Schrödinger equation | Nuclear Physics 101

Variational method for Woods-Saxon potential

August 31, 2021

GoLuckyRyan Basic Hankel, Runge-Kutta, Schrödinger equation, spherical Bessel, Variational method Leave a comment

In the past, we solved the radial wavefunction for the Woods-Saxon potential by Runge-Kutta method. But there is another method based on linear algebra, that all wave function can be expanded as a linear combination of a set of basis. In here, we use a free wave as basis, which is the spherical Bessel function.

to be annoying due to my poor memory , lets state the Schrödinger equation again,

$\displaystyle \left( -\frac{\hbar^2}{2m}\nabla^2 + V(r) \right)\Psi = E\Psi$

Separating the radial part

$\displaystyle \left( -\frac{\hbar^2}{2m} \frac{1}{r^2} \left( \frac{d}{dr} r^2 \frac{d}{dr} \right) +\frac{\hbar^2}{2m} \frac{l(l+1)}{r^2}+ V(r) \right) R(r) = E R(r)$

replace $R(r) = u(r) /r$

$\displaystyle \frac{1}{r^2} \left( \frac{d}{dr} r^2 \frac{d}{dr} \right) \frac{u(r)}{r} = \frac{1}{r}\frac{d^2}{dr^2} u(r)$

$\displaystyle \left( -\frac{\hbar^2}{2m} \frac{d^2}{dr^2} + \frac{\hbar^2}{2m} \frac{l(l+1)}{r^2} + V(r) \right) u(r) = E u(r)$

simplify

$\displaystyle \left( \frac{d^2}{dr^2} - \frac{l(l+1)}{r^2} - \frac{2m}{\hbar^2} V(r) \right) u(r) = -\frac{2m}{\hbar^2}E u(r)$

We can define a “free wave” Hamiltonian as

$\displaystyle \left( \frac{d^2}{dr^2} - \frac{l(l+1)}{r^2} \right) u(r) = H_0 u(r) = -k^2 u(r)$

In this post, we explored the orthogonality of the the Riccati-Bessel function, and it is the eigen state for $H_0$ with eigenvalue of $-k_n^2$ .

$\displaystyle b_k(r) = \sqrt{\frac{2}{pi}} \hat{j_l}\left(k r\right), ~~~ H_0 b_k(r) = -k^2 b_k(r)$

$\displaystyle \left< b_k | b_k' \right> = \delta(k-k')$

Thus, the Riccati-Bessel function can be served as the set of basis functions that span the whole space.

The total Hamiltonian is then

$\displaystyle \left( -\frac{\hbar^2}{2m} \frac{d^2}{dr^2} + \frac{\hbar^2}{2m} \frac{l(l+1)}{r^2} + V(r) \right) u(r) \\~~~~~~~~~~~~~~~~= \left( -\frac{\hbar^2}{2m} H_0 + V(r) \right) u(r) = E u(r)$

Since the solution can be expressed as a linear combination of the basis functions, then

$\displaystyle u(r) = \sum_n a_n b_n(r)$

$\displaystyle \left( -\frac{\hbar^2}{2m} H_0 + V(r) \right) \sum_n a_n b_n(r) = E\sum_n a_n b_n(r)$

Multiple $b_m(r)$ and integrate,

$\displaystyle \sum_{n} a_n \left< b_m \left| -\frac{\hbar^2}{2m}H_0 + V(r) \right|b_n \right> = E a_m$

The above is a matrix equation

$\displaystyle \left< -\frac{\hbar^2}{2m} H_0 + V(r) \right> \vec{a} = E \vec{a}, ~~ \vec{a} = (a_1, a_2,... a_N)$

Everything seems OK. But when I test the theory using Mathematica, it fail. For simplicity, we use the s-orbital, and the basis is

$\displaystyle b_n(r) = \sqrt{\frac{2}{\pi}} \sin\left( \frac{2\pi}{a} n r\right)$

The Woods-Saxon parameters are $V_0 = -50 \mbox{MeV}, V_{so} = 22 \mbox{MeV}, R_0 = R_{so} = 4 \mbox{fm}, a_0 = a_{so} = 0.67 \mbox{fm}$ .

