## Hartree-Fock method for Helium excited state

There are two excited wavefunctions, singlet and triplet state. In the singlet state, the total spin is zero, while in the triplet state, the total spin is 1.

In any case, the Slater determinant is

$\displaystyle \Psi(x,y) = \frac{1}{\sqrt{2}} \begin{vmatrix} \phi_1(x) & \phi_2(x) \\ \phi_1(y) & \phi_2(y) \end{vmatrix}$

For spin-singlet state, we set,

$\phi_1(q) = \uparrow \phi_{1s}(r), \phi_2(q) = \downarrow \phi_{2s}(r)$

For triplet state,

$\phi_1(q) = \uparrow \phi_{1s}(r), \phi_2(q) = \uparrow \phi_{2s}(r)$

Since the Hamitonian is spin independent, we only take care of the spatial part.

For the triplet state, if we put in the wavefucntion in the Hamiltonian. Using the basis set approximation, The Fock matrix is

$\displaystyle F_{ij} = H_{ij} + \sum_{hk}^M a_{\nu h}^*a_{\nu k} (G_{ij}^{hk} - \delta(\sigma_h, \sigma_k)G_{ik}^{hj})$

The factor $latex \delta(\sigma_h, \sigma_k)$ is due to the spin component.

From the beginning, the Fock matrix is

$\displaystyle \sum_{j}^M F_{ij} a_{\mu j} = \sum_{j}^M H_{ij} a_{\mu j} + \sum_{j}^M a_{\mu j} \sum_{hk}^M a_{\nu h}^*a_{\nu k} G_{ij}^{hk} - \sum_{j}^M a_{\nu j} \sum_{hk}^M a_{\nu h}^*a_{\mu k} G_{ij}^{hk}$

We can see the exchange term is difference. We can interchange $k \leftrightarrow j$

$\displaystyle \sum_{j}^M a_{\nu j} \sum_{hk}^M a_{\nu h}^*a_{\mu k} G_{ij}^{hk} = \sum_{k}^M a_{\nu k} \sum_{hj}^M a_{\nu h}^*a_{\mu j} G_{ik}^{hj}$

and pull the $a_{\mu j}$ out.

The calculation is straight forward. First, we calculate the 1s state due to the mean field created by the 2s orbital, then calculate the 2s state due to the mean field by the 1s orbital. And then repeat until converged.

Using the Hydrogen 1s, 2s, 3s, and 4s orbtial, we calculated

1s KE+PE = -2 Hartree = -54.42 eV
2s KE+PE = -0.4349 Hartree = -11.83 eV
The direct term = 0.2800 Hartree = 7.62 eV
The exchange term = 0.016 Hartree = 0.44 eV

The total energy = -2.171 Hartree = -59.07 eV.

Compare to the calculated ground state energy ( -77.11 eV), the excitation energy is 18.04 eV. And the ionization energy is 4.65 eV.

The experimental value for the triplet state excitation energy is 19.77 eV, and the ionization energy is 4.70 eV.

Consider the ground state energy is different from the experimental value by 2 eV. the calculation is fairly good.

For the singlet state, because the spin state of the two electron are opposite, the Fock matrix becomes Hartree matrix.

$\displaystyle F_{ij} = H_{ij} + \sum_{hk}^M a_{\nu h}^*a_{\nu k} G_{ij}^{hk}$

The calculated result is,

1s KE+PE = -2.000 Hartree = -54.42 eV
2s KE+PE = -0.400 Hartree = -10.89 eV
The direct term = 0.2531 Hartree = 6.89 eV
The exchange term = 0.040 Hartree = 1.09 eV

The total energy = -2.147 Hartree = -58.43 eV.

Compare to the calculated ground state energy ( -77.11 eV), the excitation energy is 18.68 eV. And the ionization energy is 4.00 eV.

The experimental value for the triplet state excitation energy is 20.55 eV, and the ionization energy is 3.92 eV.

