Hartree-Fock for Helium excited state II

This times, we will show the energy level diagram for helium atom. We already calculated the ground state, the 1s2s singlet and triplet excited states. We will calculate higher excited states, for example, 1s3s, 1s2p or 1s3p, etc, and included in the diagram.

The most difficult part is the evaluation of the mutual interaction matrix element

$\displaystyle G_{ij}^{hk} = \langle b_i(x) b_h(y) | \frac{1}{r_{xy}} | b_j(x) b_k(y) \rangle$

The angular integral was evaluated in the last post. And the radial integral is done using Mathematica.

I will use hydrogen 1s, 2s, 2p, 3s, 3p, 4s, and 4p states for basis. Thus, the diagram will contain some of the excited states from some possible combination. The 1s4s and 1s4p state will not be calculated, because there is no room for mixing with higher excited states but only for lower excited states. This will make the eigen state be unbound.

During the calculation, one of the tricky point is the identify of the n-th s-state. It is because the mixing is always there, and the mixing can be very large. In order to determine the principle quantum number, we have to check the wave function can see how many peaks in there.

Here is the energy level diagram obtained by Hartree-Fock method using limited hydrogen wave functions as basis set.

Evaluation of mutual interaction

We are going to evaluation the integral

$\displaystyle G_{ij}^{hk} = \langle b_i(x) b_h(y) | \frac{1}{r_{xy}} | b_j(x) b_k(y) \rangle$

Recalling the multi-pole expansion,

$\displaystyle \frac{1}{r_{12}} = \sum_{l=0}^{\infty} \frac{4 \pi}{2l+1} \frac{r_{<}^l}{r_{>}^{l+1}} \sum_{m=-l}^{l} Y_{lm}^{*}(\Omega_1) Y_{lm}(\Omega_2)$

and the basis

$b_{nlm}(\vec{r}) = R_{nl}(r) Y_{lm}(\Omega)$

Set an averaged basis

$\displaystyle b_{nl}(\vec{r}) = R_{nl}(r) \frac{1}{\sqrt{2l+1}} \sum_{m=-1}^{l}Y_{lm}(\Omega)$

$\displaystyle \Gamma_{ij}^{hk}(l) = \frac{4 \pi}{2l+1} \sum_{m=-l}^{l} \frac{1}{\sqrt{(2l_i+1)(2l_j+1)(2l_h+1)(2l_k+1)}} \\ \sum_{m_i,m_j}\int Y_{l_i m_i}^{*}(\Omega_1) Y_{l m}^{*}(\Omega_1) Y_{l_j m_j}(\Omega_1) d \Omega_1 \\ \sum_{m_h,m_k} \int Y_{l_h m_k}^{*}(\Omega_2) Y_{l m}(\Omega_2) Y_{l_k m_k}(\Omega_2) d \Omega_2$

In the angular integrals, using Wigner 3-j symbol and the integral

$\displaystyle\int Y_{l_1 m_1}(\Omega) Y_{l m}(\Omega) Y_{l_2 m_2}(\Omega) d \Omega \\= \frac{\sqrt{(2l_1+1)(2l+1)(2l_2+1)}}{\sqrt{4\pi}} \begin{pmatrix} l_1 & l & l_2 \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} l_1 & l & l_2 \\ m_1 & m & m_2 \end{pmatrix}$

$Y_{lm}^{*}(\Omega) = (-1)^m Y_{l(-m)}(\Omega)$

$\displaystyle \begin{pmatrix} l_1 & l & l_2 \\ m_1 & m & m_2 \end{pmatrix} = (-1)^{l_1+l+l_2} \begin{pmatrix} l_1 & l & l_2 \\ -m_1 & -m & -m_2 \end{pmatrix} \\ =(-1)^{l_1+l+l_2} \begin{pmatrix} l_1 & l_2 & l \\ m_1 & m_2 & m \end{pmatrix}$

we have

$\displaystyle\int Y_{l_i m_i}^{*}(\Omega) Y_{l m}^{*}(\Omega) Y_{l_j m_j}(\Omega) d \Omega \\= (-1)^{m_j} \frac{\sqrt{(2l_i+1)(2l+1)(2l_j+1)}}{\sqrt{4\pi}} \begin{pmatrix} l_i & l & l_j \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} l_i & l & l_j \\ m_i & m & -m_j \end{pmatrix}$

$\displaystyle\int Y_{l_h m_h}^{*}(\Omega) Y_{l m}(\Omega) Y_{l_k m_k}(\Omega) d \Omega \\= (-1)^{m_h} \frac{\sqrt{(2l_h+1)(2l+1)(2l_k+1)}}{\sqrt{4\pi}} \begin{pmatrix} l_h & l & l_k \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} l_h & l & l_k \\ -m_h & m & m_k \end{pmatrix}$

Thus,

$\displaystyle \Gamma_{ij}^{hk}(l) = \sum_{m=-l}^{l} \sum_{m_i,m_j } (-1)^{m_j} \begin{pmatrix} l_i & l_j & l \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} l_i & l_j & l \\ m_i & -m_j & m \end{pmatrix} \\ \sum_{m_h,m_k} (-1)^{m_h} \begin{pmatrix} l_h & l_k & l \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} l_h & l_k & l \\ -m_h & m_k & m \end{pmatrix}$

The 3-j symbol restricted that

$l_i+l_j+l = even$

$l_h+l_k+l = even$

$m_i-m_j + m = 0, |m_i| \leq l_i , |m_j| \leq l_j$

$-m_h + m_k +m = 0, |m_h| \leq l_h , |m_k| \leq l_k$

I guess it is the most simplified formula for the angular part.