Using Runge-Kutta method, the ground state energy is -36.09 MeV, and the wave function is

We set $a = 40~\mbox{fm}$ , and $n = 1, 2, ...., 40$ . The matrix elements is

$\displaystyle \left<n \left| -\frac{\hbar^2}{2m_n}H_0 + V(r) \right| m \right> = \delta_{nm} \frac{\hbar^2 k^2}{2m_n} + \left<V(r) \right>, ~~ k = \frac{2\pi}{a} n$

where $m_n$ is neutron mass.

After calculate the matrix, find the eigenvalue and eigenstate. The lowest eigen energy is -465.18 MeV, which indicate something wrong. And the wave function, constructed by the eigenstate look likes

Ground state constructed by basis functions

The shape of the wavefunction looks similar to the correct wavefunction, but narrower and larger amplitude.

The next eigen wave function also look like 1s-orbital. The spectrum of the eigenvalues, which are the eigen energies, has 7 bound states.

I don’t know the problems….

One possible reason could be that the basis function is always $b_n(a) = 0$ , but the real solution is not necessary to be $\psi(a) = 0$ . In fact, the 0s-wave function never be zero except $r= 0$ .

I use the same basis and extract the coefficient,

$\displaystyle \psi(r) \approx \sum_{n}^{40} C_n \frac{\sin(k_n r)}{r}, ~ C_n = \int_0^\infty \psi(r) \sin(k_n r) r dr, ~ k_n = \frac{2\pi}{a}n$

And it can reproduce the original wave function at $r < 10 \mbox{fm}$

Since the basis function is a sine wave and we only sum limited term, the long range oscillation can not be suppressed.

Another possible reason is that the basis is not normalizable. This root at the orthogonality of Bessel function that

$\displaystyle \int_0^\infty J_\nu^2(kx) x dx = \infty$

An-another possible reason is that, the basis is for unbound state. But in this post, we used a not-normalizable, unbounded basis to construct the bound state wave function. It can form the solution correctly at short-range, but it is periodic. This is the nature of series expansion, always periodic.

In this sense, we are also expanding the solution into a series, that the series only produce periodic function, therefore, a correct way to do is using the spherical Hankel transform, which is transform the Schrodinger equation into momentum space.

Hartree method for 1D infinite potential well

October 7, 2017

GoLuckyRyan theory Eigen System, Hartree-Fock, Schrödinger equation, Variational method Leave a comment

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.

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.

Reason for n>l in atomic orbit

September 14, 2017

GoLuckyRyan atomic, Basic Coulomb, hydrogen, Laguerre, rotation, Schrödinger equation Leave a comment

If we study atomic structure, we will find the principle quantum number must larger than azimuthal quantum number (the number of angular momentum), i.e.

$n > l$

But most text book give this result using mathematics that there is no solution for the radial equation when $n \leq l$ . Some text book solves the radial equation using power series, and argue that the series has to be terminated at power of $n-1$ because the number of node is $n-1$ . The exactly solution is related to Laguerre polynomial, and the polynomial is only defined for $n \geq l+1$ .

Also, in nuclear physics, there is no restriction and $n$ and $l$ can take any integers. Why these two cases are different? of course, the key point is the potential difference. Coulomb potential is used in atomic orbit. Wood-Saxon potential (or finite square well ) is used in nuclear orbit.

The Schrodinger equation for potential $V(r)$ is

$\displaystyle \left( \frac{1}{r^2} \frac{\partial}{\partial r}\left( r^2 \frac{\partial}{\partial r}\right) - \frac{1}{r^2} L^2 - \frac{2m}{\hbar^2} V(r) \right) \psi(r, \Omega) = -\frac{2mE}{\hbar^2} \psi(r,\Omega)$

Separate the radial and spherical part $\psi(r, \Omega) = R(r) Y(\Omega)$

$\displaystyle L^2 Y(\Omega) = l(l+1) Y(\Omega)$

$\displaystyle \left( \frac{1}{r^2} \frac{d}{d r}\left( r^2 \frac{d}{d r}\right) - \frac{l(l+1)}{r^2} - \frac{2m}{\hbar^2} V(r) \right) R(r) = -\frac{2mE}{\hbar^2} R(r)$

