a review on Hydrogen’s atomic structure

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I found that most of the book only talk part of it or present it separately. Now, I am going to treat it at 1 place. And I will give numerical value as well. the following context is on SI unit.

a very central idea when writing down the state quantum number is, is it a good quantum number? a good quantum number means that its operator commute with the Hamiltonian. and the eigenstate states are stationary or the invariant of motion. the prove on the commutation relation will be on some post later. i don’t want to make this post too long, and with hyperlink, it is more reader-friendly. since somebody may like to go deeper, down to the cornerstone.  but some may like to have a general review.

the Hamiltonian of a isolated hydrogen atom is given by fews terms, deceasing by their strength.

H = H_{Coul} + H_{K.E.} + H_{Rel} + H_{Darwin} + H_{s-0} + H_{i-j} + H_{lamb} + H_{vol} + O

the Hamiltonian can be separated into 3 classes.

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Bohr model

H_{Coul} = - \left(\frac {e^2}{4 \pi \epsilon_0} \right) \frac {1}{r}

is the Coulomb potential, which dominate the energy. recalled that the ground state energy is -13.6 eV. and it is equal to half of the Coulomb potential energy, thus, the energy is about 27.2 eV, for ground state.

H_{K.E.} = \frac {P^2}{ 2 m}

is the non-relativistic kinetic energy, it magnitude is half of the Coulomb potential, so, it is 13.6 eV, for ground state.

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this 2 terms are consider in the Bohr model, the quantum number, which describe the state of the quantum state, are

n = principle number. the energy level.

l = orbital angular momentum. this give the degeneracy of each energy level.

m_l = magnetic angular momentum.

it is reasonable to have 3 parameters to describe a state of electron. each parameter gives 1 degree of freedom. and a electron in space have 3. thus, change of basis will not change the degree of freedom. The mathematic for these are good quantum number and the eigenstate \left| n, l, m_l \right> is invariant of motion, will be explain in later post. But it is very easy to understand why the angular momentum is invariant, since the electron is under a central force, no torque on it. and the magnetic angular momentum is an invariant can also been understood by there is no magnetic field.

the principle quantum number n is an invariance. because it is the eigenstate state of the principle Hamiltonian( the total Hamiltonian )!

the center of mass also introduced to make more correct result prediction on energy level. but it is just minor and not much new physics in it.

Fine structure

H_{Rel} = - \frac{1}{8} \frac{P^4}{m^3 c^2}

is the 1st order correction of the relativistic kinetic energy. from K.E. = E - mc^2 = \sqrt { p^2 c^2 + m^2c^4} - mc^2 , the zero-order term is the non-relativistic kinetic energy. the 1st order therm is the in here. the magnitude is about 1.8 \times 10^{-4} eV . ( the order has to be recalculate, i think i am wrong. )

H_{Darwin} = \frac{\hbar^{2}}{8m_{e}^{2}c^{2}}4\pi\left(\frac{Ze^2}{4\pi \epsilon_{0}}\right)\delta^{3}\left(\vec r\right)

is the Darwin-term. this term is result from the zitterbewegung, or rapid quantum oscillations of the electron. it is interesting that this term only affect the S-orbit. To understand it require Quantization of electromagnetic field, which i don’t know. the magnitude of this term is about 10^{-3} eV

H_{s-o} = \left(\frac{Ze^2}{4\pi \epsilon_{0}}\right)\left(\frac{1}{2m_{e}^{2}c^{2}}\right)\frac{1}{r^3} L \cdot S

is the Spin-Orbital coupling term. this express the magnetic field generated by the proton while it orbiting around the electron when taking electron’s moving frame. the magnitude of this term is about 10^{-4} eV

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this fine structure was explained by P.M.Dirac on the Dirac equation. The Dirac equation found that the spin was automatically come out due to special relativistic effect. the quantum number in this stage are

n = principle quantum number does not affected.

l = orbital angular momentum.

m_l = magnetic total angular momentum.

s = spin angular momentum. since s is always half for electron, we usually omit it. since it does not give any degree of freedom.

m_s = magnetic total angular momentum.

at this stage, the state can be stated by \left| n, l, m_l, m_s \right> , which shown all the degree of freedom an electron can possible have.