The total integral is

$\displaystyle G_{ij}^{hk} = \sum_{l=0}^\infty \Gamma_{ij}^{hk}(l) \langle R_i(x) R_h(y) | \frac{r_<^l}{r_>^{l+1}} |R_j(x) R_k(y) \rangle$

The angular integral imposes condition for $l$.

I am not sure this is a correct way to treat the problem.

First, the averaged basis is still an energy eigen state. It is not the eigen state for the angular part. So, this averaging could introduces an error and we should reminded that this is an approximation. But in the perturbation view point, this averaged basis is still valid.

Second thing is, the sum

$\displaystyle P_{(l_i m_i) (l_j m_j)}^{(l_h m_h)(l_k m_k)}(l) = \sum_{m=-l}^{l} \langle Y_{l_i m_i}|Y_{lm}^*|Y_{l_j m_j}\rangle \langle Y_{l_h m_h}|Y_{lm}|Y_{l_k m_k}\rangle$

is not symmetry for exchange of $i, j, h,k$ in general. For example,

$\displaystyle \frac{-1}{3}=P_{(1-1) (00)}^{(11)(00)}(1) {\neq} P_{(00)(1-1)}^{(11)(00)}(1) = 0$

This is a very uprising result that the mutual interaction dependent on the magnetic quantum number. Thus, in detail, we should use $n l m m_s$ as a basis.

Third, the sum $P_{ij}^{hk}(l)$ is depend on $l$. The mutual interaction require us to sum all possible $l$.

Fourth, the coupling between 1s2p triplet state, the total spin is $S = 1$, total L is $L = 1$, and the total angular momentum can be  $J = 0, 1, 2$. In our treatment, we did not coupled the angular momentum in the calculation explicitly. In fact, in the integral of the spherical harmonic, the coordinate are integrated separately, and the coupling seem to be calculated implicitly. I am not sure how to couple two spherical harmonics with two coordinates.

Using generating function to calculate integrals of Laguerre polynomial

The generating function of Laguerre polynomial is

$\displaystyle \sum_n^\infty L_n^\alpha(x) t^n = \frac{1}{(1-t)^{\alpha+1}} \exp \left(-\frac{tx}{1-t}\right) =G_\alpha(t, x)$

We are going to evaluate the integral

$\displaystyle \int_{0}^{\infty} x^{\alpha +k} e^{-x} L_m^{\beta}(x) L_n^{\alpha}(x) dx$

Consider this integral

$\displaystyle \int_0^\infty x^{\alpha+k} e^{-x} G_\alpha(t,x) G_{\beta}(s,x) dx$

It is equal to

$\displaystyle \int_0^\infty x^{\alpha+k} e^{-x} \sum_{n,m}^{\infty} L_n^\alpha(x) L_m^\beta(x) t^n s^m dx \\ = \sum_{n,m}^{\infty} t^n s^m \int_0^\infty x^{\alpha+k} e^{-x} L_n^\alpha(x) L_m^\beta(x) dx \\ = \frac{1}{(1-t)^{\alpha+1}} \frac{1}{(1-s)^{\beta+1}} \int_0^\infty x^{\alpha+k} e^{-x} e^{-\frac{t x}{1-t} - \frac{s x}{1-s}} dx$

It is easy to calculate the last integral,

$\displaystyle \int_0^\infty x^{\alpha+k} e^{-x} e^{-\frac{t x}{1-t} - \frac{s x}{1-s}} dx = \int_0^\infty x^{\alpha+k} \exp \left(-\frac{1- t s}{(1-t)(1-s)}x \right) dx$

Replace variable,

$\displaystyle \int_0^\infty x^{\alpha+k} e^{-x} e^{-\frac{t x}{1-t} - \frac{s x}{1-s}} dx \\ = \frac{(1-t)^{(\alpha+1+k)}(1-s)^{(\alpha+1+k)}}{(1-t s)^{(\alpha+1+k)}}\int_0^\infty y^{\alpha+k} e^{-y} dy$

The integral becomes

$\displaystyle \sum_{n,m}^{\infty} t^n s^m \int_0^\infty x^{\alpha+k} e^{-x} L_n^\alpha(x) L_m^\beta(x) dx = \frac{(1-t)^{k}(1-s)^{(\alpha-\beta+k)}}{(1-t s)^{(\alpha+1+k)}} (\alpha+k)!$

Expand the right hand side,

$\displaystyle (1-t)^k = \sum_i^k (-1)^i \begin{pmatrix} k \\ i \end{pmatrix} t^i$

$\displaystyle (1-s)^{(\alpha-\beta+k)} = \sum_i^{\alpha-\beta+k} (-1)^i \begin{pmatrix} \alpha-\beta+k \\ i \end{pmatrix} s^i$

$\displaystyle (1-ts)^{-(\alpha+1+k)} = \sum_i^\infty (-1)^i \begin{pmatrix} -(\alpha+k+1) \\ i \end{pmatrix} (t s)^i = \sum_i^\infty \begin{pmatrix} \alpha+k+i\\ i \end{pmatrix} (t s)^i$

We used

$\displaystyle \begin{pmatrix} m \\ n \end{pmatrix} = \prod_{i=0}^{n-1} \frac{m-i}{n-i}$

$\displaystyle (-1)^n\begin{pmatrix} -m \\ n \end{pmatrix} = \prod_{i=0}^{n-1} \frac{m+i}{n-i} = \frac{(m+n-1)!}{n!(m-1)!}$