The radial equation further reduce by using $u(r) = r R(r)$

$\displaystyle \frac{d^2 u}{dr^2} - \frac{2m}{\hbar^2} U(r) u(r) = \frac{2mE}{\hbar^2} u(r)$

we can see, the effective potential is

$\displaystyle U(r) = V(r) + \frac{\hbar^2}{2m} \frac{l(l+1)}{r^2}$

For Coulomb potential,

$\displaystyle U(r) = -\frac{e^2}{4\pi \epsilon_0 r} + \frac{\hbar^2}{2m} \frac{l(l+1)}{r^2}$

Introduce dimensionless quantities, or Express everything using the fine structure constant. $a_0$ is Bohr radius,

$\displaystyle \rho = \frac{2r}{a_0}, \lambda = \frac{e^2}{4\pi \epsilon_0} \sqrt{ - \frac{m}{2\hbar^2 E}} = n$

The equation becomes

$\displaystyle \frac{d^2u}{d\rho^2} + \left(\frac{\lambda}{\rho} - \frac{l(l+1)}{\rho^2}\right) u = \frac{1}{4} u$

The simplified effective potential,

$\displaystyle U(\rho) = \frac{n}{\rho} - \frac{l(l+1)}{\rho^2}$

Now, we related the principle quantum number and azimuthal quantum number in one simply formula.

Since the repulsive part due to the rotation is getting larger and larger, the minimum point of the effective potential is

$\displaystyle \rho_{min} = \frac{2l(l+1)}{n}$

When $l = n$ , $\rho_{min} = 2(n+1) , r_{min} = (n+1)a_0$ .

I plot the case for $n=2$

n=2.PNG

The bottom line is $l = 0$ . We can see, when $l = 2$ , the minimize point of the potential is at $r = 3 a_0$ . The potential is vary shallow, and the bound state is very diffuses. Since the 2nd derivative of the wave function $u$ around the minimum point is

$\displaystyle \frac{d^2u}{d\rho^2} = \left( \frac{l(l+1)}{\rho^2} - \frac{n}{\rho} + \frac{1}{4} \right) u \approx \frac{1}{4} \frac{l(l+1) - n^2}{l(l+1)} u = k u$

For $l = n-1$ , $k$ is negative, that give oscillating wave function.

For $l = n$ , $k$ is positive, that gives explosive wave function that cannot be normalized.

Therefore, $l < n$ .

In fact, if we plot the effective potential,

$\displaystyle U(r) = -\frac{e^2}{4\pi \epsilon_0 r} + \frac{\hbar^2}{2m} \frac{l(l+1)}{r^2}$

we can find that when $l \geq 1$ , the potential is also very shallow. For such a picture, we can see the wave function must be oscillating that the number of node must be larger then 1, so the principle quantum number should be larger than 1.

In nuclear orbit, the effective potential using Wood-Saxon potential is

$\displaystyle U(r) = -\frac{1}{1-\exp(\frac{r-R}{a})} + \frac{\hbar^2}{2m} \frac{l(l+1)}{r^2}$

where $R \approx 1.25 A^{1/3} [fm]$ and $a = 0.6 [fm]$ . For simplicity, we use

$\displaystyle U(r) = -\frac{1}{1-\exp(\frac{r-R}{a})} + \frac{l(l+1)}{r^2}$

The position of minimum potential is no simple formula. I plot R=10, a=1

We can see, for any $l < l_{u}$ there is a “well” that can bound a nucleon.

I haven’t solve the finite spherical well, but the infinite spherical well is solved. The solution is the spherical Bessel function of first kind. that function only has 1 parameter, which is the angular momentum. To determine the energy is simply find the node position, and the number of node is related to the principle quantum number. Therefore, the principle quantum number and angular momentum is unrelated. A similar result should be applied well in the case of finite spherical well like Wood-Saxon type, that, for a fixed angular momentum, there can be many solution with different number of nodes.

