3D Harmonic oscillator

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Set x = r/\alpha The Schrodinger equation is

\displaystyle \left(-\frac{\hbar^2}{2m} \nabla^2 + \frac{1}{2} m \omega^2 r^2 \right)\Psi = E \Psi

in Cartesian coordinate, it is,

\displaystyle  -\frac{\hbar^2}{2m}\left( \frac{d^2}{dx^2}+\frac{d^2}{dy^2}+\frac{d^2}{dz^2} \right) \Psi + \frac{1}{2} m \omega^2 (x^2+y^2+z^2) \Psi = E \Psi

We can set the wave function to be \Psi(r, \Omega)  = X(x) Y(y) Z(z)

\displaystyle  \left( -\frac{\hbar^2}{2m}\frac{d^2X}{dx^2} + \frac{1}{2} m \omega^2 x^2 X \right) YZ + \\ \left( -\frac{\hbar^2}{2m}\frac{d^2Y}{dy^2} + \frac{1}{2} m \omega^2 y^2 Y \right) XZ + \\ \left( -\frac{\hbar^2}{2m}\frac{d^2Z}{dz^2} + \frac{1}{2} m \omega^2 z^2 Z \right) XY = E XYZ

we can see, there are three repeated terms, we can set

\displaystyle -\frac{\hbar^2}{2m}\frac{d^2X}{dx^2} + \frac{1}{2} m \omega^2 x^2 X = E_x X

We decoupled the X, Y, Z. Each equation is a quadratic equation with energy

\displaystyle E_x, E_y, E_z = \left(\frac{1}{2} + n \right) \hbar \omega

and

\displaystyle E_x + E_y + E_z = \left(\frac{3}{2} +  n_x + n_y + n_z  \right) \hbar \omega = \left(\frac{3}{2} +  n + n + n  \right) \hbar \omega = E

The number of states for each energy level is

\displaystyle C^{n_x+n_y+n_z+2}_2 = C^{n+2}_2 = \frac{(n+2)!}{n!2!}

The first few numbers of states are 1, 3, 6, 10, 15, 21, 28, … The accumulated numbers of states are 1, 4, 10, 20, 35, 56, 84, … Due to the spin-state, the accumulated numbers of particles are 2, 8, 20, 40, 70, 112, 168, … The few magic numbers are reproduced.

The wave function is the product of the Hermite functions H_n(x) and exponential function

\Phi(x,y,z) = N H_{n_x} (x) H_{n_y}(y) H_{n_z}(z) \exp(-r^2/2)

If we simply replace (x,y,z) \rightarrow r( \cos(\phi) \sin(\theta), \sin(\phi) \sin(\theta) , cos(\theta) ) , we can see the ground state consists of s-orbit, the 1st excited state consists of p-orbit, and the 2nd excited state consists of d-orbit.


To have a better understanding, the radial function has to be solved. The procedure is very similar to solving Coulomb potential.

\displaystyle \left(-\frac{\hbar^2}{2m}\left(\frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{d}{dr}\right)\right) + \frac{\hbar^2}{2m} \frac{L^2}{r^2} + \frac{1}{2} m \omega^2 r^2 \right)\Psi = E \Psi

separate the radial part and angular part.

\displaystyle \left( \frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{d}{dr}\right) - \frac{l(l+1)}{r^2} - \frac{m^2 \omega^2}{\hbar^2} r^2\right) R = - (2n+3) \frac{m \omega}{\hbar} R

Set \alpha^2 = \frac{\hbar}{m \omega}

\displaystyle \left( \frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{d}{dr}\right) + \frac{2n+3}{\alpha^2} - \frac{l(l+1)}{r^2} - \frac{r^2}{\alpha^4}\right) R =  0

Set x = r/\alpha

\displaystyle \left( \frac{1}{x^2}\frac{d}{dx}\left(x^2\frac{d}{dx}\right) + (2n+3) - \frac{l(l+1)}{x^2} - x^2\right) R =  0

Set u(x) = x R(x)

\displaystyle \frac{d^2 u }{d x^2} + \left( (2n+3) - x^2 - \frac{l(l+1)}{x^2} \right) u = 0

as usual, the short range behaviour is r^{l+1}, long range behaviour is \exp(-x^2/2) , as stated in the Cartesian coordinate. Thus, we set

