## Rotation operator on x, y in Matrix form

in the J.J. Sakurai’s book, the formalism of finding the matrix representation of rotation operator is general, but quite long and detail. A general treatment is necessary for understanding the topic, but i think, who will use arbitrary rotation? so,  here i give a simple and direct calculation on $J_x$ and $J_y$, for use-ready.

the method is diagonalization. because we already knew the matrix form of the angular momentum operator. which is not given in J.J.Sakurai’s book.

recall that the formalism:

$f(M) = P \cdot f(D) \cdot P^{-1}$

since $D$ is diagonal matrix, thus

$f(D)_{ij} = f(\lambda_i) \delta_{ij}$

so, we have to find out the $P$ for $J_x$ and $J_y$.

i am still trying to obtain the equation, but…..

anyway, using program can solve it without headache. ( but typing Latex is )here are some result.

$J_x(\frac{1}{2}) = \begin {pmatrix} \cos \left( \frac {\theta}{2} \right) & - i \sin \left( \frac{\theta}{2} \right) \\ -i \sin ( \frac{\theta}{2} ) & \cos (\frac {\theta}{2}) \end {pmatrix}$

$J_y(\frac{1}{2}) = \begin {pmatrix} \cos \left( \frac {\theta}{2} \right) & - \sin ( \frac{\theta}{2} ) \\ \sin ( \frac{\theta}{2} ) & \cos (\frac {\theta}{2}) \end {pmatrix}$

## 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.

## WKB approximation

I was scared by this term once before. ( the approach an explanation from J.J. Sakurai’s book is not so good)  in fact, don’t panic, it is easy. Let me explain.

i just copy what written in Introduction to Quantum Mechanics by David Griffiths (1995) Chapter 8.

The approx. can be applied when the potential is varies slowly compare the wavelength of the wave function. when it expressed in $Exp( i k x)$, wavelength = 2 π / k, when it expressed in $Exp( - \kappa x )$, wavelength = 1/κ.

in general, the wavefunction can be expressed as amplitude and phase:

$\Psi(x) = A(x)Exp(i \phi(x))$

where $A(x)$ and $\phi(x)$ are real function

sub this into the time-independent Schrödinger equation (TISE)

$\Psi '' (x) = - \frac {2 m} {\hbar^2 } ( E - V(x) ) \Psi (x)$

$\Psi ''(x) = ( A''(x)- A(x) \phi'(x)^2 + 2 i A'(x) \phi'(x)+ i A(x)\phi''(x) ) Exp(i \phi (x) )$

and separate the imaginary part and real part.

The imaginary part is can be simplified as:

$2 A'(x) \phi '(x) + A(x) \phi ''(x) = 0 = \frac {d}{dx} ( A^2(x) \phi '(x)$

$A(x) = \frac {const.} {\sqrt {\phi '(x)}}$

The real part is

$A''(x) = \left ( \phi ''(x) - \frac {2m}{\hbar^2 } ( E - V(x) ) \right) A(x)$

we use the approx. that $A''(x) = 0$ ,  since it varies slowly.

Thus,

$\phi '(x) = \sqrt { \frac {2m}{\hbar^2} (E - V(x) ) }$

$\Rightarrow \phi(x) = \int \sqrt { \frac {2m}{\hbar ^2} ( E - V(x ) )} dx$

if we set,

$p(x) = \sqrt { \frac {2m}{ \hbar^2 } ( E - V(x) )}$

for clear display and $p(x)$ is the energy different between energy and the potential. the solution is :

$\Psi(x) = \frac {const.}{\sqrt {p(x)}} Exp \left( i \int p(x) dx \right)$

Simple! but one thing should keep in mind that, the WKB approx is not OK when Energy = potential.

This tell you, the phase part of the wave function is equal the square of the area of the different of Energy and the Potential.

when the energy is smaller then the potential, than, the wavefunction is under decay.

one direct application of WKB approxi is on the Tunneling effect.

if the potential is large enough, so, the transmittance is dominated by the decay, Thus, the probability of the tunneling is equal to

$Exp \left( - 2 \sqrt { \frac {2m}{\hbar ^2 } A_{area} ( V(x) - E )} \right)$

Therefore, when we have an ugly potential, we can approx it by a rectangular potential with same area to give the similar estimation.