It is interesting to compare with other formalism. Many other textbook use singlet wavefunction,

$\displaystyle \Psi_S(x,y) = \frac{1}{\sqrt{2}}( \phi_1(x) \phi_2(y) + \phi_2(x) \phi_1(y)) \frac{1}{\sqrt{2}}(\uparrow \downarrow - \downarrow \uparrow)$

The triplet state is the same,

$\displaystyle \Psi_T(x,y) = \frac{1}{\sqrt{2}}( \phi_1(x) \phi_2(y) - \phi_2(x) \phi_1(y)) \frac{1}{\sqrt{2}}(\uparrow \uparrow)$

Since the Hamiltonian is spin independent, we can see, when we calculate the total energy of the singlet state and triplet state are

$E_S = E_0 + J + K$

$E_T = E_0 + J - K$

Thus, the energy difference between singlet and triplet state is 2 exchange term.

In our calculation, the wave function of the singlet state is difference. We directly calculate from the Slater determinant. And the Fock matrix is lack of the exchange term due to the opposite spin. However, the triplet state is the same, the exchange term is 0.44 eV. If we do not do Hartree-Fock on the singlet state, but calculate the singlet state using the exchange term, the singlet state energy would be -58.19 eV. Comparing with the Hartree-Fock result (-58.43 eV), I would say they agree. The small difference is due to the difference of the mean fields.

## 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 $k = 0$, without lost of generality, 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 $e^{-x} x^{\alpha}$.

## Hartree-Fock method for 1D infinite potential well

Following the previous post, I tested my understanding on the Hartree method. Now, I move to the Hartree-Fock method. The “mean field” energy of the Hartree-Fock is

$\displaystyle G_{\alpha \beta}= \sum_{j=i+1}^{N} \langle \psi_{\alpha}(i) \psi_{\nu} (j) | G(i,j) \left(| \phi_{\beta}(i) \phi_{\nu}(j) \rangle - |\phi_{\nu}(i) \phi_{\beta}(j) \rangle \right) \\ = \langle \alpha \nu | \beta \nu \rangle - \langle \alpha \nu | \nu \beta \rangle$

I also use the same method, in which, the trial wave function is replaced every iteration and the integration is calculated when needed.

Since the total wave function must be anti-symmetry under permutation, therefore one state can be occupied by only one particle. Thus, if we use the ground state in the mean field, the “meaningful” wave functions are the other states.

It is interesting that the mean field energy is zero when $\mu = \nu$, the consequence is no mean field for the same state. Suppose the mean field energy is constructed using the ground state, and we only use 3 states, the direct term is

$G_D = \begin{pmatrix} \langle 11|11 \rangle & \langle 11|21 \rangle & \langle 11|31 \rangle \\ \langle 21|11 \rangle & \langle 21|21 \rangle & \langle 21|31 \rangle \\ \langle 31|11 \rangle & \langle 31|21 \rangle & \langle 31|31 \rangle \end{pmatrix}$

The exchange term is

$G_E = \begin{pmatrix} \langle 11|11 \rangle & \langle 11|12 \rangle & \langle 11|13 \rangle \\ \langle 21|11 \rangle & \langle 21|12 \rangle & \langle 21|13 \rangle \\ \langle 31|11 \rangle & \langle 31|12 \rangle & \langle 31|13 \rangle \end{pmatrix}$

Due to the symmetry of the mutual interaction. We can see that some off-diagonal terms are cancelled. For example,

$\displaystyle \langle 1 1 | 3 1 \rangle = \int \psi_1^*(x) \psi_1^*(y) \cos(x-y) \psi_3(x) \psi_1(y) dy dx$

$\displaystyle \langle 1 1 | 1 3 \rangle = \int \psi_1^*(x) \psi_1^*(y) \cos(x-y) \psi_1(x) \psi_3(y) dy dx$

