Diagonal elements for Nilsson orbital in spherical-spin function

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In this post, we explain how to calculate the Nilsson orbital using perturbation method by compute the matrix elements using spherical-spin function. In that post, I said I will give the calculation for the perturbation element. Here we go for the diagonal elements

The diagonal matrix element,

\displaystyle \left<Nljk|r^2 Y_{20}|Nljk\right>

where the spherical-spin wave-function is

\displaystyle |Nljk\rangle = A r^l e^{-\frac{r^2}{2}} L_{\frac{N-l}{2}}^{l + \frac{1}{2}}(r^2) \sum_{m m_s} Y_{lm}(\theta, \phi) \chi_{\frac{1}{2} m_s} C_{lm\frac{1}{2} m_s}^{jk}

\displaystyle A = \sqrt{\frac{(\frac{N-l}{2})!(\frac{N+l}{2})! 2^{N+l+2}}{\sqrt{\pi} (N+l+1)!}}

First, the radial part is

\displaystyle R_{Nl}(r) = r^l e^{-\frac{r^2}{2}} L_{\frac{N-l}{2}}^{l + \frac{1}{2}}(r^2)

The radial integral is easy, we can use the integration formula. The radial integration is,

\displaystyle \int_0^{\infty} r^2 R_{Nl}^2(r) r^2 dr = \int_0^{\infty} r^{2l+4} e^{-r^2} \left(L_{\frac{N-l}{2}}^{l + \frac{1}{2}}(r^2)\right)^2 dr

When using the integration formula, one has to be careful when changing variable r^2 \rightarrow r . Since dr^2 = 2 r dr , we have to pull 2r out to properly do the change of variable.

\displaystyle \int_0^{\infty} R_{Nl}^2(r) r^2 dr = \frac{1}{2}\int_0^{\infty} r^{2(l+1/2 + 1)} e^{-r^2} \left(L_{\frac{N-l}{2}}^{l + \frac{1}{2}}(r^2)\right)^2 (2 r dr)

Set r^2 \rightarrow x , l + 1/2 \rightarrow \alpha , and n = \frac{N-l}{2}

\displaystyle  = \frac{1}{2}\int_0^{\infty} x^{\alpha+1} e^{-x} \left(L_{N}^{\alpha}(x)\right)^2 (dx) = \frac{1}{2}\frac{(\alpha+n)!}{n!}(2n+\alpha+1)


\displaystyle \int_0^{\infty} R_{Nl}^2(r) r^2 dr = \frac{1}{2}\frac{( \frac{N+l+1}{2})!}{(\frac{N-l}{2})!}(N+\frac{3}{2})

The angular-spin part is

\displaystyle S_{ljk}(\theta,\phi, m_s) = \sum_{m m_s} Y_{lm}(\theta, \phi) \chi_{\frac{1}{2} m_s} C_{lm\frac{1}{2} m_s}^{jk}

This contains spatial and spin part.

\displaystyle S_{ljk}^*(\theta,\phi, m_s) Y_{20} S_{ljk}(\theta,\phi, m_s) \\=\sum_{m' m'_s} \sum_{m m_s} Y_{lm'}^*(\theta,\phi) Y_{20}  Y_{lm}(\theta, \phi) \chi_{\frac{1}{2} m'_s}  \cdot \chi_{\frac{1}{2} m_s} C_{lm'\frac{1}{2} m'_s}^{jk} C_{lm\frac{1}{2} m_s}^{jk}

The dot-product of the spin part restricted the m'_s

\displaystyle = \sum_{m' m m_s} Y_{lm'}^*(\theta,\phi) Y_{20}  Y_{lm}(\theta, \phi) C_{lm'\frac{1}{2} m_s}^{jk} C_{lm\frac{1}{2} m_s}^{jk}

And since m+m_s = k and m'+m_s =k , therefore m = m',

\displaystyle = \sum_{m m_s} Y_{lm}^*(\theta,\phi) Y_{20}  Y_{lm}(\theta, \phi) \left(C_{lm\frac{1}{2} m_s}^{jk}\right)^2

And the integration of spherical harmonic gives,

\displaystyle \int Y_{lm}^*(\theta,\phi) Y_{20}  Y_{lm}(\theta, \phi) = \sqrt{\frac{5}{4\pi}} C_{20l0}^{l0} C_{20lm}^{lm}

