What is the expectation value for the ground state of $ H = \sum_i Z_i Z_{i+1} + \sum_i X_i $?

Question

What is the expectation value for the ground state of $ H = \sum_i Z_i Z_{i+1} + \sum_i X_i $ ? In Eq. 15b this provides a solution in k space. The minimum would be reached for $E = -4$. But for example here the energy is calculated numerically as -7.3 ?

Answer 1

What is the expectation value for the ground state of $ H = \sum_i Z_i Z_{i+1} + \sum_i X_i $ ?

The expectation value of the energy in the ground state is: $$ E_0 = \langle \Psi_0|\hat H|\Psi_0\rangle\;, $$ where $|\Psi_0\rangle$ is the full ground state.

OP's first reference provides notes on a standard procedure for diagonalizing the 1d Ising model. A set of transformations are performed to diagonalize the Hamiltonian in terms of creation $\gamma_k^\dagger$ and annihilation $\gamma_k$ operators in momentum space.

As with the usual quantum simple harmonic oscillator, the ground state satistfies $\gamma_k|\Psi_0\rangle = 0$, meaning there are no excitations above the ground state.

OP's first reference provides the form of the diagonalized Hamiltonian as: $$ H=\sum_k\epsilon_k(\gamma_k^\dagger\gamma_k - 1/2)\;. $$

If the above form is correct, then the ground state energy is: $$ E_0 = -\frac{1}{2}\sum_k \epsilon_k\;. $$

The single-particle spectrum ("energy band") is: $$ \epsilon_k = 2J\sqrt{1+g^2-2g\cos(k)}\;, $$ which is Eq. 15b of the first reference.

Note that there is clearly at least one typo in the above-mentioned reference, e.g., the mixed use of $j$ (a site index) and $k$ (a momentum index) in Eq. 15a. I can not vouch for the validity of all the formulas in the reference.

In Eq. 15b this provides a solution in k space. The minimum would be reached for $E = -4$.

Yes, the minimum of the single-particle spectrum (the "energy band") for $J=-1$ and $g=1$ is $-4$.

The provided reference equation for the band can be evaluated for $J=-1$ and $g=1$ to see that the single particle spectrum looks like: $$ \epsilon(k) = 2J\sqrt{1+g^2-2g\cos(k)} \to -2\sqrt{2-2\cos(k)} = -2\sqrt{4\sin^2(k/2)} = -4\sqrt{\sin^2(k/2)}\;, $$ which has a minimum value of -4.

But for example here the energy is calculated numerically as -7.3 ?

First of all, the authors of OP's second reference are calculating the full ground state energy not the bottom-of-band energy, and these are not expected to be the same (i.e., there's no reason to expect the answer is $-4$).

Second, in the linked example, they are performing a variational calculation of the energy using a trial wave function that seems to depends on just one parameter $\theta$. The variational calculation is not exact, so they do not arrive at the exact result for the ground state energy. Presumably the exact result is a bit less than the variational result.

Third, not that it really matters, the authors of the second reference are considering a different Hamiltonian: $Z_iZ_{i+1}-X_i$ rather than $Z_iZ_{i+1}+X_i$. (But, this really just changes the single-particle spectrum from $sin(k/2)$ to $\cos(k/2)$.)

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Example:

To get a feel for how to do an exact calculation, consider the case of just three qubits and "periodic boundary conditions". In this case the full Hamiltonian is: $$ H = Z_2Z_1 + Z_1Z_0 + Z_0Z_2 + X_0 + X_1+X_2\;, $$ where there are three $ZZ$ terms because of the "periodic boundary conditions."

The full Hamiltonian is, explicitly: $$ {\left(\begin{matrix} 3 & 1 & 1 & 0 & 1 & 0 & 0 & 0\\ 1 & -1 & 0 & 1 & 0 & 1 & 0 & 0\\ 1 & 0 &-1 & 1 & 0 & 0 & 1 & 0\\ 0 & 1 & 1 &-1 & 0 & 0 & 0 & 1\\ 1 & 0 & 0 & 0 &-1 & 1 & 1 & 0\\ 0 & 1 & 0 & 0 & 1 &-1 & 0 & 1\\ 0 & 0 & 1 & 0 & 1 & 0 &-1 & 1\\ 0 & 0 & 0 & 1 & 0 & 1 & 1 & 3\end{matrix}\right)}\;, $$ which has lowest eigenvalue of approximately $-3.46$.

Similarly, for 4 q-bits the ground state energy is approximately -5.23, for 5 q-bit the ground state energy is approximately -6.16, and for 6 q-bits the ground state energy is approximately -7.73, which is slightly less (i.e., more negative) than what was arrived at via the variational calculation, as expected.