These two integrals are the same. In fact,

$latex \langle \alpha \nu | \beta \nu \rangle – \langle \alpha \nu | \nu \beta \rangle =$

whenever $\alpha = \nu$

$\displaystyle \langle \nu \nu | \beta \nu \rangle = \int \psi_\nu^*(x) \psi_\nu^*(y) \cos(x-y) \psi_\beta(x) \psi_\nu(y) dy dx$

$\displaystyle \langle \nu \nu | \nu \beta \rangle = \int \psi_\nu^*(x) \psi_\nu^*(y) \cos(x-y) \psi_\nu(x) \psi_\beta(y) dy dx$

We can see, when interchange $x \leftrightarrow y$, the direct term and the exchange term are identical, and then the mean field energy is zero. Also, when $\beta = \nu$ the mean field energy is also zero.

Due to the zero mean field, the off-diagonal terms of the Hamiltonian $H_{\alpha \beta}$ with $\alpha = \nu$ or $\beta = \nu$ are zero. Then, the eigen energy is the same as the diagonal term and the eigen vector is un-change.

Back to the case, the direct matrix at the 1st trial is,

$G_D = \begin{pmatrix} 0.720506 & 0 & -0.144101 \\ 0 & 0.576405 & 0 \\ -0.144101 & 0 & 0.555819 \end{pmatrix}$

The exchange matrix is

$G_E = \begin{pmatrix} 0.720506 & 0 & -0.144101 \\ 0 & 0.25 & 0 \\ -0.144101 & 0 & 0.0288202 \end{pmatrix}$

Thus, the Fock matrix is

$F = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 4.3264 & 0 \\ 0 & 0 & 9.527 \end{pmatrix}$

Therefore, the eigen states are the basis, unchanged

$\displaystyle \psi_1(x) = \sqrt{\frac{2}{\pi}} \sin(x)$

$\displaystyle \psi_2(x) = \sqrt{\frac{2}{\pi}} \sin(2x)$

$\displaystyle \psi_3(x) = \sqrt{\frac{2}{\pi}} \sin(3x)$

Only the eigen energies are changed, as $\epsilon_1 = 1$$\epsilon_2 = 4.3264$$\epsilon_3 = 9.527$

The total wave function for 2 particles at state 1 and state-\mu is

$\Psi(x,y) = \frac{1}{\sqrt{2}} ( \psi_1(x) \psi_\mu(y) - \psi_\mu(x) \psi_1(y) )$

I found that the “mean field” function is not as trivial as in the Hartree case, because of the exchange term. In principle, the mean field for particle-i at state-$\mu$ is,

$G(i) = \int \phi_\nu^*(j) G(i,j) \phi_\nu(j) dj - \int \phi_\nu^*(j) G(i,j) \phi_\mu(j) dj$

However, the direct term is multiply with $\psi_\mu(i)$, but the exchange term is multiply with $\psi_\nu(i)$, which are two different functions. i.e., the “mean field” is affecting two functions, or the “mean field” is shared by two states.

Although the mean field for single state can be defined using exchange operator symbolically, but I don’t know how to really implement into a calculation. Thus, I don’t know how to cross-check the result.

## Method on solving differential equation

The foundation of using Hartree-Fock method to get the self-consistence wave function and potential is solving the Hartree-Fock equation, which is kind of differential equation.

The idea of the Hartree-Fock method is

1. Assume the trial wave function $\psi^{(0)}_\mu(i)$
2. put the trial wave function into the Hartree-Fock equation $\displaystyle F(\psi_{\nu}^{(0)}(j))$
3. Solve the equation and obtain a new wave function $\psi^{(1)}_\mu(i)$
4. Go to step 2. until converge.

In this post, we will discuss at least 2 method on the step 3.

One method is solve the Hartree-Fock equation using numerical method, for example, Rungu-Kutta method, or Numerov’s method. Those method can also obtain the energy when impose some conditions, for example, the behaviour of the wave function at long range.