\displaystyle u(x) = f(x) \exp\left(-\frac{x^2}{2}\right) x^{l+1}

\displaystyle x\frac{d^2f}{dx^2} + (2(l+1) -2x^2) \frac{df}{dx} + 2x(n-l) f(x) = 0

with change of variable y = x^2 , the equation becomes

\displaystyle y\frac{d^2f}{dy^2} + \left( l + \frac{1}{2} + 1 - y \right)\frac{df}{dy} + \frac{n-l}{2} f(y) = 0

This is our friend, the Laguerre polynomial! In the Laguerre polynomial, (n-l)/2 must be non-negative integer. Now we set k = (n-l)/2 , than the energy is

\displaystyle E = \hbar \omega \left( 2k + l + \frac{3}{2} \right)

In order to have n, l, k are integer, when

n = 0 \rightarrow l = 0 \\ n = 1 \rightarrow l = 1 \\ n= 2 \rightarrow l = 0, 2 \\ n = 3 \rightarrow l = 1, 3 \\ n = 4 \rightarrow l = 0,2,4

The overall solution without a normalization factor is

\displaystyle \Psi_{nlm}(r, \theta, \phi) = r^l \exp\left(-\frac{r^2}{2\alpha^2}\right) L_{k}^{l+\frac{1}{2}}\left( \frac{r^2}{\alpha^2} \right) Y_{lm}(\theta, \phi)

 

 

 

 

 

 

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Kinematics of Transfer Reaction

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Another surprise that no post on the kinematics of transfer reaction!


Suppose the reaction is A(B, 1)2 , where A is the incident particle, B is the target nucleus that is at rest, 1, 2 are the outgoing particle.

The Q-value for the reaction is the difference between initial mass and final mass

Q = m_A+ m_B - m_1-m_2

When the incident energy is too low, lower then the Q-value (we will see the relationship later) , the reaction is impossible, because the nuclei-1 and 2 cannot be created.

The equation for the reaction is

P_A + P_B = P_1 + P_2

where P_i are 4-momenta. The number of freedom is 2. The total unknown is 8 from P_1, P_2, the masses of particle-1, and -2 give 2 constrains, the equations give 4 constrains.

P_A = ( E_A , 0, 0, k_A) = ( m_A + T_A, 0, 0, \sqrt{2 m_A T_A + T_A^2})\\ P_B = (m_B, 0, 0, 0)

where T_A is the incident energy. We first transform the system into center of momentum (CM) frame. Since the total momentum is zero in the CM frame, we construct a CM 4-momentum

P_{cm} = (P_A+P_B)/2 = ( m_A+ m_B + T_A, 0, 0, k_A)

The Lorentz beta is \beta = k_A/(m_A+m_B+T_A) .

The invariance mass of the CM 4-momentum is unchanged, which is equal to the total energy of the particles in the CM frame. We can check

P_A' = ( \gamma E_A - \gamma \beta k_A, 0, 0, -\gamma \beta E_A + \gamma k_A) \\ P_B' = ( \gamma m_B , 0, 0, -\gamma \beta m_B )

The total energy in CM frame is

E_{tot} = \gamma( m_A+m_B + T_A) - \gamma \beta k_A = \sqrt{(m_A+m_B+T_A)^2 - k_A^2}


After the reaction, the momenta of the particle-1 and -2 are the same an opposite direction in the CM frame.Balancing the energy and momentum, the momentum is

\displaystyle k  = \frac{1}{2 E_{tot}} \sqrt{(E_{tot}^2 - (m_1+m_2)^2)(E_{tot}^2 - (m_1-m_2)^2)}

In the equation, if either of the bracket is negative, the momentum cannot be formed. The smaller term is

E_{tot}^2 - (m_1+m_2)^2 = (m_A+m_B+T_A)^2 - k_A^2 - (m_1+m_2)^2  \\ =(m_A+m_B+T_A)^2 - k_A^2 - (m_A+m_B-Q)^2 \\ = 2m_BT_A + 2(m_A + m_B)Q - Q^2

If we set the line for this quantity be zero,

$latex 2m_BT_A + 2(m_A + m_B)Q – Q^2 = 0 \\ T_A = \frac{Q^2 – 2(m_A+m_B)}{2m_B}  \\ = \frac{1}{2m_B} ( (Q-m_A-m_B)^2 – (m_A+m_B)^2) \\ = \frac{1}{2m_B} ( (m_1+m_2)^2 – (m_A+m_B)^2)$

Thus, if the T_A is smaller then that value, the reaction is impossible.