\displaystyle C_{20l0}^{l0} = -\frac{l+1}{\sqrt{4l^3+8l^2+l-3}} , l > 0

\displaystyle C_{20l0}^{l0} = -\frac{l(l+1)-3m^2}{\sqrt{l(4l^3+8l^2+l-3)}} , l > 0

Sum up everything,

\displaystyle \left<Nljk|r^2 Y_{20}|Nljk\right> \\ =  \frac{(\frac{N-l}{2})!(\frac{N+l}{2})! 2^{N+l+2}}{\sqrt{\pi} (N+l+1)!} \frac{1}{2}\frac{( \frac{N+l+1}{2})!}{(\frac{N-l}{2})!}(N+\frac{3}{2}) \sum_{m m_s} \sqrt{\frac{5}{4\pi}} C_{20l0}^{l0} C_{20lm}^{lm} \left(C_{lm\frac{1}{2} m_s}^{jk}\right)^2 \\ = \sqrt{\frac{5}{4}} \frac{(\frac{N+l}{2})! ( \frac{N+l+1}{2})! 2^{N+l+1}}{\pi (N+l+1)!}(N+\frac{3}{2}) \sum_{m m_s}  C_{20l0}^{l0} C_{20lm}^{lm} \left(C_{lm\frac{1}{2} m_s}^{jk}\right)^2

Note: I haven’t numerically check the formula. ( may be later )

For the off-diagonal element. The angular-spin part should be similar. The difficulty is the radial part, we have to evaluate the most general orthogonal relation of the Laguerre polynomial with weighting x^{\frac{1}{2}(\alpha+\alpha'+2)} e^{-x} .

We only knew the Laguerre polynomil is orthogonal with respect to r^\alpha e^{-r}, i.e.

\displaystyle \int_0^{\infty} L_n^\alpha L_m^\alpha r^\alpha e^{-r} dr  = \frac{(\alpha+n)!}{n!} \delta_{nm}

But not this.

\displaystyle \int_0^{\infty} L_n^\alpha L_m^\beta r^\frac{\alpha+\beta}{2} e^{-r} dr

or this

\displaystyle \int_0^{\infty} L_n^\alpha L_m^\beta r^{(\frac{\alpha+\beta}{2}+k)} e^{-r} dr

But we are quite sure the last one, with $\alpha \neq \beta$ will not give zero, otherwise, The Nilsson orbital will be very simple and boring.


Axial Harmonic Oscillator – Nilsson Orbit

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The Hamiltonian is

\displaystyle H = -\frac{\hbar^2}{2m}\nabla^2 +\frac{m}{2}(\omega_\rho^2(x^2+y^2)+\omega_z z^2 )

Use cylindrical coordinate, the Schrodinger equation is

\displaystyle \left(-\frac{\hbar^2}{2m}\left(\frac{d^2}{dz^2} + \frac{1}{\rho}\frac{d}{d\rho}\left(\rho\frac{d}{d\rho}\right) + \frac{1}{\rho^2}\frac{d^2}{d\phi^2}\right) +\frac{m}{2}(\omega_\rho^2(\rho^2)+\omega_z z^2 ) \right) \Psi = E \Psi

Where the energy is

\displaystyle E = \hbar\omega_\rho(1+ n_x + n_y) + \hbar\omega_z\left(\frac{1}{2}+ n_z\right)

Note that n_x+n_y \neq n_\rho. One of the reason is there are 2 degree of freedom, it cannot be solely expressed into 1 parameter.

Set \Psi = Z P \Phi ,

\displaystyle -\frac{\hbar^2}{2m}\left(\frac{d^2Z}{dz^2} P \Phi  + \frac{1}{\rho}\frac{d}{d\rho}\left(\rho\frac{dP}{d\rho} \right) Z \Phi + \frac{1}{\rho^2}\frac{d^2\Phi}{d\phi^2} Z P\right)  \\ +  \frac{m}{2}\omega_\rho^2\rho^2 ZP\phi+\frac{m}{2}\omega_z z^2 ZP\Phi= E ZP\Phi

\displaystyle -\frac{\hbar^2}{2m}\left(\frac{d^2Z}{dz^2} \frac{1}{Z}  + \frac{1}{\rho}\frac{d}{d\rho}\left(\rho\frac{dP}{d\rho} \right) \frac{1}{P} + \frac{1}{\rho^2}\frac{d^2\Phi}{d\phi^2} \frac{1}{\Phi}\right)  \\ +  \frac{m}{2}\omega_\rho^2\rho^2 +\frac{m}{2}\omega_z z^2 = E 

The angular part, we can set

\displaystyle \frac{d^2\Phi}{d\phi^2} \frac{1}{\Phi} = -m_\phi^2

The solution is \Phi(\phi) = \exp(i m_\phi \phi)

The z-part is the 1D Harmonic oscillator, the solution is the Hermite function.

