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I was wondering if there is anywhere a formal proof that shows that the ground state energy of a Heisenberg XXX model with periodic boundary conditions becomes equal to the ground state energy with open boundary conditions.

Basically does the ground state energy that you can analytically compute for the Heisenberg model on a cyclic chain becomes that of the straight chain in the limit to infinity? I'm assuming it does, but can't find a rigorous proof somewhere. Any help would be appreciated!

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The ground state energy for the one-dimensional spin-1/2 Heisenberg model can be obtained using Bethe ansatz methods. For periodic boundary conditions, this was done as early as 1938 by Lamek Hulthén, and the solution is now presented in textbooks such as Giamarchi's Quantum Physics in One Dimension, Oxford (2003). The open boundary case is more complicated to handle, but there is a treatment in F. C. Alcaraz, M. N. Barber, M. T. Batchelor, R. J. Baxter, and G. R. W. Guispel: Surface exponents of the quantum XXZ, Ashkin-Teller and Potts models, J. Phys. A: Math. Gen. 20, 6397 (1987). In the notation of Alcaraz et al., the XXZ Hamiltonian is given by $$ H_\mathrm{XXZ} = -\frac{1}{2} \sum_{j=1}^{L-1} \left( \sigma_j^x \sigma_{j+1}^x + \sigma_j^y \sigma_{j+1}^y + \Delta \sigma_j^z \sigma_{j+1}^z \right). $$

The conclusion you're looking for is in Eqs. (2.49, 2.50) for the ground state energy per site for $|\Delta|\leq 1$ and open boundaries in the thermodynamic limit. These equations reproduce the result for periodic boundaries [Eqs. (31a)-(31c) in C. N. Yang and C. P. Yang, Phys. Rev. 150, 327 (1966)]. Specifically, the ground state energy per site in the thermodynamic limit is given by (for both boundary conditions) $$ e_\infty (\gamma) = \frac{1}{2} \cos \gamma - 2\sin^2\gamma \int_0^\infty \frac{\mathrm{d}x}{\cosh(\pi x) \left[ \cosh \left( 2\gamma x\right) -\cos \gamma\right]}, $$ where $\Delta=-\cos\gamma$. The isotropic ferromagnetic and antiferromagnetic chains are recovered with $\Delta=+1$ and $\Delta=-1$, respectively. In the ferromagnetic case, you get $e_\infty=-1/2$, and in the antiferromagnetic case you get $$ e_\infty=\frac{1}{2}-2\ln 2, $$ which both coincide with the values familiar from the case with periodic boundary conditions.

For SU($N$) Heisenberg chains with $N>2$, Sec. III of Phys. Rev. B 97, 134420 (2018) and works cited therein may allow a similar comparison. I don't know if that's been done. There is also a related question looking for fully general arguments.

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The ground state energy of finite-size quantum spin systems depends on the boundary conditions. For example, the energy of the ground state of a ferromagnetic one-dimensional XXX chain of $N$ spins $1/2$ is equal to $E^{(o)}_0(N) = -\frac{J}{4}(N-1)$ for open boundary conditions and $E^{(p)}_0(N) = -\frac{J}{4}N$ for periodic boundary conditions.