The 4-momenta of the outgoing particles can be constructed.

P_1' = ( \sqrt{m_1^2 - k^2} , k \cos(\phi) \sin(\theta) , k \sin(\phi) \sin(\theta) , k \cos(\theta)) \\ P_2' = ( \sqrt{m_2^2 - k^2} , -k \cos(\phi) \sin(\theta) ,- k \sin(\phi) \sin(\theta) , -k \cos(\theta))

Back to Lab frame,

P_1 = ( \gamma \sqrt{m_1^2 + k^2}  + \gamma \beta k \cos(\theta), k \cos(\phi) \sin(\theta) , k \sin(\phi) \sin(\theta) , \gamma \beta \sqrt{m_1^2 - k^2}  + \gamma k \cos(\theta))

For simplicity, set \phi = 0,

P_1 = ( \gamma \sqrt{m_1^2 + k^2}  + \gamma \beta k \cos(\theta), k \cos(\phi) \sin(\theta) , 0, \gamma \beta \sqrt{m_1^2 - k^2}  + \gamma k \cos(\theta))

We can see, the locus in the momentum space is

( k cos(\phi) \sin(\theta) , 0, \gamma \beta \sqrt{m_1^2 - k^2}  + \gamma k \cos(\theta))

This is an ellipse with length k (1, \gamma) , and centered at (0, \gamma \beta \sqrt{m_1^2 + k^2}) . If the relativistic effect is small, it is a circle.

Here is an example,

transfer.PNG

Finite Spherical Square Well with spin-orbital potential

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Last time, we studied the finite spherical square well and calculated the energy levels for difference angular momentum. In that case, the good quantum numbers (quantities that is commute with the Hamiltonian, i.e. conserved, invariance with time) are the principle quantum number n, which dictated the number of node and the angular momentum l , and the spin state s . A eigen state can be denoted as |nl m_l s m_s\rangle . The degeneracy for each state is 2(2l+1).

With the spin-orbital potential

\displaystyle V_{so} = a L \cdot S

The angular momentum m_l and spin m_s are not good quantum numbers. The sum, total angular momentum j = l + s and it z-component m_j are. The new eigen state is |n j m_j l s\rangle .

The Hamiltonian inside the well is

\displaystyle H = H_0 + V_{so} = -\frac{\hbar^2}{2m} \frac{1}{r^2} \frac{d}{dr}r^2\frac{d}{dr} - |V_0| + \frac{\hbar^2}{2m} \frac{1}{r^2}L^2 + a L\cdot S

The energy for H_0 dose not change, as the L^2 is still a good quantum number. The spin-orbital coupling is evaluated like

\displaystyle L\cdot S = \frac{1}{2}(J^2 - L^2 - S^2)

then,

\displaystyle \langle n j m_j l s| L\cdot S |n j m_j l s\rangle = \frac{1}{2} ( j(j+1) - l(l+1) - s(s-1) )

Since the spin has only 2 spin states, thus,

\displaystyle \langle L\cdot S \rangle \\ = \frac{1}{2} l , j = l+\frac{1}{2} \\ = -\frac{1}{2}(l+1), j = l - \frac{1}{2}

Note that, the mean energy of the L-states are unchanged. The degeneracy for j_+ = l+1/2 is 2(l+1) , for j_- = l-1/2 is 2l. Also, the s-states are not affected, because l = 0 .

Thus, the wave vector inside the well k becomes

\displaystyle k = \sqrt{\frac{2m}{\hbar^2 }(|V_0|- \beta \langle L\cdot S\rangle) - |E|)}

the wave vector outside the well k'

\displaystyle k' = i \kappa = i \sqrt{\frac{2m}{\hbar^2 }(\beta \langle L\cdot S\rangle) + |E|)}

And the rest is matching the boundary condition.


We can also add one more term, an additional \gamma L^2 term so that the centroid of the L-state is shifted. The energy for this potential is \gamma l(l+1) .

Here is the energy levels for \beta = -1 and \gamma = -0.5.

LS coupled Spherical Square Well.png

We can see, we reproduced the shell ordering 0s1/2, 0p3/2, 0p1/2, 0d5/2, 1s1/2, 0d3/2, 0f7/2, 1p3/2, 1p1/2, 0f5/2….