At the last step. The product

$\displaystyle \frac{(1-t)^{k}(1-s)^{(\alpha-\beta+k)}}{(1-t s)^{(\alpha+1+k)}} \\ = \sum_a^k \sum_{b}^{\alpha-\beta+k} \sum_c^\infty (-1)^{(a+b)} \begin{pmatrix} k \\ a \end{pmatrix} \begin{pmatrix} \alpha-\beta+k \\ b \end{pmatrix} \begin{pmatrix}\alpha+k+c \\ c \end{pmatrix} t^{(a+c)} s^{(b+c)}$

By comparing the order of $t^n s^m$, the integral can be evaluated. With $a+c = n, b+c = m$

$\displaystyle \int_0^\infty x^{\alpha+k} e^{-x} L_n^\alpha(x) L_m^\beta(x) dx \\ = (\alpha+k)! \sum_a^k \sum_{b}^{\alpha-\beta+k} \sum_c^\infty (-1)^{(a+b)} \begin{pmatrix} k \\ a \end{pmatrix} \begin{pmatrix} \alpha-\beta+k \\ b \end{pmatrix} \begin{pmatrix} \alpha+k+c \\ c \end{pmatrix}$

When $\alpha = \beta$

$\displaystyle \int_0^\infty x^{\alpha+k} e^{-x} L_n^\alpha(x) L_m^\alpha(x) dx \\ = (\alpha+k)! \sum_a^k \sum_{b}^{k} \sum_c^\infty (-1)^{(a+b)} \begin{pmatrix} k \\ a \end{pmatrix} \begin{pmatrix} k \\ b \end{pmatrix} \begin{pmatrix} \alpha+k+c \\ c \end{pmatrix}$

For example, $k = 0$,

$\displaystyle \frac{(1-s)^{(\alpha-\beta)}}{(1-t s)^{(\alpha+1)}} = \sum_{b}^{\alpha-\beta} \sum_c^\infty (-1)^{b} \begin{pmatrix} \alpha-\beta \\ b \end{pmatrix} \begin{pmatrix} \alpha+c \\ c \end{pmatrix} t^{c} s^{(b+c)}$

When $\alpha \neq \beta, , n= m = 1$, only $b = 0, c = 1$

$\displaystyle \int_0^\infty x^{\alpha} e^{-x} L_1^\alpha(x) L_1^\beta(x) dx = (\alpha)! \begin{pmatrix} \alpha-\beta \\ 0 \end{pmatrix} \begin{pmatrix} \alpha+1 \\ 1 \end{pmatrix} = (\alpha+1)!$

When $\alpha = \beta, , n = m$, only $b = 0, c = n$

$\displaystyle \int_0^\infty x^{\alpha} e^{-x} (L_n^\alpha(x))^2 dx = (\alpha)! \begin{pmatrix} \alpha+n \\ n \end{pmatrix} = \frac{(\alpha+n)!}{n!}$

For $k = 1, \alpha = \beta, n= m$, there are two terms, $a = b = 0, c = n$  and $a = b = 1, c = n -1$

$\displaystyle \int_0^\infty x^{\alpha+1} e^{-x} (L_n^\alpha(x))^2 \\ = (\alpha+1)! \left(\begin{pmatrix} \alpha+n+1 \\ n \end{pmatrix} + \begin{pmatrix} \alpha+n \\ n-1 \end{pmatrix} \right) = \frac{(\alpha+n)!}{n!}(2n+\alpha+1)$

We recovered all formula. Using generating function, we don’t have to worry about the condition of $n, m, \alpha, \beta, k$. We only need to fulfill the sum with condition $a +c = n, b+c = m$ and $a, b, c \geq 0$.

Hartree method for Helium ground state

After long preparation, I am ready to do this problem.

The two electron in the helium ground state occupy same spacial orbital but difference spin. Thus, the total wavefunction is

$\displaystyle \Psi(x,y) = \frac{1}{\sqrt{2}}(\uparrow \downarrow - \downarrow \uparrow) \psi(x) \psi(y)$

Since the Coulomb potential is spin-independent, the Hartree-Fock method reduce to Hartree method. The Hartree operator is

$F(x) = H(x) + \langle \psi(y)|G(x,y) |\psi(y) \rangle$

where the single-particle Hamiltonian and mutual interaction are

$\displaystyle H(x) = -\frac{\hbar^2}{2m} \nabla^2 - \frac{Ze^2}{4\pi\epsilon_0 x} = -\frac{1}{2}\nabla^2 - \frac{Z}{x}$

$\displaystyle G(x,y) = \frac{e^2}{4\pi\epsilon_0|x-y|} = \frac{1}{|x-y|}$

In the last step, we use atomic unit, such that $\hbar = 1, m=1, e^2 = 4\pi\epsilon_0$. And the energy is in unit of Hartree, $1 \textrm{H} = 27.2114 \textrm{eV}$.

We are going to use Hydrogen-like orbital as a basis set.

$\displaystyle b_i(r) = R_{nl}(r)Y_{lm}(\Omega) \\= \sqrt{\frac{(n-l-1)!Z}{n^2(n+l)!}}e^{-\frac{Z}{n}r} \left( \frac{2Z}{n}r \right)^{l+1} L_{n-l-1}^{2l+1}\left( \frac{2Z}{n} r \right) \frac{1}{r} Y_{lm}(\Omega)$

I like the left the $1/r$, because in the integration $\int b^2 r^2 dr$, the $r^2$ can be cancelled. Also, the $i = nlm$ is a compact index of the orbital.