Wood-Saxon potential, the effective potential is always sloping, especially when outside of the well, that forces that wave function to be decay faster. However, the Coulomb potential is a long range force, the effective potential can be slowly changing for long distance, that allow the wave function to be oscillating for long range.

Express Hydrogen orbit using fine structure constant

September 14, 2017

GoLuckyRyan atomic, Basic hydrogen, Schrödinger equation Leave a comment

The fine structure constant is

$\displaystyle \alpha = \frac{e^2}{4\pi \epsilon_0} \frac{1}{\hbar c}$

The Bohr radius

$\displaystyle a_0 = \frac{4\pi \epsilon_0}{e^2} \frac{\hbar^2}{m} = \frac{\hbar}{m c \alpha}$

The energy

$\displaystyle E_n = -\frac{1}{2n^2} \left(\frac{e^2}{2\pi \epsilon_0}\right)^2 \frac{m}{\hbar^2} = - \frac{1}{2} m (c\alpha)^2 \frac{1}{n^2}$

We can remember that the electron is traveling at $\alpha c$ at ground state.

When solving the radial part of the schrodinger equation, we can use

$\displaystyle \rho = \frac{2r}{a_0} = r\sqrt{- \frac{8mE}{\hbar^2}}$

$\displaystyle \lambda = n = \sqrt{ -\frac{mc^2 \alpha^2}{2 E}}$

A little summery of what I am doing

August 31, 2017

GoLuckyRyan atomic, Basic Clebsch-Gordon, Fourier transform, hydrogen, Laguerre, rotation, Schrödinger equation, Wavelet Leave a comment

My current position is developing a computational program that can measure the system of polarized target automatically and repeatedly. The program needs to connect with the microwave generator, voltage supply, power meter, and oscilloscope. That is purely technical and nothing special.

Later, after the program started to collecting data, another program is needed to analysis the data. Although there are many program that can do analysis, but those program are not so easy to use, in both control and display. So, I make an analysis program. The program is simply read the 2-D time-domain signal, and then determine some parameters of a specific function that fit the signal. However, the signal could be very noisy, so FFTW and wavelet analysis were implemented. That’s why the wavelet analysis appeared in this blog.

After that, the sample of polarized target is made from various pentacene derivatives, that no known energy level exist. One way to find out the energy levels of the singlet excited states is to measure the absorption spectrum. However, the energy of the triplet state is difficult to measure. And the energy level of the triplet state is critical for Dynamic Nuclear Polarization to be happened. Thus, one way to find out is doing computation chemistry.

My last chemistry class was like 15 years ago, when I was junior high school. But the basic of computation chemistry is solving the Schrodinger equation. That is what theoretical nuclear physicists do! The variation method, the Hartree-fock method, I heard it and somehow know it, but never do it with my own hand and computer. That is why I revisit the Hartree-fock method, and found out my previous understanding is so naive.

In the course of studying Hartree-Fock method, and one problem is evaluating the overlap integral

$\displaystyle \int \psi_a(r_1) \psi_b(r_2) \frac{1}{r_{12}} \psi_c(r_1) \psi_d(r_2) dr_1dr_2$

For a 3-D system, the integral involves product of multiple spherical harmonics. That is really troublesome. Therefore, I move to study the spherical harmonics, and the related rotational invariant, Wigner D-matrix, Clebcsh-Gordon series and Fourier series. The spherical harmonics arises from solving the Laplace equation in spherical coordinate. A general theory of the solution of Lapalace equation involves Legendre polynomial, which is a special case for Hypergeometric function. And a very interesting connection is that the elliptical function of the 1st and 2nd kind are also two special cases of hypergeometric function, that, the solution of Laplace equation for elliptical boundary condition is elliptical functions. That connects spherical harmonics and elliptical function! Wow!

I am now a bit off-track, that I am very interesting on the function of all functions. We know that there are many elementary functions, such as sin, cos, and Log. And even more kind of special functions, such as

Hermite — solving 1-D harmonic oscillator
Laguerre — the radial function of hydrogen
Legendre — the solution of the “ $\theta$ ” of Laplace equation
Gamma — a continuation of factorial
Bessel — solution of 3-D infinite square well
Elliptic — magnetic field of a solenoid
Dirac delta
Gaussian or Error function

For discrete argument

Clebsch-Gordon
Factoral
Binomial

As far as I know, the Hypergemetric function is like “mother of functions”, although not all special functions can be expressed as it. I am driven by curiosity, so, I am not sure where I will go. For instance, the transformation of Hypergeometric function is very interesting.