Thus the rest is

\displaystyle -\frac{\hbar^2}{2m}\left( \frac{1}{\rho}\frac{d}{d\rho}\left(\rho\frac{dP}{d\rho} \right) \frac{1}{P} - \frac{m_\phi^2}{\rho^2}\right) +  \frac{m}{2}\omega_\rho^2\rho^2 =  \hbar\omega_\rho(1+ n_x + n_y)


\displaystyle  \frac{1}{\rho}\frac{d}{d\rho}\left(\rho\frac{dP}{d\rho} \right) - \frac{m_\phi^2}{\rho^2}P +  \frac{m^2\omega_\rho^2}{\hbar^2}\rho^2 P + \frac{m\omega_\rho}{\hbar}(1+ n_x + n_y) P = 0

Set a dimensionless constant \alpha^2 = \frac{\hbar}{m \omega_\rho} , and x = \rho/\alpha

\displaystyle \frac{1}{x}\frac{d}{dx}\left(x\frac{dP}{dx}\right) + \left( 2(1+n_x+n_y) - \frac{m_\phi^2}{x^2} - x^2 \right) P = 0

Using the normalisation formula, \int P^2 x dx = 1, set u = P \sqrt{x}

\displaystyle x^2\frac{d^2u}{dx^2}+ \left( 2(1+n_x+n_y) - \frac{\frac{1}{4}-m_\phi^2}{x^2} - x^2 \right) u = 0

Because of the long-range and short-range behavior, set

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

\displaystyle x\frac{d^2f}{dx^2} + (1+2m_\phi-2x^2)\frac{df}{dx} + 2(n_x+n_y-m_\phi)x f= 0

Set y = x^2

\displaystyle y \frac{d^2f}{dy^2} + (m_\phi+1-y)\frac{df}{dy} + \frac{n_x+n_y-m_\phi}{2} f= 0

Define n_\rho = \frac{n_x+n_y-m_\phi}{2} \rightarrow n_x+n_y = 2 n_\rho + m_\phi = N - n_z.

\displaystyle y \frac{d^2f}{dy^2} + (m_\phi+1-y)\frac{df}{dy} + n_\rho f= 0

This is our friend again! The Laguerre polynomial! The complete solution is

\displaystyle \Psi_{n_z n_\rho m_\phi}(z, \rho, \phi) \\ = \sqrt{\frac{1}{\alpha_z \alpha^2}}\sqrt{\frac{ n_\rho !}{2^n_z  n_z! (m_\phi + n_\rho)!\sqrt{\pi^3}}} H_{n_z}\left(\frac{z}{\alpha_z}\right) \\ \exp\left(- \frac{1}{2}\left(\frac{z^2}{\alpha_z^2}+\frac{\rho^2}{\alpha^2}\right)\right) \left(\frac{\rho}{\alpha}\right)^{m_\phi} L_{n_\rho}^{m_\phi} \left(\frac{\rho^2}{\alpha^2}\right) \exp(i m_\phi \phi)

where \alpha_z^2 = \hbar/m/\omega_z

The energy is

\displaystyle E = \hbar\omega_z \left(\frac{1}{2} + n_z \right) + \hbar\omega_\rho \left( 2n_\rho + m_\phi + 1 \right)

The quantum number m_\phi has same meaning as m_l in spherical case.

The notation for the state is

|Nn_z m_l \rangle

with the spin, the only good quantum number is the z-component of the total angular momentum J = L+S  long the body axis, i.e. K = m_\phi \pm 1/2 , thus, the state is

|Nn_z m_l K \rangle

Note that the total angular momentum J^2 is not a good quantum number in deformation, as the rotational symmetry is lost. However, the quantum number K is linked with the angular momentum of the Nilsson single particle orbit. It is because when a particle has m_j = K, the angular momentum must at least j \geq K .

The above is a general solution for the harmonic oscillator in cylindrical coordinate. When \omega_z = \omega_\rho , it reduce to spherical case.