Using basis set expansion, we need to calculate the matrix elements of

$\displaystyle H_{ij}=\langle b_i(x) |H(x)|b_j(x)\rangle = -\delta \frac{Z^2}{2n^2}$

$\displaystyle G_{ij}^{hk} = \langle b_i(x) b_h(y) |G(x,y) |b_j(x) b_k(y) \rangle$

Now, we will concentrate on evaluate the mutual interaction integral.

Using the well-known expansion,

$\displaystyle G(x,y) = \frac{1}{|x-y|}=\frac{1}{r_{12}} = \sum_{l=0}^{\infty} \sum_{m=-l}^{l} \frac{4\pi}{2l+1} \frac{r_<^l}{r_>^{l+1}} Y_{lm}^{*}(\Omega_1)Y_{lm}(\Omega_2)$

The angular integral

$\displaystyle \langle Y_i(x) Y_h(y)| Y_{lm}^{*}(x) Y_{lm}(y) | Y_j(x) Y_k(y) \rangle \\ = \big( \int Y_i^{*}(x) Y_{lm}^{*}(x) Y_j(x) dx \big) \big( \int Y_h^{*}(y) Y_{lm}(y) Y_k(y) dy \big)$

where the integral $\int dx = \int_{0}^{\pi} \int_{0}^{2\pi} \sin(\theta_x) d\theta_x d\phi_x$.

From this post, the triplet integral of spherical harmonic is easy to compute.

$\displaystyle \int Y_h^{*}(y) Y_{lm}(y) Y_k(y) dy = \sqrt{\frac{(2l+1)(2l_k+1)}{4\pi (2l_h+1)}} C_{l0l_k0}^{l_h0} C_{lm l_km_k}^{l_hm_h}$

The Clebsch-Gordon coefficient imposed a restriction on $l,m$.

$\displaystyle \langle R_i(x) R_h(y)| \frac{r_<^l}{r_>^{l+1}} | R_j(x) R_k(y) \rangle \\ = \int_0^{\infty} \int_{0}^{\infty} R_i(x) R_h(y) \frac{r_<^l}{r_>^{l+1}} R_j(x) R_k(y) y^2 x^2 dy dx \\ = \int_0^{\infty} R_i(x) R_j(x) \\ \left( \int_{0}^{x} R_h(y) R_k(y) \frac{y^l}{x^{l+1}} y^2dy + \int_{x}^{\infty} R_h(x)R_k(x) \frac{x^l}{y^{l+1}} y^2 dy \right) x^2 dx$

The algebraic calculation of the integral is complicated, but after the restriction of $l$ from the Clebsch-Gordon coefficient, only few terms need to be calculated.

The general consideration is done. now, we use the first 2 even states as a basis set.

$\displaystyle b_{1s}(r) = R_{10}(r)Y_{00}(\Omega) = 2Z^{3/2}e^{-Zr}Y_{00}(\Omega)$

$\displaystyle b_{2s}(r) = R_{20}(r)Y_{00}(\Omega) = \frac{1}{\sqrt{8}}Z^{3/2}(2-Zr)e^{-Zr/2}Y_{00}(\Omega)$

These are both s-state orbital. Thus, the Clebsch-Gordon coefficient

$\displaystyle C_{lm l_k m_k}^{l_h m_h} = C_{lm00}^{00}$

The radial sum only has 1 term. And the mutual interaction becomes

$\displaystyle G(x,y) = \frac{1}{|x-y|}=\frac{1}{r_{12}} = 4\pi \frac{1}{r_>} Y_{00}^{*}(\Omega_1)Y_{00}(\Omega_2)$

The angular part

$\displaystyle \langle Y_i(x) Y_h(y)| Y_{lm}^{*}(x) Y_{lm}(y) | Y_j(x) Y_k(y) \rangle = \frac{1}{4\pi}$

Thus, the mutual interaction energy is

$G_{ij}^{hk} = \displaystyle \langle b_i(x) b_h(y) |G(x,y) |b_j(x) b_k(y) \rangle = \langle R_i(x) R_h(y)| \frac{1}{r_>} | R_j(x) R_k(y) \rangle$

$G_{ij}^{hk} = \displaystyle \langle R_i(x) R_h(y)| \frac{1}{r_>} | R_j(x) R_k(y) \rangle \\ \begin{pmatrix} G_{11}^{hk} & G_{12}^{hk} \\ G_{21}^{hk} & G_{22}^{hk} \end{pmatrix} = \begin{pmatrix} \begin{pmatrix} G_{11}^{11} & G_{11}^{12} \\ G_{11}^{21} & G_{11}^{22} \end{pmatrix} & \begin{pmatrix} G_{12}^{11} & G_{12}^{12} \\ G_{12}^{21} & G_{12}^{22} \end{pmatrix} \\ \begin{pmatrix} G_{21}^{11} & G_{21}^{12} \\ G_{21}^{21} & G_{21}^{22} \end{pmatrix} & \begin{pmatrix} G_{22}^{11} & G_{22}^{12} \\ G_{22}^{21} & G_{22}^{22} \end{pmatrix} \end{pmatrix} \\= \begin{pmatrix} \begin{pmatrix} 1.25 & 0.17871 \\ 0.17871 & 0.419753 \end{pmatrix} & \begin{pmatrix} 0.17871 & 0.0438957 \\ 0.0439857 & 0.0171633 \end{pmatrix} \\ \begin{pmatrix} 0.17871 & 0.0438957 \\ 0.0438957 & 0.0171633 \end{pmatrix} & \begin{pmatrix} 0.419753 & 0.0171633 \\ 0.0171633 & 0.300781 \end{pmatrix} \end{pmatrix}$

We can easy to see that $G_{ij}^{hk} = G_{ji}^{hk} = G_{ij}^{kh} = G_{hk}^{ij} = G_{ji}^{kh}$. Thus, if we flatten the matrix of matrix, it is Hermitian, or symmetric.

