\cos(k_{n}+\theta) -\mu N, \end{array} $$

(94)

which for the special torsion Θ ≡ L𝜃 = (π/2)L reads

$$\begin{array}{@{}rcl@{}} E= 2t_{0}\sum\limits_{n=1}^{N}\sin(k_{n}) -\mu N. \end{array} $$

(95)

The chemical potential is determined by the condition $\partial \bar E/\partial N=0$, where $\bar E$ is the thermodynamical average of the eigenvalues E. At zero temperature, μ is such that the N occupied k _n satisfy the inequality − 2t cos(k _n ) ≤ μ (n = 1,…, N).

The U → ∞ limit simplifies the Lieb-Wu equations. A schematic depiction of the procedure determining the ground-state energy is presented in Fig. 13. The charge and spin degrees of freedom decouple and can be described separately. The k _n satisfy a relatively simple equation, analogous to (46) [34, 38, 39]:

$$\begin{array}{@{}rcl@{}} e^{ik_{n}L}=e^{i{\Lambda}}, \end{array} $$

(96)

where the phase Λ depends only on the spin degrees of freedom.

Fig. 13

Computation of the ground-state energy from the solution of the Lieb-Wu equations in the $U\to \infty$ limit, under TBC with the special torsion Θ = (π/2)L. L, N, and M are the lattice size, the number of electrons, and the number of ↓-electrons respectively. The ground-state energy is computed from (95), where the k _n , given by (97), can be regarded as momenta of spinless electrons. To determine the phase Λ, one starts out by considering a subsidiary gas of non-interacting particles with momenta q _m . The M integers m are chosen so that the resulting q _m , given by (98), lie in the First Brillouin Zone. Given the q _m , (100) determines the phase Λ. The next steps are depicted on the right-hand panel. We start by determining the L allowed momenta k _n . The integers n are chosen to position the k _n in the First Brillouin Zone and to minimize the energy in (94). The resulting minimum energy $E_{\mathcal {M}}$ depends on Λ and hence upon our choice of the set $\mathcal {M}$. To find the ground-state energy, we have to repeat the procedure for all possible $\mathcal {M}$s. The lowest overal $E_{\mathcal {M}}$ is the ground-state energy

Equation (96) allows momenta of the form

$$\begin{array}{@{}rcl@{}} k_{n} = \frac{2\pi n+{\Lambda}}{L}, \end{array} $$

(97)

where the n’s are integers that define the eigenstate of the Hamiltonian. The integers defining the ground state for TBC, for example, are those that minimize the sum on the right-hand side of (95).

To determine the allowed momenta, we therefore need the phase Λ and have to examine the spin degrees of freedom. Again, we let the number M of electrons with ↓ spin be smaller or equal to the number N − M of ↑ electrons. Although the Lieb-Wu equation describing the spin degrees of freedom seem unwieldy, they have been found to be identical with the equations describing a simpler system, a subsidiary gas with a Hamiltonian that can be trivially diagonalized [43]. The eigenvalues of the latter Hamiltonian determine the phase Λ, which can then be substituted on the right-hand side of (97) to yield the allowed momenta k _n .

More specifically, to determine Λ one has to find the total momentum of a subsidiary system with M particles on an N-site one-dimensional lattice. The particles in the subsidiary system occupy M distinct states labeled by their momenta q _m (where 1 ≤ m ≤ N), which lie on a flat band, with dispersion relation 𝜖 _q = 0. The subsidiary particles must satisfy either anti-periodic or periodic boundary conditions, depending on whether M is even or odd, respectively. The M allowed momenta must therefore satisfy the equalities

$$\begin{array}{@{}rcl@{}} e^{iq_{m}N}= \left\{\begin{array}{ll} -1&\qquad(M=\text{even})\\ 1 &\qquad(M=\text{odd}), \end{array}\right. \end{array} $$

(98)

which are equivalent to the expressions

$$\begin{array}{@{}rcl@{}} q_{m} = \left\{\begin{array}{ll} \frac{(2m+1)\pi}{N}&\qquad(M=\text{even})\\ \frac{2m\pi}{N}&\qquad(M=\text{odd}) \end{array}\right., \end{array} $$

(99)

with integers 1 ≤ m ≤ N that depend on the desired eigenstate of the Hamiltonian.