So, for now, as an ending, I found one article is very interesting. The History and Future of Special Functions, by Stephen Wolfram.

Variational method & Eigenvalue Problem

August 29, 2017

GoLuckyRyan Basic Eigen System, Schrödinger equation, Variational method Leave a comment

The relationship I haven’t stated explicitly, so, here.

In solving the Schrodinger equation (or a general Hamiltonian problem )

$H |\psi \rangle = E|\psi \rangle$

The variation method is a golden method to find the minimum energy. For the minimum energy solution $\psi_0$

$\displaystyle \frac{\langle \psi_0|H|\psi_0\rangle}{\langle \psi_0|\psi_0 \rangle} = E_0$

For other solution

$\displaystyle \frac{\langle \psi|H|\psi\rangle}{\langle \psi|\psi \rangle} \geq E_0$

This implies the diagonal elements of $H$ are always larger or equal the minimum energy.

If we expand the solution in to an orthogonal complete basis $\phi_i$

$\displaystyle |\psi\rangle = \sum_i a_i |\phi_i\rangle$

The Schrodinger equation

$\displaystyle \sum_i a_i H|\phi_i\rangle = \sum_i a_i E |\phi_i\rangle$

since the basis is orthogonal, multiple

$\displaystyle \sum_i a_i \langle \phi_j |H|\phi_i \rangle = \sum_i a_i E \langle \phi_j |\phi_i \rangle$

$\displaystyle \sum_i \langle \phi_j | H|\phi_i\rangle a_i = E a_j$

$\displaystyle \sum_i H_{ji} a_i = E a_j$

In matrix form

$H \cdot \vec{a} = E \vec{a}$

this is the eignevalue problem.

The eigenvalue problem does not solve the ground state, but also the excited states.

Exchange energy

June 15, 2017

GoLuckyRyan Basic Eigen System, exchange energy, perturbation, Schrödinger equation Leave a comment

In QM, when treating a system with identical particles, the exchange term raised because the system must be the same after particle exchanged.

I am reluctant to use the term “exchange interaction” or “exchange force”, “exchange potential”, because there is no interaction, no force carrier, no potential at all.

Suppose the Hamiltonian for single fermion is

$H\psi = \epsilon \psi$

The 2-particle system, the Schrodinger equation is

$H_T \Psi = H_1 + H_2 + H'= E \Psi$

where $H'$ is the interaction between the 2 particles.

Now, guessing the total wave function to be

$\Psi = a \psi_1(r_1) \psi_2(r_2) + b \psi_1(r_2)\psi_2(r_1) = a \Psi_n + b\Psi_e$

The first term is “normal”, the second term is “exchanged”. substitute to the equation

$a H_T \Psi_n + b H_T \Psi_e = a E \Psi_n + b E \Psi_e$

multiply both side with $\Psi_1$ or $\Psi_2$ , and integrate $r_1, r_2$ , we have

$\begin{pmatrix} J & K \\ K & J \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = E \begin{pmatrix} a \\ b \end{pmatrix}$

where

$J = \int \psi_1^*(r_1) \psi_2^*(r_2) H_T \psi_1(r_1) \psi_2(r_2) dr_1 dr_2$

$K = \int \psi_1^*(r_2) \psi_2^*(r_1) H_T \psi_1(r_1) \psi_2(r_2) dr_1 dr_2$

Here, we assumed $H'$ is symmetric against $r_1, r_2$ . This is a fair assumption as $H'$ is the mutual interaction, and $H_1, H_2$ are independent.

The $J$ is the energy from the interaction of the particles.

The $K$ is the energy due to EXCHANGE of the particles.

As we can see, the energy $J = \epsilon_1 + \epsilon_2 + J'$ .