According to P. Ring & P. Schuck (2004) (The Nuclear Many-Body Problem, P.68), the Hamiltonian can be expressed as a quadruple deform field by setting

\displaystyle \omega_\rho^2 = \omega^2 \left(1+\frac{2}{3} \delta\right) \\ \omega_z^2 = \omega^2 \left(1-\frac{4}{3} \delta\right)

\displaystyle H = -\frac{\hbar^2}{2m}\nabla^2 + \frac{1}{2} m \omega^2 \left(r^2- \frac{2}{3}\sqrt{\frac{16\pi}{5}} \delta r^2 Y_{20}(\theta,\phi) \right)

This Hamiltonian has similarity with the deformation

\displaystyle R(\theta, \phi) = R_0 \left(1 + \alpha_00 + \sum_{\lambda=1}^{\infty}\sum_{\mu=-\lambda}^{\lambda} \alpha_{\lambda \mu} Y_{\lambda \mu}(\theta, \phi) \right)

take the first quadruple term, and calculate R^2(\theta, \phi)

R^2(\theta, \phi) = R_0^2 (1 + 2 \beta Y_{20}(\theta,\phi) )

Compare, we have

\displaystyle  \beta = \frac{1}{3}\sqrt{\frac{16\pi}{5}}\delta \approx 1.05689 \delta


The ratio

\displaystyle \frac{\omega_z}{\omega_\rho} =  \left(\frac{\alpha_\rho}{\alpha_z} \right)  = \sqrt{\frac{1-\frac{4}{3}\delta}{1+\frac{2}{3}\delta}} = 1- \delta + \frac{1}{6}\delta^2 - \frac{5}{18}\delta^3... 

We need volume conservation

\omega_x \omega_y \omega_z = \omega_0^3


\displaystyle \omega = \omega_0 \left(\frac{1}{(1+2/3\delta)(1-4/3\delta)} \right)^{\frac{1}{6}} \approx \omega_0 \left(1+\frac{2}{9}\delta^2 + \frac{8}{81}\delta^3 \right)

The energy is

\displaystyle E = \hbar\omega \left( \sqrt{1-\frac{4}{3}\delta}\left(\frac{1}{2} + n_z \right) + \sqrt{1+\frac{2}{3}\delta} \left( N - n_z + 1 \right) \right) \\ \approx \hbar \omega_0 \left( \frac{3}{2} + \left(1 + \frac{1}{3}\delta\right) N - \delta n_z \right)


Here is some plots with various \delta = 0, 0.3, 0.5




Next time, I will add the LS coupling and L-square term to recreate the Nilsson diagram. Also I will expand the solution from cylindrical coordinate into spherical coordinate. This unitary transform is the key to understand the single particle-ness.



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


\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  \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 see more clearly, we can project the function onto spherical harmonic, which is fixed angular momentum, i.e.

\langle \Phi(x,y,z) | Y_{lm}(\theta,\phi) \rangle = C_{lm} R_{nl}(r).

where C_{lm} is coefficients and R_{nl}(r) is radial function.

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 \geq 0 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)

The normalization constant can be calculate easily using the integral formula of Laguerre polynomial.

\displaystyle \int R^2(r) r^2 dr = 1

\displaystyle N^2 \int r^{2l} \exp\left(-\frac{r^2}{2\alpha^2}\right) \left( L_{k}^{l+\frac{1}{2}}\left( \frac{r^2}{\alpha^2} \right)\right)^2 r^2 dr  = 1

change of variable y = r^2/\alpha^2 \rightarrow r=\alpha \sqrt{y}

\displaystyle dr = \frac{\alpha}{2\sqrt{y}}dy

\displaystyle N^2 \frac{\alpha^{2l+3}}{2} \int y^{l+\frac{1}{2}} \exp(-y) \left( L_k^{l+\frac{1}{2}} \right)^2 dy = 1