Now, we can start doing the Hartree method.

The general solution of the wave function is

$\psi(x) = a_1 b_{1s}(x) + a_2 b_{2s}(x)$

The Hartree matrix is

$F_{ij} = H_{ij} + \sum_{h,k} a_h a_k G_{ij}^{hk}$

The first trial wave function are the Hydrogen-like orbital,

$\psi^{(0)}(x) = b_{1s}(r)$

$F_{ij}^{(0)} = \begin{pmatrix} -2 & 0 \\ 0 & -0.5 \end{pmatrix} + \begin{pmatrix} 1.25 & 0.17871 \\ 0.17817 & 0.419753 \end{pmatrix}$

Solve for eigen system, we have the energy after 1st trial,

$\epsilon^{(1)} = -0.794702 , (a_1^{(1)}, a_2^{(1)}) = (-0.970112, 0.242659)$

After 13th trial,

$\epsilon^{(13)} = -0.880049 , (a_1^{(13)}, a_2^{(13)}) = (-0.981015, 0.193931)$

$F_{ij}^{(13)} = \begin{pmatrix} -2 & 0 \\ 0 & -0.5 \end{pmatrix} + \begin{pmatrix} 1.15078 & 0.155932 \\ 0.155932 & 0.408748 \end{pmatrix}$

Thus, the mixing of the 2s state is only 3.7%.

Since the eigen energy contains the 1-body energy and 2-body energy. So, the total energy for 2 electrons is

$E_2 = 2 * \epsilon^{(13)} - G = -2.82364 \textrm{H} = -76.835 \textrm{eV}$

In which ,

$G = \langle \psi(x) \psi(y) |G(x,y) |\psi(x) \psi(y) \rangle = 1.06354 \textrm{H} = 28.9403 \textrm{eV}$

So the energies for

From He to He++.  $E_2 = -2.82364 \textrm{H} = -76.835 \textrm{eV}$
From He+ to He++, $E_1^+ = -Z^2/2 = 2 \textrm{H} = -54.422 \textrm{eV}$.
From He to He+, is $E_1 = E_2 - E_1^+ = -0.823635 \textrm{H} = -22.4123 \textrm{eV}$

The experimental 1 electron ionization energy for Helium atom is

$E_1(exp) = -0.90357 \textrm{H} = -24.587 \textrm{eV}$
$E_1^+(exp) = -1.99982 \textrm{H} = -54.418 \textrm{eV}$
$E_2(exp) = -2.90339 \textrm{H} = -79.005 \textrm{eV}$

The difference with experimental value is 2.175 eV. The following plot shows the Coulomb potential, the screening due to the existence of the other electron, the resultant mean field, the energy, and $r \psi(x)$

Usually, the Hartree method will under estimate the energy, because it neglected the correlation, for example, pairing and spin dependence. In our calculation, the $E_2$ energy is under estimated.

From the $(a_1^{(13)}, a_2^{(13)}) = (-0.981015, 0.193931)$, we can see, the mutual interaction between 1s and 2s state is attractive. While the interaction between 1s-1s and 2s-2s states are repulsive. The repulsive can be easily understood. But I am not sure how to explain the attractive between 1s-2s state.

Since the mass correction and the fine structure correction is in order of $10^{-3} \textrm{eV}$, so the missing 0.2 eV should be due to something else, for example, the incomplete basis set.

If the basis set only contain the 1s orbit, the mutual interaction is 1.25 Hartree = 34.014 eV. Thus, the mixing reduce the interaction by 5.07 eV, just for 3.7% mixing

I included the 3s state,

$\epsilon^{(13)} = -0.888475 , (a_1^{(13)}, a_2^{(13)}, a_3^{(13)}) = (0.981096, -0.181995, -0.06579)$

The mutual energy is further reduced to 1.05415 Hartree = 28.6848 eV. The $E_2 = -77.038 \textrm{eV}$. If 4s orbital included, the $E_2 = -77.1058 \textrm{eV}$. We can expect, if more orbital in included, the $E_2$ will approach to $E_2(exp)$.

Integrations of Laguerre polynomial

Since the Laguerre polynomial is deeply connected to the hydrogen-like electron orbital. The radial solution is

$\displaystyle R_{nl}(r) = A \frac{1}{r} \exp(- \frac{x}{2}) x^{l+1}L_{n-l-1}^{2l+1}(x)$

$\displaystyle x = \frac{2Z}{na_0} r$

From the normalization condition, we can get the coefficient A

$\displaystyle \int_0^{\infty} R_{nl}^2(r) r^2 dr$

Beside of calculating the normalization factor, the general expectation value also need to evaluation of the integration of the Laguerre polynomial.