The $H_T$ in $K$ can be reduced to $H'$ when the eigen wave function $\psi_1 , \psi_2$ are in difference orbits. When $\psi_1 = \psi_2$ , $K = \epsilon_1 + \epsilon_2 + K'$ .

Thus the individual energy can be subtracted, yield,

$\begin{pmatrix} J' & K' \\ K' & J' \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = \Delta E \begin{pmatrix} a \\ b \end{pmatrix}$

The solution is

$\Delta E = J' + K', (a,b) = (1,1)$

This is the symmetric state. For fermion, the spin-part is anti-symmetric. This is the spin-singlet state.

$\Delta E = J' - K', (a,b) = (1,-1)$

This is the anti-symmetric state. For fermion, the spin-part is symmetric, or spin-triplet state.

It is worth to notice that, in the symmetric state, the energy is higher because the two particles can be as close as possible, that create large energy. In the opposite, in the anti-symmetric state, the two particle never contact each other, thus, the interaction energy reduced.

We can summarized.

The energy $J'$ is the shift of energy due to the mutual interaction.

But the system of identical particle subjects to the exchange symmetry. The exchanged wave function is also a state, that the total wave function has to include the exchanged wave function. This exchanged wave function creates the exchange term or exchange energy $K'$ .

In the system of Fermion, the exchange energy is related to Pauli exclusion principle.

Hartree-Fock for 2-body ground state

June 8, 2017

GoLuckyRyan Basic Hartree-Fock, Schrödinger equation Leave a comment

Long time ago, we talked about the mean-field calculation, an touched Hartree-Fock method. In that time, we explained excited-state approach. Now, we explain another approach by variation of the wave functions. This approach is inevitable in atomic physics, because the potential is fixed.

The Hamiltonian is

$H = H_1 + H_2 + V_{12}$

Since the spin component is anti-parallel, the space part of the total wave function is

$\Psi = \phi_1(r_1) \phi_2(r_2)$

The Schrodinger equation is

$H\Psi = E\Psi$

Integrate both side with $\phi_1(r_1)$

$\int dr_1 \phi_1(r_1) H \Psi = \int dr_1 \phi_1(r_1) E \Psi$

using the normalization and define $\left<x| 1\right> = \phi_1(r_1)$

$\left<1|H|1\right> \phi_2(r_2) = E \phi_2(r_2)$

similarly

$\left<2|H|2\right> \phi_1(r_1) = E \phi_1 (r_1)$

expand $H = H_1 + H_2 + V_{12}$

$(H_2 + \left<1|V_{12}|1\right>) \phi_2(r_2) = (E - \left<1|H_1|1\right>) \phi_2(r_2)$

$(H_1 + \left<2|V_{12}|2\right>) \phi_1(r_1) = (E - \left<2|H_2|2\right>) \phi_1(r_1)$

Now, we have 2 equations, an initial guess of $\Psi = \phi_1(r_1) \phi_2(r_2)$ ,

The difficulty is that, the $\left<1|V_{12}|1\right>$ contains 2 variables.

Solving radial SE numerically

July 8, 2016

GoLuckyRyan Basic, Math, theory bound state, Eigen System, Laplacian, Runge-Kutta, scattering, Schrödinger equation 4 Comments

The time-independent Schrödinger equation is

$\displaystyle (-\frac{\hbar^2}{2m}\nabla^2 + V ) \Psi = E \Psi$

Using the Laplacian in spherical coordinate. and Set $\Psi = R Y$

$\displaystyle \nabla^2 R Y - \frac{2m}{\hbar^2}(V-E) R Y = 0$

$\displaystyle \nabla^2 = \frac{1}{r^2}\frac{d}{dr}(r^2 \frac{d}{dr}) - \frac{1}{r^2} L^2$

The angular part,

$L^2 Y = l(l+1) Y$

The radial part,

$\displaystyle \frac{d}{dr}(r^2\frac{dR}{dr}) - l(l+1)R - \frac{2mr^2}{\hbar^2}(V-E) R = 0$

To simplify the first term,

$\displaystyle R = \frac{u}{r}$

$\displaystyle \frac{d}{dr}(r^2 \frac{dR}{dr})= r \frac{d^2u}{dr^2}$

A more easy form of the radial function is,

$\displaystyle \frac{d^2u}{dr^2} + \frac{l(l+1)}{r^2} u - \frac{2m}{\hbar^2} (V-E) u = 0$

The effective potential $U$

$U = V + \frac{\hbar^2}{m} \frac{l(l+1)}{r^2}$

$\displaystyle \frac{d^2u}{dr^2} + \frac{2m}{\hbar^2} (E - U) u = 0$

We can use Rungu-Kutta method to numerically solve the equation.