However, L_z is no longer a good quantum number. it does not commute with the Hamiltonian. so, m_l does not be the eigenstate anymore. the total angular momentum was introduced J = L + S . and J^2 and J_z commute with the Hamiltonian.  therefore,

j = total angular momentum.

m_j = magnetic total angular momentum.

an eigenstate can be stated as \left| n, l, s, j, m_j \right> . in spectroscopy, we denote it as ^{2 s+1} L _j , where L is the spectroscopy notation for l .

there are 5 degrees of freedom, but in fact, s always half, so, there are only 4 real degree of freedom, which is imposed by the spin ( can up and down).  the reason for stating the s in the eigenstate is for general discussion. when there are 2 electrons, s can be different and this is 1 degree of freedom.

Hyperfine Structure

H_{i-j} = \alpha I \cdot J

is the nuclear spin- electron total angular momentum coupling. the coefficient of this term, i don’t know. Sorry. the nuclear has spin, and this spin react with the magnetic field generate by the electron. the magnitude is 10^{-5}

H_{lamb}

is the lamb shift, which also only affect the S-orbit.the magnitude is 10^{-6}

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the hyperfine structure always makes alot questions in my mind. the immediate question is why not separate the orbital angular momentum and the electron spin angular momentum? why they first combined together, then interact with the nuclear spin?

may be i open another post to talk about.

The quantum number are:

n = principle quantum number

l = orbital angular momentum

s = electron spin angular momentum.

j = spin-orbital angular momentum of electron.

i = nuclear spin. for hydrogen, it is half.

f = total angular momentum

m_f = total magnetic angular momentum

a quantum state is $\left| n, l, s, j,i, f , m_f \right>$. but since the s and i are always a half. so, the total degree of freedom will be 5. the nuclear spin added 1 on it.

Smaller Structure

H_{vol}

this term is for the volume shift. the magnitude is 10^{-10} .

in diagram:

on angular momentum adding & rotation operator

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the angular momentum has 2 kinds – orbital angular momentum L , which is caused by a charged particle executing orbital motion, since there are 3 dimension space. and spin S , which is an internal degree of freedom to let particle “orbiting” at there.

thus, a general quantum state for a particle should not just for the spatial part and the time part. but also the spin, since a complete state should contains all degree of freedom.

\left| \Psi \right> = \left| x,t \right> \bigotimes \left| s \right>

when we “add” the orbital angular momentum and the spin together, actually, we are doing:

J = L \bigotimes 1 + 1 \bigotimes S

where the 1 with L is the identity of the spin-space and the 1 with S is the identity of the 3-D space.

the above was discussed on J.J. Sakurai’s book.

the mathematics of L and S are completely the same at rotation operator.

R_J (\theta) = Exp( - \frac {i}{\hbar} \theta J)

where J can be either L or S.

the L can only have effect on spatial state while S can only have effect on the spin-state. i.e:

R_L(\theta) \left| s \right> = \left| s\right>

R_S(\theta) \left| x \right> = \left| x\right>

the L_z can only have integral value but S_z can be both half-integral and integral. the half-integral value of Sz makes the spin-state have to rotate 2 cycles in order to be the same again.

thus, if the different of L and S is just man-made. The degree of freedom in the spin-space is actually by some real geometry on higher dimension. and actually, the orbital angular momentum can change the spin state:

L \left| s \right> = \left | s' \right > = c \left| s \right>

but the effect is so small and

R_L (\theta) \left| s\right > = Exp( - \frac {i}{\hbar} \theta c )\left| s \right>

but the c is very small, but if we can rotate the state for a very large angle, the effect of it can be seen by compare to the rotation by spin.

\left < R_L(\omega t) + R_S(\omega t) \right> = 2 ( 1+ cos ( \omega ( c -1 ) t)

the experiment can be done as follow. we apply a rotating magnetic field at the same frequency as the Larmor frequency. at a very low temperature, the spin was isolated and T_1 and T_2 is equal to \infty . the different in the c will come up at very long time measurement and it exhibit a interference pattern.

if c is a complex number, it will cause a decay, and it will be reflected in the interference pattern.

if we find out this c, then we can reveal the other spacial dimension!

