# Simple iterative method to obtain any ground state given Hamiltonian operator

So, I'm considering the operator $\frac{h_0}{\bf H}$ with $\bf H$ any hamiltonian operator and $h_0$ a small non-zero positive constant with dimensions of energy and assumed to bound by below all energy eigenvalues

The idea is that the operation of any state vector on this operator is given by

$$\frac{h_0}{\textbf{H}} | \Psi \rangle = \sum_i \frac{h_0}{E_i} c_i | \phi_i \rangle$$

Applying this operator $N$ times gives

$$\bigg(\frac{h_0}{\textbf{H}}\bigg)^N | \Psi \rangle = \sum_i \frac{h_0^N}{E_i^N} c_i | \phi_i \rangle$$

The assertion of my claim is that the limit of this iteration with $N \rightarrow \infty$ is esentially equivalent up to a normalization constant to the ground state of $\textbf{H}$

I've read that obtaining the ground state eigenvector is in general a problem with high computational complexity, so I'm suspicious of the robustness of this claim. Is this valid?

• This is the obverse of Sylvester's largest eigenvalue method. You are assuming all eigenvalues are strictly positive and $h_0$ is strictly equal to the lowest one, to get a nontrivial answer. – Cosmas Zachos Jun 12 '18 at 15:08
• I don't need to assume $h_0$ equals the lowest one, in any case all higher eigenvalues will be attenuated faster. One would have to scale up by some compensating factor on each iteration if one wants to avoid doing a full proper normalization step – lurscher Jun 12 '18 at 15:21
• Well, if $h_0$ is labile, rescaling will eventually identify it with the lowest eigenvalue, or you are projected out. This is the aggressively trivial part. Try a 2x2 example. Barring ground state degeneracy (all but impossible), you have a tautology. – Cosmas Zachos Jun 12 '18 at 15:33
• The reason finding the ground state eigenvector has high computational complexity is because the size of the matrix $\mathbf{H}$ is exponentially large in the number of particles. Your method thus requires you to take the inverse of an exponentially large matrix, and apply it to an exponentially long vector some number of times. This is fairly similar to the projector method, where you apply $(1-\tau \mathbf{H})$ to an arbitrary state over and over again. In either case, you haven't solved the problem that $\mathbf{H}$ is large. – Jahan Claes Jun 12 '18 at 16:05
• ok, that makes sense – lurscher Jun 12 '18 at 16:09

This claim is obviously false, and I have no idea why the OP suspects this could work. Suppose that the spectrum of $H$ satisfies $\inf \sigma(H)=\delta>0$. Now take $h_0<\delta$ (this is always possible, and such choice satisfies OP's assumptions).
In addition, suppose that $H$ has at least one ground state, and take $\Psi_\delta$ to be a ground state of $H$. Then $$\langle\Phi,h_0^N H^{-N}\Psi_\delta\rangle=\frac{h_0^N}{\delta^N}\langle\Phi,\Psi_ \delta\rangle=\epsilon^N\langle\Phi,\Psi_\delta\rangle\; ,$$ where $\epsilon=\frac{h_0}{\delta}<1$. Hence $\lim_{N\to \infty} \epsilon^N=0$. This means that $h_0^N H^{-N}\Psi_\delta$ converges weakly to zero in the Hilbert space. If it also converges strongly (and it does), it must converge to zero as well.
However, any (normalized) ground state cannot be zero (by definition), so $h_0^N H^{-N}\Psi_\delta$ does not converge to the ground state in general. The same reasoning applies also taking $h_0=\delta$, and choosing $\Psi$ to be any eigenstate for an eigenvalue bigger than the ground state.
• @lurscher Well, what such "normalization factor" should be? I am pretty sure that there is no general choice of normalization factor, even if you make it depend on $h_0$, $H$ and also $N$, that would give you that the claim is true for any $\Psi$. – yuggib Jun 12 '18 at 15:53
• @lurscher I can actually even prove that: take a $\Psi$ that is an eigenvector for an eigenvalue $\lambda>\delta$. Then, $\Psi$ is orthogonal to every ground state $\Psi_\delta$, and its whole linear span is orthogonal to any $\Psi_\delta$. Now, for any $\alpha(h_0,H,N)\in \mathbb{C}$, the vector $\alpha(h_0,H,N)h_0^NH^{-N}\Psi$ is in the linear span of $\Psi$ since $\Psi$ is an eigenvalue. Therefore, its limit for $N\to\infty$, provided it exists, would belong to the linear span of $\Psi$. Therefore, it is never a ground state for $H$ (it is actually orthogonal to any ground state). – yuggib Jun 12 '18 at 15:59
• right, that would make the $c_i$ factors be basically $\delta_{i \lambda}$. Unless you make a very bad choice for the initial $\Psi$ you should have no problem – lurscher Jun 12 '18 at 16:01