The initial condition of $u$ has to be 0. (home work)

I used excel to calculate a scattered state of L = 0 of energy 30 MeV. The potential is a Wood-Saxon of depth 50 MeV, radius 3.5 fm, diffusiveness 0.8 fm.

Another example if bound state of L = 0. I have to search for the energy, so that the wavefunction is flat at large distance. The outermost eigen energy is -7.27 MeV. From the radial function, we know it is a 2s orbit.

Lippmann-Schwinger equation

February 20, 2011

GoLuckyRyan Scattering, theory Fourier transform, Lippmann, scattering, Schrödinger equation, Schwinger 1 Comment

For elastic scattering, the scattered state’s energy should equal the incident state’s energy. However, the equation for them are different.

$H_0 \left| \phi \right> = E \left| \phi \right>$

$H_0 + V \left| \psi \right> = E \left| \psi \right>$

the Lippmann-Schwinger equation proposed an equation ( or solution )

$\left| \psi ^{\pm} \right> = \frac{1} { E - H_0 \pm i \epsilon } V \left|\psi^{\pm} \right> + \left| \phi \right>$

the complex energy is the key, without it, the operator will be singular. to compute the wave function, we apply the position bra on the right:

$\left< x | \psi ^{\pm} \right> = \int {dx'^3} \left< x| \frac {1}{E-H_0 \pm \epsilon } | x' \right> \left< x' | V | \psi^{\pm} \right> + \left< x | \phi \right>$

to compute the

$< x | \frac {1} {E-H_0 \pm \epsilon } |x'>$

we have to use the momentum ket, since it is commute with the Hamiltonian. then we have

$< x | \frac {1} {E-H_0 \pm \epsilon } |x'> = - \frac { 2 m } { 4 \pi \hbar^2 } \frac {1} {|x-x'| } Exp ( \pm i k |x-x'| )$

which is the Green function. We can also see this by compare the equation with the scatter equation.thus , the solution is

$\left< x | \psi ^{\pm} \right> =- \frac { 2 m } { \hbar^2 } \int {dx'^3} \frac{1}{4 \pi |x-x'| }Exp( \pm i k |x-x'|) \left< x' | V | \psi^{\pm} \right> + \left< x | \phi \right>$

for the potential is local. which means it is confined in a small space. or

$\left < x' | V | x'' \right> = V(x') \delta(x'-x'')$

and for the detector is far, plus using plane wave as incident wave the solution is

$\psi^{pm}(x) = \frac{1}{\sqrt {2 \pi \hbar}^3 } \left( Exp( \pm i k z ) + \frac { Exp( \pm i k \cdot r ) } {r} f(k, k') \right)$

$f(k,k') = - \frac { (2 \pi )^3 } { 4 \pi } \frac { 2 m } {\hbar ^2 } < k' |V | \psi ^{pm} >$

$k = p / \hbar$

since the scattered wave can be a composition of different wave-number k. thus, the matrix

$T = < k'| V | \psi^ {\pm} >$

also called the transition matrix, which mean, it is the transition through potential V to a wave-number k.

we can see that the equation is very similar to the scatter equation, since basically, they are the solution of the same Schrödinger equation.

if we further apply the Born approximation, we will get the same result that the scattered wave amplitude is a Fourier transform of the nuclear potential.

Nuclear Physics 101

Variational method for Woods-Saxon potential

Hartree method for 1D infinite potential well

Reason for n>l in atomic orbit

Express Hydrogen orbit using fine structure constant

A little summery of what I am doing

Variational method & Eigenvalue Problem

Exchange energy

Hartree-Fock for 2-body ground state

Solving radial SE numerically

Lippmann-Schwinger equation

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