## on angular momentum adding & rotation operator

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.

## Natural unit

on Size, Energy and Unit, we know that the speed of light better be equal to 1, that simplify the equation of relativity.

$c = 1$

now, we impose 1 more things, the Reduced Planck constant,  $\hbar$ also set to 1. that simplify all equations with angular momentum or spin.

$\hbar = 1$

the Angular momentum:

$J^2 \left| l,m \right > = j(j+1) \hbar ^2 \left| l,m\right >$

$J_z \left|l,m \right> = m \hbar \left|l,m\right>$

now becomes :

$J^2 \left| l,m \right > = j(j+1) \left| l,m\right >$

$J_z \left|l,m \right> = m \left|l,m\right>$

when we want to calculate the real value, we can recover the $\hbar$ by considering the dimension. the reduced Planck constant has dimension

$[\hbar] = [kg][m^2][s^{-1}] = [energy][second]$

for example,

$E = \omega \hbar$

## Parity

It is Pa-ri-ty, not Par-ty.

( it needs to clean up)

Parity is just a reflection on every space dimension.

$\begin {pmatrix} x \\ y \\ z \end {pmatrix} \rightarrow \begin{pmatrix} - x \\ - y \\-z \end {pmatrix}$

## General

This is just a mirror reflection, although mirror reflection only reflects on 1 dimension, the dimension that perpendicular to the mirror surface.

May be we start on 2-D space instead of 3-D, draw a F and flips it upside down, and left-side right. then, you have a F just rotated 180 degree, not a reflection. however, in 3-D, then there is something different.

The parity transform is taking everything reverted. For example, when you stand up, your arms place horizontal and you left arm points forward and your right arm points right. After a parity transform. You right arm point left. Your left arm point backward, and you are standing on the ceiling, upside down. The result is a mirror image of your self. If we rotate the reverted-self from the ceiling to the ground.

Thus, parity also related as mirror reflection. In physics, we like to call the right-hand system (RHS) or left-hand system (LHS).

A simple RHS and LHS are on your hands! Although our left hand and right hand has some minor different, in general, they are the mirror image of each other. And the great interesting thing is, your left hand cannot overlap the right hand. They are equal but not the same.

Another thing is spring, when a wire is rolled clockwise and going upward, it form a left-hand spring and vice aver. Thus 2 springs are not the same.

For those which keep function as before parity transform, we called it parity positive, for those who are not, we called it parity negative.

Be reminded that the chiral material that interact circularly polarized light different still the parity positive. For example, a material which only let right hand light passes through, but not let the left hand pass. After parity transform, it lets left hand light pass through but not right hand .Thus, the left hand and right hand are work equally well!

We also cannot say our left hand is more weak then our right hand, then we called it parity negative. It is because, if we reflected ourself, our left hand is as good and right hand and the right hand is as weak as left hand.

A more physical example is the polarization of light, there are lefthand rotating light and right hand rotating light, called circular polarization. And material which interact differenty with different circular polarization are called chiral material. We should stop talking about examples in here. Because in nature, there are so many things has chiral property. Never the less, potenient and drug also has chirality. One book I recommend on general science for the chirality is “right-hand, left-hand” by chris McManus.

## Physics

Physics encounters parity is because we believe if the whole world is reverted, every thing just work fine and the same. For example, if our orgasms are all reflected, left goes to right, right go to left. We still alive. In fact, there are some real cases, that some peole do have reverted orgasm. Because there should be symmetric in the world.

In normal day, parity positive never break. It is seem impossible to break. How coome some thing work differently under parity transform?

For position, linear momentum, parity just make them change

However, in mathematics, there are many parity negative things. One example is the spherical harmonic. It is can be parity positive and negative depends on the parameter.

Lets take a imaginary example in parity negative. If we use photon to hit a target, all photons are going left. Now, we reflet the whole system. But now, the photons are still going left.

The first discovery of parity negative is on beta- decay from Co-60. Whe. Applied an external magnetic field from down to up, the beta particle come out at left. When we change the magnetic field, now is from up to down, the beta particle should come out at right, if parity is positive. But it is not, it still keep coming out left!

The reason of it is beyond my understanding… Sorry.