Lets consider a spin $\frac{1}{2}$ chain with $n$ spins and an associated local hamiltonian $H= \sum_i h_{i,i+1}$. We also assume that $\|h_{i,i+1}\|_{\infty} \leq 1$. In this question, we will be interested in properties of this local hamiltonian system under local perturbation of $H$ to $H' = \sum_i h'_{i,i+1}$, where $\|h'_{i,i+1}-h_{i,i+1}\|_{\infty}\leq \varepsilon$. Following things are easy to observe.

1) One can easily find a $H,H'$ such that both have unique ground states, and fidelity between these ground states is $(1-\varepsilon)^n$. For example, take $h_{i,i+1} = |0\rangle\langle 0|_i \otimes I_{i+1}$ and $h'_{i,i+1} = |\varepsilon\rangle\langle \varepsilon|_i \otimes I_{i+1}$, where $|\varepsilon\rangle = \sqrt{1-\varepsilon}|0\rangle + \sqrt{\varepsilon}|1\rangle$. Both $H,H'$ are gapped with spectral gap $1$.

2) If $H$ has degenerate ground space, then fidelity between maximally mixed states in ground space can drastically change for any $\varepsilon > 0$. It could be of the order of $2^{-n}$, for any $\varepsilon$. An easy example is when $h_{i,i+1}$ is a zero matrix.

Question: Assuming that $H$ has constant spectral gap (that is, larger than a constant $c$ for any $n$) and unique ground state, can we show that fidelity between ground states of $H$ and $H'$ is at least $(1-f(\varepsilon))^n$, where $f(\varepsilon)\rightarrow 0$ as $\varepsilon\rightarrow 0$? Otherwise, is there a counter-example?

A supporting argument: It is not hard to show that for thermal state $\rho_H(\beta) = \frac{e^{-\beta H}}{Tr(e^{-\beta H})}$, we have $F(\rho_H(\beta), \rho_{H'}(\beta)) \geq 2^{-2\beta\cdot\varepsilon\cdot n}$, where $F(.,.)$ is fidelity. Argument goes via showing that $-\log F(\rho_H(\beta), \rho_{H'}(\beta)) \leq D(\rho_H(\beta), \rho_{H'}(\beta)) \leq 2\beta\cdot\varepsilon\cdot n$, where $D(.,.)$ is relative entropy. Now, if $H$ has constant spectral gap, then constant $\beta$ should be a good approximation to the ground state, suggesting similar property for the ground state. But this argument is far from sufficient, as fidelity between $\rho_H(\beta)$ and ground state of $H$ needs to be exponentially (in $n$) close to $1$, which may hold only when $\beta \approx n$.

Locality is important: In above argument, its important to use locality. Note that $\|H-H'\|_{\infty} \approx \varepsilon\cdot n$, which can be much larger than the spectral gap of $H$. Hence above question clearly has negative answer if $H$ is not a local hamiltonian.

  • $\begingroup$ Maybe you can refer to the work on the uncle Hamiltonian on Matrix product states. There if the MPS state is injective, then the parent Hamiltonian has gapped nondegenerated ground state and it is continuous wrt perturbations. But for non-injective, the parent Hamiltonian has degenerated ground states and perturbations will lead to abrupt change of the ground states. $\endgroup$ – XXDD Oct 2 '18 at 13:06

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