In here, we will show the calculation for 3 integrals:

$\displaystyle \int_0^{\infty} x^{\alpha+m} e^{-x} L_n^\alpha(x) dx, n,\alpha\geq0$

$\displaystyle \int_0^{\infty} x^{\alpha+k} e^{-x} (L_n^\alpha(x))^2 dx$

$\displaystyle \int_0^{\infty} x^{\alpha +k} e^{-x} L_m^{\beta}(x) L_n^\alpha(x) dx$

Using the Rodrigues formula

$\displaystyle L_n^\alpha(x) = \frac{1}{n!} x^{-\alpha} e^x \frac{d^n}{dx^n}(e^{-x}x^{n+\alpha})$

$\displaystyle \int_0^{\infty} x^{\alpha+m} e^{-x} L_n^\alpha(x) dx = \frac{1}{n!} \int_0^{\infty} x^{m} \frac{d^n}{dx^n}(e^{-x}x^{n+\alpha})dx$

For $m=0$,

$\displaystyle \int_0^{\infty} x^{\alpha+m} e^{-x} L_n^\alpha(x) dx \\ = \frac{1}{n!} \int_0^{\infty} \frac{d^n}{dx^n}(e^{-x}x^{n+\alpha})dx \\ = \frac{1}{n!} (F_{n-1}(\infty)-F_{n-1}(0) )$

$\displaystyle F_{n-1}(x) = \int_0^{\infty} \frac{d^n}{dx^n}(e^{-x}x^{n+\alpha})dx \\ = \sum_{i=0}^{n-1} (-1)^i \begin{pmatrix} n-1 \\ i \end{pmatrix} \frac{(n+\alpha)!}{(\alpha+1+i)!} x^{\alpha+1+i} e^{-x}$

It is obvious that $F(\infty) = 0$ due to the $e^{-x}$. Since $\alpha>0$, then $x^{\alpha+1+i} = 0$ at $x = 0$. Thus,

$\displaystyle \int_0^{\infty} x^{\alpha+m} e^{-x} L_n^\alpha(x) dx = 0 , m=0$

Using integration by path,

$\displaystyle \int_0^{\infty} x^{m} d \left(\frac{d^{n-1}}{dx^{n-1}}(e^{-x}x^{n+\alpha}) \right)dx \\ = [ x^{m} \frac{d^{n-1}}{dx^{n-1}}(e^{-x}x^{n+\alpha}) ]_0^{\infty} - (m) \int_0^{\infty} x^{m-1} \frac{d^{n-1}}{dx^{n-1}}(e^{-x}x^{n+\alpha})dx$

For the same reason, the first term is zero for $\alpha + m \geq 0$. Repeat the integration by path $k$ times,

$\displaystyle \int_0^{\infty} x^{m} d \left(\frac{d^{n-1}}{dx^{n-1}}(e^{-x}x^{n+\alpha}) \right)dx \\ = (-1)^k \frac{(m)!}{(m-k)!} \int_0^{\infty} x^{m-k} \frac{d^{n-k}}{dx^{n-k}}(e^{-x}x^{n+\alpha})dx$

When $m \geq n || 0 > m$, $k$ can go to $n$, then,

$\displaystyle \int_0^{\infty} x^{m} d \left(\frac{d^{n-1}}{dx^{n-1}}(e^{-x}x^{n+\alpha}) \right)dx \\ = (-1)^{n} \frac{(m)!}{(m-n)!} \int_0^{\infty} x^{m-n}(e^{-x}x^{n+\alpha})dx \\ = (-1)^{n} \frac{(m)!}{(m-n)!} (\alpha+m)!$

When $n > m > 0$, $k$ can only go to $m$. then,

$\displaystyle \int_0^{\infty} x^{m} d \left(\frac{d^{n-1}}{dx^{n-1}}(e^{-x}x^{n+\alpha}) \right)dx \\ = (-1)^{m} (m)! \int_0^{\infty} \frac{d^{n-k}}{dx^{n-k}}(e^{-x}x^{n+\alpha})dx \\ = (-1)^{m} (m)! \frac{d^{n-m-1}}{dx^{n-m-1}}(e^{-x}x^{n+\alpha}) = 0$

For the similar result as $F_{n-1}(x)$, the result is zero.

To summarize, the following formula suitable for all $m > n || 0>m$

$\displaystyle \int_0^{\infty} x^{\alpha+m} e^{-x} L_n^\alpha(x) dx = (-1)^n (\alpha+m)! \begin{pmatrix} m \\ n \end{pmatrix}$

To evaluate the integral,

$\displaystyle \int_0^{\infty} x^{\alpha+k} e^{-x} (L_n^\alpha(x))^2 dx$

We need the know that the Laguerre polynomial can be expressed as,

$\displaystyle L_n^\alpha(x) = \sum_{i=0}^{n} (-1)^i \begin{pmatrix} n+\alpha \\ n-i \end{pmatrix} \frac{x^i}{i!}$

Thus, we can evaluate

$\displaystyle \int_0^{\infty} x^{\alpha+k} e^{-x} x^i L_n^\alpha(x) dx$

Then combine everything,

$\displaystyle \int_0^{\infty} x^{\alpha+k+i} e^{-x} L_n^\alpha(x) dx = (-1)^n (\alpha+i+k)! \begin{pmatrix} i+k \\ n \end{pmatrix}$

put inside the series expression,

$\displaystyle \int_0^{\infty} x^{\alpha+k} e^{-x} (L_n^\alpha(x))^2 dx = \sum_{i=0}^{n} (-1)^{i+n} \begin{pmatrix} n+\alpha \\ n-i \end{pmatrix} \begin{pmatrix} i+k \\ n \end{pmatrix} \frac{(\alpha+k+i)!}{i!}$

In general, $i + k \geq n || 0 > i+k$.

