Laplacian in spherical coordinate

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the Momentum operator in spherical coordinate

\nabla^2 = \frac {1}{r^2}\frac {\partial } { \partial r} \left ( r^2 \frac {\partial} {\partial r} \right ) - \frac {1}{r^2} L^2

where L is the Reduced angular momentum operator. the minus sign is very important for giving a correct sign. the original angular momentum operator J is related by:

J=\hbar^2 L

by compare the Laplacian in spherical coordinate, the L is

L^2 = - \frac {1}{sin(\theta)} \frac {\partial}{\partial \theta} \left( sin(\theta) \frac {\partial}{\partial \theta} \right ) - \frac {1}{sin(\theta)} \frac{\partial^2} {\partial \phi ^2}

But this complicated form is rather useless, expect you are mathematic madman.

we can start from classical mechanic

\vec{L} = \vec {r} \times \vec{p}

L_x = y \frac {\partial} {\partial z} - z \frac {\partial}{\partial y }

L_y = z \frac {\partial} {\partial x} - x \frac {\partial}{\partial z }

L_z = x \frac {\partial} {\partial y} - y \frac {\partial}{\partial x }

with the change of coordinate

\begin {pmatrix} x \\ y \\ z \end{pmatrix} = \begin {pmatrix} r sin(\theta) cos(\phi) \\ r sin(\theta) sin(\phi) \\ r cos(\theta) \end{pmatrix}

and the Jacobian Matrix M_J , which is used for related the derivatives.


\frac {\partial}{\partial x} = \frac {\partial r}{\partial x} \frac {\partial} {\partial r} +\frac {\partial \theta}{\partial x} \frac {\partial} {\partial \theta}+\frac {\partial \phi}{\partial x} \frac {\partial} {\partial \phi}

\frac {\partial}{\partial y} = \frac {\partial r}{\partial y} \frac {\partial} {\partial r} +\frac {\partial \theta}{\partial y} \frac {\partial} {\partial \theta}+\frac {\partial \phi}{\partial y} \frac {\partial} {\partial \phi}

\frac {\partial}{\partial z} = \frac {\partial r}{\partial z} \frac {\partial} {\partial r} +\frac {\partial \theta}{\partial z} \frac {\partial} {\partial \theta}+\frac {\partial \phi}{\partial z} \frac {\partial} {\partial \phi}

which can be simplify

\nabla_{(x,y,z)} = M_J^T \nabla_{(r, \theta, \phi )}

M_J = \frac {\partial ( r, \theta, \phi) }{\partial (x,y,z)}

M_J^{\mu\nu} = \frac {\partial \mu}{\partial \nu}

then, we have

L_x = i sin(\phi) \frac {\partial }{\partial \theta} +i cot(\theta) cos(\phi) \frac { \partial }{\partial \phi}

L_y =-i cos(\phi) \frac {\partial }{\partial \theta} + i cot(\theta) sin(\phi) \frac { \partial }{\partial \phi}

L_z = - i \frac {\partial }{\partial \phi}

However, even we have the functional form, it is still not good.  we need the ladder operator

L_+ = L_x + i L_y = Exp(i \phi) \left( \frac {\partial }{\partial \theta} + i cot(\theta) \frac { \partial }{\partial \phi} \right)

L_- = L_x - i L_y = Exp(-i \phi) \left( \frac {\partial }{\partial \theta} - i cot(\theta) \frac { \partial }{\partial \phi} \right)

notice that

L_+^\dagger = L_-

so, just replacing i \rightarrow -i .

when we looking for the Maximum state of the spherical Harmonic Y_{max}(\theta, \phi)

L_+ Y_{max}(\theta,\phi) = 0 *)

use the separable variable assumption.

Y_{max}(\theta, \phi) = \Theta \Phi

L_+ \Theta \Phi = 0 = - Exp(i \phi) \left( \frac {d\Theta}{d \theta} \Phi + i cot(\theta) \frac { d\Phi}{d\phi} \right) \Theta

\frac {tan(\theta)}{\Theta} \frac { d \Theta} {d \Theta } = - \frac {i}{\Phi} \frac {d \Phi} {d \phi} = m

the solution is

Y_{max}(\theta,\phi) = sin^m(\theta) Exp(i m \phi )

L^2 Y_{max}(\theta, \phi) = m(m+1) Y_{max}(\theta,\phi)

an application on Hydrogen wave function is here.

Hydrogen Atom (Bohr Model)

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OK, here is a little off track. But that is what i were learning and learned. like to share in here. and understand the concept of hydrogen is very helpful to understand the nuclear, because many ideas in nuclear physics are borrow from it, like “shell”.

The interesting thing is about the energy level of Hydrogen atom. the most simple atomic system. it only contains a proton at the center, um.. almost center, and an electron moving around. well, this is the “picture”. the fact is, there is no “trajectory” or locus for the electron, so technically, it is hard to say it is moving!

why i suddenly do that is because, many text books said it is easy to calculate the energy level and spectrum for it. Moreover, many famous physicists said it is easy. like Feynman, Dirac, Landau, Pauli, etc… OK, lets check how easy it is.

anyway, we follow the usual say in every text book. we put the Coulomb potential in the Schrödinger equation, change the coordinate to spherical. that is better and easy for calculation because the coulomb potential is spherical symmetric. by that mean, the momentum operator (any one don’t know what is OPERATOR, the simplest explanation is : it is a function of function.) automatically separated into 2 parts : radial and angular part. The angular part can be so simple that it is the Spherical harmonic.

Thus the solution of the “wave function” of the electron, which is also the probability distribution of  the electron location, contains 2 parts as well. the radial part is not so trivial, but the angular part is so easy. and it is just Y(l,m) .

if we denote the angular momentum as L, and the z component of it is Lz, thus we have,

L^2 Y(l,m) = l(l+1) \hbar^2 Y(l,m)

L_z Y(l,m) = m \hbar Y(l,m)

as every quadratic operator, there are “ladder” operator for “up” and “down”.

L_\pm Y(l,m) =\hbar \sqrt{l(l+1) - m(m\pm 1)} Y(l,m \pm 1)

which means, the UP operator is increase the z-component by 1, the constant there does not brother us.

it is truly easy to find out the exact form of the Y(l,m) by using the ladder operator. as we know, The z component of the a VECTOR must have some maximum. so, there exist an Y(l,m) such that

L_+ Y(l,m) =0

since there is no more higher z-component.

by solve this equation, we can find out the exact form of Y(l,m) and sub this in to L2, we can knowMax(m) = l . and apply the DOWN operator, we can fins out all Y(l,m) , and the normalization constant is easy to find by the normalization condition in spherical coordinate, the normalization factor is sin(\theta) , instead of 1 in rectangular coordinate.

\int_0^\pi \int_0^{2 \pi} Y^*(l',m') Y(l,m) sin(\theta) d\theta d \psi = \delta_{l' l} \delta_{m' m}

more on here