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the problem is. How can we act the orbital angular momentum on the spin with out the effect of spin angular momentum? since L and S always coupled.

one possibility is make the S zero. in the system of electron and positron. the total spin is zero.

another possibility is act the S on the spatial part. and this will change the energy level.

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an more fundamental problem is, why L and S commute? the possible of writing this

\left| \Psi \right> = \left| x,t \right> \bigotimes \left| s \right>

is due to the operators are commute to each other. by why?

if we break down the L in to position operator x and momentum operator p, the question becomes, why x and S commute or p and S commute?

[x,S]=0 ?

[p,S]=0 ?

[p_x, S_y] \ne 0 ?

i will prove it later.

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another problem is, how to evaluate the Poisson bracket? since L and S is not same dimension. may be we can write the eigenket in vector form:

\begin {pmatrix} \left|x, t \right> \\ \left|s\right> \end {pmatrix}

i am not sure.

 

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For any vector operator, it must satisfy following equation, due to rotation symmetry.

[V_i, J_j] = i \hbar V_k   run in cyclic

Thus,

where J is rotation operator. but i am not sure is it restricted to real space rotation. any way, spin is a vector operator, thus

$latex [S_x, L_y] = i \hbar S_z = – [S_y, L_x] $

so, L, S is not commute.

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.

since

\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.

angular momentum operator in matrix form

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in order to fine out the matrix elements, we just need to know 2 equations

J_z \left| j,m\right> = m \hbar \left|j,m\right>

J_\pm \left| j, m \right>=\hbar \sqrt{ j(j+1) -m ( m\pm 1) } \left |j,m \right >

The calculation is straight forward, but be careful of the sign.

i define the coefficient :

K_j(m) = \frac{\hbar}{2} \sqrt {j(j+1) -m(m+1) }

and the matrix coefficients are:

J_z ^{\mu\nu} (j) = \hbar (j -\mu +1) \delta_{\mu \nu}

J_x ^{\mu\nu} (j) = K_j(j-\mu) \delta_{\mu (\nu-1)}+K_j(j-\nu)\delta_{(\mu-1)\nu}

J_y ^{\mu\nu} (j) = -i K_j(j-\mu) \delta_{\mu (\nu-1)}+ i K_j (j-\nu)\delta_{(\mu-1)\nu}

where

\mu , \nu = 1,2,...,2j+1

the Kronecker Delta indicated that only 1st upper and lower diagonal elements are non-zero.

δμ(ν-1) means the 1st upper diagonal elements. since ν = μ+1 to make it non-zero.

For example:

J_x (1) = \frac {\hbar }{2} \begin {pmatrix} 0 & \sqrt {2} & 0 \\ \sqrt{2} & 0 &\sqrt{2} \\ 0 & \sqrt{2} & 0\end{pmatrix}

J_x (\frac {3}{2}) = \frac {\hbar }{2} \begin {pmatrix} 0 & \sqrt {3} & 0 & 0 \\ \sqrt{3} & 0 & 2 & 0 \\ 0 & 2 & 0 & \sqrt{3} \\ 0 & 0 & \sqrt{3} & 0 \end{pmatrix}

To compute J_y , we just need to multiply the upper diagonal with i and the lower diagonal with - i .

The coefficient K_j(j-\mu)  is every interesting. if we only look at the first upper diagonal. and take the square of each element.

J_x(\frac {1}{2}) = 1

J_x(1) = \begin{pmatrix} 2 & 2 \end{pmatrix}

J_x(\frac{3}{2}) = \begin{pmatrix} 3 & 4 & 3 \end{pmatrix}

and we use this to form a bigger matrix

\begin {pmatrix} ... & 5 & 4 & 3 & 2 & 1 \\ ... & 10 & 8 & 6 & 4 & 2 \\ ... & ... & 12 & 9 & 6 & 3 \\ ... & ... & ...& 12 & 8 & 4 \\ ... & ... & ... & ... & 10 & 5 \end {pmatrix}

if we read from the top right hand corner, and take the diagonal element. we can see that they fit the 1st upper diagonal element of J_x ( j) , if we take square root of each one.

and the pattern are just the multiplication table! How nice~

so, i don’t have to compute for j = 5/2.

