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- TL;DR Summary
- Repeated acting on a quantum state, with an operator related to the Hamiltonian, can apparently allow solving of all energy eigenvalues of the system.

The classical "power method" for solving one special eigenvalue of an operator works, in a finite-dimensional vector space, as follows: suppose an operator ##\hat{A}## can be written as an ##n\times n## matrix, and its unknown eigenvectors are (in Dirac bra-ket notation) ##\left|\psi_1 \right.\rangle,\dots\left|\psi_n \right.\rangle##. Now, choose a random trial vector ##\left|\psi_t \right.\rangle##, constructed so that it's unlikely to be exactly orthogonal with any of the eigenvectors ##\left|\psi_k \right.\rangle##. Then when you operate on ##\left|\psi_t \right.\rangle## many times with operator ##\hat{A}##, the result is

##\hat{A}^m \left|\psi_t \right.\rangle = \hat{A}^m \left( c_1 \left|\psi_1 \right.\rangle + \dots + c_n \left|\psi_n \right.\rangle \right) = c_1 a_{1}^{m} \left|\psi_1 \right.\rangle + \dots + c_n a_{n}^m \left|\psi_n \right.\rangle##,

and when ##m## is a large integer, the resulting vector is very nearly parallel with the eigenvector ##\left|\psi_k \right.\rangle## for which the eigenvalue ##a_k## has the largest absolute value ##|a_k |##. Other eigenvalues can be obtained by shifting the operator ##\hat{A}## as ##\hat{A} \rightarrow \hat{A} + \lambda \hat{I}## before this calculation. There the ##\hat{I}## is the identity operator and ##\lambda \in \mathbb{R}##.

As far as I know, using this method is more problematic in an infinite-dimensional vector space, such as when calculating the eigenstates of the Hamiltonian for a quantum mechanical system. For instance, suppose you attempt to find the ground state wave function of a hydrogen atom by choosing a trial function ##\psi_t (x,y,z)## and form the function ##\hat{H}^m \psi_t (x,y,z)##, where ##\hat{H}## is the hydrogenic atom Hamiltonian. In an infinite-dimensional vector space, it's not guaranteed that a vector/state is still normalizable after acting on it with an operator, so this solution scheme doesn't necessarily work properly. This can even still be a problem if you're able to choose the initial function ##\psi_t (x,y,z)## so that it's orthogonal with all scattering state eigenfunctions that have eigenvalues ##E>0##. This problem seems to be connected to the "boundedness" of an operator in Hilbert space, as descibed on pages 453-457 of Sadri Hassani's

A new Arxiv preprint, still not peer-reviewed, claims to describe a way to calculate the ##\it{all}## energy eigenvalues of a quantum system with this computational tool, without the infinite-dimensionality being a problem:

https://arxiv.org/pdf/2211.06303.pdf

Does this seem legit to you others? It seems that this calculation only requires repeated applying of an operator on a state, similar to a matrix-vector multiplication in finite dimensions, which is much less computationally intensive than normal solution of an eigenvalue problem. I haven't been studying every possible way of solving quantum eigenvalue problems recently, to be able to see how large an improvement this is to existing methods.

##\hat{A}^m \left|\psi_t \right.\rangle = \hat{A}^m \left( c_1 \left|\psi_1 \right.\rangle + \dots + c_n \left|\psi_n \right.\rangle \right) = c_1 a_{1}^{m} \left|\psi_1 \right.\rangle + \dots + c_n a_{n}^m \left|\psi_n \right.\rangle##,

and when ##m## is a large integer, the resulting vector is very nearly parallel with the eigenvector ##\left|\psi_k \right.\rangle## for which the eigenvalue ##a_k## has the largest absolute value ##|a_k |##. Other eigenvalues can be obtained by shifting the operator ##\hat{A}## as ##\hat{A} \rightarrow \hat{A} + \lambda \hat{I}## before this calculation. There the ##\hat{I}## is the identity operator and ##\lambda \in \mathbb{R}##.

As far as I know, using this method is more problematic in an infinite-dimensional vector space, such as when calculating the eigenstates of the Hamiltonian for a quantum mechanical system. For instance, suppose you attempt to find the ground state wave function of a hydrogen atom by choosing a trial function ##\psi_t (x,y,z)## and form the function ##\hat{H}^m \psi_t (x,y,z)##, where ##\hat{H}## is the hydrogenic atom Hamiltonian. In an infinite-dimensional vector space, it's not guaranteed that a vector/state is still normalizable after acting on it with an operator, so this solution scheme doesn't necessarily work properly. This can even still be a problem if you're able to choose the initial function ##\psi_t (x,y,z)## so that it's orthogonal with all scattering state eigenfunctions that have eigenvalues ##E>0##. This problem seems to be connected to the "boundedness" of an operator in Hilbert space, as descibed on pages 453-457 of Sadri Hassani's

*Mathematical physics: A Modern Introduction to its Foundations*.A new Arxiv preprint, still not peer-reviewed, claims to describe a way to calculate the ##\it{all}## energy eigenvalues of a quantum system with this computational tool, without the infinite-dimensionality being a problem:

https://arxiv.org/pdf/2211.06303.pdf

##\bf{Abstract}##

We present a new power method to obtain solutions of eigenvalue

problems. The method can determine not only the dominant or lowest eigenvalues

but also all eigenvalues without the need for a deflation procedure. The method uses

a functional of an operator (or a matrix). The method can freely select a solution by

varying a parameter associated to an estimate of the eigenvalue. The convergence of

the method is highly dependent on how closely the parameter to the eigenvalues. In

this paper, numerical results of the method are shown to be in excellent agreement

with the analytical ones.

Does this seem legit to you others? It seems that this calculation only requires repeated applying of an operator on a state, similar to a matrix-vector multiplication in finite dimensions, which is much less computationally intensive than normal solution of an eigenvalue problem. I haven't been studying every possible way of solving quantum eigenvalue problems recently, to be able to see how large an improvement this is to existing methods.

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