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 ?

  • $\begingroup$ Ok, but for a simple model like this there should be an analytical result. I would not fix the number of qubits so the ground state energy would be a function also of n. $\endgroup$ Jan 28, 2023 at 10:25

1 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 $\cos(k/2)$.)


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.

  • 2
    $\begingroup$ Both concepts are somewhat meaningless as absolute numbers. For example, the ground state energy is "just" a constant and doesn't really affect the dynamics. (It's like the "zero point" energy of a system of harmonic oscillators in quantum field theory). For a system of non-interacting fermions, you would get the total energy by adding up all the single-particle energies associated with each fermion, and they can't all be at the bottom of the band because of the exclusion principle. $\endgroup$
    – hft
    Jan 27, 2023 at 21:32
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    $\begingroup$ Yes, except when you pass to the continuum limit (go from sum to integral) you will find you need to include a factor of the overall length of the system (just like in 3d when you get an overall factor of volume, which makes the energy extensive). This comes from counting states in phase space (usually with periodic boundary conditions). So more like: $\sim \frac{L}{2\pi}\int_{\text BZ} dk 2|sin(k/2)|$, where the integration is restricted to the "Brillouin Zone," which in 1d is just $-\pi$ to $\pi$. $L=aN$ where $a$ is the spacing between ising lattice sites, and N is the number of sites. $\endgroup$
    – hft
    Jan 27, 2023 at 21:47
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    $\begingroup$ Yes, $N$ is the number of Q-bits. In the Ising model, $N$ is the number of "lattice sites." It is the max value in the sum over "$i$": $H = \sum_{i=1}^N Z_i Z_{i+1} +X_i$. $\endgroup$
    – hft
    Jan 28, 2023 at 15:45
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    $\begingroup$ $a$ is the spacing between sites. For example, the distance between two physical "trapped ions" in the quantum computer. Don't worry about it, it is already effectively absorbed into the definitions of the interaction constants J and g, which are here already just set to 1. $\endgroup$
    – hft
    Jan 28, 2023 at 15:46
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    $\begingroup$ That result can't be right. I mean, you calculated it correctly, but it is positive, so it can't be correct. I think the issue has something to do with all the transformations in reference 1 and they might shift the overall zero level... I'm not sure... I think the easiest way to get the exact theoretical answer would be to diagonalize the 6 site hamiltonian. $\endgroup$
    – hft
    Jan 28, 2023 at 18:33

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