The magic number 2, 8, 20, and 40 are reproduced. May be, if I adjust the strength of the spin-orbit coupling, my get the magic number 28 and 50. Since the energy level of the s-states are fixed, I need to adjust the shift of the centroid of the others, and the splitting.

Here is the result of \beta = -1.2, \gamma = -0.2

LS coupled Spherical Square well 2.png

It seems that the magic number of 28, which is from the isolation of 0f7/2, is recreated. And the isolation of 0g9/2, created the magic number of 50.

 

 

1-D Harmonic Oscillator

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It is quite surprised that This is not here!


The Hamiltonian is

\displaystyle H=\frac{P^2}{2m} + \frac{m \omega^2}{2} X^2

Lets define creation operator A^\dagger and destruction operator A, or the ladder operators, such that,

\displaystyle A^\dagger =\sqrt{\frac{1}{2m}} P + i \sqrt{\frac{m \omega^2}{2}} X

\displaystyle A =\sqrt{\frac{1}{2m}} P - i \sqrt{\frac{m \omega^2}{2}} X

Then we can see

\displaystyle H = A^\dagger A + \frac{\hbar \omega}{2} = A A^\dagger - \frac{\hbar \omega}{2}

and the commutative relations

\displaystyle A^\dagger A - A A^\dagger = -\hbar \omega

\displaystyle HA^\dagger - A^\dagger H = \hbar \omega A^\dagger

\displaystyle HA - A H = -\hbar \omega A

From the last two equations, we can see

\displaystyle H |n\rangle = E_n |n\rangle \\ A H|n\rangle = (HA+\hbar \omega A) |n\rangle = E_n A|n\rangle \\  HA|n\rangle = (E_n-\hbar \omega) A|n\rangle

This shows that, if |n\rangle is a state with energy E_n, then A|n\rangle is also a state with energy E_n - \hbar \omega. Similar for A^\dagger |n\rangle , the energy will be E_n + \hbar \omega .


Suppose the ground state is |0\rangle , which is is lowest energy state. If we apply the destruction operator, it will be gone, i.e.

\displaystyle A|0 \rangle = 0 \\ A^\dagger A|0\rangle = 0 \\ \left(H- \frac{\hbar\omega}{2}\right)|0\rangle \\ H|0\rangle = \frac{\hbar\omega}{2}|0\rangle = E_0 |0\rangle

Thus, the ground state energy is \hbar \omega /2 !


We can donate

\displaystyle A^\dagger |n\rangle = U_n |n+1\rangle \\ A |n\rangle = D_n |n-1\rangle

Using \displaystyle A^\dagger A - A A^\dagger = -\hbar \omega, we have

\displaystyle (A^\dagger A - A A^\dagger ) |n \rangle \\ = A^\dagger D_n |n-1 \rangle - A U_n|n+1 \rangle \\ = (U_{n-1}D_n- U_n D_{n+1} )|n\rangle = - \hbar \omega |n\rangle

or

\displaystyle  U_n D_{n+1}  = U_{n-1}D_n + \hbar \omega 

Consider above equation on |0\rangle , such that A|0\rangle = 0, thus,

\displaystyle (A^\dagger A - A A^\dagger ) |0 \rangle = - U_0 D_{1} )|0\rangle = - \hbar \omega |0\rangle

Thus, U_0 D_1 = \hbar \omega , then, in general, we have

\displaystyle  U_n D_{n+1}  = (n+1) \hbar \omega 

Then

\displaystyle A^\dagger A |n\rangle = U_{n-1} D_n |n\rangle \\ A A^\dagger |n\rangle = U_{n} D_{n+1} |n\rangle

The simplest solution is

U_n = \sqrt{\hbar \omega}\sqrt{n+1} \\ D_n =\sqrt{\hbar \omega}\sqrt{n}

And, if we treat the state |n\rangle contains n oscillators, we have a number operator N = A^\dagger A , such that

A^\dagger A |n \rangle = n \hbar \omega |n \rangle


some people will normalized the ladder operators with \sqrt{\hbar \omega}. For me, I don’t want to make the starting point be so “artificial”. We can summarized the 1-D Harmonic oscillator as

HarmonicOsciilator.png


Now, we are going to solve the wave function. One way is use the ladder operator,