Given a set of M occupied momenta q _m , the phase Λ is the total momentum

$$\begin{array}{@{}rcl@{}} {\Lambda} = \sum\limits_{m=1}^{M}q_{m}. \end{array} $$

(100)

This explained, we are ready to find the eigenvalues of the L = 4, U → ∞ Hubbard Hamiltonian for N = 3.

First, we set M = 0, which is equivalent to letting S _z = S = 3/2. With M = 0, the number of particles in the subsidiary gas is zero and it follows from (100) that Λ = 0. As in Section A.1, we choose the k _n to lie in the first Brillouin Zone. 97 then yields the allowed momenta:

$$\begin{array}{@{}rcl@{}} k_{n} = \frac{\pi n}{2}\qquad(n=-1,0,1,2). \end{array} $$

(101)

To obtain the smallest eigenvalue of the Hamiltonian associated to the k _n in (101), we fill the three levels making the smallest contribution to the right-hand side of (95), i. e., the levels associated with k ₋₁, k ₀ and k ₂. The resulting eigenvalue coincides with the right-hand side of (87).

Consider now M = 1. With M = 1, the q _m allowed by (99) are

$$\begin{array}{@{}rcl@{}} q_{m} = \frac{2m\pi}{3}\qquad(n=-1,0,1). \end{array} $$

(102)

Equation (100) then determines Λ. Since M = 1, the sum on the right-hand side is restricted to a single q _m , namely one of the three values in (102). The resulting phases are given by the equality

$$\begin{array}{@{}rcl@{}} {\Lambda} = -\frac{2\pi}{3}, 0, \frac{2\pi}{3}. \end{array} $$

(103)

Substitution of the right-hand side of (103) for Λ in (97) yields the following allowed momenta:

$$\begin{array}{@{}rcl@{}} k = \left\{\begin{array}{ll} -\frac{2\pi}{3}, -\frac{\pi}{6}, \frac{\pi}{3}, -\frac{5\pi}{6}&\qquad\left( {\Lambda}=-\frac{2\pi}{3}\right)\\ \frac{\pi}{2}, 0, \frac{\pi}{2}, -\pi&\qquad({\Lambda}=0)\\ \frac{\pi}{3}, -\frac{5\pi}{6}, \frac{\pi}{6}, -\frac{2\pi}{3}&\qquad\left( {\Lambda}=\frac{2\pi}{3}\right) \end{array}\right.. \end{array} $$

(104)

To obtain the corresponding eigenvalues, from (95), for each Λ, we have to occupy three of the four allowed k-states, i. e., leave one level vacant. The resulting energies are given by the equality

$$\begin{array}{@{}rcl@{}} E+3\mu = \left\{\begin{array}{ll} \pm\sqrt3t_{0}, \pm t_{0}&\qquad\left( {\Lambda}=\pm\frac{2\pi}{3}\right)\\ 0, -2t_{0}, 2t_{0} &\qquad({\Lambda}=0), \end{array}\right. \end{array} $$

(105)

the eigenvalues for Λ = 2π/3 being degenerate with those for Λ = −2π/3, and the first eigenvalue for Λ = 0 being doubly degenerate. The lowest eigenvalues for Λ = ±2π/3 and for Λ = 0 are $-\sqrt 3t_{0}-3\mu$ and − 2t ₀ − 3μ, respectively.

Comparison of (105) with (87) and (92) shows that with M = 1 the phase Λ = 0 corresponds to S = 3/2, S _z = 1/2 (87), while Λ = ±2π/3 corresponds to S = S _z = 1/2 (92). This concludes our illustrative discussion.

The same procedure can be applied to other lattice lengths L and electron numbers N. We are especially interested in the minimum energies in the sectors with N = L and N = L − 1, from which we can compute the U → ∞ ground-state energy E _Ω and the energy gap E _g at half filling.

With N = L, the ground-state energy vanishes in the U → ∞ limit. Since each k-level can host at most one electron, all levels must be occupied for N = L. Particle-hole symmetry then guarantees that the positive contributions to E _Ω cancel the negative contributions. The ground-state energy is therefore zero.

