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!


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.


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.


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.



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

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


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.