## 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.

## 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 $E_2(exp)$.

## Method II (decay)

there is another way to study nuclear physics, which is by observe the decay process.

there are 3 major decays, the alpha, beta and gamma. the alpha decay is an excited nucleus go to a lower energy state by emitting an Helium ( 2 protons and 2 neutrons) nucleus. This process change the nucleus constitution and make it lighter. the beta decay is an excited nucleus go to a lower energy state by emitting an electron ( or positron, or capture an electron). This process also change the nucleus constitution but the change in mass is very small. the gamma decay is an excited nucleus go to a lower energy state by emitting a photon ( or the light particle). this process does not change the nucleus constitution.

there are many other decays, like neutron, proton, and even fission can regard as decay. in general, decay is a process that a nucleus go to a lower energy state and become more stable. But other decays has too short for lifetime, so that after the earth was formed, they are almost gone. and only alpha, beta and gamma has long left time to survive.

Via a decay process, we can know the energy level, lifetime, parity of each level of a nucleus. These informations can help us to build nuclear models and theories to understand and predict nuclear properties, like the strong and weak force.

i will present a series about decay in future posts. watch out!

On 11st, March, a M9.0 earthquake strike east Japan around 3pm. this caused malfunction of the nuclear plant, Fukushima Daiichi.

i am not going to talk about the malfunction, but on the radioactivity.

The radioactivity is caused by nucleus decay into a more stable state. like Carbon-14 will decay to Carbon-13 by beta decay. there are 3 common types of radioactivities: alpha – a Helium nucleus, beta – a electron and gamma – high energy light.

the radioactivity was measured by Bq ( Becquerel ), which is 1 decay per second. for a typical radioactivity substance, there may be 10 billion decay per second. think about how many atom in 1 mole of matter.

each radiation carry some energy, for alpha decay, energy is about 5MeV, beta decay is 1 MeV, and for gamma decay, it is around 0.1 to 1 MeV. 1 MeV is about $1.60217653 \times 10^{-13}$ Joule. for boiling a water to increase 1 degree, it take 2400Joule. you can see how small it is.

1 more thing to consider is, not every radiation will be absorbed. some will just pass through and leave and no effect. the chance for being absorbed is proportion to the density of the material, or for simplicity, how much weight of the material.

for easy understanding, we multiple these 3 factors together, to give us how much energy absorbed per kg. and we call it Gy (Gray), the absolute does.  and we can have Gy per hour, Gy per second, depends on the time scale.

Thus, for 1Gy, an alpha source, typical 5MeV per decay and a total absorption, it takes 6241 billion hits. for 10 billion decay a second, it takes 10 mins for this does.

in other system, people use rad for radioactivity, for example in the game fallout. which is 1 Gy = 100 rad.

Here is a list of different does and the health effect

• 0.25 Gy or less  –  no effect
• 0.50 Gy  – temporary reduction of while blood cells
• 1 Gy  –  feel nausea, vomit
• 1.5 Gy – 50%  more chance to have cancer, cataracts or leukemia
• 2 Gy – 5% people will die
• 4 Gy – 50% people will die in 30 days
• 6 Gy – 90% people will die in 14 days
• 7 Gy – all die

However, different part of our body have different does limit. so, for 1Gy radiation, our eye will be blinded, but our skin has no effect. therefore, we introduce another does, called Effective does Sv ( Seivert ). it is produced by a weight factor.

$Gy \times w = Sv$

different part of out body has different weight factor, for lower weight factor, the more easy the tissue will get hurt.

here is the list of weight factor

• whole body  – 1
• skin — 0.01
• lung — 0.12
• liver — 0.05
• breast — 0.05
• stomach — 0.12

we see that the weight factor for whole body is 1, therefore, in general we use Gy or Sv in equal way.

And we also observed that, when we eat some radioactive matter inside, the limit does will be decreased by 8 times in out stomach. or to say, the radioactive matter is 8 times dangerous than it is at outside. the reason is, when the matter is inside our body, we will absorb all energy.

a typical does is 3.75mSv per year. which means we received 3.75mSv in a year, 0.4uSv per hours.

for radioactive effect, there are 3 types of effect: acute ( short term ), late ( long term) and genetic effect ( permanent ) .

• short term – Erythema, hair loss, etc
• long term – Cateracts, Leukemia or cancer, etc

and the effects can be divided into 2 groups – deterministic and stochastic

• deterministic mean the chance of having those effect is increase with the radiation increased.
• stochastic is similar to deterministic, but there is a threshold. below the threshold there is no effect. – Cancer and genetic effect are belong to this group.

## Differential Cross Section II

Last time, the differential cross section discussion is based on quantum mechanics. This time, i try to explain it will out any math. so, that my mum ask me, i can tell her and make her understand. :)

in a scattering experiment, think about a target, say, a proton fixed in the center, it is positive charged. if another proton coming with some energy. it will get repelled, due to the repulsive nature of Coulomb force of same charge. it should be easy to understand, if the proton coming with high energy, it will get closer to the target, or even enter inside the target.

the repelling angle of the proton is not just depend on the energy it carry, but also on the impact parameter ( we usually call it b , but i like to call it r). the impact parameter is the shortest distance between the target and the line of the moving direction of the proton at long long away.

if the impact parameter is large, the proton miss the target. it almost cannot feel the target affection. thus, it go straight and unaffected. when the r is zero, it will hit the target head on head. and due to the repulsion. it will return back.  so, we can understand. the smaller the impact parameter, the deflection will be larger. since it can feel the force stronger.