J_x ( \frac{5}{2} ) = \frac {\hbar}{2} \begin {pmatrix} 0 & \sqrt{5} & 0 & 0& 0 & 0 \\ \sqrt{5} & 0 & \sqrt{8} & 0 & 0 & 0 \\ 0 & \sqrt{8} & 0 & \sqrt{9} & 0 & 0 \\ 0 & 0 & \sqrt{9} & 0 & \sqrt{8} & 0 \\ 0& 0 &0 & \sqrt{8} &0 & \sqrt{5} \\ 0 & 0 & 0 & 0 & \sqrt{5} & 0 \end {pmatrix}

but the physical reason for this trick , i don’t know. for the Pascal triangle, we understand the reason for each element – it is the multiplicity.

In general:

\displaystyle J_x(L) = \frac{\hbar}{2} \begin{pmatrix} 0 & \sqrt{2L} & 0 & ... & ... & ... \\ \sqrt{2L} & 0 & \sqrt{2(2L-1)} & ... & ... & ... \\ ... & ... & ... & ... & ... \\ ... & ... & ... & ... & \sqrt{r (2L-r+1)} & ... \\  .. & ... & ... & ... & ...  \end{pmatrix}

or the ij element of the matrix is

\displaystyle [J_x(L)]_{i, i>j} = \frac{\hbar}{2} \sqrt{i(2L-i+1)} \delta_{i+1,j}

Informations we can extract

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in scattering experiment,  the raw informations we can know or observe are only few things:

  1. the number of particles counted at particular solid angle. ( when you have a unit sphere, the area on the surface is called solid angle)
  2. The polarization (spin)
  3. charge
  4. energy
  5. momentum (time of flight)

Since the number of particles counted is related to the intensity of the incident beam, the density of the target, the interaction and the differential cross section.

on the other hand, the number of particles counted should be related to intensity of incident beam, density of the target and interaction potential. Thus, the differential cross section is related to the interaction potential.

The polarization can be measured by 2nd scattering of known polarization target. or directly from a polarized primary target. or by a polarized beam.

For a nucleus there are 12 properties, and we can group them into 2, 1 is static properties, another is dynamic properties.

Static properties

intrinsic:

  1. mass
  2. radius
  3. spin
  4. parity

extrinsic:

  1. relative abundance
  2. decay half-live
  3. magnetic dipole
  4. electric quadrapole

Dynamic properties

  1. decay modes
  2. reaction mode
  3. cross section
  4. excited state

Of course, the final goal of nuclear physics is find the Hamiltonian for governing the motion of nuclear matter. Thus, we can base on the intrinsic static properties, to deduced the extrinsic and dynamic properties.

think about in atomic physics, we know the potential, the spin, then we can give out every things, like the radius, parity, decay half-live, cross section, etc….

so, the nuclear Hamiltonian is the KEY to open the door of understanding of nuclear matter.

Larmor Precession (quick)

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Magnetic moment (\mu ) :

this is a magnet by angular momentum of charge or spin. its value is:

\mu = \gamma J

where J is angular momentum, and \gamma is the gyromagnetic rato

\gamma = g \mu_B

Notice that we are using natural unit.

the g is the g-factor is a dimensionless number, which reflect the environment of the spin, for orbital angular momentum, g = 1.

\mu_B is Bohr magneton, which is equal to

\mu_B = \frac {e} {2 m} for positron

since different particle has different mass, their Bohr magneton value are different. electron is the lightest particle, so, it has largest value on Bohr magneton.

Larmor frequency:

When applied a magnetic field on a magnetic moment, the field will cause the moment precess around the axis of the field. the precession frequency is called Larmor frequency.

the precession can be understood in classical way or QM way.

Classical way:

the change of angular momentum is equal to the applied torque. and the torque is equal to the magnetic moment  cross product with the magnetic field. when in classical frame, the angular momentum, magnetic moment, and magnetic field are ordinary vector.