The integration is

\displaystyle N^2 \frac{\alpha^{2l+3}}{2} \frac{(k+l+\frac{1}{2})!}{k!} = 1

\displaystyle N^2 = \frac{2}{\alpha^{2l+3}} \frac{k!}{(k+l+\frac{1}{2})!}

we can use

\displaystyle \left(n+\frac{1}{2}\right) = \frac{(2n+1)!}{2^{2n+1}n!}\sqrt{\pi}


\displaystyle N^2 = \frac{1}{\sqrt{\pi}\alpha^{2l+3}} \frac{k! (k+l)! 2^{2k+2l+3}}{(2k+2l+1)!}

replace k = (n-l)/2 . The total wave function is

\displaystyle \Psi_{nlm}(r, \theta, \phi) \\ =\sqrt{ \frac{1}{\sqrt{\pi}\alpha^{2l+3}} \frac{(\frac{n-l}{2})! (\frac{n+l}{2})! 2^{n+l+2}}{(n+l+1)!}} 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)

Here are some drawing of the square of the wave functions. From the below is n = 0, 1, 2, 3, from left to right, are s-orbit, p-orbit, d-orbit, f-orbit.

3-D Harmonic Oscillator.png

With the LS coupling, the spatial function does not affected, unless the coupling has spatial dependence. With the LS coupling, the good quantum numbers are n, l, j, m_j



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})


\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)


\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(k) 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 .




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,






Complete derivation from Schrodinger equation to Laguerre equation for Coulomb potential

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The Hamiltonian is

\displaystyle H = -\frac{\hbar^2}{2m}\nabla^2 - \frac{Ze^2}{4\pi\epsilon_0r}

Separate the radial and angular part. The radial equation is

\displaystyle  \left( -\frac{\hbar^2}{2m}\frac{1}{r^2}\left(\frac{d}{dr} r^2 \frac{d}{dr} \right) - \frac{Ze^2}{4\pi\epsilon_0 r} + \frac{\hbar^2}{2m} \frac{l(l+1)}{r^2} \right) R(r) = E R(r)

rearrange, using Atomic unit

\displaystyle  \left( -\frac{1}{2}\frac{1}{r^2}\left(\frac{d}{dr} r^2 \frac{d}{dr} \right) - E -\frac{Z}{ r} + \frac{1}{2} \frac{l(l+1)}{r^2} \right) R(r) = 0

Since the normalization condition of R(r) is \int_0^\infty R(r)R(r) r^2 dr = 1, it is natural to define u(r) = r R(r)

\displaystyle R = \frac{u}{r}, \frac{d}{dr}\left(r^2\frac{dR}{dr}\right) = r \frac{d^2u}{dr^2}

The radial equation becomes

\displaystyle -\frac{1}{2}\frac{d^2u}{dr^2} - \left( E + \frac{Z}{ r} - \frac{l(l+1)}{2r^2} \right) u(r)= 0

Substitute the eigen energy E_n = - \frac{Z^2}{2n^2}

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


\displaystyle x = \frac{2Z}{n} r,   \frac{d}{dr} = \frac{d}{dx} \frac{2Z}{n}

The radial equation becomes

\displaystyle \left(\frac{2Z}{n}\right)^2\frac{d^2u(x)}{dx^2} - \left( - \frac{Z^2}{n^2} + \frac{4Z^2}{n x} - \frac{4Z^2}{n^2}\frac{l(l+1)}{x^2} \right) u(x)= 0

Take out the 4Z^2/n^2

\displaystyle \frac{d^2u(x)}{dx^2} - \left(\frac{n}{ x} - \frac{1}{4}  - \frac{l(l+1)}{x^2} \right) u(x)= 0

Now, we know the short and long-range behaviour of u(x) , assume it to be

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

Then the equation of y(x) is

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

This is the Laguerre equation

\displaystyle x \frac{d^2y}{dx^2} + \left( \alpha+1 - x\right) \frac{dy}{dx} + m y(x) = 0

where \alpha = 2l+1 and m = n-l -1. Therefore, the solution of the radial equation is,

R(r) = \frac{1}{r} A \exp\left(-\frac{x}{2} \right) x^{l+1} L_{n-l-1}^{2l+1}(x)

where A is normalization factor.

Notice that the Laguerre polynomial is only defined for m \geq 0, thus, n > l .

Using generating function to calculate integrals of Laguerre polynomial

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The generating function of Laguerre polynomial is

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

We are going to evaluate the integral

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

Consider this integral

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

It is equal to

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

It is easy to calculate the last integral,

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

Replace variable,

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

The integral becomes

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

Expand the right hand side,

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

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

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

We used

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

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

At the last step. The product

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

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

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

When \alpha = \beta

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

For example, k = 0,

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

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

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

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

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

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

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

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

Hartree method for Helium ground state

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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.

The radial part,

\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

The radial part

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).


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