\displaystyle A|0\rangle = 0 \\ \left( \sqrt{\frac{1}{2m}} P - i \sqrt{\frac{m \omega^2}{2}} X \right) \phi_0 = 0 \\ \frac{d\phi_0}{dx} = - \frac{m \omega}{\hbar} x \phi_0 = 0

The solution is

\displaystyle \phi_0 = N \exp\left(-\frac{m \omega}{2\hbar}x^2\right)

Another way is the Schrodinger equation is

\displaystyle -\frac{\hbar^2}{2m}\frac{d^2\phi}{dx^2} + \frac{1}{2}m\omega^2 x^2 \phi = E \phi = \frac{2n+1}{2}\hbar \omega \phi

\displaystyle \frac{d^2\phi}{dx^2} - \frac{m^2 \omega^2}{\hbar^2} x^2 \phi = (2n+1)\frac{m \omega}{\hbar} \phi

Define \alpha = \sqrt{\frac{\hbar}{m \omega}} , the dimension of \alpha is length. Set u = x/ \alpha , then

\displaystyle \frac{d^2\phi}{du^2} + ((2n+1) - u^2) \phi = 0

Since we know that the function must decay outside the well, set \phi = f(y) \exp(-y^2/2) , then

\displaystyle \frac{d^2f}{du^2} -2 u \frac{df}{du} +2 nf = 0

The solution is Hermite polynomial H_n(u) of order n.

Assume

\displaystyle H_n(x) = \sum_{i = 0}^{\infty} a_i x^i

sub into the Hermite equation,

\displaystyle a_{i+2} = \frac{2(i-n)}{(i+1)(i+2)} a_i

First,  the even and odd series are separated. And either the odd-series or the even-series are converge, as the ratio

\displaystyle \lim_{i->\infty} \frac{a_{i+2}}{a_i}  = 2

However, the series can be terminated when i = n. Thus, the Hermite polynomial has either even terms or odd terms, but not mixed.

When n = 0, a_2 = 0 and all later even terms are zero. The equation is f''(x) - 2 x f'(x) = 0 , which only support even function. Thus, H_0(x) = a_0 is a constant. Thus, we can see, when n is even (odd), Hermite polynomial is even (odd) with only even (odd) n.

Hermite polynomial are orthogonal with \exp(-x^2), thus, the wave function \phi_n(x) = H_n(x) \exp(-x^2/2) are orthogonal.

 

Finite Spherical Square Well

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Hello, everyone, in order to calculate deuteron by Hartree-Fock method, I need a basis. The basis of infinite spherical square well is too “rigid”, that it has to “extension” to non-classical region. Beside of the basis of Wood-Saxon potential. The finite spherical square well is a good alternative. The radial equation is basically the same as infinite spherical square well.

The potential is

\displaystyle V(r) = -|V_0|, r\leq a

\displaystyle V(r) = 0, r > a

Within the well, the wave vector is

\displaystyle k = \sqrt{\frac{2m}{\hbar^2} (|V_0 |-|E|) }

, outside the well, the wave vector is

\displaystyle k' = i \kappa = i \sqrt{\frac{2m}{\hbar^2} |E| }

The solution is spherical Bessel function. Since the Bessel function of the first and second kind are oscillating like sin or cosine function. To form a decay function when r > a, we need the Hankel function with complex argument.

\displaystyle h_n( i \kappa r) = h_n(x) = - (i x)^n \left( \frac{1}{x} \frac{d}{dx}\right)^n \left(\frac{\exp{(-x)}}{x} \right)

To make it real, we need a factor i^n .

The boundary conditions are continuity and differential continuity.

\displaystyle j_l(ka) = A i^l h_l(i\kappa a)

\displaystyle \left(\frac{d}{dr}j_l(kr)\right)_{r=a} = A i^l \left(\frac{d}{dr}h_l(i\kappa a) \right)_{r=a}

These two conditions solved for two parameters A and E. However, I cannot find an analytical solution to the energy E.


If the potential depth is 60 MeV for proton. Radius is 1 fm for a light nuclei. Set \hbar = 1, m = 1.

The result is follow,

FiniteSquareWell

The 1st column is 1s, 1p, 1d, 1f, and 1g. The 2nd column is 2s, 2p, 2d, 2f, and 2g. The 3rd column is 3s, 3p, and 3d.

By compare with infinite square well,

compare

The energies are lower in finite well, because the wave functions can spread-out to non-classical region, so that the wave length is longer and energy is lower.