For same impact parameter, the higher energy proton will have less deflection, since it travel faster, spend less time by the force, and the deflection get less.

Thus, we have an idea that the angle of deflection is high for small impact parameter and high energy. And most important, it only depends on these 2 factors and the effect from the target.

since our detector can only detect some small angle over some small area. So, we can place out detector on some angle, get the yield, and this is the name –  differential cross section come from.

Now, we have a uniform flow of particle with energy E. they will be deflected by the target and go to some angles. If we detect at the deflection angle, see how many particles ( the yield ) we can get in each angle. we can calculate back the effect of the target. For example, for a small angle, the particle get little defected, and this means the particle is from large impact parameter. for a large angle, the particles are from small impact parameter.

In some cases, the number of particle detected will be very high at some particular angle then others angles and this means, the cross section is large. and this means something interesting.

Moreover, don’t forget we can change the energy of the beam. for some suitable energy, the particle will being absorbed or resonance with the target. that given us low or high cross section on the energy spectrum.

( the graph is an unauthorized from the link: http://www.astm.org/Standards/E496.htm )

The above diagram is the differential cross section obtained from a Deuteron to a Tritium ( an isotope of Hydrogen with 2 neutrons and 1 proton) target, and the reaction change the Tritium into Helium and a neutron get out.

The reaction notation is

$X(a,b)Y$

where a is incident particle, X is target, b is out come particle, and Y is the residual particle.

the horizontal axis is detector angle at lab-frame. and the vertical axis is energy of Tritium. we always neglect the angle 0 degree, because it means no deflection and the particle does not “see” the target. at low energy, the d.c.s. is just cause by Coulomb force. but when the energy gets higher and higher, there is a peak around 60 degree. this peak is interesting, because it penetrated into the Helium and reveal the internal structure of it. it tells us, beside of the Coulomb force, there are another force inside. that force make the particle deflects to angle 60 degree. for more detail analysis, we need mathematic. i wish someday, i can explain those mathematics in a very simple way.

Therefore, we can think that, for higher energy beam, the size we can “see” will be smaller. if we think a particle accelerator is a microscope. higher energy will have larger magnification power. That’s why we keep building large and larger machines.

## density, flux & luminosity

density is a consideration factor for scattering experiment. In low density, both for target and the beam, then the probability of collision will be small and experiment will be time consuming and uncertainly increase. Remember that the size of nuclear is 1000 times less then the atom. the cross section area of it will be 1000 x 1000 times lesser. the chance for a nucleus-nucleus collision is very small. for example, if there is only 1 particle in the area on 1 atom, the chance for hitting the nucleus is $1 / ( \pi 10^6)$ = 0.000003, 3 in 1 million. it is just more than nothing. thus, in order to have a hit, we have to send more then 3 million particles for 1 atom. in some case, the beam density is small, say, 0.3 million particles per second on an area of 1 atom. then we have to wait 10 second for 1 hit.

density is measured in particle per area for target .

for beam, since particle is moving in it, time is included in the unit. there are 2 units – flux and luminosity. flux is particle per second, and luminosity is energy per second per area. since energy of the beam is solely by the number of particle. so, density of beam is particle per second per area. but in particle physics, the energy of particle was stated. thus, the luminosity is equally understood as density of beam, and their units are the same as particle per second per area.

In daily life, density is measured by mass per volume. although the unit are different, they are the same thing – ” how dense is it? ”

in solid, the density is highest compare to other state of matter. from wiki, we can check the density. and coveted it in to the unit we want. for example, copper has density $8.94 g / cm^3$. its molar mass is $29 g / mole$. thus, it has $0.31 N_A$ copper atom in $1 cm^3$. and $N_A = 6.022 \times 10^{23}$, which is a huge number, so, the number of atom on $1 cm^2$ is $3.25 \times 10^{15}$ .

how about gas? the density depends on temperature and pressure, at $0^o C$ and 1 atm pressure, helium has density  $1.79 \times 10^{-4} g / cm^3$ and the molar mass is 2. thus, the number of Helium atom in $1 cm^2$ is $1.42 \times 10^{13}$. when the temperature go to -100 degree, the density will increase.

beside of the number of atom per area. we have to consider the thickness of the target. think about a target is a layer structure, each layer has certain number of atom per area. if the particle from the beams miss the 1st layer, there will be another layer and other chance for it to hit. thus. more the thickness, more chance to hit.

For a light beam, the power $P$ and the wavelength $\lambda$ determine the flux of photon. power is energy per second. and energy of single photon is inversely proportional to its wavelength. the density of a light beam is given by :

$L = n/area = P \frac { \lambda} { h c} = P \lambda \times 5 \times 10^{15} [W^{-1}][nm^{-1}][s^{-1}][m^{-2}]$

where, $L$ is the luminosity and $n$ is the flux. for typical green class 4 laser, which has power more then 0.5 W and wavelength is about 500nm. the flux is about  $n=1.3 \times 10^18 [s^{-1}]$ photons per second per unit area. if the laser spot light is about 5 mm in diameter. thus, the density of the beam is $L=1.7^{18} [s^{-1}][cm^2]$.

for laser pointer in office, which is class 1 laser. the power is less then0.4 mW, say, 0.1 mW. for same spot size, the luminosity  is still as high as $6.6^{14} [s^{1}][cm^2]$.

on LHC, the beam flux can be $10^{34} [s^{-1}][cm^2]$. by compare the the density of solid copper. it is much denser. thus, a collision in LHC is just like smashing 2 solid head to head and see what is going on.