The 2nd method is basis expansion. We assume the wave function is a superposition of some basis,

$\displaystyle \psi_\mu(i) = \sum_{\alpha}^{N_\alpha} a_{\mu\alpha} \phi_\alpha(i)$

Substitute into the Hartree equation (Hartree-Fock equation is similar),

$\displaystyle F \left(\sum_{j=i+1}^{N}\psi_\nu(j) \right) \psi_\mu(i) = H(i) \psi_\mu(i) + \sum_{j=i+1}^{N}\langle \psi_\nu(j) | G(i,j) | \psi_\nu(j) \rangle \psi_\mu(i)$

$\displaystyle \sum_{\alpha} a_{\mu\alpha} F |\phi_{\alpha}(i) \rangle \\ = \sum_{\alpha} a_{\mu\alpha} \left( H(i) |\phi_{\alpha}(i) \rangle + \sum_{k,l} \sum_{j=i+1}^N a_{\nu k}^{*} a_{\nu l}\langle \phi_{\nu k}(j) | G(i,j) | \phi_{\nu l} (j) \rangle |\phi_{\mu \alpha}(i) \rangle \right) \\ = \sum_{\alpha} a_{\mu\alpha} \left( H(i) + G(i) \right) |\phi_{\alpha}(i) \rangle \\ = \sum_{\alpha} a_{\mu \alpha} E |\phi_{\alpha} \rangle$

Acting $\langle \phi_{\mu \beta}(i)|$

$\displaystyle \sum_{\alpha} a_{\mu \alpha} \langle \phi_{\mu \beta} (i) | F | \phi_{\mu \alpha} (i) \rangle = \sum_{\alpha} F_{\beta \alpha} a_{\mu \alpha} = E a_{\mu \beta}$

Thus, we can solve the coefficient $a_{\mu \alpha}$, which is the eigen vector. Then, new trial functions with the new coefficients are used.

In the 2nd method, we can see the integral

$H_{\beta \alpha} = \langle \phi_\beta(i) | H(i)| \phi_\alpha (i) \rangle$

$G_{\beta \alpha}^{k l} = \langle \phi_\beta(i) \phi_k(j) | G(i,j) | \phi_l(j) \phi_\alpha(i) \rangle$

can be pre-calculated. The calculation becomes iterating the coefficient $a_{\mu \alpha}$.

In method 2, the basis does not change, but only the coefficients. The advantage, also the disadvantage of this method is that, all possible integrals are pre calculated, even though some of them will never be used. Once the integrals are calculated, it can be used for solving many similar problems. But the calculation time for the integrals may be long and consume memory.

In previous post, I did not pre-calculate the integral, I simply put the new trial function and integrate again.

This method is very similar to the method 2. However, the basis is the wave function itself. After one calculation, a new basis, which is the wave function, will be used. And the old basis will be replaced. In method 2, only the coefficient is replaced.

This method may somehow faster then method 2, because only nesscary integrals are calculated. In Mathematica, method 2 is generally slower, especially for larger basis.

## Some comments on Hartree-Fock method

Most confusing thing for the method is that, weather the index is represent particle or state.

Look back the method. The total wave function is product of the wave function of each particle.

$\Psi = \phi_1(1) \phi_2(2) .. \phi_N(N)$

Here, we fixed our notation that the subscript is state, the parentheses is particle ID.