\vec {\Gamma}= \frac { d \vec{J}}{dt} = \vec{\mu} \times \vec{B} = \gamma \vec {J} \times \vec{B}

solving gives the procession frequency is :

\omega = - \gamma B

the minus sign is very important, it indicated that the J is precessing by right hand rule when \omega >0 .

QM way:

The Tim dependent Schrödinger equation (TDSE) is :

i \frac {d}{d t} \left| \Psi\right> = H \left|\Psi\right>

H is the Hamiltonian, for the magnetic field is pointing along the z-axis.

H = -\mu \cdot B = - \gamma J\cdot B = -gamma B J_z = \omega J_z

the solution is

\left|\Psi(t) \right> = Exp( - i \omega t J_z) \left| \Psi(0) \right>

Thus, in QM point of view, the state does not “rotate” but only a phase change.

However, the rotation operator on z-axis is

R_z ( \theta ) = Exp( - i \frac {\theta}{\hbar} J_z )

Thus, the solution can be rewritten as:

\left|\Psi (t)\right> = R_z( \omega t) \left|\Psi(0)\right>

That makes great analogy on rotation on a real vector.

on the sum of 4 momentum and excited mass

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when we have a decay process, there are many fragments, we can measure their momentum and energy and construct the 4-momentum

\vec {P_i} = ( E_i , p_i )

we use the c = 1 unit as usual.

to find out the mass before the decay, we can use

\sum E_i^2 - \sum p_i^2 = m_{excite}^2

the reason for the term “excited mass”, we can see by the following illustration.

consider a head on collision of 2 particles in C.M. frame, with momentum p and energy E1 and E2.

the mass for each one is given by

m_1 = \sqrt {E_1^2 - p^2 }

m_2 =\sqrt {E_2^2 - p^2 }

but if we use the sum of the 4 momentum and calculate the mass,

\sqrt { (E_1 +E_2)^2 - (p - p)^2} = E_1+E_2

which is not equal to

\sqrt { E_1^2 - p^2} + \sqrt{E_2^2 - p^2 }

in fact, it is larger.

the reason for its larger is, when using the sum of 4 momentum, we actually assumed the produce of collision is just 1 particles, and the collision is inelastic. Thus, if we think about the time-reverse process, which is a decay, thus, some of the mass will convert to K.E. for the decay product.

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

Differential Cross Section

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In nuclear physics, cross section is a raw data from experiment. Or more precisely differential cross section, which is some angle of the cross section, coz we cannot measure every scatter angle and the differential cross section gives us more detail on how the scattering going on.

The differential cross section (d.s.c.) is the square of the scattering amplitude of the scatter spherical wave, which is the Fourier transform of the density.

d.s.c = |f(\theta)|^2 = Fourier ( \rho (r), \Delta p , r )

Where the angle θ come from the momentum change. So, sometime we will see the graph is plotted against momentum change instead of angle.

By measuring the yield of different angle. Yield is the intensity of scattered particle. We can plot a graph of the Form factor, and then find out the density of the nuclear or particle.

However, the density is not in usual meaning, it depends on what kind of particle we are using as detector. For example, if we use electron, which is carry elected charge, than it can feel the coulomb potential by the proton and it reflected on the “density”, so we can think it is kind of charge density.

Another cross section is the total cross section, which is sum over the d.s.c. in all angle. Thus, the plot always is against energy. This plot give us the spectrum of the particle, like excitation energy, different energy levels.

Special Relativity II

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We are going to talk about coordinate transform from “center of momentum frame” to “Laboratory frame”.

At the center of momentum [C.M.] frame, the total momentum is zero.

Refer here on Google Docs. for the Mathematica 7 code and calculation steps.

anyone want to have the .nb file, feel free to ask.

i am just discuss on the result.

  1. In the C.M. frame, energy of each particle reserved, there is no exchange in energy and momentum. after collision, they just change the moving direction.
  2. the energy is C.M frame is always smaller then Lab frame, or other fame. which is also from the face that, at the CM frame, total momentum is zero and the corresponding energy is the Proper Energy.
  3. In Lab frame, the scatter angle is always smaller than 90 degree for incident particle’s mass > target particle.
  4. In Lab frame, The larger the scatter angle, the smaller the momentum and larger the momentum transfer.

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