For $k = 0$, the sum only has 1 term for $i = n$

$\displaystyle \int_0^{\infty} x^{\alpha} e^{-x} (L_n^\alpha(x))^2 dx = \frac{(\alpha+n)!}{n!}$

Using the formula, for $k=1$, we have

$\displaystyle \int_0^{\infty} x^{\alpha+1} e^{-x} (L_n^\alpha(x))^2 dx = \frac{(\alpha+n)!}{n!} (2n+\alpha +1)$

To evaluate the expectation value of radial function

$\displaystyle \langle R_{nl} | \frac{1}{r^m} | R_{nl} \rangle$

We have to calculate the integral,

$\displaystyle \int_0^\infty e^{-x} x^{2l+2-m} \left( L_{n-l-1}^{2l+1} (x) \right)^2 dx$

For $m = 0$, we get the normalization constant,

$\displaystyle A^2 \frac{na_0}{2Z} \int_0^{\infty} e^{-x} x^{2l+2} \left( L_{n-l-1}^{2l+1}(x) \right)^2 dx = 1$

$\displaystyle \langle \frac{1}{r} \rangle = \frac{1}{n^2} \frac{Z}{a_0}$

$\displaystyle \langle \frac{1}{r^2} \rangle = \frac{2}{n^3(2l+1)} \frac{Z^2}{a_0^2}$

In general,

$\displaystyle \langle \frac{1}{r^m} \rangle = \left(\frac{Z}{a_0}\right)^m \frac{2^{m-1}(n-l-1)!}{n^{m+1}(n+l)!} G_{nlm}$

$\displaystyle G_{nlm} = \sum_{i=0}^{n-l-1} (-1)^{n-l-1+i} \begin{pmatrix} n+l \\ n-l-1+i \end{pmatrix} \begin{pmatrix} 1-m+i \\ n-l-1 \end{pmatrix} \frac{(2l+2-m+i)!}{i!}$

For the integral, for $m \neq n$

$\displaystyle \int_0^{\infty} x^{\alpha +k} e^{-x} L_m^{\beta}(x) L_n^\alpha(x) dx$

When $k = 0$, when $m < n$, the integral is zero.

When $k \neq 0, m

$\displaystyle \int_0^{\infty} x^{\alpha +k} e^{-x} L_m^{\beta}(x) L_n^\alpha(x) dx \\ = \sum_{i=0}^{m} \frac{(-1)^i}{i!}\begin{pmatrix} m+\beta \\ m-i \end{pmatrix} \int_0^{\infty} x^{\alpha+k+i} e^{-x} L_n^\alpha(x) dx \\ = \sum_{i=0}^{m} \frac{(-1)^{i+n}}{i!}\begin{pmatrix} m+\beta \\ m-i \end{pmatrix} \begin{pmatrix} k+i \\ n \end{pmatrix} (\alpha+k+i)!$

The sum only non-zero when $m\geq i > n-k$ or $-k > i \geq 0$

This shows that the Laguerre polynomial is orthogonal with weighting $e^{-x} x^{\alpha}$.

Algorithm of Wavelet Transform (with Qt class)

There are many kind of wavelet transform, and I think the names are quite confusing.

For instance, there are continuous and discrete wavelet transforms, in which, the “continuous” and “discrete” are for the wavelet parameters, not for the “data” itself. Therefore, for discrete data, there are “continuous” and “discrete” wavelet transforms, and for function, there are also “continuous” and “discrete” wavelet transforms.

In here, we will focus on discrete wavelet transform for function first. This discrete wavelet transform is also called as wavelet series, which express a compact support function into series of wavelet.

For simplicity, we also focus on orthonormal wavelet.

As the wavelet span the entire space, any compact function can be expressed as

$\displaystyle f(t) = \sum_{j,k} \left \psi_{j,k}(t)$

$\psi_{j,k}(t) = 2^{j/2} \psi(2^j t - k)$

where $j, k$ are integer.

Now, we move to discrete data discrete wavelet transform. The data is discrete, we can imagine only $t_n = t_0 + n \Delta$ points are known with finite $n$.

$\displaystyle f_n = f(t_n) = \sum_{j,k} \left \psi_{j,k}(t_n)$

the integration becomes a finite sum.

Without loss of generality, we can set $t_0 = 0, \Delta = 1$, and then the time axis becomes an integer number axis. We found that $j \leq 0$ as the wavelet can only be expand, not shrink. Because there are finite number of data point, i.e. $n < \infty$, $-Log_2(n) < j \leq 0$.

However, this double summation for each $f_n$ is very time consuming. There is a Fast Discrete Wavelet Transform. Before we continuous, we must study the wavelet.

From the last post, we know that the scaling function that generate a MRA must be:

$\displaystyle \phi(t) = \sum_{k} g_0(k) \phi(2t-k)$

$\left<\phi(t-k) | \phi(t-k') \right> = \delta_{kk'}$

, where $k$ are integer. The set of shifted scaling function span a space $V_0$. For the wavelet,

$\displaystyle \psi(t) = \sum_{k} g_1(k) \psi(2t-k)$

$\left<\psi(t-k) | \psi(t-k') \right> = \delta_{kk'}$

The set of shifted wavelet span a space $W_0$, so that $W_0 \perp V_0$, so that

$\left<\phi(t-k)|\psi(t-k') \right> = 0$

Since the wavelet is generated from the scaling function, we expect the coefficient of $g_0(k)$ and $g_1(k)$ are related. In fact, the relationship for orthonormal scaling function and wavelet is

$g_1(k) = (-1)^k g_0(1-k)$

For discrete data $x_i$, it can be decomposed into the MRA space. We start by the largest $V_0$ space, where the wavelet is most shrunken.