But this question is probably about the ground state energy per spin in the $N\to\infty$ limit. An exact solution is not required to prove the following equality $$ \lim_{N\to\infty} \frac{E_0^{(o)}(N)}{N} = \lim_{N\to\infty} \frac{E_0^{(p)}(N)} {N} \tag{*} $$ for a system with a finite interaction radius and a finite number of states. Suppose that $\hat{A}$, $\hat{B}$, $\hat{C} = \hat{A} +\hat{B}$ are Hermitian operators and $a$, $b$, $c $ are their smallest eigenvalues, and $b'$ is the largest eigenvalue of $\hat{B}$. It is not difficult to proof the following inequalities $$ a+b \leq c \leq a+b'\tag{**} $$ In the case of one-dimensional spin $1/2$ XXX model we have $$ \hat{A} = -J\sum_{i=1}^{N-1}\hat{\vec{S}}_i\hat{\vec{S}}_{i+1}, \quad \hat{B} = -J\hat{\vec{S}}_N\hat{\vec{S}}_1,\quad \hat{C} = -J\sum_{i=1}^{N}\hat{\vec{S}}_i\hat{\vec{S}}_{i+1}, $$ $$ a = E_0^{(o)}(N),\quad c = E_0^{(p)}(N), \quad b = \mbox{min}\left(-\frac{J}4,\frac{3J}{4}\right), \quad b'= \mbox{max}\left(-\frac{J}4,\frac{3J}{4}\right) $$ and $$ E_0^{(o)}(N) + \mbox{min}\left(-\frac{J}4,\frac{3J}{4}\right) \leq E_0^{(p)}(N) \leq E_0^{(o)}(N) + \mbox{max}\left(-\frac{J}4,\frac{3J}{4}\right). $$ From the last inequalities it obviously follows (*).

Update on TobiasFunke's comment. To obtain $(**)$, it is sufficient to take the corresponding eigenvector $|\psi\rangle$ of the operator $\hat{C}$ and consider the matrix elements. Let $\hat{C}|\psi\rangle = c|\psi\rangle$ and $\langle\psi|\psi\rangle = 1$, then we have $$ c = \langle\psi|\hat{C}|\psi\rangle = \langle\psi|\hat{A}|\psi\rangle + \langle\psi|\hat{B}|\psi\rangle \geq a + b $$ The last inequality is due to the well known fact that for any $|\phi\rangle$, $\langle\phi|\phi\rangle = 1$ the following inequalities are valid $\langle\phi|\hat{A}|\phi\rangle \geq a$, $\langle\phi|\hat{B}|\phi\rangle \geq b$. To obtain the second inequality in (**), it is enough to write the first one with $\hat{A}_1 = \hat{C}$, $\hat{B}_1 = -\hat{B}$, $\hat{C}_1 = \hat{A} = \hat{A}_1 + \hat{B}_1$,

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    $\begingroup$ For reference: The inequality cited is a special instance of Weyl's inequality. If it is "not difficult to prove" depends on what exactly you mean with difficult. I think it is requires much non-trivial matrix algebra (and probably some other results/theorems). $\endgroup$ Nov 12, 2023 at 8:26
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    $\begingroup$ @TobiasFünke thanks for the reference. I didn't know about the general case. But for this special case it is really easy to prove the inequalities. I'll edit my answer. $\endgroup$
    – Gec
    Nov 12, 2023 at 8:50
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    $\begingroup$ Yes, I agree. Another way to see this is using Rayleigh-Ritz: Defining $\langle\cdot\rangle_\psi:=\langle\psi,\cdot \psi\rangle/\langle \psi,\psi\rangle$, it basically says that $l=\min \langle L\rangle_\psi$ and $l^\prime=\max \langle L\rangle_\psi$ for $\psi \neq 0$ and a linear hermitian operator $L$. Hence: $\langle C\rangle_\psi=\langle A\rangle_\psi + \langle B\rangle_\psi$ and $\langle C\rangle_\psi \geq \langle C\rangle_{\psi_c}=c\geq a+b$. OTOH, taking $\psi=\psi_a$, we have $\langle C\rangle_{\psi_a}\leq a+\langle B\rangle_{\psi_a}\leq a+b^\prime$. $\endgroup$ Nov 12, 2023 at 13:26
  • $\begingroup$ ...sorry, it obviously should read $\langle C\rangle_{\psi_a}=a+\langle B\rangle_{\psi_a}$ (always a pain to use Math in comments...). To be clear, I denoted by $\psi_l$ a vector for which $L$ assumes its minimum Ritz quotient. $\endgroup$ Nov 12, 2023 at 13:38

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