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$\def\braket#1{\langle#1\rangle}$

I am attempting to solve a particular Hamiltonian by variational method. The wavefunction that I have selected is as follows:

$$ \Psi = Ne^{\frac{-kr}{2}}\sum_{i=0}^{m-1}\alpha_i(kr)^i $$

$$ N=\left(\frac{k^3}{\sum_{i,j}^{m-1}\alpha_i\alpha_j(i+j+2)!}\right)^{\frac{1}{2}} $$

Where $N$ is the normalization condition, $k$ and the set of $\alpha$ are my parameters that I wish to vary for some value of $m$. The expectation value for any operator is then:

$$ \braket{O} = |N|^2\sum_{i,j}\alpha_i\alpha_jO_{ij} = |N|^2 \braket{O}' $$

Where $O_{ij}$ is simply the value of the integral over the $i$ and $j$ terms. In order to minimize this operator we are trying to find the following:

$$ argmin_\alpha\frac{\partial\braket{O}}{\partial \alpha} = N\frac{\partial N}{\partial \alpha} \braket{O}' + |N|^2 \frac{\partial \braket{O}'}{\partial \alpha}. $$

Using gradient descent with some non-zero starting condition for the set $\alpha$, I can pretty quickly find solutions. (I have tried random starting conditions, unitary, and a few others, but they all converge to similar results)

As a test of this method, I want to solve a known Hamiltonian, namely the Hydrogen atom. If I use the Hamiltonian for a hydrogen atom, and set $k$ and $m$ to match the known solution ($k=\frac{2Z}{na_0}$,$ m=1$; see eg Griffiths, Wikipedia, Hyperphysics, etc), then I get the expected $-13.6$ eV for the $n=1, \ell=0$ state.

Using the same Hamiltonian, but for larger values of $m$*, I find that the expectation values of my operators don't seem to yield the sensible results that I had expected. In fact, in order to achieve something close the the true eigenvalue I have to scale my Hamiltonian as so:

$$ \braket{H} = (2m-1)\braket{T} + m\braket{V} $$

Where our definitions for these expectation values are as follows:

$$ \braket{T} = -\frac{1}{2}|N|^2\sum_{i,j}^{m-1}\alpha_i\alpha_j(i^2+j(j-1)-i(1+2j)-2)(i+j)!\frac{1}{4k} $$ $$ \braket{V} = -Z|N|^2\sum_{i,j}^{m-1}\alpha_i\alpha_j(i+j+1)!\frac{1}{k^2}. $$

(Note that since we are assuming this is the ground state of Hydrogen, we are taking $\ell=0$).

The scaling factors of $(2m-1)$ and $m$ were discovered through simple trial-and-error for a wide range of $m$, however they achieve almost exactly the same values for $\braket{H}$ for all of the $m$ that I tested (up to $m=12$, as I recall).

Question: Why do I have to use these scaling factors in my equation? Shouldn't the expectation value of the Hamiltonian approach the same value regardless of the value of $m$*?

*Note: I realize that for $m=1$ this is the eigenfunction of the Hydrogen atom, but for $m>1$, as I understand the variational principal, we should get a result that is close to the $m=1$ value once $\alpha$ have been determined.

Edit: It was pointed out that I might be missing a negative sign, however this was an error transposing the equation into SE, my code has the correct sign and the weird factor still remains.

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It seems to me you might have a wrong sign in the variational energy. I was able to reproduce same minimal value $E_0 = -0.25$ of the function $E(\alpha, k) = -T - V$ at $k = 1$, $\alpha_1 = \dots = \alpha_{m-1} = 0$ for different values of m. Here $T$ and $V$ are exactly your expressions.

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  • $\begingroup$ That's a good catch, but unfortunately, that was an error in transposing the equation into SE, my actual code has the correct factors (there are also some factors of hbar, c, m_e etc that I omitted). $\endgroup$ – Dace Jan 11 at 17:49
  • $\begingroup$ Also the issue was that if I start with m>1 the coefficients will tend towards non-zero values and the equation will require scaling of T and V $\endgroup$ – Dace Jan 11 at 17:53

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