The single particle energy is,

$\displaystyle \langle \Psi |\sum_{i}^{N} H(i)|\Psi \rangle = \sum_{i}^{N} \langle \phi_\mu(i) | H(i)|\phi_\mu(i) \rangle$

Here, the $i$-particle in state $\mu$

The mutual interaction energy is

$\displaystyle \langle \Psi |\sum_{i}^{N-1}\sum_{j=i+1}^{N} G(i,j)|\Psi \rangle = \sum_{i}^{N-1}\sum_{j=i+1}^{N} \langle \phi_\mu(i) \phi_\nu(j) | G(i,j)|\phi_\mu(i) \phi_\nu(j) \rangle$

In both energy terms, we can factor out the sum of particle, we get the Hartree equation

$\displaystyle \left( H(i) + \sum_{j=i+1}^{N} \langle \phi_\nu(j) | G(i,j)| \phi_\nu(j) \rangle \right) |\phi_\mu(i) \rangle = \epsilon_\mu | \phi_\mu(i) \rangle$

Now, we can see, the sum is sum over the number of particle, not state.

When solving the Hartree equation using a given basis $b_k(x)$, we set the first trial of the solution to be

$| \phi_\mu(i) \rangle = | b_\mu(i) \rangle$

Then, in the first trial, the Hartree matrix is

$\displaystyle \hat{F}_{\alpha \beta} = \langle b_\alpha(i) | \left( H(i) + \sum_{j=i+1}^{N} \langle b_\nu(j) |G(i,j) | b_\nu(j) \rangle \right) | b_\beta(i) \rangle$

In here, we keep the particle ID. Inside the mean field integral, the trial state $b_\nu(j)$ is used for the particle-j.

After a trial, we will get the eigen vector for each state $v_\mu$, then we construct a new trial function and iterate until converge.

We can see, for multi-particle wave function, the sum is sum over the particle, even thought there are particles share the same state. In fact, since each particle should be at a state, the state should be a function of particle ID. So, the general Hartree matrix is

$\displaystyle \hat{F}_{\alpha \beta} = \langle \psi_\alpha(i)|\left( H(i) + \sum_{j=i+1}^{N} \langle \psi_{\nu(j)}(j) | G(i,j)| \psi_{\nu(j)}(j) \rangle \right) |\psi_\beta(i) \rangle$

where $\alpha, \beta = 1,2,3,..., k$, $k$ is number state to use.

In Hartree-Fock method, the sum is also over all other particle, not state.

## Hartree method for 1D infinite potential well

In order to test my understanding on the Hartree method. I played a simple case. The potential is

$V(x) = \left\{ \begin{matrix} 0 , 0 \leq x \leq \pi \\ \infty , else \end{matrix} \right\}$

The mutual interaction between 2 particles is, for simple calculation,

$G(x_1, x_2) = g_0 \cos(x_1 - x_2)$

This mutual interaction is repulsive, we can expect the eigen wave function will be more spread out.

We use the eigen state of the $V(x)$ potential to be the basis,

$\displaystyle \phi_n(x) = \sqrt{\frac{2}{\pi}} \sin(nx)$

The Hartree method is basically solving the Hartree equation

$\displaystyle \left( -\frac{\hbar^2}{2m} \frac{d^2}{dx_1^2} + V(x_1) + \int G(x_1, x_2) (\psi_k(x_2))^2 dx_2 \right) \psi_n(x_1) = \epsilon_n \psi_n(x_1)$

We assume the two particles are in the same ground state, thus, $k = n = 1$. We assume,

$\psi_1(x) = \sum a_i \phi_i(x)$

The algorithm is that, for the 1 particles matrix

$\displaystyle F_{ij} = \int \psi_i(x) \left( -\frac{\hbar^2}{2m} \frac{d^2}{dx^2} + V(x) \right) \psi_j(x) dx$

Define

$\displaystyle G_0(x) = \int G(x, x_2) (\psi_1(x_2))^2 dx_2$

The 2-particle matrix

$\displaystyle G_{ij} = \int \psi_i(x) G_0(x) \psi_j(x) dx$

For the 1st trial, we set

$\psi_i^{(0)}(x) = \phi_{(2i+1)}(x)$

We only use “even” function with respect to $x = \pi/2$, because mixing parity will result non-reasonable result. ( why ? )

Then we get the matrix of the Hartree Hamiltonian $H_{ij} = F_{ij} + G_{ij}$, and solve for the eigen value and eigen vector,

$\vec{v}^{(0)} = (v_1^{(0)}, v_2^{(0)}, .... v_n^{(0)})$

The next trial is

$\displaystyle \psi_i^{(1)}(x) = \sum_i v_i^{(0)} \psi_i^{(0)}$

The process repeated until converge. The convergent can be exam using the Hartree Hamiltonian that the non-diagonal terms are vanished, i.e.