With N = L − 1, except for the special length L = 2, the ground-state energy is negative. For fixed Λ, (97) defines the allowed momenta. In the ground state all levels are filled, except for the highest one, with energy 𝜖 _max . The ground-state energy is − 𝜖 _max . With Θ = (πL)/2, provided that the momentum k _n = π/2 be allowed, the highest allowed energy is $\epsilon _{max}=\epsilon _{k_{n}=\pi /2}=2t_{0}$. If k _n = π/2 is not allowed, the ground-state energy will be $-2t_{0}\sin (\bar k)$, where $\bar k$ is the allowed momentum closest to π/2.

For lengths L that are multiples of four, one of the momenta allowed by (97) is k _n = π/2 + Λ/L. The phase Λ = 0 is always allowed, since we can always choose M = 0. The momentum k _n = π/2 is therefore allowed, and the ground-state energy is − 2t ₀.

The ground-state energy is also − 2t ₀ if N = L − 1 is a multiple of four. Given Λ, the momentum k _{n = 0} = Λ is always allowed by (97). We choose M = 1. According to (99), the subsidiary momentum q _N/4 = π/2 is allowed, and hence the phase can take the value Λ = π/2. It follows that k _n = π/2 is allowed, and that the ground-state energy is − 2t ₀.

If neither L nor N are multiples of four, k _n cannot equal π/2, and the ground-state energy E _Ω is positive. To compute it, we must first let M run from zero to N, consider all subsidiary momenta q _m momenta compatible with (99) for each M and obtain the resulting phases Λ from (100). Once the Λ are computed, the allowed k _n are given by (97). The ground-state energy under TBC is given by the set of N momenta k _n thus determined that minimizes the right-hand side of (95).

1.3 B.3 Ground-State Energy for L → ∞

As L → ∞, the quantum numbers k _n and λ _m characterizing the Bethe-Ansatz solution form continua. When the ground state is considered, the Lieb-Wu equations reduce to two coupled integral equations for the densities of the k _n and λ _n . For the special case 2M = N = L, i.e., for the spin-unpolarized half-filled band, Lieb and Wu were able to solve the integral equations and derive closed expressions for the ground-state energy E _Ω and chemical potential [34, 38, 39]. Their expression for the ground-state energy, which excludes the contribution from the term proportional to μ on the right-hand side of (1), reads

$$\begin{array}{@{}rcl@{}} E^{LW}_{\Omega,N=L} = -4L{\int}_{0}^{\infty} \frac{J_{0}(\omega)J_{1}(\omega)}{\omega\left( 1+e^{\omega U/2}\right)} \,\mathrm{d}\omega \end{array} $$

(106)

where J _ν denotes the ν-th order Bessel function.

The chemical potential, defined as the energy difference $E^{LW}_{\Omega ,N+1}-E^{LW}_{\Omega ,N}$ needed to add a particle to the ground state, is given by the equality

$$\begin{array}{@{}rcl@{}} \mu_{+} = \frac{U}{2} - 2 + 4{\int}_{0}^{\infty}\frac{J_{1}(\omega)}{\omega\left( 1+e^{\omega U/2}\right)}\,\mathrm{d}\omega. \end{array} $$

(107)

1.4 B.4 Energy Gap for L → ∞

The subscript + on the left-hand side of (107) is necessary, because the chemical potential is discontinuous for U ≠ 0. The chemical potential μ ₋, equal to the energy $E^{LW}_{\Omega ,N}-E^{LW}_{\Omega ,N-1}$ needed to add a particle to the N − 1-electron ground state, can be obtained from the particle-hole transformation in Section 3.2.5:

$$\begin{array}{@{}rcl@{}} \mu_{-} = U -\mu_{+}. \end{array} $$

(108)

The energy gap E _g = μ ₊ − μ ₋ is therefore given by the closed expression

$$\begin{array}{@{}rcl@{}} E_{g} = U - 4 + 8{\int}_{0}^{\infty}\frac{J_{1}(\omega)}{\omega\left( 1+e^{\omega U/2}\right)}\,\mathrm{d} \omega, \end{array} $$

(109)

the right-hand side of which vanishes as U → 0.