In this example, the 3d and 2g orbits are bounded (of course, all orbits in infinite well are bound). This is not because of the depth of the well, but the boundary of the well. In other word, to bring down an orbit, the wave function has to spread out, that is connected with the neutron-halo.

 

Wave function in momentum space

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The wave function often calculated in spatial coordinate. However, in experimental point of view, the momentum distribution can be extracted directly from the experimental data.

The conversion between momentum space and position space is the Fourier transform

\displaystyle \phi(\vec{k}) = \frac{1}{\sqrt{2\pi}^3} \int Exp\left(-i \vec{k}\cdot \vec{r} \right) \phi(\vec{r}) d\vec{r}

Using the plane wave expansion

\displaystyle Exp(i k\cdot r) = \sum_{l=0}^\infty (2l+1) i^l j_l(kr) P_l(\hat{k}\cdot\hat{r})

or

\displaystyle Exp(i k\cdot r) = 4\pi \sum_{l=0}^\infty \sum_{m=-l}^{l} i^l j_l(kr) Y_{lm}(\Omega_k) Y_{lm}^{*}(\Omega_r)

Thus,

\displaystyle \phi(\vec{k}) = \frac{4\pi}{\sqrt{2\pi}^3} \sum_{l=0}^\infty (-i)^l \sum_{m=-l}^{l} \int j_l(k r) Y_{lm}(\Omega_k) Y_{lm}^*(\Omega) \phi(\vec{r}) r^2 dr d\Omega

where j_l (x) is spherical Bessel function. Usually, due to conservation of angular momentum, the angular part can be separated from the spatial part. Let assume the wave function in position space is

\phi(\vec{r}) = \psi(r) Y_{l_r m_r}(\Omega)

\phi(\vec{k}) = \psi(p) Y_{l_k m_k}(\Omega_k)

Then we have

\displaystyle \psi(k) = \frac{4\pi}{\sqrt{(2\pi)^3}} \int j_l(k r) \psi(r) r^2 dr

\displaystyle \phi(\vec{k}) = \psi(k) (-i)^l Y_{lm}(\Omega_k)

where l = l_r = l_k, m=m_r = m_k , due to the orthogonality of spherical harmonics.

For s, p, d, f-state, the spherical Bessel function is

\displaystyle j_0(x) = \frac{\sin(x)}{x}

\displaystyle j_1(x) = \frac{\sin(x)}{x^2} - \frac{\cos(x)}{x}

\displaystyle j_2(x) = \sin(x)\frac{2-x^2}{x^3} - \cos(x)\frac{3}{x^2}

\displaystyle j_3(x) = \sin(x)\frac{15-6x^2}{x^4} - \cos(x)\frac{15-x^2}{x^3}

For Hydrogen-like wave function, the non-normalized momentum distribution is

\displaystyle \psi_{10}(k) =  \frac{4 Z^{5/2}}{(k^2 + Z^2)^2}

\displaystyle \psi_{20}(k) = \frac{16 \sqrt{2}Z^{5/2}(4k^2-Z^2)}{(4k^2 + Z^2)^3}

\displaystyle \psi_{21}(k) =\sqrt{\frac{2}{3}}\frac{64 k Z^{7/2}}{(4k^2 + Z^2)^3}

\displaystyle \psi_{30}(k) =  \frac{36 \sqrt{3} Z^{5/2} (81k^3-30k^2Z^2+Z^4)}{(9k^2 + Z^2)^4}

\displaystyle \psi_{31}(k) = \frac{144 \sqrt{6} Z^{7/2} (9k^3-kZ^2)}{(9k^2 + Z^2)^4}

Here is the plot for momentum distribution (\psi_{nl}(k))^2 k^2 .

k_1.PNG

k_2.PNG

k_3.PNG

It is interesting that, the number of node decrease with higher angular momentum. But be-aware that it is only in atomic case, not a universal.

The higher the principle quantum number, the smaller of the spread of momentum. This is because, the spread of position wave function getting larger, and the uncertainty in momentum space will be smaller. This is a universal principle.

We also plot the Hydrogen radial function in here \psi(r)^2 x^2 , for reference,

r_1.PNG

r_2.PNG

r_3.PNG

 

 

Hartree-Fock for Helium excited state II

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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.

He_excite.PNG

 

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