$\displaystyle x_i = \sum_{k} v_{0,k} \phi(i-k)$

to decompose to the $V_{-1}$ and $W_{-1}$ space. We can use the nested property of the MRA space, $\phi(2t)$ can be decomposed into $\phi(t-k)$ and $\psi(t-k)$,

$\displaystyle \psi(2t-l) = \sum_{k} h_0(2k-l) \phi(t-k) + h_1(2k-l) \psi(t-k)$

where (given that $\phi(t)$ and $\latex \psi(t)$ are orthonormal ),

$h_0(2k-l) = \left< \phi(2t-l) | \phi(t-k) \right>$

$h_1(2k-l) = \left< \phi(2t-l) | \psi(t-k) \right>$

Therefore, using the coefficient of $h_0$ and $h_1$, the wavelet coefficient $v_{0,k}$ can be decomposed to

$\displaystyle v_{s-1,k} = \sum_{l} h_0(2k-l) v_{s,l}$

$\displaystyle w_{s-1,k} = \sum_{l} h_1(2k-l) v_{s,l}$

in graphic representation

This is a fast discrete wavelet transform.

Due to the nested space of MRA, we also expect that the coefficient $h_0$ and $h_1$ are related to $g_0$. For orthonormal wavelet,

$\displaystyle h_0(k) = \frac{1}{2} g_0(-k)$

$\displaystyle h_1(k) = \frac{1}{2} (-1)^{k} g_0 (k+1)$

Since the $g_0$ is finite, the $g_1, h_0, h_1$ are all finite. That greatly reduce the computation cost of the discrete wavelet transform.

To reconstruct the discrete data $x_i$, we don’t need to use

$\displaystyle v_{s+1,l} = \sum_{k} v_{s,k} \phi(l - k) + w_{s,k} \psi(l-k)$

using the nested space of MRA, $\psi(t) = \sum_{k} g_1(k) \psi(2t-k)$,

$\displaystyle v_{s+1,l} = \sum_{k} g_0(l-2k) v_{s,k} + g_1(l-2k) w_{s,k}$

in graphical representation,

I attached the wavelet transfrom class for Qt, feel free to modify.

in the code, the data did not transform to MRA space. The code treats the data already in the MRA space. Some people said this is a “crime”. But for the seek of “speed”, it is no need to map the original discrete data into MRA space. But i agree, for continuous function, we must map to MRA space.

Wavelet Analysis or MRA

Although the Fourier transform is a very powerful tool for data analysis, it has some limit due to lack of time information. From physics point of view, any time-data should live in time-frequency space. Since the Fourier transform has very narrow frequency resolution, according to  uncertainty principle, the time resolution will be very large, therefore, no time information can be given by Fourier transform.

Usually, such limitation would not be a problem. However, when analysis musics, long term performance of a device, or seismic survey, time information is very crucial.

To over come this difficulty, there a short-time Fourier transform (STFT) was developed. The idea is the applied a time-window (a piecewise uniform function, or Gaussian) on the data first, then FT. By applying the time-window on difference time of the data (or shifting the window), we can get the time information. However, since the frequency range of the time-window  always covers the low frequency, this means the high frequency  signal is hard to extract.

To improve the STFT, the time-window can be scaled (usually by 2). When the time window is shrink by factor of 2, the frequency range is expanded by factor of 2. If we can subtract the frequency ranges for the time-window and the shrink-time-window, the high frequency range is isolated.

To be more clear, let say the time-window function be

$\phi_{[0,1)}(t) = 1 , 0 \leq t < 1$

its FT is

$\hat{\phi}(\omega) = sinc(\pi \omega)$

Lets also define a dilation operator

$Df(t) = \sqrt{2} f(2t)$

the factor $\sqrt{2}$ is for normalization.

The FT of $D\phi(t)$ has smaller frequency range, like the following graph.

We can subtract the orange and blue curve to get the green curve. Then FT back the green curve to get the high-frequency time-window.

We can see that, we can repeat the dilation, or anti-dilation infinite time. Because of this, we can drop the FT basis $Exp(-2\pi i t \omega)$, only use the low-pass time-window to see the low-frequency behaviour of the data, and use the high-pass time-window to see the high-frequency behaviour of the data. Now, we stepped into the Multi-resolution analysis (MRA).

In MRA, the low-pass time-window is called scaling function $\phi(t)$, and the high-pass time-window is called wavelet $\psi(t)$.

Since the scaling function is craetd by dilation, it has the property

$\phi(t) = \sum_{k} g_{0}(k) \phi(2t-k)$

where $k$ is integer. This means the vector space span by ${\phi(t-k)}_{k}=V_0$ is a subspace of the dilated space $DV_0 =V_1$. The dilation can be go one forever, so that the whole frequency domain will be covered by $V_{\infty}$.

Also, the space span by the wavelet, ${\psi(t-k)}=W_0$, is also a subspace of $V_1$. Thus, we can imagine the structure of MRA is:

Therefore, any function $f(t)$ can also be expressed into the wavelet spaces. i.e.

$f(t) = \sum_{j,k} w_{j,k} 2^{j/2}\psi(2^j t - k)$

where $j, k$ are integers.

I know this introduction is very rough, but it gives a relatively smooth transition from FT to WT (wavelet transform), when compare to the available material on the web.