$F_{ij} + G_{ij} = 0, i\neq j$

I tried with 2 basis and 6 basis, I set $\hbar^2/2m = 1$ for simplicity, the Hartree ground state is 1.7125 for both basis. The eigen wavefunction for the 2-basis is

$\psi(x) = 0.797752 \sin(x) + 0.0145564 \sin(3x)$

with the mean field

$V_m(x) = 0.842569 \sin(x)$

For 6 basis, the wave function is

$\psi(x) = 0.797751 \sin(x) + 0.0145729 \sin(3 x) + 0.000789143 \sin(5 x) + 0.000125871 \sin(7 x) + 0.0000336832 \sin(9 x) + 0.0000119719 \sin(11 x)$

we can see, the contribution of the higher energy states are getting smaller and smaller. The higher energy states only contributes 0.03%. Following is the plot compare $\phi_1(x)$ (orange) and $\psi(x)$ (blue). We can see that the wave function becomes little spread out.

If we increase the strength of the mutual interaction by 10 times, the wave function becomes more spread out.

To cross check the result,  we can use the mean field and solve the Schrodinger equation using variational method. The method gives the ground state energy be 1.7125 and converge.

This is interesting that the variational method needs 4 or 5 basis to converge, while Hartree method only need 2.

At last, we substitute the wave functon back to the Hartree equation, it turns out the result is not satisfied 100%. The total wave function is

$\Psi(x_1, x_2) = \psi(x_1) \psi(x_2)$

The Schrodinger equation is

$\displaystyle H \Psi = -\frac{\hbar^2}{2m}\left(\frac{d^2}{dx_1^2} + \frac{d^2}{dx_2^2}\right) \Psi(x_1,x_2) + G(x_1,x_2)\Psi(x_1,x_2) = E \Psi(x_1,x_2)$

However, the left side and the right side are not equal, but the integration

$\langle \Psi |H|\Psi\rangle = \langle |\Psi|\Psi \rangle$

is equal.

I plot the $\Psi H\Psi$ (orange) and $E \Psi^2$ (blue), and there difference is the following

We can imagine, the integration of the bottom plot is zero. I don’t know the reason, but I guess it is not because of the number of basis, because the higher energy states only contributes for 0.03%. I guess, it is because of the energy is the quantity that is optimized rather then the wave function. And I am not sure that the reverse of the following is true.

$H|\Psi\rangle = E |\Psi\rangle \rightarrow \langle \Psi|H|\Psi \rangle = E \langle \Psi|\Psi \rangle$

For the case of many particles, the $G_0$ has to sum over all other particle. When all particles are in the same states, the sum is simple. For particles distribute among different states, in general, the sum is

$\displaystyle G_0 =\sum_{k\neq i} \int G(x_1,x_2) \psi_k(x_2)^2 dx$

Here, the sum is excluded the particle $x_1$, which is in state $i$. For example, if the system consists of 2 particles on state-1, and 2 particles on state-2. The mean field for the particle in state-1 is constructed using 1 particles on state-1 and 2 particles on state-2. Thus the energy for the state-2 may be not correct.

The mean field for the particle on state-2, the mean field should be constructed by 2 particles on state-1 and 1 particle on state-2.

For higher states that without particle, the mean field is built by 2 particles on state-1 and 2 particles on state-2.

In general, the mean field is constructed using other particles.

This fixed mean field method is simple but we have to define the mean field for each state and calculate again and again.